Solucionario.de.Matematicas.para.administracion.y.economia.pdf

Table of Contents Chapter 0

1

Chapter 1

35

Chapter 2

54

Chapter 3

89

Chapter 4

132

Chapter 5

160

Chapter 6

177

Chapter 7

231

Chapter 8

295

Chapter 9

333

Chapter 10

357

Chapter 11

378

Chapter 12

423

Chapter 13

469

Chapter 14

539

Chapter 15

614

Chapter 16

658

Chapter 17

670

Chapter 0 Problems 0.1

7. True;

1. True; –13 is a negative integer.

⎛ b ⎞ ab 8. True, because a ⎜ ⎟ = . ⎝c⎠ c

2. True, because −2 and 7 are integers and 7 ≠ 0. 3. False, because the natural numbers are 1, 2, 3, and so on.

9. False; the left side is 5xy, but the right side is 5 x 2 y.

0 4. False, because 0 = . 1

10. True; by the associative and commutative properties, x(4y) = (x ⋅ 4)y = (4 ⋅ x)y = 4xy.

5 5. True, because 5 = . 1

11. distributive 12. commutative

6. False, since a rational number cannot have 7 is not a number denominator of zero. In fact, 0 at all because we cannot divide by 0. 7. False, because integer. 8. True;

13. associative 14. definition of division

25 = 5, which is a positive

15. commutative and distributive 16. associative

2 is an irrational real number.

17. definition of subtraction

9. False; we cannot divide by 0.

18. commutative

10. False, because the natural numbers are 1, 2, 3, and so on, and 3 lies between 1 and 2.

19. distributive 20. distributive

11. True

21. 2x(y − 7) = (2x)y − (2x)7 = 2xy − (7)(2x) = 2xy − (7 · 2)x = 2xy − 14x

12. False, since the integer 0 is neither positive nor negative.

22. (a − b) + c = [a + (−b)] + c = a + (−b + c) = a + [c + (−b)] = a + (c − b)

Problems 0.2 1. False, because 0 does not have a reciprocal. 2. True, because

x+2 x 2 x = + = + 1. 2 2 2 2

23. (x + y)(2) = 2(x + y) = 2x + 2y

7 3 21 ⋅ = = 1. 3 7 21

24. 2[27 + (x + y)] = 2[27 + (y + x)] = 2[(27 + y) + x] = 2[(y + 27) + x] 25. x[(2y + 1) + 3] = x[2y + (1 + 3)] = x[2y + 4] = x(2y) + x(4) = (x · 2)y + 4x = (2x)y + 4x = 2xy + 4x

3. False; the negative of 7 is −7 because 7 + (−7) = 0. 4. False; 2(3 · 4) = 2(12) = 24, but (2 · 3)(2 · 4) = 6 · 8 = 48.

26. (1 + a)(b + c) = 1(b + c) + a(b + c) = 1(b) + 1(c) + a(b) + a(c) = b + c + ab + ac

5. False; –x + y = y + (–x) = y – x. 6. True; (x + 2)(4) = (x)(4) + (2)(4) = 4x + 8. 1

Chapter 0: Review of Algebra

ISM: Introductory Mathematical Analysis

27. x(y − z + w) = x[(y − z) + w] = x(y − z) + x(w) = x[y + (−z)] + xw = x(y) + x(−z) + xw = xy − xz + xw

51. X(1) = X

28. –2 + (–4) = –6

53. 4(5 + x) = 4(5) + 4(x) = 20 + 4x

29. –6 + 2 = –4

54. –(x – 2) = –x + 2

30. 6 + (–4) = 2

55. 0(–x) = 0

31. 7 – 2 = 5

⎛ 1 ⎞ 8 ⋅1 8 = 56. 8 ⎜ ⎟ = ⎝ 11 ⎠ 11 11

52. 3(x – 4) = 3(x) – 3(4) = 3x – 12

32. 7 – (–4) = 7 + 4 = 11 33. −5 − (−13) = −5 + 13 = 8

57.

5 =5 1

58.

14 x 2 ⋅ 7 ⋅ x 2 x = = 21 y 3 ⋅ 7 ⋅ y 3 y

59.

3 3 3 = =− −2 x −(2 x) 2x

60.

2 1 2 ⋅1 2 ⋅ = = 3 x 3 ⋅ x 3x

61.

a a(3b) 3ab (3b) = = c c c

34. −a − (−b) = −a + b 35. (–2)(9) = –(2 · 9) = –18 36. 7(–9) = –(7 · 9) = –63 37. (–2)(–12) = 2(12) = 24 38. 19(−1) = (−1)19 = −(1 · 19) = −19 39.

−1 ⎛ 9⎞ = −1⎜ − ⎟ = 9 1 −9 ⎝ 1⎠

40. –(–6 + x) = –(–6) – x = 6 – x ⎛ 7 ⎞ 62. (5a ) ⎜ ⎟ = 7 ⎝ 5a ⎠

41. –7(x) = –(7x) = –7x 42. –12(x – y) = (–12)x – (–12)(y) = –12x + 12y (or 12y – 12x)

63.

−aby −a ⋅ by by = = −ax −a ⋅ x x

−3 3 1⋅ 3 1 =− =− =− 44. −3 ÷ 15 = 15 15 5⋅3 5

64.

7 1 7 ⋅1 7 ⋅ = = y x y ⋅ x xy

45. −9 ÷ (−27) =

−9 9 9 ⋅1 1 = = = −27 27 9 ⋅ 3 3

65.

2 5 2 ⋅ 5 10 ⋅ = = x y x ⋅ y xy

46. (−a ) ÷ (−b) =

−a a = −b b

66.

1 1 3 2 3+ 2 5 + = + = = 2 3 6 6 6 6

67.

5 3 5 9 5 + 9 14 2 ⋅ 7 7 + = + = = = = 12 4 12 12 12 12 2 ⋅ 6 6

68.

3 7 9 14 9 − 14 −5 5 ⋅1 1 − = − = = =− =− 10 15 30 30 30 30 5⋅6 6

43. –[–6 + (–y)] = –(–6) – (–y) = 6 + y

47. 2(–6 + 2) = 2(–4) = –8 48. 3[–2(3) + 6(2)] = 3[–6 + 12] = 3[6] = 18 49. (–2)(–4)(–1) = 8(–1) = –8 50. (−12)(−12) = (12)(12) = 144 2


69.

70.

4 6 4 + 6 10 + = = =2 5 5 5 5

X 5

−

Y 5

=

2 3 16 15 16 − 15 1 − = − = = 5 8 40 40 40 40

74.

= 6÷

l 3

m

=

a3⋅7

=

(b 4 )5

b 4⋅5

75.

−x y2 z xy

76.

7 is not defined (we cannot divide by 0). 0

2

=

11.

w4 s 6 y4

x9 x5

= x 9 −5 = x 4

⎛ 2a 4 12. ⎜ ⎜ 7b5 ⎝

z x xy x2 =− ÷ =− ⋅ =− yz y 2 xy y2 z x

6

⎞ (2a 4 )6 ⎟ = ⎟ (7b5 )6 ⎠ 26 ( a 4 ) 6 = 76 (b5 )6 =

0 77. =0 7

=

0 78. is not defined (we cannot divide by 0). 0

13.

( x3 )6 3

=

x( x )

79. 0 · 0 = 0 Problems 0.3

14.

1. (23 )(22 ) = 23+ 2 = 25 (= 32)

64a 4⋅6 117, 649b5⋅6 64a 24 117, 649b30

x3⋅6 1+3

( x 2 )3 ( x 3 ) 2 ( x3 ) 4

3. w4 w8 = w4+8 = w12

16.

4

81 = 3

4. z 3 zz 2 = z 3+1+ 2 = z 6

17.

7

−128 = −2

9 5

y y

=

x y

9+5 12⋅4

6. ( x ) = x 12 4

=

x

8

19.

48

0.04 = 0.2

3

4

4

= x18− 4 = x14

x 2⋅3 x3⋅2

25 = 5

18.

y14

=x

=

x18 x

x12−12 = x 0 = 1

15.

x x

=

x

2. x6 x9 = x6+9 = x15

5.

b 20

⎛ w2 s 3 ⎞ ( w2 s 3 ) 2 ( w2 ) 2 ( s3 ) 2 w2⋅2 s3⋅2 = = = 10. ⎜ ⎟ ⎜ y2 ⎟ ( y 2 )2 y 2⋅2 y4 ⎝ ⎠

x y 6y = 6⋅ = y x x

3+ 5

a 21

9. (2 x 2 y 3 )3 = 23 ( x 2 )3 ( y 3 )3 = 8 x 2⋅3 y 3⋅3 = 8 x6 y9

l m l 1 l ÷ = ⋅ = 3 1 3 m 3m

3 5

=

5

72.

x y

(a3 )7

⎛ x2 ⎞ ( x 2 )5 x 2⋅5 x10 8. ⎜ ⎟ = = = ⎜ y3 ⎟ ( y 3 )5 y 3⋅5 y15 ⎝ ⎠

5

3 1 1 18 3 2 18 − 3 + 2 17 − + = − + = = 2 4 6 12 12 12 12 12

6

7.

X −Y

71.

73.

Section 0.3

4 1 1 1 = = 4 16 16 2

x3⋅4

=

x6 x6 x12

=

x12 x12


20.

–8 = 27

3

3

–8

3

27

=


−2 2 =− 3 3

21. (49)1/ 2 = 49 = 7

37. (9 z 4 )1/ 2 = 9 z 4 = 32 ( z 2 )2 = 32 ( z 2 ) 2 = 3z 2

22. (64)1/ 3 = 3 64 = 4 3/ 2

23. 9

=

3 3 13 39 39 39 = ⋅ = = = 2 13 13 13 13 13 132

36.

3

( 9)

3

3

38. (16 y8 )3 / 4 = ⎡ 4 16 y8 ⎤ = ⎡ 4 (2 y 2 )4 ⎤ = (2 y 2 )3 ⎢⎣ ⎣⎢ ⎦⎥ ⎦⎥

= (3) = 27 3

= 8y 6

1

24. (9) −5 / 2 =

(9)

1

25. (32) –2 / 5 =

(32)

26. (0.09) =

1 3 10

–1/ 2

=

⎛ 1 ⎞ 27. ⎜ ⎟ ⎝ 32 ⎠

=

5/ 2

=

2/5

1

( 9)

5

=

1

=

5

3

1

( 5 32 )

1

=

2

1 243 1

=

(2)

2

=

⎛ 27t 3 ⎞ 39. ⎜ ⎟ ⎜ 8 ⎟ ⎝ ⎠

1 4

1 = = 0.3 0.09

10 3

41. 4/5

⎛ 64 ⎞ 28. ⎜ − ⎟ ⎝ 27 ⎠

4

⎛ 1 ⎞ 1 ⎛1⎞ = ⎜⎜ 5 ⎟⎟ = ⎜ ⎟ = 16 ⎝2⎠ ⎝ 32 ⎠

2/3

42.

−3 / 4

2/3

c2

= a5 ⋅ b −3 ⋅

1 c2

2

9t 2 ⎡ 3t ⎤ =⎢ ⎥ = 4 ⎣2⎦

−3 / 4

= a5 ⋅

⎡4⎤ =⎢ ⎥ ⎣ x3 ⎦

1

⋅

1

b3 c 2

= x 2 / 5 y 3 / 5 z –10 / 5 =

5 2 3 –10

x y z

−3

=

=

4−3 ( x3 )−3

a5 b3 c 2

x2 / 5 y3 / 5 z2

2

2 ⎛ 64 ⎞ 16 ⎛ 4⎞ = ⎜⎜ 3 − ⎟⎟ = ⎜ − ⎟ = 27 3 9 ⎝ ⎠ ⎝ ⎠

30.

3

54 = 3 27 ⋅ 2 = 3 27

31.

3

2 x3 = 3 2

3 3

x =x

3

2 = 33 2

3

2

32.

4x = 4 x = 2 x

33.

16 x = 16 x = 4 x 4

a5b −3

4

43. 5m−2 m−7 = 5m −2+ ( −7) = 5m −9 =

50 = 25 ⋅ 2 = 25 ⋅ 2 = 5 2

29.

⎛ ⎡ 3t ⎤ 3 ⎞ = ⎜⎢ ⎥ ⎟ ⎜⎣ 2 ⎦ ⎟ ⎝ ⎠

⎛ ⎡ 4 ⎤4 ⎞ = ⎜⎢ ⎥ ⎟ ⎜ ⎣ x3 ⎦ ⎟ ⎝ ⎠ −3 9 9 4 x x = = = −9 3 64 x 4

⎛ 256 ⎞ 40. ⎜ ⎟ ⎝ x12 ⎠

1

(0.09)1/ 2

2/3

4

44. x + y –1 = x + 1

45. (3t ) –2 =

(3t )

46. (3 − z ) –4 =

2

5 m9

1 y =

1 9t 2

1 (3 − z )4

2

47.

4 4 x x x 34. 4 = = 4 16 2 16

5

5 x 2 = (5 x 2 )1/ 5 = 51/ 5 ( x 2 )1/ 5 = 51/ 5 x 2 / 5

48. ( X 3Y −3 )−3 = ( X 3 )−3 (Y −3 )−3 = X −9Y 9

35. 2 8 − 5 27 + 3 128 = 2 4 ⋅ 2 − 5 9 ⋅ 3 + 3 64 ⋅ 2

=

= 2 ⋅ 2 2 − 5 ⋅ 3 3 + 43 2 = 4 2 − 15 3 + 4 3 2 4

Y9 X9


x − y = x1/ 2 − y1/ 2

49. 50.

Section 0.3

u −2 v −6 w3 vw−5

w3−( −5)

=

=

u 2 v1−( −6)

w8

x

9/ 4 3/ 4 1/ 2

y

52.

= a −3 / 4b −1/ 2 a5b −4

63.

= a17 / 4 b −9 / 2 =

a17 / 4 b

=

9/ 2

53. (2a − b + c) 2 / 3 = 3 (2a − b + c)2

64.

4 3 6 9

54. (ab 2 c3 )3 / 4 = 4 (ab 2 c3 )3 = a b c 55. x

=

–4 / 5

1

=

x4 / 5

1

3

1

−

=

x 2 /15

59.

6

60.

3

5

4

]

8

=

=

=

= [x

−4 / 5 1/ 6

]

=x

–4 / 30

51/ 2 3 81/ 4

(3 x)1/ 3

3x 3

2

=

33 y 2

3y2 / 3

66.

18 = 9 =3 2

=x

5

2

=

(2 y )

y 2y 2y

=

1(3 x) 2 / 3

=

(3x)1/ 3 (3 x)2 / 3

3

(3x) 2 3x

=

2 ⋅ y1/ 3 3 y 2 / 3 ⋅ y1/ 3

=

2 y1/ 3 2 3 y = 3y 3y

5

=

68.

6 ⋅ 51/ 2 51/ 2 ⋅ 51/ 2 3 ⋅ 21/ 4 81/ 4 ⋅ 21/ 4

=

=

6 5 5 34 2 4

16

=

34 2 2

2

5

=

(24 a10b15 )1/ 20 = ab 2

=

2 31/ 3

=

20

31/ 3 ⋅ 32 / 3

3

3

=

(2334 )1/ 6 6 648 = 3 3

5

3 u 5 / 2 v1/ 2

=

16a10 b15 ab

21/ 2 ⋅ 32 / 3

69. 2 x 2 y –3 x 4 = 2 x6 y –3 =

70.

2 ⋅ a1/ 2 b3 / 4

2 / 4 1/ 4 a1/ 2b1/ 4 ⋅ a1/ 2 b3 / 4 a b a b 21/ 5 a1/ 2b3 / 4 24 / 20 a10 / 20b15 / 20 = = ab ab 4 2

=

15 2

=

(2 y )

1/ 2

12 = 4=2 3

=

3

1

=

1/ 2

9 x2 3x

12

–2 /15

x

6

1

=

65.

67.

w3 / 5 (3w)3 / 5 3 1 3 1 = − = − 5 3 5 3 5 3 5 w (3w) w 27 w3

1

1/ 2

x

57. 3w−3 / 5 − (3w) −3 / 5 =

58. [( x )

(2 x)

y (2 y )1/ 2

=

(2 y )

2

5 4

56. 2 x1/ 2 − (2 y )1/ 2 = 2 x − 2 y

–4 1/ 5 1/ 6

y

=

1 3

(2 x)

4 2x 2x

=

1/ 2

2y 2

=

a −3b −2 a5b −4 = (a −3b −2 )1/ 4 a5b −4

4

1/ 2

2 2x x

2y

z

1/ 2

4(2 x)1/ 2

=

(2 x)

y

62.

4

=

2x =

u 2 v7

51. x 2 4 xy –2 z 3 = x 2 ( xy –2 z 3 )1/ 4 = x 2 x1/ 4 y –2 / 4 z 3 / 4 =

4

61.

=

23 / 634 / 6 3

2 x6 y3

3 ⋅ u1/ 2 v1/ 2 u 5 / 2 v1/ 2 ⋅ u1/ 2 v1/ 2

=

3u1/ 2 v1/ 2 u 3v


243

71.

243 = 81 = 9 3

=

3


= [(5k 2 ) 2 3]1/ 2 = 5k 2 31/ 2

72. {[(3a3 ) 2 ]−5 }−2 = {[32 a 6 ]−5 }−2 = {3−10 a −30 }−2

83.

= 320 a 60

73.

20

=

64 y ⋅ x 6

x

74.

s

3/ 2

5

=

3 2

s

75.

76.

=

(2 –2 x1/ 2 y –2 )3 1/ 2

⋅x

s

5/ 2

s

2/3

1 2 –6 x3 / 2 y –6

x3 / 2

85.

64 y x x

=

=

84.

26 y 6

6 1/ 2

=

1/ 2

15 / 6

s

s

4/6

75k 4 = (75k 4 )1/ 2 = [(25k 4 )(3)]1/ 2

82.

(ab −3c)8 (a −1c 2 ) −3

3

= s11/ 6 = x2 ÷

= (31/ 4 )8 = 38 / 4 = 32 = 9

87. − 77. 32 (32) −2 / 5 = 32 (25 ) −2 / 5

78. ⎛⎜ 5 x 2 y ⎞⎟ ⎝ ⎠

2/5

b 24

2

x6

= x 2 ÷ x6−12 = x 2 ÷ x −6

12

x 1

x6

= x 2 ⋅ x 6 = x8

(–6)(–6) = 36 = 6

86.

8s –2 2s

88.

a5 c14

⎡ x3 ⎤ x6 ( x3 ) 2 ÷⎢ = ÷ ⎥ 3 2 x4 x 4 ( x6 )2 ⎣⎢ ( x ) ⎦⎥

= x2 ÷

= 32 (2−2 ) 1 = 32 ⋅ 22 9 = 4

=

3

(–6)2 ≠ −6 since –6 < 0.

Note that

8

a 3 c −6

( x 2 )3

3 2

(4 3)

a8b −24 c8

= 7 ⋅ 7 2 = 73 = 7

3 7(49)

2

x yz 3 3 xy 2 = 3 ( x 2 yz 3 )( xy 2 ) = 3 x3 y 3 z 3 = xyz

=

3

4

=−

3 2

=−

s s

( a5b−3 c )

3

4

s5

= (a5 )3 (b −3 )3 (c1/ 2 )3 = a15b −9 c3 / 2 =

= [( x 2 y )1/ 5 ]2 / 5 = ( x 2 y )2 / 25

a15 c3 / 2 b9

⎛ 3 x3 y 2 ⎞ 89. (3x3 y 2 ÷ 2 y 2 z −3 ) 4 = ⎜ ⎟ ⎜ 2 y 2 z −3 ⎟ ⎝ ⎠

= x 4 / 25 y 2 / 25

4

4

79. (2 x –1 y 2 )2 = 22 x –2 y 4 =

80.

3 3

y4 x

=

3

y1/ 3 x1/ 4

=

4y

⎛ 3 x3 z 3 ⎞ =⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠ 3 3 4 (3x z ) = (2)4

4

x2 3 ⋅ y 2 / 3 x3 / 4

y1/ 3 x1/ 4 ⋅ y 2 / 3 x3 / 4

=

24 81x12 z12 = 16

3 x3 / 4 y 2 / 3 = xy

81.

x x 2 y3 xy 2 = x1/ 2 ( x 2 y 3 )1/ 2 ( xy 2 )1/ 2 =x

1/ 2

( xy

3/ 2

1/ 2

)( x

34 x12 z12

90.

y) = x y

2 5/ 2

6

(

1 –2

2x 16 x3

)

2

=

1

( )

1/ 2 2

2

= –2 2

(x )

(161/ 2 )2 ( x3 ) 2

1 –4

2x 16 x6

=

1 1 8 x10

= 8 x10


Section 0.4 18. −{−6a − 6b + 6 + 10a + 15b − a[2b + 10]} = −{4a + 9b + 6 − 2ab − 10a} = −{−6a + 9b + 6 − 2ab} = 6a − 9b − 6 + 2ab

Problems 0.4 1. 8x – 4y + 2 + 3x + 2y – 5 = 11x – 2y – 3 2. 6 x 2 − 10 xy + 2 + 2 z − xy + 4

19. x 2 + (4 + 5) x + 4(5) = x 2 + 9 x + 20

= 6 x 2 − 11xy + 2 z + 6

20. u 2 + (5 + 2)u + 2(5) = u 2 + 7u + 10

3. 8t 2 − 6s 2 + 4 s 2 − 2t 2 + 6 = 6t 2 − 2 s 2 + 6 4. 5.

x +2 x + x +3 x = 7 x

21. ( w + 2)( w − 5) = w2 + (−5 + 2) x + 2(−5) = w2 − 3w − 10

a + 2 3b − c + 3 3b = a + 5 3b − c

22. z 2 + (–7 − 3) z + (–7)(–3) = z 2 − 10 z + 21

6. 3a + 7b − 9 − 5a − 9b − 21 = −2a − 2b − 30

23. (2 x)(5 x ) + [(2)(2) + (3)(5)]x + 3(2) = 10 x 2 + 19 x + 6

7. 6 x 2 –10 xy + 2 − 2 z + xy − 4 = 6 x 2 − 9 xy − 2 z + 2 − 4

24. (t)(2t) + [(1)(7) + (−5)(2)]t + (−5)(7) = 2t 2 − 3t − 35

8.

x +2 x − x −3 x = − x

9.

x + 2 y − x − 3z = 2 y − 3z

25. X 2 + 2( X )(2Y ) + (2Y )2 = X 2 + 4 XY + 4Y 2 26. (2 x)2 − 2(2 x)(1) + 12 = 4 x 2 − 4 x + 1

10. 8z – 4w – 3w + 6z = 14z – 7w 11. 9x + 9y – 21 – 24x + 6y – 6 = –15x + 15y – 27

27. x 2 − 2(5) x + 52 = x 2 − 10 x + 25

12. u − 3v − 5u − 4v + u − 3 = −3u − 7v − 3 13. 5 x − 5 y + xy − 3 x − 8 xy − 28 y 2

2

2

28. (1 ⋅ 2)

2

2

+ [(1)(5) + (–1)(2)] x + (–1)(5)

= 2x + 3 x − 5

= 2 x 2 − 33 y 2 − 7 xy

29.

14. 2 – [3 + 4s – 12] = 2 – [4s – 9] = 2 – 4s + 9 = 11 – 4s

(

3x

)

2

+2

(

)

3 x (5) + (5)2

= 3x + 10 3x + 25

15. 2{3[3x 2 + 6 − 2 x 2 + 10]} = 2{3[ x 2 + 16]}

30.

= 2{3x + 48} = 6 x + 96 2

( x)

2

( y)

2

− 32 = y − 9

31. (2 s )2 − 12 = 4 s 2 − 1

16. 4{3t + 15 – t[1 – t – 1]} = 4{3t + 15 – t[–t]} = 4{3t + 15 + t 2 } = 4t 2 + 12t + 60

32. ( z 2 )2 − (3w)2 = z 4 − 9w2

17. −5(8 x3 + 8 x 2 − 2( x 2 − 5 + 2 x))

33. x 2 ( x + 4) − 3( x + 4)

= −5(8 x3 + 8 x 2 − 2 x 2 + 10 − 4 x)

= x3 + 4 x 2 − 3x − 12

= −5(8 x3 + 6 x 2 − 4 x + 10) = −40 x3 − 30 x 2 + 20 x − 50

34. x( x 2 + x + 3) + 1( x 2 + x + 3) = x3 + x 2 + 3 x + x 2 + x + 3 = x3 + 2 x 2 + 4 x + 3 7



35. x 2 (3 x 2 + 2 x − 1) − 4(3x 2 + 2 x − 1)

46.

= 3x 4 + 2 x3 − x 2 − 12 x 2 − 8 x + 4 = 3x 4 + 2 x3 − 13x 2 − 8 x + 4

47.

36. 3 y (4 y + 2 y − 3 y ) − 2(4 y + 2 y − 3 y ) 3

2

3

2

2 x3 7 x 4 4 − + = 2 x2 − 7 + x x x x 6 x5 2x

= 12 y 4 + 6 y 3 − 9 y 2 − 8 y 3 − 4 y 2 + 6 y

48.

= 12 y 4 − 2 y 3 − 13 y 2 + 6 y

37. x{2( x 2 − 2 x − 35) + 4[2 x 2 − 12 x]} = x{2 x 2 − 4 x − 70 + 8 x 2 − 48 x} = x{10 x 2 − 52 x − 70} = 10 x3 − 52 x 2 − 70 x

2

+

4 x3 2x

2

1

−

2x

2

= 3 x3 + 2 x −

3y − 4 − 9 y − 5 3y −6 y − 9 = 3y −6 y 9 = − 3y 3y 3 = −2 − y

38. [(2 z )2 − 12 ](4 z 2 + 1) = [4 z 2 − 1](4 z 2 + 1) x

= (4 z 2 ) 2 − 12 = 16 z 4 − 1

49. x + 5 x 2 + 5 x − 3 x2 + 5x

39. x(3x + 2y – 4) + y(3x + 2y – 4) + 2(3x + 2y – 4)

−3

= 3x 2 + 2 xy − 4 x + 3xy + 2 y 2 − 4 y + 6 x + 4 y − 8 = 3x 2 + 2 y 2 + 5 xy + 2 x − 8

Answer: x +

40. [ x 2 + ( x + 1)]2 = ( x ) + 2 x ( x + 1) + ( x + 1) 2 2

2

−3 x+5

x −1 50. x − 4 x 2 − 5 x + 4

2

x2 − 4 x –x + 4 –x + 4 0 Answer: x – 1

= x 4 + 2 x3 + 2 x 2 + x 2 + 2 x + 1 = x 4 + 2 x3 + 3 x 2 + 2 x + 1

41. (2a )3 + 3(2a )2 (3) + 3(2a )(3) 2 + (3)3 = 8a3 + 36a 2 + 54a + 27

3 x 2 − 8 x + 17 51. x + 2 3x3 − 2 x 2 + x − 3

42. (3 y )3 − 3(3 y )2 (2) + 3(3 y )(2)2 − (2)3 = 27 y3 − 54 y 2 + 36 y − 8

3 x3 + 6 x 2 –8 x 2 + x

43. (2 x)3 − 3(2 x)2 (3) + 3(2 x)(3)2 − 33

–8 x 2 − 16 x 17 x − 3 17 x + 34 – 37

= 8 x3 − 36 x 2 + 54 x − 27

44. x3 + 3x 2 (2 y ) + 3 x(2 y ) 2 + (2 y )3 = x3 + 6 x 2 y + 12 xy 2 + 8 y 3

Answer: 3x 2 − 8 x + 17 + 45.

z 2 18 z − = z − 18 z z

8

–37 x+2

1 2x2


Section 0.5 z+2 56. z 2 − z + 1 z 3 + z 2 + z

x3 + x 2 + 3 x + 3 52. x − 1 x 4 + 0 x3 + 2 x 2 + 0 x + 1

z3 − z 2 + z

x 4 − x3 x3 + 2 x 2

2z2

x3 – x 2

2z2 − 2z + 2 2z − 2

3x 2 + 0 x 3x 2 − 3x 3x + 1 3x − 3 4

Answer: x3 + x 2 + 3 x + 3 +

Answer: z + 2 +

z − z +1

Problems 0.5 1. 2(ax + b)

4 x −1

2. 2y(3y – 2) 3. 5x(2y + z)

x2 − 2 x + 4

53. x + 2 x + 0 x + 0 x + 0 3

2z − 2 2

2

4. 3x 2 y (1 − 3xy 2 )

x3 + 2 x 2 −2 x 2 + 0

5. 4bc(2a3 − 3ab 2 d + b3cd 2 )

−2 x − 4 x 4x + 0 4x + 8 −8 2

Answer: x 2 − 2 x + 4 −

6. 6u 2 v(uv 2 + 3w4 − 12v 2 ) 7. z 2 − 7 2 = ( z + 7)( z − 7)

8 x+2

8. (x + 2)(x − 3) 9. ( p + 3)( p + 1)

3 x − 12

54. 2 x + 3 6 x 2 + 8 x + 1

10. (s – 4)(s – 2)

6x + 9x −x +1 − x − 32 2

11. (4 x)2 − 32 = (4 x + 3)(4 x − 3) 12. (x + 6)(x – 4)

5 2

13. (a + 7)(a + 5)

5

1 Answer: 3x − + 2 2 2x + 3

14. (2t )2 − (3s )2 = (2t + 3s )(2t − 3s )

x−2 55. 3x + 2 3 x 2 − 4 x + 3

15. x 2 + 2(3)( x) + 32 = ( x + 3) 2 16. (y – 10)(y – 5)

3x2 + 2 x –6 x + 3 –6 x − 4 7

Answer: x − 2 +

17. 5( x 2 + 5 x + 6) = 5( x + 3)( x + 2)

7 3x + 2

18. 3(t 2 + 4t − 5) = 3(t − 1)(t + 5) 9



35. y 2 ( y 2 + 8 y + 16) − ( y 2 + 8 y + 16)

19. 3( x 2 − 12 ) = 3( x + 1)( x − 1)

= ( y 2 + 8 y + 16)( y 2 − 1)

20. (3y − 4)(3y − 2)

= ( y + 4)2 ( y + 1)( y − 1)

21. 6 y 2 + 13 y + 2 = (6 y + 1)( y + 2) 36. xy ( x 2 − 4) + z 2 ( x 2 − 4) = ( x 2 − 4)( xy + z 2 )

22. (4x + 3)(x – 1)

= ( x + 2)( x − 2)( xy + z 2 )

23. 2s (6 s 2 + 5s − 4) = 2 s (3s + 4)(2s − 1)

37. b3 + 43 = (b + 4)(b 2 − 4(b) + 42 ) = (b + 4)(b 2 − 4b + 16)

24. (3z ) 2 + 2(3 z )(5) + 52 = (3 z + 5) 2

38. x3 − 13 = ( x − 1)[ x 2 + 1( x) + 12 ]

25. u 3 / 5 v(u 2 − 4v 2 ) = u 3 / 5 v(u + 2v)(u − 2v)

= ( x − 1)( x 2 + x + 1)

26. (3x

) − 1 = (3 x

2/7 2

2

2/7

+ 1)(3x

2/7

− 1)

39. ( x3 )2 − 12 = ( x3 + 1)( x3 − 1) 27. 2 x( x + x − 6) = 2 x( x + 3)( x − 2) 2

= ( x + 1)( x 2 − x + 1)( x − 1)( x 2 + x + 1)

28. ( xy ) 2 − 2( xy )(2) + 22 = ( xy − 2)2

40. 33 + (2 x)3 = (3 + 2 x)[32 − 3(2 x) + (2 x)2 ] = (3 + 2 x)(9 − 6 x + 4 x 2 )

29. [2(2 x + 1)]2 = 22 (2 x + 1)2 = 4(2 x + 1)2

41. ( x + 3) 2 ( x − 1)[( x + 3) + ( x − 1)] = ( x + 3)2 ( x − 1)[2 x + 2]

30. 2 x 2 [2 x(1 − 2 x)]2 = 2 x 2 (2 x)2 (1 − 2 x) 2

= ( x + 3)2 ( x − 1)[2( x + 1)]

= 2 x 2 (4 x 2 )(1 − 2 x)2

= 2( x + 3)2 ( x − 1)( x + 1)

= 8 x 4 (1 − 2 x)2

42. (a + 5)2 (a + 1) 2 [(a + 5) + (a + 1)] 31. x( x 2 y 2 − 14 xy + 49) = x[( xy )2 − 2( xy )(7) + 7 2 ]

= (a + 5)2 (a + 1)2 (2a + 6) = 2(a + 5)2 (a + 1) 2 (a + 3)

= x( xy − 7)2

43. [P(1 + r)] + [P(1 + r)]r = [P(1 + r)](1 + r)

32. x(5x + 2) + 2(5x + 2) = (5x + 2)(x + 2)

= P (1 + r )2

33. x( x − 4) + 2(4 − x ) 2

2

44. (3 X + 5 I )[( X − 3I ) − ( X + 2 I )] = (3 X + 5I )(−5I ) = −5I (3 X + 5I )

= x( x 2 − 4) − 2( x 2 − 4) = ( x 2 − 4)( x − 2) = (x + 2)(x – 2)(x – 2) = ( x + 2)( x − 2)

45. ( x 2 )2 − 42 = ( x 2 + 4)( x 2 − 4)

2

= ( x 2 + 4)( x + 2)( x − 2)

34. (x + 1)(x – 1) + (x – 2)(x + 1) = (x + 1)[(x – 1) + (x – 2)] = (x + 1)(2x – 3)

46. (9 x 2 )2 − ( y 2 )2 = (9 x 2 + y 2 )(9 x 2 − y 2 ) = (9 x 2 + y 2 )(3 x + y )(3 x − y )

10


Section 0.6

47. ( y 4 ) 2 − 12 = ( y 4 + 1)( y 4 − 1)

8.

= ( y 4 + 1)( y 2 + 1)( y 2 − 1) = ( y 4 + 1)( y 2 + 1)( y + 1)( y − 1)

9.

48. (t 2 ) 2 − 22 = (t 2 + 2)(t 2 − 2) ⎡ = (t 2 + 2) ⎢t 2 − ⎣

(

= (t 2 + 2) t +

( 2 ) ⎤⎥⎦ 2 )( t − 2 ) 2

10.

49. ( X + 5)( X − 1) = ( X + 5)( X + 1)( X − 1) 2

2

2

(t + 3)(t − 3) t 2 t (t + 3)(t − 3)

11.

51. y ( x 4 − 2 x 2 + 1) = y ( x 2 − 1)2 = y[( x + 1)( x − 1)]2 = y ( x + 1)2 ( x − 1) 2

52. 2 x(2 x 2 − 3 x − 2) = 2 x(2 x + 1)( x − 2)

2.

3.

4.

5.

6.

7.

a 2 − 3a

=

(a − 3)(a + 3) a + 3 = a(a − 3) a

x − 3x − 10 2

x −4 2

x 2 − 9 x + 20 x + x − 20 2

=

=

( x − 5)( x − 4) x − 5 = ( x + 5)( x − 4) x + 5

2 x − 16 x + 14 x 3( x − 8) = 2 x( x − 7) 3

2

6 x2 + x − 2 2 x + 3x − 2 2

=

12 x − 19 x + 4 2

6 x − 17 x + 12 2

12.

( x + 2)( x − 5) x − 5 = ( x + 2)( x − 2) x − 2

3 x 2 − 27 x + 24

=

3( x − 8)( x − 1) 2 x( x − 7)( x − 1)

(3x + 2)(2 x − 1) 3x + 2 = ( x + 2)(2 x − 1) x+2 =

t t −3

(ax − b)(c − x) (ax − b)(−1)( x − c) = ( x − c)(ax + b) ( x − c)(ax + b) (ax − b)(−1) = ax + b b − ax = ax + b ( x + y )( x − y )( x + y )2 ( x − y )( x + y ) 2 = ( x + y )( y − x) (–1)( x − y )

2( x − 1) ( x + 4)( x + 1) ⋅ ( x − 4)( x + 2) ( x + 1)( x − 1) =

2( x − 1)( x + 4)( x + 1) ( x − 4)( x + 2)( x + 1)( x − 1)

=

2( x + 4) ( x − 4)( x + 2)

Problems 0.6 a2 − 9

=

= −( x + y ) 2

50. ( x 2 − 9)( x 2 − 1) = ( x + 3)( x − 3)( x + 1)( x − 1)

1.

2

(4 x − 1)(3x − 4) 4 x − 1 = (2 x − 3)(3x − 4) 2 x − 3

=

x( x + 2)( x − 2)2 3( x − 4)( x − 2)( x − 3)( x + 2)

=

x( x − 2) 3( x − 4)( x − 3)

13.

X 2 4 4X 2 X ⋅ = = 8 X 8X 2

14.

3x 2 14 3 x 14 3(14) x ⋅ = ⋅ = =6 7x x 7 x 7x

15.

2m n3 2mn3 n ⋅ = = n 2 6m 6mn 2 3

16.

c + d 2c 2c(c + d ) 2(c + d ) ⋅ = = c c−d c (c − d ) c−d

17.

4x 4x 1 4x 2 ÷ 2x = ⋅ = = 3 3 2x 6x 3

18.

4 x 2 x 4 x(2 x) 8 x 2 ⋅ = = 1 3 3 3

y 2 (–1) y2 =− ( y − 3)( y + 2) ( y − 3)( y + 2)

11

x ( x + 2) ( x − 2)2 ⋅ 3( x − 4)( x − 2) ( x − 3)( x + 2)


19.


–9 x3 3 –27 x3 ⋅ = = −27 x 2 1 x x

27.

−12Y 4 −12Y 3 1 −12Y 3 ÷4 = ⋅ = = −3Y 3 20. Y 1 4 4

21.

22.

23.

24.

25.

26.

x −3 x−4 x −3 1 x−3 ⋅ = ⋅ = =1 1 ( x − 3)( x − 4) 1 x−3 x−3

28.

(2 x + 3)(2 x − 3)(1 + x)(1 − x) ( x + 4)( x − 1)(2 x − 3)

=

(2 x + 3)(1 + x)(–1)( x − 1) ( x + 4)( x − 1)

2

=

( x − 3)( x + 2) ( x + 3)( x − 1) ⋅ ( x + 3)( x − 3) ( x + 2)( x − 2) x + 2 ( x + 3)( x − 1) = ⋅ x + 3 ( x + 2)( x − 2) ( x + 2)( x + 3)( x − 1) = ( x + 3)( x + 2)( x − 2) x −1 = x−2

(2 x + 3)(1 + x) x+4

y (6 x 2 + 7 x − 3) x( y − 1) + 4( y − 1) ⋅ x( y – 1) + 5( y − 1) x 2 y ( x + 4) =

10 x x +1 10 x ( x + 1) 2x ⋅ = = ( x + 1)( x − 1) 5 x 5 x( x + 1)( x − 1) x − 1 3

=

=−

( x + 3) 2 ( x + 3)2 1 ÷ ( x + 3) = ⋅ x x x+3 2 ( x + 3) x+3 = = x ( x + 3) x 3

(2 x + 3)(2 x − 3) (1 + x)(1 − x) ⋅ ( x + 4)( x − 1) 2x − 3

y (3 x − 1)(2 x + 3)( y − 1)( x + 4) ( y − 1)( x + 5) x 2 y ( x + 4) (3 x − 1)(2 x + 3) x 2 ( x + 5)

29.

x 2 + 5 x + 6 ( x + 3)( x + 2) = = x+2 x+3 x+3

30.

2+ x x+2 = =1 x+2 x+2

31. LCD = 3t 2 1 6 1 6 +1 7 + = + = = t 3t 3t 3t 3t 3t

( x + 2)( x + 5) ( x − 4)( x + 1) ⋅ ( x + 5)( x + 1) ( x − 4)( x + 2) x + 2 x +1 = ⋅ x +1 x + 2 ( x + 2)( x + 1) = ( x + 1)( x + 2) =1

32. LCD = X 3 9 1 9 X 9− X − = − = 3 2 3 3 X X X X X3 33. LCD = x3 − 1 1−

( x + 3) 2 (3 + 4 x)(3 − 4 x) ⋅ 4x − 3 7( x + 3)

x3 x3 − 1

= =

( x + 3)2 (3 + 4 x)(3 − 4 x) = 7(4 x − 3)( x + 3) ( x + 3)(3 + 4 x)(−1)(4 x − 3) = 7(4 x − 3) ( x + 3)(3 + 4 x) =− 7

= =

x3 − 1

−

x3

x3 − 1 x3 − 1 x 3 − 1 − x3 x3 − 1 −1 x3 − 1 1 1 − x3

34. LCD = s + 4 4 4 s ( s + 4) 4 + s ( s + 4) +s= + = s+4 s+4 s+4 s+4 = 12

s 2 + 4s + 4 ( s + 2)2 = s+4 s+4


Section 0.6

35. LCD = (2x – 1)(x + 3) 4 x 4( x + 3) x(2 x − 1) + = + 2 x − 1 x + 3 (2 x − 1)( x + 3) ( x + 3)(2 x − 1) =

39. LCD = (x – 1)(x + 5) −3x 2 4 −3+ x −1 −( x − 1)( x + 5)

4( x + 3) + x(2 x − 1) 2 x 2 + 3x + 12 = (2 x − 1)( x + 3) (2 x − 1)( x + 3)

36. LCD = (x – 1)(x + 1) x + 1 x − 1 ( x + 1)( x + 1) ( x − 1)( x − 1) − = − x − 1 x + 1 ( x − 1)( x + 1) ( x − 1)( x + 1) =

( x + 1) − ( x − 1) ( x + 1)( x − 1) 2

4( x + 5) 3( x − 1)( x + 5) 3x 2 − + ( x − 1)( x + 5) ( x − 1)( x + 5) ( x − 1)( x + 5)

=

4 x + 20 − 3( x 2 + 4 x − 5) + 3x 2 ( x − 1)( x + 5)

=

35 − 8 x ( x − 1)( x + 5)

2

40. LCD = (2x – 1)(x + 6)(3x – 2) 2x − 3 3x + 1 1 − + (2 x − 1)( x + 6) (3x − 2)( x + 6) 3 x − 2

x 2 + 2 x + 1 − ( x 2 − 2 x + 1) 4x = = ( x + 1)( x − 1) ( x + 1)( x − 1)

37. LCD = ( x − 3)( x + 1)( x + 3) 1 1 + ( x − 3)( x + 1) ( x + 3)( x − 3) x+3 x +1 = + ( x − 3)( x + 1)( x + 3) ( x − 3)( x + 1)( x + 3) ( x + 3) + ( x + 1) = ( x − 3)( x + 1)( x + 3) 2x + 4 = ( x − 3)( x + 1)( x + 3) 2( x + 2) = ( x − 3)( x + 1)( x + 3)

=

(2 x − 3)(3 x − 2) − (3 x + 1)(2 x − 1) + (2 x − 1)( x + 6) (2 x − 1)( x + 6)(3 x − 2)

=

6 x 2 − 13x + 6 − (6 x 2 − x − 1) + 2 x 2 + 11x − 6 (2 x − 1)( x + 6)(3 x − 2)

=

2x2 − x + 1 (2 x − 1)( x + 6)(3 x − 2) 2

2

2

x2 + 2 x + 1 ⎛ 1⎞ ⎛ x 1⎞ ⎛ x +1⎞ 41. ⎜1 + ⎟ = ⎜ + ⎟ = ⎜ ⎟ = ⎝ x⎠ ⎝ x x⎠ ⎝ x ⎠ x2 2

2

2

–1

⎛ 1 − xy ⎞ =⎜ ⎟ ⎝ x ⎠

⎛1 1⎞ ⎛ y ⎛ y+x⎞ x ⎞ 42. ⎜ + ⎟ = ⎜ + ⎟ = ⎜ ⎟ ⎝x y⎠ ⎝ xy xy ⎠ ⎝ xy ⎠

38. LCD = (x − 4)(2x + 1)(2x − 1) 4 x − ( x − 4)(2 x + 1) ( x − 4)(2 x − 1) 4(2 x − 1) x(2 x + 1) = − ( x − 4)(2 x + 1)(2 x − 1) ( x − 4)(2 x + 1)(2 x − 1) 4(2 x − 1) − x(2 x + 1) = ( x − 4)(2 x + 1)(2 x − 1) =

=

=

y 2 + 2 xy + x 2 x2 y 2 –1

⎛1 ⎞ ⎛ 1 xy ⎞ 43. ⎜ − y ⎟ `= ⎜ − ⎟ ⎝x ⎠ ⎝x x ⎠

−2 x 2 + 7 x − 4 ( x − 4)(2 x + 1)(2 x − 1)

2

2

1⎞ ⎛ ⎛ ab 1 ⎞ ⎛ ab + 1 ⎞ 44. ⎜ a + ⎟ = ⎜ + ⎟ = ⎜ ⎟ b⎠ ⎝ ⎝ b b⎠ ⎝ b ⎠ a 2b 2 + 2ab + 1 = b2

–1

=

x 1 − xy

2

45. Multiplying the numerator and denominator of 7x +1 the given fraction by x gives . 5x 46. Multiplying numerator and denominator by x x+3 x+3 1 = = gives . 2 + − −3 ( x 3)( x 3) x x −9 13



47. Multiplying numerator and denominator by 2x(x + 2) gives 3(2 x)( x + 2) − 1( x + 2) ( x + 2)[3(2 x) − 1] = x(2 x)( x + 2) + x(2 x) 2 x 2 [( x + 2) + 1] =

( x + 2)(6 x − 1) 2 x 2 ( x + 3)

.

56. = =

−12 12 = =− . ( x + 3)( x + 2)[9 + ( x − 7)] ( x + 3)( x + 2)2

3

x+h

3 3

=

x =

33 x

5+ a

1

+

a

= =

51.

1

2− 3

⋅

2+ 3 2− 3

57.

x+h x x+h x 3 3 3 x − x+h 3

3

(

3

3

)

x + h3 x

3

50. LCD = 5 + a a a a

33 x + h

−

a a

58.

( a ) + 1(

5+a a a + 5+a

5+a

)

59.

5+ a a

2

a 5+a =

2+ 3

⋅

=

2+ 3

2 2

(

2+ 3

)

2−3

4+2 6 = −4 − 2 6 −1 2 5 3+ 7 ⋅ 3− 7 3+ 7 2 5 3+ 7

(

2

)

3−7 15 + 35

(

)

−4 15 + 35 =− 2

49. LCD = 3 x + h ⋅ 3 x −

2− 3 =

48. Multiplying numerator and denominator by 3(x + 3)(x + 2) gives 3( x − 1) − 1(3)( x + 3) 3(3)( x + 3)( x + 2) + ( x − 7)( x + 3)( x + 2)

3

2 2

55.

2− 3 = 2− 3 4−3

3

⋅

t− 7

t+ 7 t− 7 ( x − 3) + 4 x −1

(

3t − 3 7

t2 − 7

x +1

=

5 2− 3

=

x −1

)

x +1

⋅

x +1

=

( x + 1)

)

x +1

x −1

(

4 1+ 2

−

(

)

( 2 + 3 )( 2 − 3 ) (1 − 2 )(1 + 2 ) 5 ( 2 − 3 ) 4 (1 + 2 ) = − =

(

4−3

5 2− 3

(

1

1− 2

) − 4 (1 + 2 ) ) (

−1

)

= 5 2 − 3 + 4 1 + 2 = 4 2 − 5 3 + 14

1+ 2

1+ 2 1+ 2 52. ⋅ = = = −1 − 2 1− 2 −1 1− 2 1+ 2 1

60. 2

53.

3− 6 2

=

(

6+ 7 =

3+ 6

3−6

5

(

⋅

)=

6 + 12 6+2 3 =− −3 3

6− 7 6− 7

6− 7 −1

3

3+ 6

3+ 6

5

54.

⋅

) =5

(

=

5

(

6− 7

)

6−7

7− 6

)

14

(

4 x2 x +2

=

) 3(

4x2

(

x +2

x −2

)(

)

x −2

)

=

4 x2

(

x −2

3( x − 4)

)


Section 0.7 Set x = –4:

Problems 0.7

2(–4) + (–4)2 − 8 ⱨ 0 –8 + 16 – 8 ⱨ 0 0=0 Thus, 2 and –4 satisfy the equation.

2

1. 9 x − x = 0 Set x = 1: 9(1) − (1) 2 ⱨ 0 9 −1 ⱨ 0 8≠0 Set x = 0:

5. x(6 + x) – 2(x + 1) – 5x = 4 Set x = –2: (–2)(6 – 2) – 2(–2 + 1) – 5(–2) ⱨ 4 –2(4) – 2(–1) + 10 ⱨ 4 –8 + 2 + 10 ⱨ 4 4=4 Set x = 0: 0(6) – 2(1) – 5(0) ⱨ 4 –2 ≠ 4 Thus, –2 satisfies the equation, but 0 does not.

9(0) − (0)2 ⱨ 0 0–0ⱨ0 0=0 Thus, 0 satisfies the equation, but 1 does not.

2. 12 − 7 x = − x 2 ; 4, 3 Set x = 4: 12 − 7(4) ⱨ − (4) 2 12 − 28 ⱨ − 16 −16 = −16 Set x = 3:

6. x( x + 1)2 ( x + 2) = 0 Set x = 0: 0(1) 2 (2) ⱨ 0 0=0 Set x = –1:

12 − 7(3) ⱨ − (3)2 12 − 21 ⱨ − 9 −9 = −9 Thus, 4 and 3 satisfy the equation.

3. z + 3( z − 4) = 5;

(–1)(0)2 (1) ⱨ 0 0=0 Set x = 2: 2(3)2 (4) ⱨ 0 72 ≠ 0 Thus, 0 and –1 satisfy the equation, but 2 does not.

17 ,4 4

17 : 4 17 ⎛ 17 ⎞ + 3⎜ − 4 ⎟ ⱨ 5 4 ⎝ 4 ⎠ 17 51 + − 12 ⱨ 5 4 4 5=5 Set z = 4: 4 + 3(4 − 4) ⱨ 5 4+0ⱨ5 4≠5 17 Thus, satisfies the equation, but 4 does not. 4

Set z =

7. Adding 5 to both sides; equivalence guaranteed 8. Dividing both sides by 8; equivalence guaranteed 9. Raising both sides to the third power; equivalence not guaranteed. 10. Dividing both sides by 2; equivalence guaranteed 11. Dividing both sides by x; equivalence not guaranteed

4. 2 x + x 2 − 8 = 0 Set x = 2:

12. Multiplying both sides by x – 2; equivalence not guaranteed

2 ⋅ 2 + 22 − 8 ⱨ 0 4+4–8ⱨ0 0=0

13. Multiplying both sides by x – 1; equivalence not guaranteed 14. Dividing both sides by (x + 3); equivalence not guaranteed.

15


15. Multiplying both sides by

ISM: Introductory Mathematical Analysis 26. 4s + 3s − 1 = 41 7 s − 1 = 41 7 s = 42 42 s= =6 7

2x − 3 ; equivalence 2x

not guaranteed 16. Adding 9 – x to both sides and then dividing both sides by 2; equivalence guaranteed

27. 5( p − 7) − 2(3 p − 4) = 3 p 5 p − 35 − 6 p + 8 = 3 p − p − 27 = 3 p −27 = 4 p 27 p=− 4

17. 4x = 10 10 5 x= = 4 2 18. 0.2x = 7 7 x= = 35 0.2

28. t = 2 – 2[2t – 3(1 – t)] t = 2 – 2[2t – 3 + 3t] t = 2 – 2[5t – 3] t = 2 – 10t + 6 11t = 8 8 t= 11

19. 3y = 0 0 y= =0 3 20. 2x – 4x = –5 –2x = –5 –5 5 x= = –2 2

29.

21. −8 x = 12 − 20 −8 x = −8 −8 x= =1 −8 22. 4 − 7 x = 3 −7 x = −1 −1 1 = x= −7 7

30.

23. 5x – 3 = 9 5x = 12 12 x= 5 24.

x = 2x − 6 5 x = 5(2x – 6) x = 10x – 30 30 = 9x 30 10 x= = 9 3 5y 6 − = 2 − 4y 7 7 5y – 6 = 14 – 28y 33y = 20 20 y= 33 4x x = 9 2 Multiplying both sides by 9 · 2 gives 9 · 2 · 7 + 2(4x) = 9(x) 126 + 8x = 9x x = 126

31. 7 +

2x + 3 = 8 2x = 5 5 ⎛ 5 2⎞ x= ⎜ or ⎟ 2 ⎟⎠ 2 ⎜⎝

32.

25. 7x + 7 = 2(x + 1) 7x + 7 = 2x + 2 5x + 7 = 2 5x = –5 –5 x= = −1 5 16

x x −4 = 3 5 5x – 60 = 3x 2x = 60 x = 30


Section 0.7

w w w + − = 120 2 6 24 Multiplying both sides by 24 gives 24 w − 12 w + 4 w − w = 2880 15w = 2880 2880 w= = 192 15

4 r −5 3 Multiplying both sides by 3 gives 3r = 4r − 15 −r = −15 r = 15

39. w −

33. r =

34.

3x 5 x + =9 5 3 9 x + 25 x = 135 34 x = 135 135 x= 34

40.

x 1 − 5 = + 5x 5 5 Multiplying both sides by 5 gives 15x + x – 25 = 1 + 25x 16x – 25 = 1 + 25x –9x = 26 26 x=− 9

35. 3x +

41.

y y y y + − = 2 3 4 5 60y – 30y + 20y – 15y = 12y 35y = 12y 23y = 0 y=0

36. y −

37.

38.

42.

2y − 3 6y + 7 = 4 3 Multiplying both sides by 12 gives 3(2y – 3) = 4(6y + 7) 6y – 9 = 24y + 28 –18y = 37 37 y=− 18

43.

t 5 7 + t = (t − 1) 4 3 2 Multiplying both sides by 12 gives 3t + 20t = 42(t − 1) 23t = 42t − 42 42 = 19t 42 t= 19

44.

17

7 + 2( x + 1) 6 x = 3 5 35 + 10(x + 1) = 18x 35 + 10x + 10 = 18x 45 = 8x 45 x= 8 x+2 2− x − = x−2 3 6 Multiplying both sides by 6 gives 2(x + 2) – (2 – x) = 6(x – 2) 2x + 4 – 2 + x = 6x – 12 3x + 2 = 6x – 12 2 = 3x – 12 14 = 3x 14 x= 3 x 2( x − 4) + =7 5 10 2x + 2(x – 4) = 70 2x + 2x – 8 = 70 4x = 78 78 39 x= = 4 2 9 3 (3 − x) = ( x − 3) 5 4 Multiplying both sides by 20 gives 36(3 – x) = 15(x – 3) 108 – 36x = 15x – 45 153 = 51x x=3 2 y − 7 8y − 9 3y − 5 + = 3 14 21 14(2y – 7) + 3(8y – 9) = 2(3y – 5) 28y – 98 + 24y – 27 = 6y – 10 46y = 115 115 5 = y= 46 2


45.

46.


4 (5 x − 2) = 7[ x − (5 x − 2)] 3 4(5 x − 2) = 21( x − 5 x + 2) 20 x − 8 = −84 x + 42 104 x = 50 50 25 x= = 104 52

52.

53.

(2 x − 5)2 + (3x − 3) 2 = 13x 2 − 5 x + 7 4 x 2 − 20 x + 25 + 9 x 2 − 18 x + 9 = 13x 2 − 5 x + 7 13x 2 − 38 x + 34 = 13x 2 − 5 x + 7 −33 x = −27 −27 9 x= = −33 11

47.

48.

54.

5 = 25 x Multiplying both sides by x gives 5 = 25x 5 x= 25 1 x= 5

55.

4 =2 x −1 4 = 2(x – 1) 4 = 2x – 2 6 = 2x x=3

56.

49. Multiplying both sides by 3 – x gives 7 = 0, which is false. Thus there is no solution, so the solution set is ∅. 50.

57.

3x − 5 =0 x −3 3x − 5 = 0 3x = 5 5 x= 3

58.

3 7 51. = 5 − 2x 2 3(2) = 7(5 − 2 x ) 6 = 35 − 14 x 14 x = 29 29 x= 14 18

x+3 2 = x 5 5(x + 3) = 2x 5x + 15 = 2x 3x = –15 x = –5 q 1 = 5q − 4 3 3q = 5q – 4 –2q = –4 q=2 4p =1 7− p 4p = 7 – p 5p = 7 7 p= 5 1 2 = p −1 p − 2 p – 2 = 2(p – 1) p – 2 = 2p – 2 p=0 2x − 3 =6 4x − 5 2x – 3 = 24x – 30 27 = 22x 27 x= 22

1 1 3 + = x 7 7 1 3 1 = − x 7 7 1 2 = x 7 7 x= 2 2 3 = x −1 x − 2 2( x − 2) = 3( x − 1) 2 x − 4 = 3x − 3 −x = 1 x = −1


59.

Section 0.7

3x − 2 3x − 1 = 2x + 3 2x +1 (3x – 2)(2x + 1) = (3x – 1)(2x + 3)

64.

6 x2 − x − 2 = 6 x2 + 7 x − 3 1 = 8x 1 x= 8

60.

−2 x 2 + 5 x − 2 − 3(–2 x 2 + 7 x − 3) = 4( x 2 − 5 x + 6) 4 x 2 − 16 x + 7 = 4 x 2 − 20 x + 24 4x = 17 17 x= 4

x + 2 x +1 + =0 x −1 3 − x (x + 2)(3 – x) + (x + 1)(x – 1) = 0

65.

3x − x 2 + 6 − 2 x + x 2 − 1 = 0 x+5=0 x = –5

61.

y−6 6 y+6 − = y y y−6

Multiplying both sides by y(y − 6) gives

66.

( y − 6)2 − 6( y − 6) = y ( y + 6) y 2 − 12 y + 36 − 6 y + 36 = y 2 + 6 y

y−2 y−2 = y+2 y+3

x+5 = 4

(

2

y + y−6 = y −4 y=2

63.

x x 3x − 4 − = x + 3 x − 3 x2 − 9 x(x – 3) – x(x + 3) = 3x – 4

67.

(y − 2)(y + 3) = (y − 2)(y + 2) 2

9 3x = x −3 x −3 9 = 3x x=3 But the given equation is not defined for x = 3, so there is no solution. The solution set is ∅.

x 2 − 3x − x 2 − 3x = 3x − 4 –6x = 3x – 4 –9x = –4 4 x= 9

y 2 − 18 y + 72 = y 2 + 6 y 72 = 24y y=3

62.

1 3 4 − = x − 3 x − 2 1− 2x (x – 2)(1 – 2x) – 3(x – 3)(1 – 2x) = 4(x – 3)(x – 2)

x+5

)

2

= 42

x + 5 = 16 x = 11

−5 7 11 = + 2 x − 3 3 − 2 x 3x + 5 Multiplying both sides by (2x − 3)(3x + 5) gives −5(3 x + 5) = −7(3 x + 5) + 11(2 x − 3) −15 x − 25 = −21x − 35 + 22 x − 33 −15 x − 25 = x − 68 −16 x = −43 43 x= 16

z−2 =3

68.

(

z−2

)

2

= 32

z–2=9 z = 11 3x − 4 − 8 = 0 3x − 4 = 8

69.

(

3x − 4

)

2

3x − 4 = 64 3x = 68 68 x= 3

19

= (8)2



70. 4 − 3x + 1 = 0 4 = 3x + 1 2

4 =

(

3x + 1

)

2

⎛ y 2 − 9 ⎞ = (9 − y ) 2 ⎜ ⎟ ⎝ ⎠

2

16 = 3 x + 1 15 = 3 x x=5

y 2 − 9 = 81 − 18 y + y 2 18y = 90 90 y= =5 18

x 2 +1 = 2 3

71.

2

⎛ x ⎞ ⎛2⎞ + 1 ⎟⎟ = ⎜ ⎟ ⎜⎜ ⎝3⎠ ⎝ 2 ⎠ x 4 +1 = 2 9 x 5 =− 2 9 10 ⎛ 5⎞ x = 2⎜ − ⎟ = − 9 ⎝ 9⎠

y + 2 = 3− y

(

(6 y )

(

) ( ) x

78.

4 + 3x

= 72

x − x +1 = 1 2

)

x +1 +1

2

x = x +1+ 2 x +1 +1

2

−2 = 2 x + 1 −1 = x + 1 , which is impossible because a ≥ 0 for all a. Thus there is no solution. The solution set is ∅.

) =( 2

2x + 5

)

79.

2

z2 + 2z = 3 + z 2

⎛ z 2 + 2 z ⎞ = (3 + z )2 ⎜ ⎟ ⎝ ⎠

4 + 3x = 2 x + 5 x =1

75.

2

( x) = (

4 + 3x = 2 x + 5

(

2

x = x +1 +1

4x – 6 = x 3x = 6 x=2 74.

2

36y = 49 49 y= 36

4x − 6 = x =

) = (3 − y )

6 y =7

[( x + 6)1/ 2 ]2 = 7 2 x + 6 = 49 x = 43

2

y+2

y+2 = 9−6 y + y

72. ( x + 6)1/ 2 = 7

4x − 6

y + y+2 =3

77.

2

73.

y2 − 9 = 9 − y

76.

z2 + 2z = 9 + 6z + z2 –9 = 4z 9 z=− 4

( x − 5)3 / 4 = 27 [( x − 5)3 / 4 ]4 / 3 = 27 4 / 3 x − 5 = 81 x = 86

20


80.

Section 0.7

1 2 − =0 w 5w − 2

87. A =

1 2 = w 5w − 2

R=

2

⎛ 1⎞ ⎛ 2 ⎞ ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟ ⎝ w⎠ ⎝ 5w − 2 ⎠ 1 2 = w 5w − 2 5w – 2 = 2w 3w = 2 2 w= 3

P=

1 − (1 + i )− n

2

88. S =

R[(1 + i )n − 1] i

Si = R[(1 + i ) n − 1] R=

Si (1 + i ) n − 1

d 1 − dt r(1 – dt) = d r – rdt = d –rdt = –r + d d −r r −d t=− = rd rd

89. r =

81. I = Prt I r= Pt p ⎞ ⎛ 82. P ⎜ 1 + ⎟−R = 0 ⎝ 100 ⎠ p ⎞ ⎛ P ⎜1 + ⎟=R ⎝ 100 ⎠

R[1 − (1 + i ) − n ] i Ai

90. R p

1 + 100

83. p = 8q – 1 p + 1 = 8q p +1 q= 8

x −a x −b = b−x a−x Multiplying both sides by (b – x)(a – x) gives (x – a)(a – x) = (x – b)(b – x) (x – a)(a – x)(–1) = (x – b)(b – x)(–1) (x – a)(x – a) = (x – b)(x – b) x 2 − 2ax + a 2 = x 2 − 2bx + b 2 a 2 − b 2 = 2ax − 2bx (a + b)(a – b) = 2x(a – b) a + b = 2x (for a ≠ b) a+b =x 2

84. p = –3q + 6 p – 6 = –3q p−6 6− p q= = −3 3

91. r =

85. S = P(1 + rt) S = P + Prt S – P = r(Pt) S–P r= Pt

2mI B (n + 1)

2mI B 2mI n +1 = rB 2mI −1 n= rB r (n + 1) =

2mI B (n + 1) r[ B (n + 1)] =I 2m rB (n + 1) I= 2m

86. r =

21


92.


1 1 1 + = p q f

98.

1 1 1 = − q f p 1 p− f = q pf q=

pf p− f

93. P = 2l + 2 w 660 = 2l + 2(160) 660 = 2l + 320 340 = 2l 340 l= = 170 2 The length of the rectangle is 170 m. 94.

vf 334.8 v(2500) 495 = 334.8 165, 726 = 2500v 165, 726 = 66.2904 v= 2500 Since the car is traveling at 66.2904 mi/h on a 65 mi/h highway, the officer can claim that you were speeding. F=

99. Bronwyn’s weekly salary for working h hours is 27h + 18. Steve’s weekly salary for working h hours is 35h. 1 (27h + 18 + 35h) = 550 5 62h + 18 = 2750 62h = 2732 2732 h= ≈ 44.1 62 They must each work 44 hours each week.

V = πr 2 h 355 = π(2) 2 h 355 = 4πh 355 h= 4π The height of the can is 355 ≈ 28.25 centimeters. 4π

100. y = a(1 – by)x y = ax(1 – by) y = ax – abxy y + abxy = ax y(1 + abx) = ax ax y= 1 + abx

95. c = x + 0.0825x = 1.0825x

1.4 x 1 + 0.09 x With y = 10 the equation is 1.4 x 10 = 1 + 0.09 x 10(1 + 0.09x) = 1.4x 10 + 0.9x = 1.4x 10 = 0.5x x = 20 The prey density should be 20.

101. y =

96. Revenue equals cost when 450x = 380x + 3500. 450x = 380x + 3500 70x = 3500 x = 50 50 toddlers need to be enrolled. n⎞ ⎛ 97. V = C ⎜ 1 − ⎟ ⎝ N⎠ ⎛ n⎞ 2000 = 3200 ⎜ 1 − ⎟ ⎝ 8⎠ 2000 = 3200 – 400n 400n = 1200 n=3 The furniture will have a value of $2000 after 3 years.

102. Let x = the maximum number of customers. 8 10 = x − 92 x − 46 8(x − 46) = 10(x − 92) 8x − 368 = 10x − 920 552 = 2x x = 276 The maximum number of customers is 276.

22


Section 0.8

d r −c t(r – c) = d tr – tc = d tr − d = tc tr − d d c= =r− t t

109. −

103. t =

110.

Problems 0.8 1. x 2 − 4 x + 4 = 0

theorem, x 2 + 1002 = ( x + 1)2 .

( x − 2)2 = 0 x–2=0 x=2

x 2 + 10, 000 = x 2 + 2 x + 1 10, 000 = 2 x + 1 9999 = 2 x 9999 x= = 4999.5 2 The distance from the top of the tower to the house is x + 1 = 4999.5 + 1 = 5000.5 meters.

45 = 24d

(

24d

)

2

2025 = 24d 2025 675 3 d= = = 84 ≈ 84 ft 24 8 8

106. Let P be the amount in the account one year ago. Then the interest earned is 0.073P and P + 0.073P = 1257. 1.073P = 1257 1257 P= ≈ 1171.48 1.073 The amount in the account one year ago was $1171.48, and the interest earned is $1171.48(0.073) = $85.52.

or t + 2 = 0 or t = –2

t 2 − 8t + 15 = 0 (t − 3)(t − 5) = 0 t−3=0 t=3

or t − 5 = 0 or t = 5

4. (x – 2)(x + 5) = 0 x–2=0 x=2

or x + 5 = 0 or x = –5

5. x 2 − 2 x − 3 = 0 (x – 3)(x + 1) = 0 x–3=0 x=3

or x + 1 = 0 or x = –1

6. (x – 4)(x + 4) = 0 x–4=0 x=4

or x + 4 = 0 or x = –4

7. u 2 − 13u = −36 u 2 − 13u + 36 = 0 (u − 4)(u − 9) = 0 u–4=0 u=4

107. Let e be Tom’s expenses in Nova Scotia before the HST tax. Then the HST tax is 0.15e and the total receipts are e + 0.15e = 1.15e. The percentage of the total that is HST is 0.15e 0.15 15 3 = = = or approximately 1.15e 1.15 115 23 13%. 108.

2. (t + 1)(t + 2) = 0 t+1=0 t = –1 3.

105. s = 30 fd Set s = 45 and (for dry concrete) f = 0.8. 45 = 30(0.8)d (45) =

14 is a root. 61

111. 0 is a root.

104. Let x = the horizontal distance from the base of the tower to the house. By the Pythagorean

2

1 is a root. 2

8. 3( w2 − 4 w + 4) = 0 3( w − 2)2 = 0 w–2=0 w=2

1 1 and − are roots. 8 14

23

or u – 9 = 0 or u = 9


9. x 2 − 4 = 0 (x – 2)(x + 2) = 0 x–2=0 x=2

17. − x 2 + 3 x + 10 = 0 or x + 2 = 0 or x = –2

10. 3u (u − 2) = 0 u=0 u=0

or u − 2 = 0 or u = 2

11. t 2 − 5t = 0 t (t − 5) = 0 t=0 t=0

or t – 5 = 0 or t = 5

12. x 2 + 9 x + 14 = 0 (x + 7)(x + 2) = 0 x+7=0 x = –7


18.

x 2 − 3x − 10 = 0 (x – 5)(x + 2) = 0 x–5=0 x=5

or x + 2 = 0 or x = –2

1 2 3 y − y=0 7 7 1 y ( y − 3) = 0 7 y=0 y=0

or y – 3 = 0 or y = 3

19. 2 p 2 = 3 p 2 p2 − 3 p = 0 p(2p – 3) = 0 p=0

or x + 2 = 0 or x = –2

2

13. 4 x + 1 = 4 x

p=0

2

4x − 4x + 1 = 0 (2 x − 1)2 = 0 2x – 1 = 0 1 x= 2

20. r 2 + r − 12 = 0 (r – 3)(r + 4) = 0 r–3=0 r=3

22. ( w − 3)2 ( w + 1)2 = 0 w−3=0 or w + 1 = 0 w=3 or w = −1

15. v(3v − 5) = −2 3v 2 − 5v = −2

23.

2

16.

or r + 4 = 0 or r = –4

21. x(x + 4)(x – 1) = 0 x = 0 or x + 4 = 0 or x – 1 = 0 x = 0 or x = –4 or x = 1

14. 2 z 2 + 9 z − 5 = 0 (2z – 1)(z + 5) = 0 2z – 1 = 0 or z + 5 = 0 1 z= or z = –5 2

3v − 5v + 2 = 0 (3v − 2)(v − 1) = 0 3v – 2 = 0 2 v= 3

or 2p – 3 = 0 3 or p = 2

t 3 − 49t = 0 t (t 2 − 49) = 0 t (t + 7)(t − 7) = 0 t = 0 or t + 7 = 0 or t − 7 = 0 t = 0 or t = −7 or t=7

or v − 1 = 0 or v = 1

24. x( x 2 − 4 x − 5) = 0 x(x – 5)(x + 1) = 0 x = 0 or x – 5 = 0 or x + 1 = 0 x = 0 or x = 5 or x = –1

−6 x 2 + x + 2 = 0 6 x2 − x − 2 = 0 (2 x + 1)(3x − 2) = 0 2x +1 = 0 or 3x − 2 = 0 1 2 x=− or x= 2 3

25. 6 x3 + 5 x 2 − 4 x = 0 x(6 x 2 + 5 x − 4) = 0 x(2x – 1)(3x + 4) = 0

24


Section 0.8

x = 0 or 2x – 1 = 0 or 3x + 4 = 0 4 1 or x = − x = 0 or x = 3 2

32. x 2 − 2 x − 15 = 0 a = 1, b = –2, c = –15 x=

26. x 2 + 2 x + 1 − 5 x + 1 = 0 x2 − 3x + 2 = 0 (x – 1)(x – 2) = 0 x–1=0 x=1

27.

=

or x – 2 = 0 or x = 2

( x − 3)( x 2 − 4) = 0 ( x − 3)( x − 2)( x + 2) = 0 x − 3 = 0 or x−2=0 x=3 or x=2

or or

28. 5(x + 4)(x − 3)(x − 8) = 0 x + 4 = 0 or x − 3 = 0 or x = −4 or x = 3 or

–(–2) ± 4 − 4(1)(–15) 2(1)

2 ± 64 2 2±8 = 2 2+8 x= =5 2

=

x+2=0 x = −2 x −8 = 0 x=8

29. p( p − 3) 2 − 4( p − 3)3 = 0 =

( p − 3) 2 [ p − 4( p − 3)] = 0 ( p − 3) 2 (12 − 3 p ) = 0

)( x − 2 ) = 0

−(−12) ± 144 − 4(4)(9) 2(4)

−b ± b 2 − 4ac 2a 5 ± 25 − 4(1)(0) = 2(1) 5 ± 25 = 2 5±5 = 2 5+5 q= =5 or 2

q=

31. x 2 + 2 x − 24 = 0 a = 1, b = 2, c = –24

=

−b ± b 2 − 4ac 2a

34. q 2 − 5q = 0 a = 1, b = −5, c = 0

x + 1 = 0 or x − 1 = 0 or x + 2 = 0 or x − 2 = 0 x = −1 or x = 1 or x = − 2 or x = 2

x=

2−8 = −3 2

12 ± 0 8 12 ± 0 = 8 3 = 2

3( p − 3) (4 − p) = 0 p–3=0 or 4 – p = 0 p=3 or p = 4

(

x=

=

2

( x + 1)( x − 1) x + 2

or

33. 4 x 2 − 12 x + 9 = 0 a = 4, b = –12, c = 9 x=

30. ( x 2 − 1)( x 2 − 2) = 0

–b ± b 2 − 4ac 2a

–b ± b 2 − 4ac 2a

–2 ± 4 − 4(1)(–24) 2(1)

–2 ± 100 2 −2 ± 10 = 2 −2 + 10 −2 − 10 x= = 4 or x = = −6 2 2

=

25

q=

5−5 =0 2



35. p 2 − 2 p − 7 = 0

39. 4 x 2 + 5 x − 2 = 0 a = 4, b = 5, c = –2

a = 1, b = −2, c = −7

−b ± b 2 − 4ac 2a −5 ± 25 − 4(4)(−2) = 2(4) −5 ± 57 = 8 −5 + 57 −5 − 57 x= or x = 8 8

−b ± b 2 − 4ac p= 2a

x=

−(−2) ± (−2)2 − 4(1)(−7) 2(1) 2 ± 32 = 2 = 1± 2 2 =

p = 1+ 2 2

or

p = 1− 2 2

40. w2 − 2w + 1 = 0 a = 1, b = −2, c = 1

36. 2 − 2 x + x 2 = 0 x2 − 2 x + 2 = 0 a = 1, b = –2, c = 2 –(–2) ± 4 − 4(1)(2) x= 2(1)

w=

−(−2) ± (−2) 2 − 4(1)(1) 2(1) 2± 0 = 2 =1 =

2 ± −4 2 no real roots =

37. 4 − 2n + n 2 = 0

41. 0.02w2 − 0.3w = 20

n 2 − 2n + 4 = 0 a = 1, b = –2, c = 4 n=

=

0.02 w2 − 0.3w − 20 = 0 a = 0.02, b = –0.3, c = –20

–b ± b 2 − 4ac 2a

w=

−(−2) ± 4 − 4(1)(4) 2(1)

=

2 ± −12 2 no real roots =

−b ± b 2 − 4ac 2a

−1 ± 1 − 4(2)(–5) 2(2)

−1 ± 41 4 −1 + 41 x= 4

−(–0.3) ± 0.09 − 4(0.02)(–20) 2(0.02)

0.3 ± 1.69 0.04 0.3 ± 1.3 = 0.04 0.3 + 1.3 w= = 0.04 0.3 − 1.3 w= = 0.04

2x2 + x − 5 = 0 a = 2, b = 1, c = –5

=

−b ± b 2 − 4ac 2a

=

38. 2 x 2 + x = 5

x=

−b ± b 2 − 4ac 2a

=

or

x=

−1 − 41 4

26

1.6 = 40 or 0.04 –1.0 = −25 0.04


Section 0.8

42. 0.01x 2 + 0.2 x − 0.6 = 0 a = 0.01, b = 0.2, c = –0.6 x=

44. −2 x 2 − 6 x + 5 = 0 a = –2, b = –6, c = 5

−b ± b 2 − 4ac 2a

x=

=

–0.2 ± 0.04 − 4(0.01)(–0.6) 2(0.01)

=

=

–0.2 ± 0.064 0.02

=

=

–0.2 ± (0.0064)(10) 0.02

=

–0.2 ± 0.08 10 0.02

x = −10 + 4 10 or x = −10 − 4 10

45. ( x 2 )2 − 5( x 2 ) + 6 = 0

43. 2 x 2 + 4 x = 5

Let w = x 2 . Then

2x2 + 4 x − 5 = 0 a = 2, b = 4, c = –5

w2 − 5 w + 6 = 0 (w – 3)(w – 2) = 0 w = 3, 2

–b ± b 2 − 4ac 2a

Thus x 2 = 3 or x 2 = 2, so x = ± 3, ± 2 .

–4 ± 16 − 4(2)(–5) = 2(2) –4 ± 56 4 –4 ± 2 14 = 4 –2 ± 14 = 2 −2 + 14 x= 2

–(–6) ± 36 − 4(–2)(5) 2(−2)

6 ± 76 −4 6 ± 2 19 = −4 −3 ± 19 = 2 −3 + 19 −3 − 19 x= or x = 2 2

= −10 ± 4 10

x=

−b ± b 2 − 4ac 2a

46. ( X 2 ) 2 − 3( X ) 2 − 10 = 0 Let w = X 2 . Then

=

w2 − 3w − 10 = 0 ( w − 5)( w + 2) = 0 w = 5, −2

Thus X 2 = 5 or X 2 = −2, so the real solutions or

x=

are X = ± 5.

−2 − 14 2

2

⎛1⎞ ⎛1⎞ 47. 3 ⎜ ⎟ − 7 ⎜ ⎟ + 2 = 0 ⎝ x⎠ ⎝ x⎠ 1 Let w = . Then x

3w2 − 7 w + 2 = 0 (3w − 1)( w − 2) = 0 1 w= , 2 3 1 Thus, x = 3, . 2

27



48. ( x −1 )2 + x −1 − 12 = 0

2

⎛ 1 ⎞ ⎛ 1 ⎞ 53. ⎜ ⎟ − 12 ⎜ ⎟ + 35 = 0 ⎝ x−2⎠ ⎝ x−2⎠ 1 , then Let w = x−2

Let w = x –1. Then w2 + w − 12 = 0 (w + 4)(w – 3) = 0 w = –4, 3 1 1 Thus, x = − , . 4 3

w2 − 12w + 35 = 0 (w – 7)(w – 5) = 0 w = 7, 5 1 1 = 5. = 7 or Thus, x−2 x−2 15 11 x= , . 7 5

49. ( x –2 )2 − 9( x –2 ) + 20 = 0 Let w = x –2 . Then w2 − 9 w + 20 = 0 (w – 5)(w – 4) = 0 w = 5, 4 1 1 1 1 = 5 or = 4, so x 2 = or x 2 = . Thus, 2 2 5 4 x x 5 1 x=± ,± . 5 2

2

⎛ 1 ⎞ ⎛ 1 ⎞ 54. 2 ⎜ ⎟ + 7⎜ ⎟+3= 0 ⎝ x+4⎠ ⎝ x+4⎠ 1 Let w = . Then x+4 2 w2 + 7 w + 3 = 0 (2w + 1)(w + 3) = 0 1 w = − , −3 2 1 1 1 = −3 . = − or Thus, x+4 2 x+4 13 x = −6, − 3

2

⎛ 1 ⎞ ⎛ 1 ⎞ 50. ⎜ ⎟ − 9 ⎜ ⎟ + 8 = 0 2 ⎝x ⎠ ⎝ x2 ⎠ 1 . Then Let w = x2 w2 − 9 w + 8 = 0 (w – 8)(w – 1) = 0 w = 8, 1 1 1 1 = 8 or = 1, so x 2 = or x 2 = 1. Thus, 2 2 8 x x 2 x=± , ± 1. 4

55. x 2 =

x+3 2

2 x2 = x + 3 2x2 − x − 3 = 0 (2x – 3)(x + 1) = 0 3 Thus, x = , − 1. 2

51. ( X − 5)2 + 7( X − 5) + 10 = 0 Let w = X − 5. Then w2 + 7 w + 10 = 0 ( w + 2)( w + 5) = 0 w = −2, −5 Thus, X − 5 = −2 or X − 5 = −5, so X = 3, 0.

56.

x 7 5 = − 2 x 2 Multiplying both sides by the LCD, 2x, gives x 2 = 14 − 5 x x 2 + 5 x − 14 = 0 (x – 2)(x + 7) = 0 Thus, x = 2, –7.

52. (3x + 2) 2 − 5(3x + 2) = 0 Let w = 3x + 2. Then w2 − 5 w = 0 w( w − 5) = 0 w = 0, 5 2 Thus 3x + 2 = 0 or 3x + 2 = 5, so x = − , 1. 3 28


57.

58.

Section 0.8

3 x −3 + =2 x−4 x Multiplying both sides by the LCD, x(x – 4), gives 3x + (x – 3)(x – 4) = 2x(x – 4)

61.

3x + x 2 − 7 x + 12 = 2 x 2 − 8 x

2r + 8 − (r 2 − r − 2) = 0

x 2 − 4 x + 12 = 2 x 2 − 8 x

−r 2 + 3r + 10 = 0

0 = x 2 − 4 x − 12 0 = (x – 6)(x + 2) Thus, x = 6, –2.

r 2 − 3r − 10 = 0 (r – 5)(r + 2) = 0 Thus, r = 5, –2.

2 6 − =5 2x +1 x −1 Multiplying both sides by the LCD, (2x + 1)(x − 1), gives 2( x − 1) − 6(2 x + 1) = 5(2 x + 1)( x − 1)

62.

2x − 3 2x + =1 2 x + 5 3x + 1 Multiplying both sides by the LCD, (2x + 5)(3x + 1), gives (2x – 3)(3x + 1) + 2x(2x + 5) = (2x + 5)(3x + 1)

−10 x − 8 = 10 x 2 − 5 x − 5

6 x 2 − 7 x − 3 + 4 x 2 + 10 x = 6 x 2 + 17 x + 5

0 = 10 x 2 + 5 x + 3 a = 10, b = 5, c = 3

10 x 2 + 3 x − 3 = 6 x 2 + 17 x + 5 4 x 2 − 14 x − 8 = 0

2

b − 4ac = 25 − 4(10)(3) = −95 < 0, thus there are no real roots.

59.

2 r +1 − =0 r−2 r+4 Multiplying both sides by the LCD, (r – 2)(r + 4), gives 2(r + 4) – (r – 2)(r + 1) = 0

2x2 − 7 x − 4 = 0 (2x + 1)(x – 4) = 0 1 Thus, x = − , 4. 2

3x + 2 2 x + 1 − =1 x +1 2x Multiplying both sides by the LCD, 2x(x + 1), gives 2 x(3 x + 2) − (2 x + 1)( x + 1) = 2 x( x + 1)

63.

6 x 2 + 4 x − (2 x 2 + 3x + 1) = 2 x 2 + 2 x 4 x2 + x − 1 = 2 x2 + 2 x 2 x2 − x − 1 = 0 (2 x + 1)( x − 1) = 0

t +1 t + 3 t +5 + = t + 2 t + 4 t 2 + 6t + 8 Multiplying both sides by the LCD, (t + 2)(t + 4), gives (t + 1)(t + 4) + (t + 3)(t + 2) = t + 5 t 2 + 5t + 4 + t 2 + 5t + 6 = t + 5 2t 2 + 10t + 10 = t + 5

1 Thus, x = − , 1. 2

2t 2 + 9t + 5 = 0 a = 2, b = 9, c = 5 −b ± b 2 − 4ac 2a −9 ± 81 − 4(2)(5) = 2(2) −9 ± 41 = 4 −9 + 41 −9 − 41 Thus t = , . 4 4

w 6( w + 1) + =3 60. 2−w w −1 Multiplying both sides by the LCD, (2 – w)(w – 1), gives 6(w + 1)(w – 1) + w(2 – w) = 3(2 – w)(w – 1)

t=

6( w2 − 1) + 2 w − w2 = 3(– w2 + 3w − 2) 5w2 + 2 w − 6 = −3w2 + 9 w − 6 8 w2 − 7 w = 0 w(8w – 7) = 0 7 Thus, w = 0, . 8

29


64.


2 3 4 + = x +1 x x + 2 Multiplying both sides by the LCD, x(x + 1)(x + 2), gives 2 x( x + 2) + 3( x + 1)( x + 2) = 4 x( x + 1)

68.

x2 + 9 x + 6 = 0 a = 1, b = 9, c = 6

70.

( x) w=

+2

( x)−5 = 0

−b ± b 2 − 4ac 2a

−2 ± 4 − 4(1)(−5) 2(1) −2 ± 24 = 2 −2 ± 2 6 = 2 = −1 ± 6

w=

2 x 2 − x ( x + 1) = 2( x + 1)( x − 1) 2 x2 − x2 − x = 2 x2 − 2 x2 − x = 2 x2 − 2 0 = x2 + x − 2 0 = (x + 2)(x − 1) x = –2 or x = 1 But x = 1 does not check. The solution is –2.

Since w = x and −1 − 6 < 0, w = −1 − 6 does not check. Thus w = −1 + 6, so

(

x = −1 + 6

3 1− x = . x x Multiplying both sides by x gives 5x – 3 = 1 – x 6x = 4 2 x= 3

66. If x ≠ –3, the equation is 5 −

2

2

Let w = x , then w2 + 2 w − 5 = 0 a = 1, b = 2, c = −5

x 2 ( x + 1)( x − 1), gives

)

2

q 2 − 12q + 32 = 0 (q – 4)(q – 8) = 0 Thus, q = 4, 8.

−

2x − 3

)

q 2 + 4q + 4 = 16q − 28

1 2 = 2 x − 1 x( x − 1) x 2 Multiplying both sides by the LCD,

(

= ( x − 6)2

(

−9 ± 92 − 4(1)(6) = 2(1) −9 ± 57 = 2 −9 + 57 −9 − 57 Thus, x = , . 2 2

67.

2

69. (q + 2) 2 = 2 4q − 7

−b ± b 2 − 4ac 2a

2

)

0 = x 2 − 21x 0 = x(x – 21) x = 0 or x = 21 Only x = 21 checks.

5 x 2 + 13 x + 6 = 4 x 2 + 4 x

65.

x+4

9 x + 36 = x 2 − 12 x + 36

2 x2 + 4 x + 3x2 + 9 x + 6 = 4 x2 + 4 x

x=

(3

)

2

= 7 − 2 6.

z + 3 = 3z + 1

71.

(

z +3

) =( 2

)

3z + 1

2

z + 3 = 3z + 2 3z + 1 −2 z + 2 = 2 3 z − z + 1 = 3z

= ( x − 3)2

(− z + 1) 2 =

2 x − 3 = x2 − 6 x + 9

(

3z

)

z 2 − 2 z + 1 = 3z

2

z 2 − 5z + 1 = 0 a = 1, b = −5, c = 1

0 = x − 8 x + 12 0 = (x – 6)(x – 2) x = 6 or x = 2 Only x = 6 checks.

30

2


z=

Section 0.8

−b ± b 2 − 4ac 2a

75.

(

(

x −2

) =(

2x − 8

)

x+3

76.

= ( x − 12)2

(

t +2

2

2x + 1

)

2

t=

= x2

0 = x2 − 4x 0 = x(x – 4) Thus, x = 0, 4.

(

y−2 +2

) =( 2

)

2

2

2

= (3t − 1)2

2y + 3

−b ± b 2 − 4ac 2a

)

–(−2.7) ± (−2.7) 2 − 4(0.04)(8.6) 2(0.04) ≈ 64.15 or 3.35

2

4 y − 2 = y +1 y−2

)

77. x =

y − 2 + 4 y − 2 + 4 = 2y + 3

(4

3t + 1

−(−7) ± (−7)2 − 4(9)(1) 2(9) 7 ± 13 = 18 7 + 13 Only checks. 18

4x = x 2

74.

(

=

2 x=x 2

=

0 = 9t 2 − 7t + 1 a = 9, b = −7, c = 1

x + 2 x + 1 = 2x + 1

(2 x )

2

t = 9t 2 − 6t + 1

x +1 = 2x + 1

) =(

)

( t)

0 = x 2 − 40 x + 144 0 = (x – 4)(x – 36) x = 4 or x = 36 Only x = 4 checks.

x +1

= (4 x − 2) 2

t + 2 = 3t + 1 t = 3t − 1

16 x = x − 24 x + 144

(

2

1 or x = 1 16 Only x = 1 checks.

2

73.

)

x=

2

x − 4 x + 4 = 2x − 8

( −4 x )

2

0 = 16 x 2 − 17 x + 1 0 = (16 x − 1)( x − 1)

−4 x = x − 12 2

2

x + 3 = 16 x 2 − 16 x + 4

x − 2 = 2x − 8 2

) = (3 x )

x + 3 +1

x + 3 + 2 x + 3 + 1 = 9x 2 x + 3 = 8x − 4 x + 3 = 4x − 2

−(−5) ± (−5)2 − 4(1)(1) = 2(1) 5 ± 21 = 2 5 − 21 Only z = checks. 2

72.

(

–0.2 ± (0.2) 2 − 4(0.01)(–0.6) 2(0.01) ≈ 2.65 or –22.65

78. x =

= ( y + 1) 2

16 y − 32 = y 2 + 2 y + 1 0 = y 2 − 14 y + 33 0 = (y – 11)(y – 3) Thus, y = 11, 3.

31



79. Let l be the length of the picture, then its width is l – 2. l(l – 2) = 48

83.

l 2 − 2l − 48 = 0 (l – 8)(l + 6) = 0 l–8=0 or l + 6 = 0 l=8 or l = –6 Since length cannot be negative, l = 8. The width of the picture is l – 2 = 8 – 2 = 6 inches. The dimensions of the picture are 6 inches by 8 inches.

24 A = A2 + 13 A + 12 0 = A2 − 11A + 12 From the quadratic formula, 11 ± 121 − 48 11 ± 73 A= = . 2 2 11 + 73 11 – 73 ≈ 10 or A = A= ≈ 1. 2 2 The doses are the same at 1 year and 10 years. A +1 c = d in Cowling’s rule when = 1, which 24 occurs when A = 23. Thus, adulthood is achieved at age 23 according to Cowling’s rule. A = 1, which is c = d in Young’s rule when A + 12 never true. Thus, adulthood is never reached according to Young’s rule.

80. The amount that the temperature has risen over the X days is (X degrees per day)(X days) = X 2 degrees. X 2 + 15 = 51 X 2 = 36 X = ± 36 X = 6 or X = –6 The temperature has been rising 6 degrees per day for 6 days.

81. M =

A A +1 d= d. A + 12 24 Dividing both sides by d and then multiplying both sides by 24(A + 12) gives 24A = (A + 12)(A + 1)

Q(Q + 10) 44

1

44 M = Q 2 + 10Q 0 = Q 2 + 10Q – 44 M From the quadratic formula with a = 1, b = 10, c = −44 M , Q= =

0

0

25

Young’s rule prescribes less than Cowling’s for ages less than one year and greater than 10 years. Cowling’s rule prescribes less for ages between 1 and 10.

–10 ± 100 – 4(1)(–44M ) 2(1)

84. a.

–10 + 2 25 + 44M 2

= −5 ± 25 + 44M

Thus, −5 + 25 + 44M is a root. 82. g = −200 P 2 + 200 P + 20 Set g = 60. 60 = −200 P 2 + 200 P + 20

(2n − 1)v 2 − 2nv + 1 = 0 From the quadratic formula with a = 2n – 1, b = –2n, c = 1, v=

–(–2n) ± 4n 2 − 4(2n − 1)(1) 2(2n − 1)

v=

2n ± 4n 2 − 8n + 4 2(2n − 1)

2 2n ± 2 n 2 − 2n + 1 n ± (n − 1) = 2(2n − 1) 2n − 1 Because of the condition that n ≥ 1, it follows that n – 1 is nonnegative. Thus,

v=

200 P 2 − 200 P + 40 = 0 5P 2 − 5P + 1 = 0 From the quadratic formula with a = 5, b = −5, c = 1, 5 ± 25 − 4(5)(1) 5 ± 5 P= = 2(5) 10

(n − 1)2 = n − 1 and we have v=

P ≈ 0.28 or P ≈ 0.72 28% and 72% of yeast gave an average weight gain of 60 grams.

n ± (n − 1) . 2n − 1

v = 1 or v = 32

1 . 2n − 1


Mathematical Snapshot Chapter 0 87. By a program, roots are 1.5 and 0.75. Algebraically:

nv 2 − (2n + 1)v + 1 = 0 From the quadratic formula with a = n, b = –(2n + 1), and c = 1,

b.

v=

−[−(2n + 1)] ± [−(2n + 1)]2 − 4(n)(1) 2n

v=

2n + 1 ± 4n 2 + 1 2n

8 x 2 − 18 x + 9 = 0 (2x – 3)(4x – 3) = 0 Thus, 2x – 3 = 0 or 4x – 3 = 0. 3 3 So x = = 1.5 or x = = 0.75. 2 4

88. By a program, roots are −0.762 and 0.262.

Because 4n 2 + 1 is greater than 2n, choosing the plus sign gives a numerator greater than 2n + 1 + 2n, or 4n + 1, so v is 4n + 1 1 = 2 + . Thus v is greater than 2n 2n greater than 2. This contradicts the restriction on v. On the other hand, because

89. By a program, there are no real roots. 90.

4n 2 + 1 is greater than 1, choosing the minus sign gives a numerator less than 2n, 2n = 1. This meets the so v is less than 2n condition on v. Thus we choose v=

85. a.

9 2 z z − 6.3 = (1.1 − 7 z ) 2 3 9 2 1.1 7 z − 6.3 = z − z2 2 3 3 ⎛ 9 7 ⎞ 2 1.1 z − 6.3 = 0 ⎜ + ⎟z − 3 ⎝2 3⎠ Roots: 0.987, –0.934

91. (πt − 4)2 = 4.1t − 3 π2t 2 − 8πt + 16 = 4.1t − 3

2n + 1 − 4n 2 + 1 . 2n

π2t 2 + (–8π − 4.1)t + 19 = 0 Roots: 1.999, 0.963

When the object strikes the ground, h must be 0, so

Mathematical Snapshot Chapter 0

0 = 39.2t − 4.9t 2 = 4.9t (8 − t ) t = 0 or t = 8 The object will strike the ground 8 s after being thrown.

1.

b. Setting h = 68.2 gives 68.2 = 39.2t − 4.9t 2 4.9t 2 − 39.2t + 68.2 = 0

2. The procedure works because multiplying a list by a number is the same as multiplying each element in the list by the number, adding a number to a list has the effect of adding the number to each element of the list, and subtracting one list from another is the same as subtracting corresponding elements. The plots match.

2

39.2 ± ( −39.2) − 4(4.9)(68.2) 2(4.9) 39.2 ± 14.1 ≈ 9.8 t ≈ 5.4 s or t ≈ 2.6 s. t=

86. By a program, roots are 4.5 and –3. Algebraically:

3.

2

2 x − 3 x − 27 = 0 (2x – 9)(x + 3) = 0 Thus, 2x – 9 = 0 or x + 3 = 0 9 So x = = 4.5 or x = –3. 2

The results agree. 33



4. The smaller quadratic residuals indicate a better fit. The fairly random pattern suggests that the model cannot be improved any further. The slight deviations from the quadratic model are presumably due to random measurement errors.

34

Chapter 1 5. Let n = number of ounces in each part. Then we have 2n + 1n = 16 3n = 16 16 n= 3 Thus the turpentine needed is 16 1 (1)n = = 5 ounces. 3 3

Problems 1.1 1. Let w be the width and 2w be the length of the plot. 2w

w

w 2w

Then area = 800. (2w)w = 800

6. Let w = width (in miles) of strip to be cut. Then the remaining forest has dimensions 2 – 2w by 1 – 2w.

2w2 = 800 w2 = 400 w = 20 ft Thus the length is 40 ft, so the amount of fencing needed is 2(40) + 2(20) = 120 ft.

2 – 2w w

1 – 2w w

1

w

2. Let w be the width and 2w be the length.

2

2w

w

w

Considering the area of the remaining forest, we have 3 (2 − 2 w)(1 − 2 w) = 4 3 2 − 6 w + 4 w2 = 4

w 2w

The perimeter P = 2w + 2l = 2w + 2(2w) = 6w. Thus 6w = 300. 300 w= = 50 ft 6 Thus the length is 2(50) = 100 ft. The dimensions are 50 ft by 100 ft.

8 − 24w + 16 w2 = 3 16w2 − 24w + 5 = 0 (4w – 1)(4w – 5) = 0 1 5 5 Hence w = , . But w = is impossible since 4 4 4 one dimension of original forest is 1 mi. Thus 1 mi. the width of the strip should be 4

3. Let n = number of ounces in each part. Then we have 4n + 5n = 145 9n = 145 1 n = 16 9 4 ⎛ 1⎞ Thus there should be 4 ⎜ 16 ⎟ = 64 ounces of 9 ⎝ 9⎠ 1 5 ⎛ ⎞ A and 5 ⎜ 16 ⎟ = 80 ounces of B. 9 ⎝ 9⎠

7. Let w = width (in meters) of pavement. The remaining plot for flowers has dimensions 8 – 2w by 4 – 2w. 8 – 2w w

4 – 2w w

w w

4. Let n = number of cubic feet in each part. Then we have 1n + 3n + 5n = 765 9n = 765 n = 85 Thus he needs 1n = 1(85) = 85 ft3 of portland cement, 3n = 3(85) = 255 ft3 of sand, and 5n = 5(85) = 425 ft3 of crushed stone.

8

Thus (8 – 2w)(4 – 2w) = 12 32 − 24w + 4 w2 = 12 4w2 − 24 w + 20 = 0 w2 − 6 w + 5 = 0 (w – 1)(w – 5) = 0 35

4

Chapter 1: Applications and More Algebra


Hence w = 1, 5. But w = 5 is impossible since one dimension of the original plot is 4 m. Thus the width of the pavement should be 1 m.

13. Let p = selling price. Then profit = 0.2p. selling price = cost + profit p = 3.40 + 0.2p 0.8p = 3.40 3.40 p= = $4.25 0.8

8. Since diameter of circular end is 140 mm, the radius is 70 mm. Area of circular end is π(radius)2 = π(70)2 . Area of square end is x 2 .

14. Following the procedure in Example 6 we obtain the total value at the end of the second year to be

Equating areas, we have x 2 = π(70)2 . Thus x = ± π(70)2 = ±70 π . Since x must be

1, 000, 000(1 + r ) 2 . So at the end of the third year, the accumulated

positive, x = 70 π ≈ 124 mm.

amount will be 1, 000, 000(1 + r ) 2 plus the

9. Let q = number of tons for $560,000 profit. Profit = Total Revenue – Total Cost 560, 000 = 134q − (82q + 120, 000) 560, 000 = 52q − 120, 000 680, 000 = 52q 680, 000 =q 52 q ≈ 13, 076.9 ≈ 13, 077 tons.

interest on this, which is 1, 000, 000(1 + r )2 r. Thus the total value at the end of the third year will be 1, 000, 000(1 + r ) 2 + 1, 000, 000(1 + r )2 r = 1, 000, 000(1 + r )3 . This must equal $1,125,800. 1, 000, 000(1 + r )3 = 1,125,800 1,125,800 (1 + r )3 = = 1.1258 1, 000, 000 1 + r ≈ 1.04029 r ≈ 0.04029 Thus r ≈ 0.04029 ≈ 4%.

10. Let q = required number of units. Profit = Total Revenue – Total Cost 150, 000 = 50q − (25q + 500, 000) 150, 000 = 25q − 500, 000 650, 000 = 25q, from which q = 26, 000

15. Following the procedure in Example 6 we obtain 3, 000, 000(1 + r ) 2 = 3, 245, 000 649 (1 + r ) 2 = 600 649 1+ r = ± 600 649 r = −1 ± 600 r ≈ −2.04 or 0.04 We choose r ≈ 0.04 = 4%.

11. Let x = amount at 6% and 1 20,000 – x = amount at 7 %. 2 x(0.06) + (20,000 – x)(0.075) = 1440 –0.015x + 1500 = 1440 –0.015x = –60 x = 4000, so 20,000 – x = 16,000. Thus the investment should be $4000 at 6% and $16,000 1 at 7 %. 2

16. Total revenue = variable cost + fixed cost 100 q = 2q + 1200

12. Let x = amount at 6% and 20,000 – x = amount at 7%. x(0.06) + (20,000 – x)(0.07) = 20,000(0.0675) –0.01x + 1400 = 1350 –0.01x = –50, so x = 5000 The investment consisted of $5000 at 6% and $15,000 at 7%.

50 q = q + 600 2500q = q 2 + 1200q + 360, 000 0 = q 2 − 1300q + 360, 000 0 = (q – 400)(q – 900) q = 400 or q = 900

36


Section 1.1 23. Let v = total annual vision-care expenses (in dollars) covered by program. Then 35 + 0.80(v – 35) = 100 0.80v + 7 = 100 0.80v = 93 v = $116.25

17. Let n = number of room applications sent out. 0.95n = 76 76 n= = 80 0.95 18. Let n = number of people polled. 0.20p = 700 700 p= = 3500 0.20

24. a.

0.031c

b. c – 0.031c = 600,000,000 0.969c = 600,000,000 c ≈ 619,195, 046 Approximately 619,195,046 bars will have to be made.

19. Let s = monthly salary of deputy sheriff. 0.30s = 200 200 s= 0.30 ⎛ 200 ⎞ Yearly salary = 12s = 12 ⎜ ⎟ = $8000 ⎝ 0.30 ⎠

25. Revenue = (number of units sold)(price per unit) Thus ⎡ 80 − q ⎤ 400 = q ⎢ ⎥ ⎣ 4 ⎦

20. Yearly salary before strike = (7.50)(8)(260) = $15,600 Lost wages = (7.50)(8)(46) = $2760 Let P be the required percentage increase (as a decimal). P(15,600) = 2760 2760 P= ≈ 0.177 = 17.7% 15, 600

1600 = 80q − q 2 q 2 − 80q + 1600 = 0 (q − 40)2 = 0 q = 40 units

26. If I = interest, P = principal, r = rate, and t = time, then I = Prt. To triple an investment of P at the end of t years, the interest earned during that time must equal 2P. Thus 2P = P(0.045)t 2 = 0.045t 2 t= ≈ 44.4 years 0.045

21. Let q = number of cartridges sold to break even. total revenue = total cost 21.95q = 14.92q + 8500 7.03q = 8500 q ≈ 1209.10 1209 cartridges must be sold to approximately break even.

27. Let q = required number of units. We equate incomes under both proposals. 2000 + 0.50q = 25,000 0.50q = 23,000 q = 46,000 units

22. Let n = number of shares of stock to be bought. total investment = 4000 + 15n total yield (goal) = 6% of total investment = 0.06(4000 + 15n) total yield = bond yield + stock yield = 0.07(4000) + 0.60n Thus, 0.06(4000 + 15n) = 0.07(4000) + 0.60n 240 + 0.9n = 280 + 0.6n 0.3n = 40 1 n = 133 3

28. Let w = width of strip. The original area is 80(120) and the new area is (120 + w)(80 + w). w

120 80

80 + w w 120 + w

37



Thus (120 + w)(80 + w) = 2(80)(120)

31. 10, 000 = 800 p − 7 p 2

9600 + 200w + w2 = 19, 200

7 p 2 − 800 p + 10, 000 = 0

w2 + 200 w − 9600 = 0 (w + 240)(w – 40) = 0 w = –240 or w = 40 We choose w = 40 ft.

800 ± 640, 000 − 280, 000 14 800 ± 360, 000 800 ± 600 = = 14 14 800 + 600 For p > 50 we choose p = = $100. 14 p=

29. Let n = number of $20 increases. Then at the rental charge of 400 + 20n dollars per unit, the number of units that can be rented is 50 – 2n. The total of all monthly rents is (400 + 20n)(50 – 2n), which must equal 20,240. 20,240 = (400 + 20n)(50 – 2n)

32. Let p be the percentage increase in market value. Then ⎛ P ⎞ (1 + p) P 1.1⎜ ⎟ = ⎝ E ⎠ (1.2) E

20, 240 = 20, 000 + 200n − 40n 2 40n 2 − 200n + 240 = 0

1+ p 1.2 1.32 = 1 + p p = 0.32 = 32% 1.1 =

2

n − 5n + 6 = 0 (n – 2)(n – 3) = 0 n = 2, 3 Thus the rent should be either $400 + 2($20) = $440 or $400 + 3($20) = $460.

33. To have supply = demand, 2 p − 10 = 200 − 3 p 5 p = 210 p = 42

30. Let x = original value of the blue-chip investment, then 3,100,000 – x is the original value of the glamour stocks. Then the current 1 11 x. value of the blue-chip stock is x + x, or 10 10 For the glamour stocks the current value is 1 (3,100, 000 − x) − (3,100, 000 − x ), which 10 9 simplifies to (3,100, 000 − x). 10 Thus for the current value of the portfolio, 11 9 x + (3,100, 000 − x) = 3, 240, 000 10 10 11x + 27,900,000 – 9x = 32,400,000 2x = 4,500,000 x = 2,250,000 Thus the current value of the blue chip 11 (2, 250, 000) or $2,475,000. investment is 10

2 p 2 − 3 p = 20 − p 2

34. 2

3 p − 3 p − 20 = 0 a = 3, b = −3, c = −20

p=

−b ± b 2 − 4ac 2a

−(−3) ± (−3) 2 − 4(3)(−20) 2(3) 3 ± 249 = 6 p ≈ 3.130 or p ≈ −2.130 The equilibrium price is p ≈ 3.13. =

35. Let w = width (in ft) of enclosed area. Then length of enclosed area is 300 – w – w = 300 – 2w.

38


Section 1.1

300 – 2w w

(10 – x)(5 – x)2 = 72 (10 – x)(5 – x) = 36 x 2 − 15 x + 50 = 36

w

AREA

x 2 − 15 x + 14 = 0 (x – 1)(x – 14) = 0 x = 1 or 14 Because of the length and width of the original bar, we reject x = 14 and choose x = 1. The new bar has length 10 – x = 10 – 1 = 9 cm and width is 5 – x = 5 – 1 = 4 cm.

PLANT

38. Volume of old style candy

150

Thus w(300 – 2w) = 11,200 2w(150 – w) = 11,200 w(150 – w) = 5600

= π(7.1)2 (2.1) − π(2)2 (2.1) = 97.461π mm3 Let r = inner radius (in millimeters) of new style candy. Considering the volume of the new style candy, we have

0 = w2 − 150w + 5600 0 = (w – 80)(w – 70) Hence w = 80, 70. If w = 70, then length is 300 – 2w = 300 – 2(70) = 160. Since the building has length of only 150 ft, we reject w = 70. If w = 80, then length is 300 – 2w = 300 – 2(80) = 140. Thus the dimensions are 80 ft by 140 ft.

π(7.1)2 (2.1) − πr 2 (2.1) = 0.78(97.461π) 29.84142π = 2.1πr 2 14.2102 = r 2 r ≈ ±3.7696 Since r is a radius, we choose r = 3.77 mm.

39. Let x = amount of loan. Then the amount actually received is x − 0.16x. Hence, x − 0.16 x = 195, 000 0.84 x = 195, 000 x ≈ 232,142.86 To the nearest thousand, the loan amount is $232,000. In the general case, the amount received from a loan of L with a compensating p L. balance of p% is L − 100 p L− L=E 100 100 − p L=E 100 100 E L= 100 − p

36. Let s = length in inches of side of original square. s 3

3

3

3

s–6

s

3

3

3

3

s–6

Considering the volume of the box, we have (length)(width)(height) = volume (s – 4)(s – 4)(2) = 50 ( s − 4) 2 = 25 s − 4 = ± 25 = ±5 s=4±5 Hence s = –1, 9. We reject s = –1 and choose s = 9. The dimensions are 9 in. by 9 in.

40. Let n = number of machines sold over 600. Then the commission on each of 600 + n machines is 40 + 0.04n. Equating total commissions to 30,800 we obtain (600 + n)(40 + 0.04n) = 30,800

37. Original volume = (10)(5)(2) = 100 cm3 Volume cut from bar = 0.28(100) = 28 cm3 Volume of new bar = 100 – 28 = 72 cm3 Let x = number of centimeters that the length and width are each reduced. Then

24, 000 + 24n + 40n + 0.04n 2 = 30,800 0.02n 2 + 32n − 3400 = 0 n=

39

–32 ± 1024 + 272 −32 ± 36 = 0.04 0.04


ISM: Introductory Mathematical Analysis 43. Let q = number of units of B and q + 25 = number of units of A produced. 1000 , and each unit of A Each unit of B costs q

–32 + 36 = 100. Thus the 0.04 number of machines that must be sold is 600 + 100 = 700.

We choose n =

costs

41. Let n = number of acres sold. Then n + 20 acres 7200 were originally purchased at a cost of n + 20 each. The price of each acre sold was ⎡ 7200 ⎤ 30 + ⎢ ⎥ . Since the revenue from selling n ⎣ n + 20 ⎦ acres is $7200 (the original cost of the parcel), we have 7200 ⎤ ⎡ n ⎢30 + = 7200 n + 20 ⎥⎦ ⎣

1500 1000 = +2 q + 25 q 1500q = 1000(q + 25) + 2(q)(q + 25) 0 = 2q 2 − 450q + 25, 000 0 = q 2 − 225q + 12,500 0 = (q – 100)(q – 125) q = 100 or q = 125 If q = 100, then q + 25 = 125; if q = 125, q + 25 = 150. Thus the company produces either 125 units of A and 100 units of B, or 150 units of A and 125 units of B.

⎡ 30n + 600 + 7200 ⎤ n⎢ ⎥ = 7200 n + 20 ⎣ ⎦ n(30n + 600 + 7200) = 7200(n + 20)

Principles in Practice 1.2

30n 2 + 7800n = 7200n + 144, 000 30n 2 + 600n − 144, 000 = 0

1. 200 + 0.8S ≥ 4500 0.8S ≥ 4300 S ≥ 5375 He must sell at least 5375 products per month.

2

n + 20n − 4800 = 0 (n + 80)(n – 60) = 0 n = 60 acres (since n > 0), so 60 acres were sold.

2. Since x1 ≥ 0, x2 ≥ 0, x3 ≥ 0, and x4 ≥ 0, we have the inequalities 150 − x4 ≥ 0 3x4 − 210 ≥ 0 x4 + 60 ≥ 0 x4 ≥ 0

42. Let q = number of units of product sold last year and q + 2000 = the number sold this year. Then the revenue last year was 3q and this year it is 3.5(q + 2000). By the definition of margin of profit, it follows that 7140 4500 = + 0.02 3.5(q + 2000) 3q 2040 1500 = + 0.02 q + 2000 q 2040q = 1500(q + 2000) + 0.02q(q + 2000)

Problems 1.2 1. 3x > 12 12 x> 3 x>4 (4, ∞)

2040q = 1500q + 3, 000, 000 + 0.02q 2 + 40q 0 = 0.02q 2 − 500q + 3, 000, 000 q=

1500 . Therefore, q + 25

500 ± 250, 000 − 240, 000 0.04

4

500 ± 10, 000 0.04 500 ± 100 = 0.04 = 10,000 or 15,000 So that the margin of profit this year is not greater than 0.15, we choose q = 15,000. Thus 15,000 units were sold last year and 17,000 this year. =

2. 4x < –2 –2 x< 4 1 x<− 2

40


Section 1.2 8. 4s – 1 < –5 4s < –4 s < –1 (–∞, –1)

1⎞ ⎛ ⎜ −∞, − 2 ⎟ ⎝ ⎠ –1 2

–1

3. 5 x − 11 ≤ 9 5 x ≤ 20 x≤4 (−∞, 4]

9. 3 < 2y + 3 0 < 2y 00 (0, ∞)

4

0

4. 5 x ≤ 0 0 x≤ 5 x≤0 (−∞, 0]

10.

0

5. –4x ≥ 2 2 x≤ −4 1 x≤− 2 1⎤ ⎛ ⎜ – ∞, − 2 ⎥ ⎝ ⎦

4 ≤ 3− 2y 1 ≤ −2 y 1 − ≥y 2 1 y≤− 2 1⎤ ⎛ ⎜ – ∞, − 2 ⎥ ⎝ ⎦ –1 2

11. x + 5 ≤ 3 + 2 x − x ≤ −2 x≥2 [2, ∞)

–1 2

2

6. 2y + 1 > 0 2y > –1 1 y>− 2 ⎛ 1 ⎞ ⎜− 2 , ∞⎟ ⎝ ⎠

12. –3 ≥ 8(2 – x) –3 ≥ 16 – 8x 8x ≥ 19 19 x≥ 8 ⎡ 19 ⎞ ⎢ 8 , ∞⎟ ⎣ ⎠

–1 2

19 8

7. 5 – 7s > 3 –7s > –2 2 s< 7 2⎞ ⎛ ⎜ −∞, 7 ⎟ ⎝ ⎠

13. 3(2 – 3x) > 4(1 – 4x) 6 – 9x > 4 – 16x 7x > –2 2 x>− 7 ⎛ 2 ⎞ ⎜− 7 , ∞⎟ ⎝ ⎠

2 7

–2 7

41



14. 8(x + 1) + 1 < 3(2x) + 1 8x + 9 < 6x + 1 2x < –8 x < –4 (–∞, –4)

2 x>6 3 –x > 9 x < –9 (–∞, –9)

20. −

–4

–9

15. 2(4 x − 2) > 4(2 x + 1) 8x − 4 > 8x + 4 −4 > 4, which is false for all x. Thus the solution set is ∅.

21.

16. 4 − ( x + 3) ≤ 3(3 − x ) 1 − x ≤ 9 − 3x 2x ≤ 8 x≤4 (−∞, 4]

–5

22.

4

17. x + 2 < 3 − x 2x < 3 − 2 x< ⎛ ⎜⎜ – ∞, ⎝

3 −2 2

23. −3x + 1 ≤ −3( x − 2) + 1 −3x + 1 ≤ −3x + 7 1 ≤ 7, which is true for all x. The solution is –∞ < x < ∞. (–∞, ∞)

2 ( x + 2) > 8(3 − x) 2 ( x + 2) > 2 2 (3 − x) x + 2 > 2(3 – x) x + 2 > 6 – 2x 3x > 4 4 x> 3 ⎛4 ⎞ ⎜ 3 , ∞⎟ ⎝ ⎠

24. 0x ≤ 0 0 ≤ 0, which is true for all x. The solution is –∞ < x < ∞. (–∞, ∞)

25.

4 3

19.

3y − 2 1 ≥ 3 4 12 y − 8 ≥ 3 12 y ≥ 11 11 y≥ 12 11 ⎡ ⎞ ⎢ , ∞⎟ ⎣ 12 ⎠ 11 12

3 −2⎞ ⎟ 2 ⎟⎠ 3−2 2

18.

9 y +1 ≤ 2 y −1 4 9y + 1 ≤ 8y – 4 y ≤ –5 (–∞, –5]

5 x < 40 6 5x < 240 x < 48 (–∞, 48)

1 − t 3t − 7 < 2 3 3(1 − t ) < 2(3t − 7) 3 − 3t < 6t − 14 −9t < −17 17 t> 9 ⎛ 17 ⎞ ⎜ , ∞⎟ ⎝ 9 ⎠ 17 9

48

42


26.

Section 1.2

3(2t − 2) 6t − 3 t > + 2 5 10 15(2t – 2) > 2(6t – 3) + t 30t – 30 > 13t – 6 17t > 24 24 t> 7 ⎛ 24 ⎞ ⎜ 7 , ∞⎟ ⎝ ⎠

31.

0

2 − 0.01x 0.2 1.8 – 0.02x ≤ 2 – 0.01x –0.01x ≤ 0.2 x ≥ –20 [–20, ∞)

24 17

32. 9 − 0.1x ≤

1 x−7 3 6 x + 39 ≥ x − 21 5 x ≥ −60 x ≥ −12 (−12, ∞)

27. 2 x + 13 ≥

–20

33. 0.1(0.03x + 4) ≥ 0.02x + 0.434 0.003x + 0.4 ≥ 0.02x + 0.434 –0.017x ≥ 0.034 x ≤ –2 (–∞, –2]

–12

28.

1 5 3x − ≤ x 3 2 18 x − 2 ≤ 15 x 3x ≤ 2 2 x≤ 3 2⎤ ⎛ ⎜ −∞, ⎥ 3⎦ ⎝

–2

34.

2 3

29.

2 5 r< r 3 6 4r < 5r 00 (0, ∞)

3 y − 1 5( y + 1) < −3 −3 3y −1 > 5 y + 5 −6 > 2 y −3 > y y < −3 (−∞, −3) –3

35. 12(50) < S < 12(150) 600 < S < 1800 36. 2 0

30.

y y y + > y+ 2 3 5 15y + 10y > 30y + 6y 25y > 36y 0 > 11y 0>y y<0 (–∞, 0)

1 ≤x≤4 2

37. The measures of the acute angles of a right triangle sum to 90°. If x is the measure of one acute angle, the other angle has measure 90 – x. x < 3(90 – x) + 10 x < 270 – 3x + 10 4x < 280 x < 70 The measure of the angle is less than 70°.

7 8 t>− t 4 3 21t > –32t 53t > 0 t>0 (0, ∞) 0

43



38. Let d be the number of disks. The stereo plus d disks will cost 219 + 18.95d. 219 + 18.95d ≤ 360 18.95d ≤ 141 141 d≤ ≈ 7.44 18.95 The student can buy at most 7 disks. Problems 1.3 1. Let q = number of units sold. Profit > 0 Total revenue – Total cost > 0 20q – (15q + 600,000) > 0 5q – 600,000 > 0 5q > 600,000 q > 120,000 Thus at least 120,001 units must be sold. 2. Let q = number of units sold. Total revenue – Total cost = Profit We want Profit > 0. 7.40q – [(2.50 + 4)q + 5000] > 0 0.9q – 5000 > 0 0.9q > 5000 5000 5 q> = 5555 0.9 9 Thus at least 5556 units must be sold. 3. Let x = number of miles driven per year. If the auto is leased, the annual cost is 12(420) + 0.06x. If the auto is purchased, the annual cost is 4700 + 0.08x. We want Rental cost ≤ Purchase cost. 12(420) + 0.06x ≤ 4700 + 0.08x 5040 + 0.06x ≤ 4700 + 0.08x 340 ≤ 0.02x 17,000 ≤ x The number of miles driven per year must be at least 17,000. 4. Let N = required number of shirts. Then Total revenue = 3.5N and Total cost = 1.3N + 0.4N + 6500. Profit > 0 3.5 N − (1.3 N + 0.4 N + 6500) > 0 1.8 N − 6500 > 0 1.8 N > 6500 N > 3611.1 At least 3612 shirts must be sold. 5. Let q be the number of magazines printed. Then the cost of publication is 0.55q. The number of magazines sold is 0.90q. The revenue from dealers is (0.60)(0.90q). If fewer than 30,000 magazines are sold, the only revenue is from the sales to dealers, while if more than 30,000 are sold, there are advertising revenues of 0.10(0.60)(0.90q − 30,000). Thus,

44


Section 1.3

if 0.9q ≤ 30, 000 ⎧0.6(0.9)q Revenue = ⎨ ⎩0.6(0.9)q + 0.1(0.6)(0.9q − 30, 000) if 0.9q > 30, 000 q ≤ 33,333 ⎧0.54q =⎨ − > 33,333 q q 0.594 1800 ⎩ Profit = Revenue − Cost q ≤ 33,333 ⎧0.54q − 0.55q =⎨ ⎩0.594q − 1800 − 0.55q q > 33,333 q ≤ 33,333 ⎧−0.01q =⎨ 0.044 q − 1800 q > 33,333 ⎩ Clearly, the profit is negative if fewer than 33,334 magazines are sold. 0.044q − 1800 ≥ 0 0.044q ≥ 1800 q ≥ 40,910 Thus, at least 40,910 magazines must be printed in order to avoid a loss.

6. Let q = number of clocks produced during regular work week, so 11,000 – q = number produced in overtime. Then 2q + 3(11,000 – q) ≤ 25,000 –q + 33,000 ≤ 25,000 8000 ≤ q At least 8000 clocks must be produced during the regular workweek. 3 7. Let x = amount at 6 % and 30,000 – x = amount at 5%. Then 4 3 1 interest at 6 % + interest at 5% ≥ interest at 6 % 4 2 x(0.0675) + (30,000 – x)(0.05) ≥ (0.065)(30,000) 0.0175x + 1500 ≥ 1950 0.0175x ≥ 450 x ≥ 25,714.29 3 Thus at least $25,714.29 must be invested at 6 %. 4

8. Let L be current liabilities. Then current assets Current ratio = current liabilities 570, 000 3.8 = L 3.8L = 570,000 L = $150,000 Let x = amount of money they can borrow, where x ≥ 0. 570, 000 + x ≥ 2.6 150, 000 + x 570,000 + x ≥ 390,000 + 2.6x 180,000 ≥ 1.6x 112,500 ≥ x Thus current liabilities are $150,000 and the maximum amount they can borrow is $112,500.

45



9. Let q be the number of units sold this month at $4.00 each. Then 2500 – q will be sold at $4.50 each. Then Total revenue ≥ 10,750 4q + 4.5(2500 – q) ≥ 10,750 –0.5q + 11,250 ≥ 10,750 500 ≥ 0.5q 1000 ≥ q The maximum number of units that can be sold this month is 1000.

2.

4.

−4 − 6 −10 = = −5 = 5 2 2

5.

⎛ 7⎞ 2 ⎜ − ⎟ = −7 = 7 ⎝ 2⎠

6. |3 − 5| − |5 − 3| = |−2| − |2| = 2 − 2 = 0

100 + q > 5000 q > 4900 At least 4901 units must be sold. 11. For t < 40, we want income on hourly basis > income on per-job basis 9t > 320 + 3(40 − t ) 9t > 440 − 3t 12t > 440 t > 36.7 hr

7.

x < 4 , –4 < x < 4

8.

x < 10, –10 < x < 10

9. Because 2 − 5 < 0 ,

(

)

2− 5 = − 2− 5 = 5 −2.

10. Because

12. Let s = yearly sales. With the first method, the salary is 35,000 + 0.03s, and with the second method it is 0.05s. 35, 000 + 0.03s > 0.05s 35, 000 > 0.02 s 1, 750, 000 > s The first method is better for yearly sales less than $1,750,000.

5 − 2 > 0,

11. a.

x−7 < 3

b.

x−2 <3

c.

x−7 ≤ 5

d.

x−7 = 4

e.

x+4 < 2

f.

x <3

g.

x >6

h.

x − 105 < 3

i.

x − 850 < 100

13. Let x = accounts receivable. Then 450, 000 + x Acid test ratio = 398, 000 450, 000 + x 1.3 ≤ 398, 000 517,400 ≤ 450,000 + x x ≥ 67, 400 The company must have at least $67,400 in accounts receivable. Principles in Practice 1.4

12. |f(x) − L| < ε

w − 22 ≤ 0.3

13.

Problems 1.4 1.

1 1 = 2 2

3. 8 − 2 = 6 = 6

10. Revenue = (no. of units)(price per unit) ⎛ 100 ⎞ + 1⎟ > 5000 q⎜ ⎝ q ⎠

1.

2 –1 =

−13 = 13 46

p1 − p2 ≤ 9

5 − 2 = 5 − 2.


14.

Section 1.4

x − µ ≤ 2σ

23. 7 − 4 x = 5 7 – 4x = ±5 –4x = –7 ± 5 –4x = –2 or –12 1 x = or x = 3 2

–2σ ≤ x – µ ≤ 2σ µ – 2 σ ≤ x ≤ µ + 2σ 15.

x =7 x = ±7

16.

−x = 2 –x = 2 or –2 x = ±2

17.

18.

19.

20.

24. 5 − 3 x = 2 5 − 3 x = ±2 −3 x = −5 ± 2 −3 x = −3 or − 7 7 x = 1 or x = 3

x =7 5 x = ±7 5 x = ±35

25.

x
−M < x < M (–M, M) Note that M > 0 is required.

5 = 12 x 5 = ±12 x 5 x=± 12

26.

−x < 3 x <3

–3 < x < 3 (–3, 3)

x −5 = 8 x – 5 = ±8 x=5±8 x = 13 or x = –3

27.

x >2 4 x x < −2 or >2 4 4 x < –8 or x > 8, so the solution is (–∞, –8) ∪ (8, ∞).

4 + 3x = 6

4 + 3x = ±6 3x = –4 ± 6 3x = –10 or 2 10 2 or x = x= − 3 3

28.

x 1 > 3 2 x 1 <− or 3 2 3 x<− or 2 3⎞ ⎛3 ⎛ ⎞ ⎜ −∞, − 2 ⎟ ∪ ⎜ 2 , ∞ ⎟ . ⎝ ⎠ ⎝ ⎠

21. 5 x − 2 = 0 5x – 2 = 0 2 x= 5 22. 7 x + 3 = x Here we must have x ≥ 0. 7x + 3 = x or –(7x + 3) = x 6x = –3 –7x – 3 = x 3 1 x=− <0 x=− <0 8 2 There is no solution.

29.

x 1 > 3 2 3 x > , so the solution is 2

x+9 < 5 −5 < x + 9 < 5 −14 < x < −4 (−14, −4)

30. |2x − 17| < −4 Because –4 < 0, the solution set is ∅.

47


31.

x−


1 1 > 2 2

1 1 <− 2 2 x<0 (–∞, 0) ∪ (1, ∞) x−

32. 1 − 3x > 2 1 – 3x > 2 –3x > 1 1 x<– 3

36. 1 1 > 2 2 or x > 1

or

x−

37. |d − 35.2 m| ≤ 20 cm or |d − 35.2| ≤ 0.20

or 1 – 3x < –2 or –3x < –3

38. Let T1 and T2 be the temperatures of the two chemicals. 5 ≤ T1 – T2 ≤ 10

or x > 1

1⎞ ⎛ The solution is ⎜ −∞, − ⎟ ∪ (1, ∞). 3⎠ ⎝

39.

40.

1. The bounds of summation are 12 and 17; the index of summation is t. 2. The bounds of summation are 3 and 450; the index of summation is m.

4 x − 1 ≥ 0 is true for all x because a ≥ 0 for all

3.

7

∑ 6i i =1

3x − 8 ≥4 2 3x − 8 ≤ −4 2 3x – 8 ≤ –8 3x ≤ 0

x − 0.01 ≤ 0.005

Problems 1.5

a. Thus –∞ < x < ∞, or (–∞,∞). 35.

x − µ > hσ

Either x – µ < –hσ, or x – µ > hσ. Thus either x < µ – hσ or x > µ + hσ, so the solution is (–∞, µ – hσ) ∪ (µ + hσ, ∞).

33. 5 − 8 x ≤ 1 –1 ≤ 5 – 8x ≤ 1 –6 ≤ –8x ≤ –4 3 1 ≥ x ≥ , which may be rewritten as 4 2 1 3 ≤x≤ . 2 4 ⎡1 3⎤ The solution is ⎢ , ⎥ . ⎣2 4⎦ 34.

x−7 ≤5 3 x−7 −5 ≤ ≤5 3 −15 ≤ x − 7 ≤ 15 −8 ≤ x ≤ 22 [−8, 22]

= 6(1) + 6(2) + 6(3) + 6(4) + 6(5) + 6(6) + 6(7) = 6 + 12 + 18 + 24 + 30 + 36 + 42 = 168

3x − 8 ≥4 2 or 3x – 8 ≥ 8 or 3x ≥ 16 16 x≤0 or x ≥ 3 ⎡ 16 ⎞ The solution is (– ∞, 0] ∪ ⎢ , ∞ ⎟ . ⎣3 ⎠

or

4.

4

∑ 10 p = 10(0) + 10(1) + 10(2) + 10(3) + 10(4)

p =0

= 0 + 10 + 20 + 30 + 40 = 100

48


5.

9

∑ (10k + 16) = [10(3) + 16] + [10(4) + 16] + [10(5) + 16] + [10(6) + 16] + [10(7) + 16] + [10(8) + 16] + [10(9) + 16]

k =3

6.

Section 1.5

= 46 + 56 + 66 + 76 + 86 + 96 + 106 = 532

11

∑ (2n − 3) = [2(7) − 3] + [2(8) − 3] + [2(9) − 3] + [2(10) − 3] + [2(11) − 3]

n =7

= 11 + 13 + 15 + 17 + 19 = 75

7. 36 + 37 + 38 + 39 + " + 60 =

60

∑i

i =36

8. 1 + 4 + 9 + 16 + 25 =

5

∑ k2

k =1

8

9. 53 + 54 + 55 + 56 + 57 + 58 = ∑ 5i i=3

16

10. 11 + 15 + 19 + 23 + " + 71 = ∑ (7 + 4i ) i =1

11. 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =

8

∑ 2i i=1 8

12. 10 + 100 + 1000 + " + 100, 000, 000 = ∑ 10 j j=1

13.

14.

15.

43

43

k =1

k =1

135

135

101

k =35

k =35

i =1

∑ 10 = 10 ∑ 1 = 10(43) = 430 ∑ 2 = 2 ∑ 1 = 2∑1 = 2(101) = 202 n

⎛

1⎞

⎛

1⎞

k =1

16.

n

⎛

1⎞

∑ ⎜⎝ 5 ⋅ n ⎟⎠ = ⎜⎝ 5 ⋅ n ⎟⎠ ∑ 1 = ⎜⎝ 5 ⋅ n ⎟⎠ (n) = 5 k =1

200

200

200

k =1

k =1

k =1

∑ (k − 100) =

∑ k − 100 ∑ 1 =

200(201) − 100(200) = 20,100 − 20, 000 = 100 2

49


17.

100


50

∑ 10k = 10∑ (i + 50)

k =51

i =1

50

23.

50

= 10∑ i + (10)(50)∑1 i =1

50(51) + 500(50) = 12, 750 + 25, 000 2 = 37,750 n

n

19.

k =1

n n(n + 1)(2n + 1) = ⋅ 6 n +1 n 2 (2n + 1) = 6

20

20

24.

k =1

k =1

20(21)(41) 20(21) +3 6 2 = 5(2870) + 3(210) = 14,980

21.

3k 2 – 200k 3 100 2 200 100 = ∑ 101 ∑ k − 101 ∑ k 101 k =1 k =1 k =1 3 100(101)(201) 200 100 ⋅101 = ⋅ − ⋅ 101 6 101 2 = 10,050 – 10,000 = 50 50

k =51 50

i =1

25.

50

∑ k 2 = ∑ (i + 50)2 = ∑ (i 2 + 100i + 2500) i =1 50

50

i =1

50

i =1

50

∑ (k + 50)2 = ∑ (k 2 + 100k + 2500)

k =1 50

=

k =1 50

50

k =1

k =1

26.

=

100 1 1 ⎛ 1 ⎞ 100 2 (4) ∑ 1 − ⎜ ⎟∑k 50 k =1 50 ⎝ 2500 ⎠ k =1

⎧⎪ ⎡ ⎛ 3 ⎞2 ⎤ 3 ⎫⎪ ∑ ⎨⎢⎢5 − ⎜⎝ n ⋅ k ⎟⎠ ⎥⎥ n ⎬ k =1 ⎪ ⎦ ⎪⎭ ⎩⎣ n 3 ⎛ 9 2⎞ k ⎟ = ∑ ⎜5 − n k =1 ⎝ n2 ⎠ n

k2

n

=

50(51)(101) 50(51) + 100 + 2500(50) 6 2 = 42,925 + 127,500 + 125,000 = 295,425 =

50

1

n

∑ (n + 1)(2n + 1) = (n + 1)(2n + 1) ∑ k 2

k =1

∑ k 2 + 100 ∑ k + 2500 ∑ 1

k =1

1 100 ⎛ 1 ⎞ k2 ⎟ ∑ ⎜4− 50 k =1 ⎝ 2500 ⎠

=

50(51)(101) 50(51) = + 100 + 2500(50) 6 2 = 42,925 + 127,500 + 125,000 = 295,425

22.

=

n 3 3⎛ 9 ⎞ n (5) ∑ 1 − ⎜ ⎟ ∑ k 2 n k =1 n ⎝ n 2 ⎠ k =1 15 27 n(n + 1)(2n + 1) = ( n) − ⋅ 6 n n3 9(n + 1)(2n + 1) = 15 − 2n 2

= ∑ i 2 + 100∑ i + 2500∑ 1 i =1

⎛ 2 ⎞ ⎫⎪ ⎥ ⎜ ⎟⎬ ⎥⎦ ⎝ 100 ⎠ ⎭⎪

2⎤

=

100

100

⎪ ⎛ 2 ⎞ ∑ ⎨⎢⎢ 4 − ⎜⎝ 100 k ⎟⎠ ⎪⎣ k =1 ⎩

2 1 100(101)(201) (100) − ⋅ 25 125, 000 6 1 6767 = 8− ⋅ 338,350 = 8 − 125, 000 2500 13, 233 733 = =5 2500 2500

= 5⋅

20.

100 ⎧ ⎡

20

∑ (5k 2 + 3k ) = 5 ∑ k 2 + 3∑ k

k =1

k =1 ⎪ ⎩⎣

4 1 10(11)(21) 1 (10) − ⋅ = 8− ⋅ 385 5 125 6 125 77 123 23 = 8− = =4 25 25 25

∑ n +1 k 2 = n +1 ∑ k 2

k =1

2 10 1 ⎛ 2k ⎞ ⎤ ⎛ 2 ⎞ ⎫⎪ 1 ⎛ ⎞ ⎥ ⎜ ⎟⎬ = ∑ ⎜ 4 − k 2 ⎟ 10 5 25 ⎠ ⎥⎦ ⎝ ⎠ ⎪⎭ k =1 ⎝

=

n

n

⎧⎪ ⎡

10 1 1 ⎛ 1 ⎞ 10 = (4) ∑ 1 − ⎜ ⎟ ∑ k 2 5 k =1 5 ⎝ 25 ⎠ k =1

i =1

= 10 ⋅

18.

10

∑ ⎨⎢⎢4 − ⎜⎝ 10 ⎟⎠

k =1

1 n(n + 1)(2n + 1) n ⋅ = (n + 1)(2n + 1) 6 6


Chapter 1 Review

Chapter 1 Review Problems

7.

1. 5 x − 2 ≥ 2( x − 7) 5 x − 2 ≥ 2 x − 14 3 x ≥ −12 x ≥ −4 [−4, ∞) 2. 2x – (7 + x) ≤ x 2x – 7 – x ≤ x –7 ≤ 0, which is true for all x, so –∞ < x < ∞, or (–∞, ∞).

8.

3. –(5x + 2) < –(2x + 4) –5x – 2 < –2x – 4 –3x < –2 2 x> 3 ⎛2 ⎞ ⎜ 3 , ∞⎟ ⎝ ⎠

9.

4. –2(x + 6) > x + 4 –2x – 12 > x + 4 –3x > 16 16 x<− 3 16 ⎞ ⎛ ⎜ −∞, − 3 ⎟ ⎝ ⎠

10.

5. 3 p (1 − p ) > 3(2 + p ) − 3 p 2 3 p − 3 p2 > 6 + 3 p − 3 p2 0 > 6, which is false for all x. The solution set is ∅.

x+5 1 − ≤2 3 2 2(x + 5) – 3(1) ≤ 6(2) 2x + 10 – 3 ≤ 12 2x ≤ 5 5 x≤ 2 5⎤ ⎛ ⎜ −∞, 2 ⎥ ⎝ ⎦ x x x − > 3 4 5 20 x − 15 x > 12 x 5 x > 12 x 0 > 7x 0>x (−∞, 0) 1 1 s − 3 ≤ (3 + 2 s ) 4 8 2s – 24 ≤ 3 + 2s 0 ≤ 27, which is true for all s. Thus –∞ < s < ∞, or (–∞,∞). 1 1 (t + 2) ≥ 3 4 4(t + 2) ≥ 3t + 48 4t + 8 ≥ 3t + 48 t ≥ 40 [40, ∞)

11. 3 − 2 x = 7 3 – 2x = 7 –2x = 4 x = –2

7 ⎞ ⎛ 6. 3 ⎜ 5 − q ⎟ < 9 3 ⎠ ⎝ 15 − 7q < 9 −7q < −6 6 q> 7 6 ⎛ ⎞ ⎜ 7 , ∞⎟ ⎝ ⎠

12.

5x − 6 =0 13 5x − 6 =0 13 5x – 6 = 0 6 x= 5

13. |2z − 3| < 5 −5 < 2z − 3 < 5 −2 < 2z < 8 −1 < z < 4 (−1, 4) 51

or 3 – 2x = –7 or –2x = –10 or x = 5


14. 4 <

ISM: Introductory Mathematical Analysis 19. Let x be the number of issues with a decline, and x + 48 be the number of issues with an increase. Then x + (x + 48) = 1132 2x = 1084 x = 542

2 x+5 3

2 x + 5 < −4 3 2 x < −9 3 27 x<− 2

2 x+5 > 4 3 2 or x > −1 3 3 or x > − 2 27 ⎛ ⎞ ⎛ 3 ⎞ The solution is ⎜ −∞, − ⎟ ∪ ⎜ − , ∞ ⎟ . 2 ⎠ ⎝ 2 ⎝ ⎠

or

20. Let x = purchase amount excluding tax. x + 0.065 x = 3039.29 1.065 x = 3039.29 x = 2853.79 Thus tax is 3039.29 − 2853.79 = $185.50. 21. Let q units be produced at A and 10,000 – q at B. Cost at A + Cost at B ≤ 117,000 [5q + 30,000] + [5.50(10,000 – q) + 35,000] ≤ 117,000 –0.5q + 120,000 ≤ 117,000 –0.5q ≤ –3000 q ≥ 6000 Thus at least 6000 units must be produced at plant A.

15. 3 − 2 x ≥ 4 3 – 2x ≥ 4 –2x ≥ 1 1 x≤− 2

or 3 – 2x ≤ –4 or –2x ≤ –7 7 or x ≥ 2 1⎤ ⎡7 ⎛ ⎞ The solution is ⎜ – ∞, − ⎥ ∪ ⎢ , ∞ ⎟ . 2⎦ ⎣2 ⎝ ⎠

16.

5

5

i =1

i =1 5

∑ (i + 2)3 = ∑ (i3 + 6i 2 + 12i + 8)

22. Total volume of old tanks

5

5

5

i =1

i =1

i =1

= π(10)2 (25) + π(20)2 (25) = 2500π + 10, 000π

= ∑ i3 + 6∑ i 2 + 12∑ i + 8∑ 1 i =1 2

2

= 12,500π ft 3 Let r be the radius (in feet) of the new tank. Then 4 3 πr = 12,500π 3 r 3 = 9375

5 (6) 5(6)(11) 5(6) +6 + 12 + 8(5) 4 6 2 = 225 + 330 + 180 + 40 = 775 =

17.

7

7

2

i =3

i =1 2

i =1

r = 3 9375 ≈ 21.0858 The radius is approximately 21.0858 feet.

∑ i3 = ∑ i3 − ∑ i3

7 (8)2 22 (3)2 = − 4 4 = 784 − 9 = 775 This uses Equation (1.9). By Equation (1.8), 7

5

i =3

i =1

23. Let c = operating costs c < 0.90 236, 460 c < $212,814

∑ i3 = ∑ (i + 2)3 .

Mathematical Snapshot Chapter 1 1 1. Here m = 120 and M = 2 (60) = 150. For LP, 2 t r = 2, so the first t minutes take up of the 120 2 available minutes. For SP, r = 1, so the 150 − t of the remaining 150 − t minutes take up 1 120 available.

18. Let p = selling price, c = cost. Then p – 0.40p = c 0.6p = c c 5c ⎛2⎞ p= = = c + ⎜ ⎟c 0.6 3 ⎝3⎠ Thus the profit is

2 2 , or 66 %, of the cost. 3 3 52


Mathematical Snapshot Chapter 1 5.

t 150 − t + = 120 2 1 t + 300 − 2t = 240 −t = −60 t = 60 Switch after 1 hour.

x = 600

1 2. Here m = 120 and M = 2 (60) = 150. For EP, 2 t of the r = 3, so the first t minutes will take up 3 120 available minutes. For SP, r = 1, so the 150 − t of the remaining 150 − t minutes take up 1 120 available. t 150 − t + = 120 3 1 t + 450 − 3t = 360 −2t = −90 t = 45 Switch after 45 minutes.

x = 310 6. Both equations represent audio being written onto 74-minute CDs. In the first equation, 18 hours (1080 minutes) are being written to a CD using a combination of 12-to-1 and 20-to-1 compression ratios. Here, x gives the maximum amount of audio (600 minutes or 10 hours) that can be written using the 12-to-1 compression ratio. In the second equation, 26.5 hours (1590 minutes) is being written using 15-to-1 and 24to-1 compression ratios. A maximum of 310 minutes can be written at 15-to-1.

3. Use the reasoning in Exercise 1, with M unknown and m = 120. t M −t + = 120 2 1 t + 2 M − 2t = 240 −t = 240 − 2 M t = 2 M − 240 The switch should be made after 2M − 240 minutes.

t of the m available R M −t minutes, the remaining M − t minutes use r of the m available. t M −t + =m R r t M t + − =m R r r M ⎛ 1 1⎞ t⎜ − ⎟ = m− r ⎝R r⎠ ⎛ r − R ⎞ mr − M t⎜ ⎟= r ⎝ rR ⎠ R (mr − M ) t= r−R

7. The first t minutes use

4. Use the reasoning in Exercise 2, with M unknown and m = 120. t M −t + = 120 3 1 t + 3M − 3t = 360 −2t = 360 − 3M 1 t = (3M − 360) 2 The switch should be made after 1 (3M − 360) minutes. 2

53

Chapter 2 b. If 200 large pizzas are being sold each week, q = 200. 200 p = 26 − 40 p = 26 – 5 p = 21 The price is $21 per pizza if 200 large pizzas are being sold each week.

Principles in Practice 2.1 1. a.

The formula for the area of a circle is πr 2 , where r is the radius. a(r ) = πr 2

b. The domain of a(r) is all real numbers. c.

2. a.

Since a radius cannot be negative or zero, the domain for the function, in context, is r > 0.

c.

The formula relating distance, time, and speed is d = rt where d is the distance, r is the speed, and t is the time. This can also be d written as t = . When d = 300, we have r 300 t (r ) = . r

Problems 2.1 1. The functions are not equal because f(x) ≥ 0 for all values of x, while g(x) can be less than 0. For

b. The domain of t(r) is all real numbers except 0. c.

example, f (−2) = (−2) 2 = 4 = 2 and

Since speed is not negative, the domain for the function, in context, is r > 0.

g(−2) = −2, thus f(−2) ≠ g(−2). 2. The functions are different because they have different domains. The domain of G(x) is [−1, ∞) (all real numbers ≥ −1) because you can only take the square root of a non-negative number, while the domain of H(x) is all real numbers.

300 . x x ⎛ x ⎞ 300 600 . Replacing r by : t ⎜ ⎟ = = x 2 x ⎝2⎠ 2

d. Replacing r by x: t ( x) =

Replacing r by e.

3. a.

x : 4

To double the number of large pizzas sold, use q = 400. 400 p = 26 − 40 p = 26 – 10 p = 16 To sell 400 large pizzas each week, the price should be $16 per pizza.

3. The functions are not equal because they have different domains. h(x) is defined for all nonzero real numbers, while k(x) is defined for all real numbers.

⎛ x ⎞ 300 1200 t⎜ ⎟ = . = x x ⎝4⎠ 4

When the speed is reduced (divided) by a constant, the time is scaled (multiplied) by ⎛ r ⎞ 300c . the same constant; t ⎜ ⎟ = r ⎝c⎠

4. The functions are equal. For x = 3 we have f(3) = 2 and g(3) = 3 − 1 = 2, hence f(3) = g(3). For x ≠ 3, we have x 2 − 4 x + 3 ( x − 3)( x − 1) = = x − 1. f ( x) = x−3 x−3 Note that we can cancel the x − 3 because we are assuming x ≠ 3 and so x − 3 ≠ 0. Thus for x ≠ 3 f(x) = x − 1 = g(x). f(x) = g(x) for all real numbers and they have the same domains, thus the functions are equal.

If the price is $18.50 per large pizza, p = 18.5. q 18.5 = 26 − 40 q −7.5 = − 40 300 = q At a price of $18.50 per large pizza, 300 pizzas are sold each week.

5. The denominator is zero when x = 0. Any other real number can be used for x. Answer: all real numbers except 0 54


Section 2.1

6. Any real number can be used for x. Answer: all real numbers

16. r 2 + 1 is never 0. Answer: all real numbers

7. For x − 3 to be real, x – 3 ≥ 0, so x ≥ 3. Answer: all real numbers ≥ 3

17. f(x) = 2x + 1 f(0) = 2(0) + 1 = 1 f(3) = 2(3) + 1 = 7 f(–4) = 2(–4) + 1 = –7

8. For z − 1 to be real, z − 1 ≥ 0, so z ≥ 1. We exclude values of z for which z − 1 = 0, so z − 1 = 0, thus z = 1. Answer: all real numbers > 1

18. H ( s ) = 5s 2 − 3 H (4) = 5(4)2 − 3 = 80 − 3 = 77

9. Any real number can be used for z. Answer: all real numbers

H

10. We exclude values of x for which x+8=0 x = –8 Answer: all real numbers except –8

2

− 3 = 10 − 3 = 7

2

20 7 ⎛2⎞ ⎛2⎞ H ⎜ ⎟ = 5⎜ ⎟ − 3 = −3 = − 3 3 9 9 ⎝ ⎠ ⎝ ⎠

19. G ( x) = 2 − x 2

11. We exclude values of x where 2x + 7 = 0 2x = –7 7 x=− 2

G (–8) = 2 − (−8)2 = 2 − 64 = −62 G (u ) = 2 − u 2 G (u 2 ) = 2 − (u 2 )2 = 2 − u 4

7 Answer: all real numbers except − 2

20. F(x) = −5x F(s) = −5s F(t + 1) = −5(t + 1) = −5t − 5 F(x + 3) = −5(x + 3) = −5x − 15

12. For 4 x + 3 to be real, 4x + 3 ≥ 0 4x ≥ –3 3 x≥− 4 Answer: all real numbers ≥ −

( 2 ) = 5( 2 )

21. γ (u ) = 2u 2 − u

3 4

γ (−2) = 2(−2) 2 − (−2) = 8 + 2 = 10 γ (2v) = 2(2v) 2 − (2v) = 8v 2 − 2v

13. We exclude values of y for which

γ ( x + a ) = 2( x + a) 2 − ( x + a)

y 2 − 4 y + 4 = 0. y 2 − 4 y + 4 = ( y − 2) 2 , so we

= 2 x 2 + 4ax + 2a 2 − x − a

exclude values of y for which y − 2 = 0, thus y = 2. Answer: all real numbers except 2.

22. h(v) =

14. We exclude values of x for which 2

x + x−6 = 0 ( x + 3)( x − 2) = 0 x = −3, 2 Answer: all real numbers except −3 and 2

h(16) =

v 1

⎛1⎞ h⎜ ⎟ = ⎝4⎠

15. We exclude all values of s for which

h(1 − x) =

2s 2 − 7 s − 4 = 0 (s – 4)(2s + 1) = 0 1 s = 4, − 2

Answer: all real numbers except 4 and −

1

1 2 55

16 1 1 4

=

=

1 4

1 1 2

1 1− x

=2

Chapter 2: Functions and Graphs

23.


f ( x) = x 2 + 2 x + 1

28. g ( x) = x 2 / 5

( 5 32 ) = (2)2 = 4 2 g (−64) = (−64)2 / 5 = ( 5 −64 ) 2 2 = ( 5 −32 5 2 ) = ( –2 5 2 ) = 4 5 4

f (1) = 12 + 2(1) + 1 = 1 + 2 + 1 = 4

g (32) = 322 / 5 =

f (–1) = (–1) 2 + 2(–1) + 1 = 1 − 2 + 1 = 0 f ( x + h) = ( x + h) 2 + 2( x + h) + 1 = x 2 + 2 xh + h 2 + 2 x + 2h + 1

24. H ( x) = ( x + 4) 2

g (t10 ) = (t10 )2 / 5 = t 4

H (0) = (0 + 4) 2 = 16

29. f(x) = 4x – 5

H (2) = (2 + 4)2 = 62 = 36

a.

H (t − 4) = [(t − 4) + 4]2 = t 2

25. k ( x) = k (5) =

b.

x−7 x2 + 2 5−7 2

=−

2 27

5 +2 3x − 7 3x − 7 k (3 x) = = 2 (3x) + 2 9 x 2 + 2 ( x + h) − 7 x+h−7 k ( x + h) = = 2 2 ( x + h) + 2 x + 2 xh + h 2 + 2

30.

26. k ( x) = x − 3 k (4) = 4 − 3 = 1 = 1 k (3) = 3 − 3 = 0 = 0

31.

k ( x + 1) − k ( x) = ( x + 1) − 3 − x − 3 = x −2 − x −3

27.

f(x + h) = 4(x + h) – 5 = 4x + 4h – 5 f ( x + h) − f ( x ) h (4 x + 4h − 5) − (4 x − 5) 4h = = =4 h h

f ( x) =

x 2 x+h 2

a.

f ( x + h) =

b.

f ( x + h) − f ( x ) = h

x+h 2

− 2x

h

=

h 2

h

=

1 2

f ( x) = x 2 + 2 x

a.

f ( x + h) = ( x + h) 2 + 2( x + h) = x 2 + 2 xh + h 2 + 2 x + 2h

f ( x) = x 4 / 3 f (0) = 04 / 3 = 0

( )

f (64) = 644 / 3 = 3 64

⎛1⎞ ⎛1⎞ f ⎜ ⎟=⎜ ⎟ ⎝8⎠ ⎝8⎠

2

4/3

4

b.

= (4) 4 = 256 4

4 ⎛ 1⎞ 1 ⎛1⎞ = ⎜3 ⎟ = ⎜ ⎟ = ⎜ 8⎟ 16 ⎝2⎠ ⎝ ⎠

32.

f ( x + h) − f ( x ) h =

( x 2 + 2 xh + h 2 + 2 x + 2h) − ( x 2 + 2 x) h

=

2 xh + h 2 + 2h = 2x + h + 2 h

f ( x) = 3x 2 − 2 x − 1

a.

f ( x + h) = 3( x + h) 2 − 2( x + h) − 1 = 3( x 2 + 2 xh + h 2 ) − 2 x − 2h − 1 = 3 x 2 + 6 xh + 3h 2 − 2 x − 2h − 1

56


b.

33.

Section 2.1

f ( x + h) − f ( x) (3x 2 + 6 xh + 3h 2 − 2 x − 2h − 1) − (3 x 2 − 2 x − 1) = h h 6 xh + 3h 2 − 2h = h = 6 x + 3h − 2

f ( x) = 3 − 2 x + 4 x 2

a.

f ( x + h) = 3 − 2( x + h) + 4( x + h)2 = 3 − 2 x − 2h + 4( x 2 + 2 xh + h 2 )

b.

34.

35.

f ( x + h) − f ( x) 3 − 2 x − 2h + 4 x 2 + 8 xh + 4h 2 − (3 − 2 x + 4 x 2 ) = h h −2h + 8 xh + 4h 2 = h = −2 + 8 x + 4h

f ( x ) = x3

a.

f ( x + h) = ( x + h)3 = x3 + 3 x 2 h + 3 xh 2 + h3

b.

f ( x + h) − f ( x) ( x3 + 3 x 2 h + 3xh 2 + h3 ) − x3 3x 2 h + 3 xh 2 + h3 = = = 3x 2 + 3xh + h 2 h h h

f ( x) =

1 x 1 x+h

a.

f ( x + h) =

b.

1 1 −h f ( x + h) − f ( x ) x + h − x 1 x( x + h) = = = =− h h h x ( x + h) h x ( x + h)

x −( x + h )

36.

f ( x) =

x +8 x ( x + h) + 8 x + h + 8 = x+h x+h

a.

f ( x + h) =

b.

x + h + 8 − x +8 x( x + h) x +x +h h+8 − f ( x + h) − f ( x ) x = x+ h = h h x ( x + h) h

(

=

2

2

x +8 x

) = x( x + h + 8) − ( x + h)( x + 8)

−8h x + xh + 8 x − x − hx − 8 x − 8h 8 = =− x ( x + h) h x ( x + h) x ( x + h) h

57

x ( x + h) h


37.

38.


f (3 + h) − f (3) [5(3 + h) + 3] − [5(3) + 3] = h h [15 + 5h + 3] − [15 + 3] = h 18 + 5h − 18 = h 5h = h =5

42. x 2 + y 2 = 1 Solving for y we have y = ± 1 − x 2 . If x = 0, then y = ±1, so y is not a function of x. Solving for x gives x = ± 1 − y 2 . If y = 0, then x = ±1, so x is not a function of y. 43. Yes, because corresponding to each input r there is exactly one output, πr 2 .

f ( x) − f (2) 2 x 2 − x + 1 − (8 − 2 + 1) = x−2 x−2 2 2x − x + 1 − 7 = x−2 2x2 − x − 6 = x−2 = 2x + 3

44. a. b.

f ( a ) = a 2 a 3 + a 3 a 2 = a 5 + a 5 = 2a 5

f (ab) = a 2 (ab)3 + a3 (ab) 2 = a 2 a 3b 3 + a 3 a 2 b 2 = a 5 b3 + a 5 b 2 = a5b 2 (b + 1)

45. Weekly excess of income over expenses is 6500 − 4800 = 1700. After t weeks the excess accumulates to 1700t. Thus the value of V of the business at the end of t weeks is given by V = f(t) = 25,000 + 1700t.

39. 9y – 3x – 4 = 0 3x + 4 shows that for The equivalent form y = 9 3x + 4 each input x there is exactly one output, . 9 Thus y is a function of x. Solving for x gives 9y − 4 x= . This shows that for each input y 3 9y − 4 . Thus x is a there is exactly one output, 3 function of y.

46. Depreciation at the end of t years is 0.02t(30,000), so value V of machine is V = f(t) = 30,000 – 0.02t(30,000), or V = f(t) = 30,000(1 – 0.02t). 47. Yes; for each input q there corresponds exactly one output, 1.25q, so P is a function of q. The dependent variable is P and the independent variable is q.

40. x 2 + y = 0

48. Charging $600,000 per film corresponds to p = 600,000. 1, 200, 000 600,000 = q q=2

The form y = − x 2 shows that for each input x there is exactly one output, − x 2 . Thus y is a function of x. Solving for x gives x = ± − y . If, for example, y = –1, then x = ±1, so x is not a function of y.

The actor will star in 2 films per year. To star in 4 films per year the actor should charge 1, 200, 000 p= = $300, 000 per film. 4

2

41. y = 7 x For each input x, there is exactly one output 7 x 2 . Thus y is a function of x. Solving for x

49. The function can be written as q = 48p. At $8.39 per pound, the coffee house will supply q = 48(8.39) = 402.72 pounds per week. At $19.49 per pound, the coffee house will supply q = 48(19.49) = 935.52 pounds per week. The amount the coffee house supplies increases as the price increases.

y . If, for example, y = 7, then 7 x = ±1, so x is not a function of y.

gives x = ±

58

ISM: Introductory Mathematical Analysis 50. a.

f(0) = 1 – 1 = 0 3

b.

c.

Section 2.2

1 1 1 52. P (1) = 1 − (1 − 0.344)0 = 1 − (1) = 2 2 2 1 1 P (2) = 1 − (1 − 0.344)1 = 1 − (0.656) = 0.672 2 2

3

27 ⎛ 300 ⎞ ⎛3⎞ f (100) = 1 − ⎜ = 1− ⎜ ⎟ = 1− ⎟ 64 ⎝ 400 ⎠ ⎝4⎠ 37 = 64 ⎛ 300 ⎞ f (900) = 1 − ⎜ ⎟ ⎝ 1200 ⎠ ⎛1⎞ = 1− ⎜ ⎟ ⎝4⎠ 1 = 1− 64 63 = 64

53. a.

3

b. Domain: 10, 12, 17, 20 g(10) = 3000, g(17) = 2300

3

54. a.

c. 55. a.

3

c. 56. a.

3

⎛ 300 ⎞ ⎜ 300 + t ⎟ = 0.5 ⎝ ⎠ 300 = 3 0.5 300 + t 300 = 300 3 0.5 + t 3 0.5 t=

300 − 300 3 0.5 3

0.5

)

4

c. 57. a.

f (1000) =

(

3

1000 2500

1,997,723.57

2,964,247.40 7.89

1.21

Principles in Practice 2.2 =

104 10, 000 = =4 2500 2500

(

)

⎡ 3 1000(2) ⎤ 10 3 2 ⎦ = f (2000) = ⎣ 2500 2500

1. a.

4

Let n = the number of visits and p(n) be the premium amount. p(n) = 125

b. The premiums do not change regardless of the number of doctor visits. c.

3

10, 000 24 3 = 4 23 ⋅ 2 = 83 2 2500

f (2 I 0 ) =

–17.43

b. 63.85

≈ 77.98

c.

4

c.

–5.13

b. 1,287,532.35

78 days

=

–18.51

b. 2.64

⎛ 300 ⎞ 0.500 = 1 − ⎜ ⎟ ⎝ 300 + t ⎠

b.

–18.97

b. –581.77

d. We solve

51. a.

Domain: 3000, 2900, 2300, 2000 f(2900) = 12, f(3000) = 10

2. a.

(2 I 0 ) 4 / 3 24 / 3 I 04 / 3 = 2500 2500

This is a constant function. d (t ) = 3t 2 is a quadratic function.

b. The degree of d (t ) = 3t 2 is 2.

⎡ I4/3 ⎤ = 23 2 ⎢ 0 ⎥ = 23 2 f ( I0 ) ⎢⎣ 2500 ⎥⎦

c.

The leading coefficient of d (t ) = 3t 2 is 3.

3. The price for n pairs of socks is given by ⎧ 3.5n 0 ≤ n ≤ 5 ⎪ c(n) = ⎨ 3n 5 < n ≤ 10 . ⎪2.75n 10 < n ⎩

Thus f (2 I 0 ) = 2 3 2 f ( I 0 ) , which means that doubling the intensity increases the response by a factor of 2 3 2. 59



4. Think of the bookshelf having 7 slots, from left to right. You have a choice of 7 books for the first slot. Once a book has been put in the first slot, you have 6 choices for which book to put in the second slot, etc. The number of arrangements is 7 · 6 · 5 · 4 · 3 · 2 · 1 = 7! = 5040.

18. g ( x) = x − 3 g (10) = 10 − 3 = 7 = 7 g (3) = 3 – 3 = 0 = 0 g (–3) = –3 − 3 = −6 = 6

19. F(10) = 1

Problems 2.2

(

)

F − 3 = −1

1. yes

F(0) = 0 ⎛ 18 ⎞ F ⎜ − ⎟ = −1 ⎝ 5⎠

x3 + 7 x − 3 1 3 7 = x + x − 1, which is a 2. f ( x) = 3 3 3 polynomial function.

20. f(3) = 4 f(–4) = 3 f(0) = 4

3. no 4. yes

21. G(8) = 8 − 1 = 7 G(3) = 3 − 1 = 2

5. yes 6. yes

G (−1) = 3 − (−1)2 = 2

7. no

G (1) = 3 − (1)2 = 2

8. g ( x) = 4 x –4 =

4 x4

22. F (3) = 32 − 3(3) + 1 = 1

, which is a rational function.

9. all real numbers

F(−3) = 2(−3) − 5 = −11 F(2) is not defined.

10. all real numbers

23. 6! = 6 · 5 · 4 · 3 · 2 · 1 = 720

11. all real numbers

24. 0! = 1

12. all x such that 1 ≤ x ≤ 3

25. (4 – 2)! = 2! = 2 · 1 = 2

13. a.

26. 6! ⋅ 2! = (6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)(2 ⋅1) = (720)(2) = 1440

3

b. 7 14. a.

1 27.

b. 7 15. a.

7 28.

b. 1 16. a.

0

b. 9 17. f(x) = 8 f(2) = 8 f(t + 8) = 8

(

)

f − 17 = 8

60

n! n ⋅ (n − 1)! = =n (n − 1)! (n − 1)!

8! 8! = 5!(8 − 5)! 5! ⋅ 3! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)(3 ⋅ 2 ⋅1) 8⋅7⋅6 = 3 ⋅ 2 ⋅1 = 8⋅7 = 56


Section 2.3

29. Let i = the passenger’s income and c(i) = the cost for the ticket. c(i) = 4.5 This is a constant function.

b.

30. Let w = the width of the prism, then w + 3 = the length of the prism, and 2w – 1 = the height of the prism. The formula for the volume of a rectangular prism is V = length · width · height.

38. a.

V ( w) = ( w + 3)( w)(2w − 1) = 2w3 + 5w2 − 3w This is a cubic function.

31. a. b.

1600 = 850 + 3q 750 = 3q q = 250

36. P (5) = 37. a.

1

2!(1!)

( 14 ) ( 34 )

5!

5

5!(0!)

39. a.

1182.74

40. a.

=

0

=

6

( 161 )( 43 ) = 2(1)

c. 41. a.

5!(1)

19.12

57.69 2.21

b. 9.98 c.

–14.52

Principles in Practice 2.3 1. The customer’s price is (c D s )( x) = c( s ( x)) = c( x + 3) = 2( x + 3) = 2x + 6 2. g ( x) = ( x + 3) 2 can be written as g ( x) = a(l ( x)) = (a D l )( x) where a( x) = x 2 and l(x) = x + 3. Then l(x) represents the length of the sides of the square, while a(x) is the area of a square with side of length x.

9 64

1 (1) ( 1024 ) =

5!

252.15

b. –62.94

34. For a committee of four, there are 4 choices for who will be member A. For each choice of member A, there are 3 choices for member G. Once members A and G have been chosen, there are two choices for member M, then one choice for member S once members A, G, and M have been chosen. Thus, there are 4 ⋅ 3 ⋅ 2 ⋅ 1 = 4! = 24 ways to label the members. Similarly, a committee of five can label itself with five labels in 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 5! = 120 ways.

35. P (2) =

1218.60

c.

33. The cost for buying n tickets is ⎧ 9.5n 0 ≤ n < 12 c ( n) = ⎨ 12 ≤ n ⎩8.75n

2

c.

b. 4985.27

32. The interest is Prt, so principal and interest amount to f(t) = P + Prt, or f(t) = P(1 + rt). Since f(t) = at + b where a (= Pr) and b (= P) are constants, f is a linear function of t.

( 14 ) ( 43 )

742.50

b. −20.28

C = 850 + 3q

3!

1 11 5 11 16 (30) + = + = =4 24 4 4 4 4 1 11 6 11 17 (36) + = + = f (36) = 24 4 4 4 4 4 175 175 33 f (39) = (39) − = 52 − = 3 4 4 4 f (30) =

Problems 2.3 1. f(x) = x + 3, g(x) = x + 5

1 1024

a.

all T such that 30 ≤ T ≤ 39

( f + g )( x) = f ( x) + g ( x) = ( x + 3) + ( x + 5) = 2x + 8

b. (f + g)(0) = 2(0) + 8 = 8

61

Chapter 2: Functions and Graphs c.

( f − g )( x) = f ( x) − g ( x) = ( x + 3) − ( x + 5) = −2

d.

( fg )( x) = f ( x) g ( x) = ( x + 3)( x + 5)


= x 2 + 8 x + 15

e.

g.

( fg )(–2) = (−2) + 8(−2) + 15 = 3 f f ( x) x + 3 ( x) = = g g ( x) x + 5

h.

( g D f )( x) = g ( f ( x)) = g (2 x) = 6 + 2 x

i.

( g D f )(2) = 6 + 2(2) = 6 + 4 = 10

( f D g )(3) = 3 + 8 = 11

i.

( g D f )( x) = g ( f ( x)) = g ( x + 3) = ( x + 3) + 5 = x+8

f ( x) = x 2 + 1, g ( x) = x 2 − x

a.

= 2x2 − x + 1

b.

b.

c.

c.

1 1 ⎛ 1⎞ ( f − g) ⎜ − ⎟ = − +1 = 2 2 2 ⎝ ⎠

d.

( fg )( x) = f ( x) g ( x) = ( x 2 + 1)( x 2 − x)

( g D f )(3) = 3 + 8 = 11

= x 4 − x3 + x 2 − x

( f + g )( x) = f ( x) + g ( x) = (2 x) + (6 + x) = 3x + 6

e.

f f ( x) x 2 + 1 ( x) = = g g ( x) x 2 − x

f.

− 12 + 1 f ⎛ 1⎞ 5 − = = 2 g ⎜⎝ 2 ⎟⎠ 3 − 12 − − 12

g.

( f D g )( x) = f ( g ( x)) = f ( x 2 − x)

(f – g)(4) = (4) – 6 = –2 ( fg )( x) = f ( x) g ( x) = 2 x(6 + x) = 12 x + 2 x

e.

f f ( x) 2x = ( x) = g g ( x) 6 + x

f.

f 2(2) 4 1 (2) = = = g 6+2 8 2

( ) ( ) ( ) 2

( f − g )( x) = f ( x) − g ( x) = (2 x) − (6 + x ) = x−6

d.

( f − g )( x) = f ( x) − g ( x) = ( x 2 + 1) − ( x 2 − x) = x +1

2. f(x) = 2x, g(x) = 6 + x a.

( f + g )( x) = f ( x) + g ( x) = ( x 2 + 1) + ( x 2 − x )

( f D g )( x) = f ( g ( x)) = f ( x + 5) = ( x + 5) + 3 = x+8

h.

j.

( f D g )( x) = f ( g ( x)) = f (6 + x) = 2(6 + x) = 12 + 2 x

2

3. f.

g.

= ( x 2 − x) 2 + 1

2

= x 4 − 2 x3 + x 2 + 1

h.

( g D f )( x) = g ( f ( x)) = g ( x 2 + 1) = ( x 2 + 1) 2 − ( x 2 + 1) = x4 + x2

i.

62

( g D f )(−3) = (−3) 4 + (−3)2 = 90


4.

Section 2.3 7. ( F D G )(t ) = F (G (t )) ⎛ 2 ⎞ = F⎜ ⎟ ⎝ t −1 ⎠

f ( x) = x 2 + 1, g ( x) = 5

a.

( f + g )( x) = f ( x) + g ( x)

2

= ( x 2 + 1) + 5

⎛ 2 ⎞ ⎛ 2 ⎞ =⎜ ⎟ + 7⎜ t –1 ⎟ +1 ⎝ t –1⎠ ⎝ ⎠ 4 14 = + +1 (t − 1) 2 t − 1 (G D F )(t ) = G ( F (t ))

= x2 + 6 2

b. c.

58 ⎛2⎞ ⎛2⎞ ( f + g) ⎜ ⎟ = ⎜ ⎟ + 6 = 3 3 9 ⎝ ⎠ ⎝ ⎠

= G (t 2 + 7t + 1) 2 = 2 (t + 7t + 1) –1 2 = 2 t + 7t

( f − g )( x) = f ( x) − g ( x) = ( x 2 + 1) – 5 = x2 – 4

d.

( fg )( x) = f ( x) g ( x) = ( x 2 + 1)(5)

8. ( F D G )(t ) = F (G (t ))

2

= 5x + 5

= F (3t 2 + 4t + 2) = 3t 2 + 4t + 2 (G D F )(t ) = G ( F (t ))

e.

( fg )(7) = 5(7 2 ) + 5 = 245 + 5 = 250

f.

f f ( x) x 2 + 1 ( x) = = g g ( x) 5

=G

g.

( f D g )( x) = f ( g ( x)) = f (5) = 52 + 1 = 26

= 3t + 4 t + 2

h.

( f D g )(12, 003) = 26

i.

( g D f )( x) = g ( f ( x)) = g ( x 2 + 1) = 5

( t) 2 = 3( t ) + 4 ( t ) + 2

9. ( f D g )(v) = f ( g (v)) = f =

5. f(g(2)) = f(4 – 4) = f(0) = 0 + 6 = 6 g(f(2)) = g(12 + 6) = g(18) = 4 – 36 = –32

v+2

)

1

(

v+2

)

2

+1

1 v + 2 +1 1 = v+3 ( g D f )(v) = g ( f (v)) ⎛ 1 ⎞ = g⎜ ⎟ ⎝ v2 + 1 ⎠ 1 = +2 2 v +1 =

6. ( f D g )( p) = f ( g ( p )) ⎛ p–2⎞ = f⎜ ⎟ ⎝ 3 ⎠ 4 = p –2 3

=

(

12 p–2

4 ⎛ 4 ⎞ p – 2 4 – 2p ( g D f )( p) = g ( f ( p )) = g ⎜ ⎟ = = 3 3p ⎝ p⎠

= =

63

1 + 2(v 2 + 1) v2 + 1 2v 2 + 3 v2 + 1


ISM: Introductory Mathematical Analysis 19. ( g D f )(m) = g ( f (m)) ⎛ 40m – m2 ⎞ = g⎜ ⎟ ⎜ ⎟ 4 ⎝ ⎠ ⎛ 40m – m 2 ⎞ = 40 ⎜ ⎟ ⎜ ⎟ 4 ⎝ ⎠

10. ( f D f )( x) = f ( f ( x)) = f ( x 2 + 2 x − 1) = ( x 2 + 2 x − 1)2 + 2( x 2 + 2 x − 1) − 1 = x 4 + 4 x3 + 4 x 2 − 2

11. Let g(x) = 11x and f(x) = x − 7. Then h(x) = g(x) − 7 = f(g(x))

= 10(40m − m2 ) = 400m − 10m2 This represents the total revenue received when the total output of m employees is sold.

12. Let g(x) = x 2 – 2 and f ( x) = x . Then h( x) = x 2 – 2 = g ( x) = f ( g ( x))

20. ( f D g )( E ) = f ( g ( E ))

1 13. Let g ( x) = x – 2 and f ( x) = . Then x 1 1 h( x ) = = = f ( g ( x)) 2 g ( x) x –2

= f (7202 + 0.29 E 3.68 )

2

= 0.45(7202 + 0.29 E 3.68 – 1000)0.53 = 0.45(6202 + 0.29 E 3.68 )0.53 This represents status based on years of education.

14. Let g ( x) = 9 x3 – 5 x and f ( x) = x3 – x 2 + 11.

21. a.

Then h( x) = [ g ( x)]3 – [ g ( x)]2 + 11 = f ( g ( x))

b. 1169.64

x2 − 1 15. Let g ( x) = and f ( x) = 4 x . x+3

22. a.

Then h( x) = 4 g ( x) = f ( g ( x)). 16. Let g(x) = 3x − 5 and f ( x) = h( x ) =

17. a.

2 − (3x − 5) (3x − 5) 2 + 2

2− x x2 + 2

−0.13

b. 18.85 23. a.

. Then

194.47

b. 0.29

= f ( g ( x)).

24. a.

0.45

b. 1.61

The revenue is $9.75 per pound of coffee sold, so r(x) = 9.75x.

Problems 2.4

b. The expenses are e(x) = 4500 + 4.25x. c.

14.05

f −1 ( x) =

x 7 − 3 3

2. g −1 ( x) =

x 1 − 2 2

1.

Profit = revenue – expenses. (r – e)(x) = 9.75x – (4500 + 4.25x) = 5.5x – 4500.

18. v( x) = (4 x – 2)3 can be written as

3. F −1 ( x) = 2 x + 14

v( x) = f (l ( x)) = ( f D l )( x) where f ( x) = x3 and l(x) = 4x – 2. Then l(x) represents the length of the sides of the cube, while f(x) is the volume of a cube with sides of length x.

4.

5. r ( A) =

64

x 5 + 4 4

f −1 ( x) = A π


6. r (V ) = 3

Section 2.5

⎛ 1, 200, 000 ⎞ p(q( p )) = p ⎜ ⎟ p ⎝ ⎠ 1, 200, 000 =

3V 4π

7. f(x) = 5x + 12 is one-to-one, for if f ( x1 ) = f ( x2 ) then 5 x1 + 12 = 5 x2 + 12, so 5 x1 = 5 x2 and thus x1 = x2 .

1,200,000 p

= 1, 200, 000 ⋅

=p Similarly, q(p(q)) = q, so the functions are inverses.

8. g ( x) = (5 x + 12) 2 is not one-to-one, because g ( x1 ) = g ( x2 ) does not imply x1 = x2 . For ⎛ 11 ⎞ example, g ⎜ − ⎟ = ⎝ 5⎠

⎛ 13 ⎞ g ⎜ − ⎟ = 1. ⎝ 5⎠

9. h( x) = (5 x + 12)2 , for x ≥ −

q , we get q = 48p. Since q > 0, p is 48 also greater than 0, so q as a function of p is q = q(p) = 48p, p > 0. q ⎛ q ⎞ q( p (q )) = q ⎜ ⎟ = 48 ⋅ =q 48 ⎝ 48 ⎠ 48 p p(q ( p )) = p (48 p ) = =p 48 Thus, p(q) and q(p) are inverses.

14. From p =

5 , is one-to-one. 12

If h( x1 ) = h( x2 ) then (5 x1 + 12)2 = (5 x2 + 12)2 . Since x ≥ −

5 we have 5x + 12 ≥ 0, and thus 12

(5 x1 + 12)2 = (5 x2 + 12)2 only if 5 x1 + 12 = 5 x2 + 12, and hence x1 = x2 .


10. F ( x) = x − 9 is not one-to-one, because

1. Let y = the amount of money in the account. Then, after one month, y = 7250 – (1 · 600) = $6650, and after two months y = 7250 – (2 · 600) = $6050. Thus, in general, if we let x = the number of months during which Rachel spends from this account, y = 7250 – 600x. To identify the x-intercept, we set y = 0 and solve for x. y = 7250 – 600x 0 = 7250 – 600x 600x = 7250 1 x = 12 12 ⎛ 1 ⎞ The x-intercept is ⎜12 , 0 ⎟ . ⎝ 12 ⎠ Therefore, after 12 months and approximately 2.5 days Rachel will deplete her savings. To identify the y-intercept, we set x = 0 and solve for y. y = 7250 – 600x y = 7250 – 600(0) y = 7250 The y-intercept is (0, 7250). Therefore, before any months have gone by, Rachel has $7250 in her account.

F ( x1 ) = F ( x2 ) does not imply x1 = x2 . For example, F(8) = F(10) = 1.

11. The inverse of f ( x) = (4 x − 5) 2 for x ≥ f −1 ( x ) =

5 is 4

x 5 + , so to find the solution, we 4 4

find f −1 (23). f −1 (23) =

23 5 + 4 4

The solution is x =

23 5 + . 4 4

12. The inverse of V (r ) =

4 3 3V πr is r (V ) = 3 , so 3 4π

the solution is r (100) = 3

p 1, 200, 000

3(100) . 4π

1, 200, 000 1, 200, 000 , we get q = . q p Since q > 0, p is also greater than 0, so q as a 1, 200, 000 , p > 0. function of p is q = q ( p) = p

13. From p =

65


ISM: Introductory Mathematical Analysis Problems 2.5

2. Let y = the cost to the customer and let x = the number of rides he or she takes. Since the cost does not change, regardless of the number of rides taken, the equation y = 24.95 represents this situation. The graph of y = 24.95 is a horizontal line whose y-intercept is (0, 24.95). Since the line is parallel to the x-axis, there is no x-intercept.

1.

0

1

2

2.5

3

4

5

y

0

12

24

30

24

12

0

( 2.

Q. II

3. a.

Miles

(2.5, 30)

Q. I (1, 1)

x 10

f(0) = 1, f(2) = 2, f(4) = 3, f(–2) = 0

Range: all real numbers

d. f(x) = 0 for x = –2. So a real zero is –2. (5, 0)

1 2 3 4 5

4. a. x hours

f(0) = 2, f(2) = 0

b. Domain: all x ≥ 0 c.

Range: all y ≥ 2

d. f(x) = 0 for x = 2. So a real zero is 2. 5. a.

f(0) = 0, f(1) = 1, f(–1) = 1

b. Domain: all real numbers c.

0

10

30

50

70

80

90

100

x

0

5.3

15.9

26.5

37.1

44.5

51.9

59.3

y

Cost (dollars)

y

b. Domain: all real numbers

x

40

10

(0, –6)

4. The monthly cost of x therms of gas is ⎧0.53x, if 0 ≤ x ≤ 70 y=⎨ ⎩0.53(70) + 0.74( x – 70), if x >70 or if 0 ≤ x ≤ 70 ⎧0.53x, y=⎨ ⎩0.74 x − 14.7, if x > 70

60

(8, –3)

(3, 0)

24

(0, 0)

x 10

Q. IV

(–4, 5)

c.

12

(

– 1 , –2 2

Q. III

y 36

y (2, 7) Q. I (0, 0)

3. The formula relating distance, time, and speed is d = rt, where d is the distance, r is the speed, and t is the time. Let x = the time spent biking (in hours). Then, 12x = the distance traveled. Brett bikes 12 · 2.5 = 30 miles and then turns around and bikes the same distance back to the rental shop. Therefore, we can represent the distance from the turn-around point at any time x as 30 – 12x . Similarly, the distance from the rental shop at any time x can be represented by the function y = 30 – 30 – 12 x . x

10

(100, 59.3)

Range: all nonnegative real numbers

d. f(x) = 0 for x = 0. So a real zero is 0. 6. a.

f(0) = 0, f(2) = 1, f(3) = 3, f(4) = 2

b. Domain: all x such that 0 ≤ x ≤ 4

(70, 37.1)

c.

Range: all y such that 0 ≤ y ≤ 3

20

(0, 0)

20 40 60 80 100

d. f(x) = 0 for x = 0. So a real zero is 0.

x therms

66


Section 2.5

7. y = 2x If y = 0, then x = 0. If x = 0, then y = 0. Intercept: (0, 0) y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers 5

10. y = 3 – 2x If y = 0, then 0 = 3 – 2x, x =

3 . 2

⎛3 ⎞ If x = 0, then, y = 3. Intercepts: ⎜ , 0 ⎟ , (0, 3) ⎝2 ⎠ y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers

y

y

x 5

3

3 2

x 5

8. y = x + 1 If y = 0, then x = –1. If x = 0, then y = 1. Intercepts: (–1, 0), (0, 1) y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers 5

11. y = x 4 If y = 0, then 0 = x 4 , x = 0. If x = 0, then y = 0. Intercept: (0, 0) y is a function of x. Not one-to-one. Domain: all real numbers Range: all real numbers ≥ 0

y

1

x

–1

y

5

5

x 5

9. y = 3x – 5 If y = 0, then 0 = 3x – 5, x =

5 . 3

⎛5 ⎞ If x = 0, then y = –5. Intercepts: ⎜ , 0 ⎟ , (0, –5) ⎝3 ⎠ y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers 10

12. y =

2 x2

If y = 0, then 0 =

, which has no solution. x2 Thus there is no x-intercept. Because x ≠ 0 , Not one-to-one. Domain: all real numbers except 0 Range: all real numbers > 0

y

x –5

5 3

2

5

y

10 (–1, 2)

(–2, 12 )

(1, 2)

(2, 12 )

x 5

67


ISM: Introductory Mathematical Analysis 16. x = −9 If y = 0 then x = −9. Because x cannot be 0, there is no y-intercept. Intercept: (−9, 0). y is not a function of x.

13. x = 0 If y = 0, then x = 0. If x = 0, then y can be any real number. Intercepts: every point on y-axis y is not a function of x. 5

y

y

10

x 5

17. x = −|y| If y = 0, then x = 0. If x = 0, then 0 = −|y|, y = 0. Intercept: (0, 0) y is not a function of x.

14. y = 4 x 2 – 16 If y = 0, then 0 = 4 x 2 – 16 = 4( x 2 − 4) , 0 = 4( x + 2)( x − 2) , x = ±2. If x = 0, then y = –16. Intercepts: (±2, 0), (0, –16) y is a function of x. Not one-to-one. Domain: all real numbers Range: all real numbers ≥ –16 20

x 10

ñ 10

5

x 5

y

18. x 2 = y 2

x 2

–2

5

If y = 0, then x 2 = 0, x = 0 . If x = 0, then 0 = y 2 , y = 0 . Intercept: (0, 0) y is not a function of x.

–16

5

15. y = x3 If y = 0, then 0 = x3 , x = 0. If x = 0, then y = 0. Intercept: (0, 0). y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers 5

y

y

x 5

y

19. 2x + y – 2 = 0 If y = 0, then 2x – 2 = 0, x = 1. If x = 0, then y – 2 = 0, y = 2. Intercepts: (1, 0), (0, 2) Note that y = 2 – 2x. y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers

x 5

68


5

Section 2.5

y

⎛ 10 ⎞ Intercepts: ⎜ ± , 0 ⎟ , (0,5) ⎜ ⎟ 2 ⎝ ⎠ Domain: all real numbers Range: all real numbers ≤ 5

2 x 1

5

5

f(x)

– √10

√10 2

2

20. x + y = 1 If y = 0, then x = 1. If x = 0, then y = 1. Intercepts: (1, 0), (0, 1) Note that y = 1 – x. y is a function of x. One-to-one. Domain: all real numbers Range: all real numbers 5

23. y = h(x) = 3 Because y cannot be 0, there is no x-intercept. If x = 0, then y = 3. Intercept: (0, 3) Domain: all real numbers Range: 3

y

1

x 1

x 5

5

y

5 3 x 5

21. s = f (t ) = 4 – t 2 If s = 0, then 0 = 4 – t 2 0 = (2 + t)(2 – t) t = ±2. If t = 0, then s = 4. Intercepts: (2, 0), (–2, 0), (0, 4) Domain: all real numbers Range: all real numbers ≤ 4 5

24. g(s) = –17 Because g(s) cannot be 0, there is no s-intercept. If s = 0, then g(s) = –17. Intercept: (0, –17) Domain: all real numbers Range: –17

s

20

4

y

t –2

2

5

x 20

–20

–20

22.

f ( x) = 5 – 2 x 2 . If f(x) = 0, then 0 = 5 – 2x 2

25. y = h( x) = x 2 – 4 x + 1

2

2x = 5 5 x2 = 2 5 10 . x=± =± 2 2 If x = 0, then f(x) = 5.

If y = 0, then 0 = x 2 – 4 x + 1 , and by the 4 ± 12 = 2 ± 3 . If 2 x = 0, then y = 1. Intercepts: (2 ± 3, 0), (0,1) Domain: all real numbers Range: all real numbers ≥ –3

quadratic formula, x =

69



y

5 2–

1

3

1

x (2, – 3)

2+

3

t 2 − 9 to be a real number,

Note that for

t 2 − 9 ≥ 0, so t 2 ≥ 9, and |t| ≥ 3. If s = 0, then

0 = t 2 − 9, 0 = t 2 − 9, or t = ±3. Because |t| ≥ 3, we know t ≠ 0, so no s-intercept exists. Intercepts: (−3, 0), (3, 0) Domain: all real numbers t ≤ −3 and ≥ 3 Range: all real numbers ≥ 0

y

x 10

10

y

–8

(1, –9)

–3

27.

5

29. s = f (t ) = t 2 − 9

If y = 0, then 0 = x 2 + 2 x – 8 0 = (x + 4)(x – 2), so x = –4, 2. If x = 0, then y = –8. Intercepts: (–4, 0), (2, 0), (0, –8). Domain: all real numbers Range: all real numbers ≥ –9

–4

q

–1

26. y = f ( x) = x 2 + 2 x – 8

10

p

3

x 10

f (t ) = – t 3

If f(t) = 0, then 0 = – t 3 , t = 0 . If t = 0, then f(t) = 0. Intercept: (0, 0) Domain: all real numbers Range: all real number 5

30. F (r ) = –

f(t)

1 r

1 , which has no solution. r Because r ≠ 0, there is no vertical-axis intercept. Intercept: none. Domain: all real numbers ≠ 0 Range: all real numbers ≠ 0

If F(r) = 0, then 0 = –

t 5

5

F(r)

28. p = h(q) = 1 + 2q + q 2

r

2

2

5

If p = 0, then 1 + 2q + q = 0, (1 + q) = 0, so q = −1. If q = 0 then p = 1. Intercepts: (−1, 0), (0, 1) Domain: all real numbers Range: all real numbers ≥ 0

31.

f ( x) = 2 x –1

If f(x) = 0, then 0 = 2 x –1 , 2 x –1 = 0, so 1 . 2 If x = 0, then f ( x) = –1 = 1 . x=

70


Section 2.5

2 x–4 Note that the denominator is 0 when x = 4. Thus 2 , which has no x ≠ 4. If y = 0, then 0 = x–4 1 solution. If x = 0, then y = – . 2 1 ⎛ ⎞ Intercept: ⎜ 0, – ⎟ 2⎠ ⎝ Domain: all real numbers except 4 Range: all real numbers except 0

⎛1 ⎞ Intercepts: ⎜ , 0 ⎟ , (0,1) ⎝2 ⎠ Domain: all real numbers Range: all real numbers ≥ 0 5

34. y = f ( x) =

f(x)

1

x 1 2

5

10

32. v = H (u ) = u – 3

y

If v = 0, then 0 = u – 3 , u – 3 = 0, so u = 3. If u = 0, then v = –3 = 3 . Intercepts: (3, 0), (0, 3). Domain: all real numbers Range: all real numbers ≥ 0 10

v

x 10

4

–1 2

35. Domain: all real numbers ≥ 0 Range: all real numbers 1 ≤ c < 8

3 3

10 8

u 10

c

p 6

33. F (t ) =

10

16 t2

If F(t) = 0, then 0 =

36. Domain: all real numbers ≥ –1 Range: all real numbers ≤ 11

16

, which has no solution. t2 Because t ≠ 0, there is no vertical-axis intercept. No intercepts Domain: all nonzero real numbers Range: all positive real numbers 10

14

␾(x)

F(t)

x 10

t 10

37. Domain: all real numbers Range: all real numbers ≥ 0 10 9

g(x)

x 3

71

5


ISM: Introductory Mathematical Analysis y

38. Domain: all positive real numbers Range: all real numbers > 1

20

f(x)

Cost (dollars)

10

4 1

16 12 8 4 10 12 2 4 6 8 10 A.M. P.M.

x 3

5

10

x

43. As price increases, quantity supplied increases; p is a function of q.

39. From the vertical-line test, the graphs that represent functions of x are (a), (b), and (d).

p

50

40. From the horizontal line test, the graphs which represent one-to-one functions of x are (c) and (d). 41. Let y = the amount that is owed and let x = the number of monthly payments made. Then, the amount Tara owes is represented by the equation y = 2400 − 275x. To determine the x-intercept, we set y = 0 and solve for x. y = 2400 − 275 x 0 = 2400 − 275 x 275 x = 2400 8 x=8 11 ⎛ 8 ⎞ The x-intercept is ⎜ 8 , 0 ⎟ . Therefore, Tara will ⎝ 11 ⎠ have paid off her debt after 9 months. To determine the y-intercept, we set x = 0 and solve for y. y = 2400 − 275x y = 2400 − 275(0) y = 2400 The y-intercept is (0, 2400). Therefore, before any payments are made, Tara owes $2400.

10 q 30

210

44. As price decreases, quantity increases; p is a function of q. p

25

5 q 25

5

45.

1000

y

300

42. The cost of an item as a function of the time of day, x is ⎧9, if 10:30 A.M. ≤ x < 2 : 30 P.M. ⎪8, if 2:30 P.M. ≤ x < 4 : 30 P.M. ⎪ y = ⎨13, if 4:30 P.M. ≤ x < 6 : 00 P.M. ⎪18, if 6:00 P.M. ≤ x < 8 : 00 P.M. ⎪13, if 8:00 P.M. ≤ x ≤ 10:00 P.M. ⎩

7

46.

14

x 21

y 4

4 5

47. 0.39 48. −0.50, 0.57 72

x 12


Section 2.5 range: (– ∞, ∞)

49. –0.61, –0.04

a.

50. 0.62, 1.73, 4.65

b. intercepts: (–1.73, 0), (0, 4)

51. –1.12

35

59.

52. No real zeros 53. −1.70, 0 54. –0.49, 0.52, 1.25

–5

25

55.

a.

maximum value of f(x): 28

b. range: (−∞, 28]

4

1

c.

–15

a.

5

–5

5

60.

maximum value of f(x): 19.60

real zeros: −4.02, 0.60

b. minimum value of f(x): –10.86 5

–5

56.

4

–5 1

–1

a.

b. intercepts: (0, 0.29), (−1.03, 0)

–2

a.

c.

maximum value of f(x): 3.94 61.

b. minimum value of f(x): –1.94 57.

range: (−∞, ∞)

real zero: −1.03

35

6

2 3

3

a.

a.

5

c.

range: [18.68, 34.21]

d. no intercept

10

5

–5

maximum value of f(x): 34.21

b. minimum value of f(x): 18.68

maximum value of f(x): 5

b. minimum value of f(x): 4 58.

5

15

–5

73


ISM: Introductory Mathematical Analysis line y = x: (a, b) on graph, then

Problems 2.6

2a 2 + b 2 a 4 = 8 − b, but

1. y = 5x Intercepts: If y = 0, then 5x = 0, or x = 0; if x = 0, then y = 5 · 0 = 0. Testing for symmetry gives: x-axis: –y = 5x y = –5x y-axis: y = 5(–x) = –5x origin: –y = 5(–x) y = 5x line y = x: (a, b) on graph, then b = 5a, and 1 a = b ≠ 5b for all b, so (b, a) is not 5 on the graph.

2b 2 + a 2b 4 = 8 − a will not necessarily be true, so (b, a) is not on the graph. Answer: (±2, 0), (0, 8); symmetry about y-axis

4. x = y 3 Intercepts: If y = 0, then x = 0; if x = 0, then 0 = y3 , so y = 0. Testing for symmetry gives:

x-axis:

x = (– y )3 = – y 3

y-axis:

– x = y3

Answer: (0, 0); symmetry about origin

x = – y3 – x = (– y )3

origin:

2. y = f ( x) = x 2 – 4 Intercepts: If y = 0, then

x = y3

2

0 = x – 4 = ( x + 2)( x – 2) , or x = ±2; if x = 0,

line y = x: (a, b) on graph, then a = b3 , and

then y = 02 – 4 = –4 . Testing for symmetry gives:

b = 3 a ≠ a3 for all a, so (b, a) is not on the graph. Answer: (0, 0); symmetry about origin

x-axis:

– y = x2 – 4 y = – x2 + 4

5. 16 x 2 − 9 y 2 = 25

y-axis:

y = (– x)2 – 4 = x 2 – 4

origin:

– y = (– x)2 – 4

Intercepts: If y = 0, then 16 x 2 = 25, x 2 = 5 so x = ± ; 4

y = – x2 + 4

line y = x: (a, b) on graph, then b = a 2 − 4, and

if x = 0, then −9 y 2 = 25, y 2 = −

2

a = ± b + 4 ≠ b − 4 for all b, so (b, a) is not on the graph. Answer: (±2, 0), (0, –4); symmetry about y-axis 2

x-axis:

2 4

y-axis:

16(− x) 2 − 9 y 2 = 25

16 x 2 − 9 y 2 = 25 origin: Since the graph has symmetry about x- and y-axes, there is also symmetry about the origin. line y = x: (a, b) on graph, then

2 x 2 + (– y ) 2 x 4 = 8 – (– y ) 2 x2 + y2 x4 = 8 + y

16a 2 − 9b 2 = 25, and 1 a 2 = (9b 2 + 25). (b, a) on graph, 16

2(– x) 2 + y 2 (– x)4 = 8 – y 2 x2 + y2 x4 = 8 − y

origin:

16 x 2 − 9(− y ) 2 = 25 16 x 2 − 9 y 2 = 25

2 x 2 = 8, x 2 = 4, or x = ±2; if x = 0, then 0 = 8 – y, so y = 8. Testing for symmetry gives:

y-axis:

25 , which has 9

no real root. Testing for symmetry gives:

3. 2 x + y x = 8 – y Intercepts: If y = 0, then

x-axis:

25 , 16

2(– x) 2 + (– y )2 (– x) 4 = 8 – (– y )

then 16b 2 − 9a 2 = 25 and 1 1 a 2 = (16b 2 − 25) ≠ (9b 2 + 25) 9 16

2 x2 + y2 x4 = 8 + y

74


Section 2.6

for all b, so (b, a) and (a, b) are not always both on the graph. ⎛ 5 ⎞ Answer: ⎜ ± , 0 ⎟ ; symmetry about x-axis, ⎝ 4 ⎠ y-axis, and origin.

9. x = – y –4 Intercepts: Because y ≠ 0, there is no 1 , which has x-intercept; if x = 0, then 0 = – y4 no solution. Testing for symmetry gives:

6. y = 57 Intercepts: Because y ≠ 0, there is no x-intercept; if x = 0, then y = 57. Testing for symmetry gives: x-axis: (–y) = 57 y = –57 y-axis: y = 57 origin: (–y) = 57 y = –57 line y = x: (a, b) on graph, then b = 57, but a can be any value, so (b, a) = (57, a) is not necessarily on the graph. Answer: (0, 57); symmetry about y–axis

x-axis:

x = – y –4

y-axis: origin:

– x = –(– y ) –4 x = y –4

line y = x: (a, b) on graph, then a = −b −4 and b = (−a )−1/ 4 ≠ −a −4 for all a, so (b, a) is not on the graph. Answer: no intercepts; symmetry about x-axis

10. y = x 2 – 25 Intercepts: If y = 0, then

x 2 – 25 = 0,

x 2 – 25 = 0, x 2 = 25, so x = ±5;

if x = 0, then y = –25 , which has no real root. Testing for symmetry gives: x-axis:

– y = x 2 – 25 y = – x 2 – 25

y-axis:

8. y = 2 x – 2

y = (– x)2 – 25 y = x 2 – 25

Intercepts: If y = 0, then 2 x = 2, 2 x = 2, x = 1, so x = ±1; if x = 0, then y = –2. Testing for symmetry gives: – y = 2x – 2 x-axis:

origin:

– y = (– x)2 – 25 y = – x 2 – 25 .

line y = x: (a, b) on graph, then b = a 2 − 25 or

y = – 2x + 2

b 2 = a 2 − 25 and

y = 2(− x) – 2

a 2 = b 2 + 25 ≠ b 2 − 25 for all b, so (b, a) is not on the graph. Answer: (±5, 0); symmetry about y-axis

y = 2x – 2

origin:

– x = – y –4 x = y –4

7. x = –2 Intercepts: If y = 0, then x = –2; because x ≠ 0, there is no y-intercept. Testing for symmetry gives: x-axis: x = –2 y-axis: –x = –2 x=2 origin: –x = –2 x=2 line y = x: (a, b) on graph, then a = −2, but b can be any value, so (b, a) = (b, −2) is not necessarily on the graph. Answer: (–2, 0); symmetry about x-axis

y-axis:

x = –(– y ) –4

– y = 2(– x) – 2 y = – 2x + 2

11. x – 4 y – y 2 + 21 = 0 Intercepts: If y = 0, then x + 21 = 0, so x = –21;

line y = x: (a, b) on graph, then b = 2a − 2 and b+2 ≠ 2b − 2 for all b, so 2 (b, a) is not on the graph. Answer: (±1, 0), (0, –2); symmetry about y-axis a=±

if x = 0, then –4 y – y 2 + 21 = 0, y 2 + 4 y – 21 = 0, (y + 7)(y – 3) = 0, so y = –7 or y = 3. 75



Testing for symmetry gives: x-axis:

y=

y-axis:

2

x – 4(– y ) – (– y ) + 21 = 0 2

x + 4 y – y + 21 = 0

y-axis:

y=

(– x) – 4 y – y 2 + 21 = 0 – x – 4 y – y 2 + 21 = 0

origin:

− x3 − 2 x 2 − x x2 + 1

(– x) – 4(– y ) – (– y ) 2 + 21 = 0 y=

a = b 2 + 4b − 21, but b = a 2 + 4a − 21 will not necessarily be true, so (b, a) is not on the graph. Answer: (–21, 0), (0, –7), (0, 3); no symmetry

a=

Intercepts: If y = 0, then x 2 = 0, so x = 0;

x3 + 2 x 2 + x

b3 − 2b 2 + b

14. x 2 + xy + y 2 = 0

if x = 0, then y 3 = 0, so y = 0. Testing for symmetry gives:

Intercepts: If y = 0, then x 2 = 0, so x = 0;

x-axis: x 2 + x(− y ) + (− y )3 = 0

if x = 0, then y 2 = 0, so y = 0. Testing for symmetry gives:

3

x − xy − y = 0 2

y-axis: (− x) + (− x) y + y 3 = 0

x 2 + x(– y ) + (– y ) 2 = 0

x-axis:

x 2 − xy + y 3 = 0

x 2 – xy + y 2 = 0

(− x) 2 + (− x)(− y ) + (− y )3 = 0 2

(− x)2 + 1

is not necessarily b2 + 1 true, so (b, a) is not on the graph. Answer: (1, 0), (0, 0); no symmetry of the given types

12. x 2 + xy + y3 = 0

2

(− x)3 − 2(− x) 2 + (− x)

x2 + 1 line y = x: (a, b) on graph, then a 3 − 2a 2 + a b= , but a2 + 1

a − 4b − b 2 + 21 = 0 and

(– x)2 + (– x) y + y 2 = 0

y-axis:

3

x + xy − y = 0 line y = x: (a, b) on graph, then

x 2 – xy + y 2 = 0 (– x) 2 + (– x)(– y ) + (– y ) 2 = 0

origin:

a 2 + ab + b3 = 0, but

x 2 + xy + y 2 = 0

b 2 + ab + a3 = 0 will not necessarily be true, so (b, a) is not on the graph. Answer: (0, 0); no symmetry.

13. y = f ( x) =

(− x) 2 + 1

−y =

origin:

– x + 4 y – y 2 + 21 = 0 line y = x: (a, b) on graph, then

origin:

(− x)3 − 2(− x) 2 + (− x)

line y = x: (a, b) on graph, then a 2 + ab + b 2 = 0 and b 2 + ba + a 2 = 0, so (b, a) is on the graph. Answer: (0, 0); symmetry about origin, symmetry about y = x

x3 − 2 x 2 + x

x2 + 1 Intercepts: If y = 0, then x3 − 2 x 2 + x x( x − 1) 2 = = 0, so x = 0, 1; x2 + 1 x2 + 1 if x = 0, then y = 0. Testing for symmetry gives: x-axis: Because f is not the zero function, there is no x-axis symmetry

15. y =

3 3

x +8 3

= 0 , which has x +8 3 no solution; if x = 0, then y = . 8 Testing for symmetry gives:

Intercepts: If y = 0, then

76

3


x-axis:

–y = y=–

y-axis:

y=

3

line y = x: (a, b) on graph, then b =

x3 + 8 3

a4 b4 , but a + b = will not b a necessarily be true, so (b, a) is not on the graph. Answer: no intercepts; symmetry about origin

x +8 3 3 3

17. 3x + y 2 = 9 Intercepts: If y = 0, then 3x = 9, so x = 3;

–x + 8 3 –y = (– x )3 + 8 –y = y=

if x = 0, then y 2 = 9, so y = ±3. Testing for symmetry gives:

3 3

–x + 8 3

3x + y 2 = 9

x3 – 8 3 a3 + 8

−3 x + y 2 = 9

and

−3 x + y 2 = 9

line y = x: (a, b) on graph, then 3a + b 2 = 9 and 1 1 a = (9 − b 2 ), but b = (9 − a 2 ) will 3 3 not necessarily be true, so (b, a) is not on the graph. Answer: (3, 0), (0, ±3); symmetry about x-axis 5

4

x = 0 , which has no x 0 solution; if x = 0, then y = , which has no y solution. Testing for symmetry gives: x4 x-axis: –y = x + (– y )

y=

(– x )4 (– x) + y

x 3

18. x − 1 = y 4 + y 2 or x = y 4 + y 2 + 1 Intercepts: If y = 0, then x = 1; if x = 0, then y 4 + y 2 = −1, so no y-intercept Testing for symmetry gives:

x-axis:

x − 1 = (– y )4 + (− y )2 x −1 = y4 + y2

y-axis:

− x = y4 + y2 + 1 x = − y4 − y2 −1

(– x) 4 –y = (– x) + (– y ) y=

5

–3

x4 y= –x + y

origin:

y

3

Intercepts: If y = 0, then

y-axis:

3(− x) + (− y ) 2 = 9

origin:

x4 x+ y

x4 –x + y

3(− x) + y 2 = 9

y-axis:

3 3 a = 3 −8 ≠ for all b, so 3 b b +8 (b, a) is not on the graph. ⎛ 3⎞ Answer: ⎜0, ⎟ ; no symmetry of the given types ⎝ 8⎠

y=

3x + (− y )2 = 9

x-axis:

line y = x: (a, b) on graph, then b =

16. y =

a4 , and a+b

a+b =

3

(– x)3 + 8

y=

origin:

Section 2.6

origin:

− x = ( − y ) 4 + (− y ) 2 + 1 x = − y4 − y2 − 1

x4 x+ y 77



line y = x: (a, b) on graph, then a = b 4 + b 2 + 1 and b ≠ a 4 + a 2 + 1 for all a so (b, a) is not on the graph. Answer: (1, 0); symmetry about x-axis. 5

3 y = 5(– x) – (– x)3

y-axis:

3 y = –5 x + x3 3(– y ) = 5(– x) – (– x)3

origin:

3 y = 5 x – x3 .

y

line y = x: (a, b) on graph, then 3b = 5a − a3 , but 3a = 5b − b3 will not necessarily be true so (b, a) is not on the graph.

x 5

(

)

Answer: (0, 0), ± 5, 0 ; symmetry about origin 5

19. y = f ( x) = x3 – 4 x Intercepts: If y = 0, then x3 – 4 x = 0, x(x + 2)(x – 2) = 0, so x = 0 or x = ±2; if x = 0, then y = 0. Testing for symmetry gives: x-axis: Because f is not the zero function, there is no x-axis symmetry.

x –5

y = (– x) – 4(– x)

21.

y = – x3 + 4 x – y = (– x ) – 4(– x)

x = 0, then – y = 0, so y = 0. Testing for symmetry gives: x – –y = 0 x-axis:

3

y = x – 4x

line y = x: (a, b) on graph, then b = a3 − 4a, but

x − y =0

a = b3 − 4b will not necessarily be true, so (b, a) is not on the graph. Answer: (0, 0), (±2, 0); symmetry about origin. 5

–x – y = 0

y-axis:

x – y =0

y

origin:

Since there is symmetry about the x- and y-axes, symmetry about origin exists. line y = x: (a, b) on graph, then a − b = 0, thus

x –2

x – y =0

Intercepts: If y = 0, then x = 0, so x = 0; if

3

origin:

5

–5

3

y-axis:

y

2

5

a = b , and b − a = 0, so (b, a) is on the graph. Answer: (0, 0); symmetry about x-axis, y-axis, origin, line y = x.

20. 3 y = 5 x – x3

5

y

Intercepts: If y = 0, then 5 x – x3 = 0, x

(

5+x

)(

)

5 − x = 0, so x = 0 or x = ± 5; if

x

x = 0, then y = 0. Testing for symmetry gives: x-axis:

5

3(– y ) = 5 x – x3 3 y = –5 x + x3

78


Section 2.6 graph. 5⎞ ⎛ 5 ⎞ ⎛ Answer: ⎜ ± , 0 ⎟ , ⎜ 0, ± ⎟ ; symmetry about 2⎠ ⎝ 3 ⎠ ⎝ x-axis, y-axis, origin

22. x 2 + y 2 = 16 Intercepts: If y = 0, then x 2 = 16, so x = ±4; if x = 0, then y 4 = 16, so y = ±4. Testing for symmetry gives: 2

y

5

2

x + (– y ) = 16

x-axis:

x 2 + y 2 = 16

y-axis:

(– x )2 + y 2 = 16

origin:

x 2 + y 2 = 16 Since there is symmetry about x- and y-axes, symmetry about origin exists.

x 5

24. x 2 − y 2 = 4

line y = x: (a, b) on graph, then a 2 + b 2 = 16

Intercepts: If y = 0, then x 2 = 4, so x = ±2;

and b 2 + a 2 = 16, so (b, a) is on the graph. Answer: (±4, 0), (0, ±4); symmetry about x-axis, y-axis, origin, line y = x. 5

if x = 0, then − y 2 = 4, y 2 = −4, which has no real roots. Testing for symmetry gives:

y

x 2 − (− y )2 = 4

x-axis:

x2 − y 2 = 4 x

y-axis:

(− x) 2 − y 2 = 4

origin:

x2 − y 2 = 4 Since there is symmetry about x-and y-axes, symmetry about origin exists.

5

line y = x: (a, b) on graph, then a 2 − b 2 = 4 and a 2 = 4 + b 2 ≠ b 2 − 4 for all b, so (b, a) is not on the graph. Answer: (±2, 0); symmetry about x-axis, y-axis, origin.

23. 9 x 2 + 4 y 2 = 25 25 , so 9 5 5 x = ± ; if x = 0, then 4 y 2 = 25, so y = ± . 3 2 Testing for symmetry gives:

Intercepts: If y = 0, then 9 x 2 = 25, x 2 =

x-axis:

5

y

9 x 2 + 4(− y )2 = 25 2

x

9 x + 4 y = 25

y-axis:

9(− x)2 + 4 y 2 = 25

origin:

9 x 2 + 4 y 2 = 25 Since there is symmetry about x- and y-axes, symmetry about origin exists.

2

−2

2

25.

6 –6

line y = x: (a, b) on graph, then 9a 2 + 4b 2 = 25 1 and b 2 = (25 − 9a 2 ). (b, a) on 4

5

6 –6

y = f ( x) = 5 − 1.96 x 2 − πx 4 . Replacing x by –x

graph, then 9b 2 + 4a 2 = 25 and 1 b 2 = (25 − 4a 2 ), so (a, b) and 9 (b, a) are not always both on the

gives y = 5 − 1.96(− x)2 − π(− x) 4 or y = 5 − 1.96 x 2 − πx 4 , which is equivalent to 79



original equation. Thus the graph is symmetric about the y-axis. a.

2.

5

y

f(x) = x2

Intercepts: (±0.99, 0), (0, 5)

x

b. Maximum value of f(x): 5 c.

5 y = –x 2

Range: (–∞, 5]

26.

8

3.

10

y

f(x) = 1 x

4

–4

x 10

–3

y= 1

x–2

4

2

y = f ( x) = 2 x – 7 x + 5 . Replacing x by –x

gives y = 2(– x)4 – 7(– x) 2 + 5 or

4.

10

y

y = 2 x 4 – 7 x 2 + 5 , which is equivalent to original equation. Thus the graph is symmetric about y-axis. Real zeros of f: ±1, ±1.58

27.

5

y = √x + 2 x 10 f(x) = √x

y

y

5

5.

x

2

f(x) = 1 x

–11

1 –1

Problems 2.7 1.

–2 5

y= 2 3x

y y = x3 – 1

6.

x f(x) = x3

x

5

10

y f(x) = |x| y = |x| – 2 x 10

80


7.

Section 2.7

y

10

13. Translate 3 units to the left, stretch vertically away from the x-axis by a factor of 2, reflect about the x-axis, and move 2 units upward.

f(x) = |x| x 10

y = |x + 1| – 2

14. Translate 3 units to the left and 4 units downward. 15. Reflect about the y-axis and translate 5 units downward.

8.

5

y f(x) =

x

16. Shrink horizontally toward the y-axis by a factor of 3.

5x

17.

5

y = – 13 x

5

–5

9.

10

y

–5

f(x) = x2

Compared to the graph for k = 0, the graphs for k = 1, 2, and 3 are vertical shifts upward of 1, 2, and 3 units, respectively. The graphs for k = –1, –2, and –3 are vertical shifts downward of 1, 2, and 3 units, respectively.

x 10 y = 1 – (x – 1)2

3

18.

y

10.

y = (x – 1)2 + 1 f(x) = x 2

5

–5

x 10

–3

11.

10

Compared to the graph for k = 0, the graphs for k = 1, 2, and 3 are horizontal shifts to the left of 1, 2, and 3 units, respectively. The graphs for k = –1, –2, and –3 are horizontal shifts to the right of 1, 2, and 3 units, respectively.

y f(x) = √x

y = √–x

x

19.

10

5

–5

12.

10

5

y

f(x) = 1

–5

x

Compared to the graph for k = 1, the graphs for k = 2 and 3 are vertical stretches away from the x-axis by factors of 2 and 3, respectively. The 1 graph for k = is a vertical shrinking toward 2 the x-axis.

x 10 y= 1

2–x

81




5–3 2 = 5+ 4 9 ( x + 3) – 3 x = F ( x + 3) = ( x + 3) + 4 x + 7 F (5) =

1. Denominator is 0 when x2 − 6 x + 5 = 0 ( x − 1)( x − 5) = 0 x = 1, 5 Domain: all real numbers except 1 and 5.

2. all real numbers

5+ 4 9 3 = = 5 5 5

h(5) =

3. all real numbers 4. all real numbers

h(–4) =

5. For x to be real, x must be nonnegative. For the denominator x – 1 to be different from 0, x cannot be 1. Both conditions are satisfied by all nonnegative numbers except 1. Domain: all nonnegative real numbers except 1.

h( x ) =

–4 + 4 0 = =0 –4 –4 x+4 x

h(u – 4) =

6. s – 5 ≥ 0 s≥5 Domain: all real numbers s such that s ≥ 5. 7.

u+4 u

11. h(u ) =

12. H ( s ) =

f (0) = 3(0) 2 – 4(0) + 7 = 7

H (7) =

f (–3) = 3(–3) 2 – 4(–3) + 7 = 27 + 12 + 7 = 46

u–4

=

u u–4

( s – 4) 2 3

H (−2) =

f ( x) = 3x 2 – 4 x + 7

(u – 4) + 4

(–2 – 4)2 36 = = 12 3 3

(7 – 4)2 9 = =3 3 3

( )

2 2 ⎡ 1 ⎤ 49 − 72 49 ⎛ 1 ⎞ ⎣ 2 − 4⎦ = = 4 = H⎜ ⎟= 2 3 3 3 12 ⎝ ⎠

2

f (5) = 3(5) – 4(5) + 7 = 75 – 20 + 7 = 62 f (t ) = 3t 2 – 4t + 7

( )

H x2 =

8. h(x) = 7; all function values are 7. Answer: 7, 7, 7, 7

( )

( x 2 – 4)2 x 4 – 8 x 2 + 16 = 3 3

13. f(4) = 4 + 16 = 20 f(−2) = −3 f(0) = −3 f(1) is not defined.

4

9. G ( x) = x − 3 G (3) = 4 3 − 3 = 4 0 = 0 G (19) = 4 19 − 3 = 4 16 = 2

14.

G (t + 1) = 4 (t + 1) − 3 = 4 t − 2 4

G ( x3 ) = x3 − 3

1 3 ⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = −⎜ − ⎟ +1 = +1 = 2 2 2 2 ⎝ ⎠ ⎝ ⎠ f (0) = 02 + 1 = 1 2

1 5 ⎛1⎞ ⎛1⎞ f ⎜ ⎟ = ⎜ ⎟ +1 = +1 = 4 4 ⎝2⎠ ⎝2⎠

x–3 10. F ( x) = x+4 –1 – 3 4 F (–1) = =– –1 + 4 3 0−3 3 F (0) = =− 0+4 4

f (5) = 53 − 99 = 125 − 99 = 26 f (6) = 63 − 99 = 216 − 99 = 117

15. a.

82

f(x + h) = 3 – 7(x + h) = 3 – 7x – 7h


b. 16. a.

Chapter 2 Review

f ( x + h) – f ( x) (3 − 7 x − 7 h) − (3 − 7 x) –7 h = = = –7 h h h f ( x + h) = 11( x + h) 2 + 4 = 11x 2 + 22 xh + 11h 2 + 4

b. 17. a. b.

18. a.

b.

f ( x + h) – f ( x) (11x 2 + 22 xh + 11h 2 + 4) – (11x 2 + 4) 22 xh + 11h 2 = = = 22 x + 11h h h h f ( x + h) = 4( x + h)2 + 2( x + h) – 5 = 4 x 2 + 8 xh + 4h 2 + 2 x + 2h – 5 f ( x + h) – f ( x) (4 x 2 + 8 xh + 4h 2 + 2 x + 2h – 5) – (4 x 2 + 2 x – 5) = h h 2 8 xh + 4h + 2h = h = 8 x + 4h + 2 f ( x + h) =

7 7 = ( x + h) + 1 x + h + 1

f ( x + h) – f ( x ) = h

7 x + h +1

–

7 x +1

h

=

7( x +1)–7( x + h +1) ( x + h +1)( x +1)

h

=

−7 h −7 = ( x + h + 1)( x + 1)h ( x + h + 1)( x + 1)

19. f(x) = 3x – 1, g(x) = 2x + 3 a.

(f + g)(x) = f(x) + g(x) = (3x – 1) + (2x + 3) = 5x + 2

b. (f + g)(4) = 5(4) + 2 = 22

20.

c.

(f – g)(x) = f(x) – g(x) = (3x – 1) – (2x + 3) = x – 4

d.

( fg )( x) = f ( x) g ( x) = (3 x –1)(2 x + 3) = 6 x 2 + 7 x – 3

e.

( fg )(1) = 6(1)2 + 7(1) – 3 = 10

f.

f f ( x) 3 x –1 ( x) = = g g ( x) 2 x + 3

g.

( f D g )( x) = f ( g ( x)) = f (2 x + 3) = 3(2 x + 3) –1 = 6 x + 8

h.

( f D g )(5) = 6(5) + 8 = 38

i.

( g D f )( x) = g ( f ( x)) = g (3x –1) = 2(3 x –1) + 3 = 6 x + 1

f ( x) = − x 2 , g(x) = 3x − 2

a.

( f + g )( x) = f ( x) + g ( x) = − x 2 + 3 x − 2

b.

( f − g )( x) = f ( x) − g ( x) = − x 2 − (3x − 2) = − x 2 − 3x + 2 83



c.

( f − g )(−3) = −(−3) 2 − 3(−3) + 2 = 2

d.

( fg )( x) = f ( x) g ( x)

23. f(x) =

( f D g )( x) = f ( g ( x)) = f ( x3 ) = x3 + 2

= (− x 2 )(3x − 2)

( g D f )( x) = g ( f ( x)) = g

= −3 x3 + 2 x 2

e.

f f ( x) − x2 ( x) = = g g ( x) 3x − 2

f.

f −(2) 2 (2) = = −1 g 3(2) − 2

g.

x-axis:

y-axis:

= −3 x 2 − 2 ( g D f )(−4) = −3(−4) 2 − 2 = −48 − 2 = −50

origin:

, g(x) = x + 1

line y = x: (a, b) on graph, then b = 3a − a3 , but

1

a = 3b − b3 is not necessarily true, so (b, a) is not on the graph.

( x + 1)2

(

⎞ 1 1 + x2 = + = 1 ⎟ x2 ⎠ x2

origin 26.

( x) =

x +1 4 x +1 ⎛ x +1 ⎞ ( g D f )( x) = g ( f ( x)) = g ⎜ ⎟= 4 4 ⎝ ⎠ =

)

Answer: (0, 0), ± 3, 0 ; symmetry about

x +1 , g ( x) = x 4

( f D g )( x) = f ( g ( x)) = f

− y = 3(− x) − (− x)3 y = 3x − x3 , which is the original equation.

⎛ 1 ( g D f )( x) = g ( f ( x)) = g ⎜ ⎝ x2 f ( x) =

y = 3(− x) − (− x)3 y = −3 x + x3

( f D g )( x) = f ( g ( x)) = f ( x + 1) =

22.

− y = 3 x − x3 y = −3 x + x3 , which is not the original equation.

= 3(− x 2 ) − 2

x2

3

x(3 − x 2 ) = 0, x = 0, ± 3. If x = 0, then y = 0. Testing for symmetry gives:

( g D f )( x) = g ( f ( x))

f ( x) =

)

Intercepts: If y = 0, then 0 = 3 x − x3 ,

= g (− x 2 )

21.

x+2

25. y = 3x − x3

( f D g )( x) = f ( g ( x)) = f (3 x − 2)

1

) (

x+2 =

24. f(x) = 2, g(x) = 3 ( f D g )( x) = f ( g ( x)) = f (3) = 2 ( g D f )( x) = g ( f ( x)) = g (2) = 3

= −9 x 2 + 12 x − 4

i.

(

= ( x + 2)3 / 2

= −(3x − 2)2

h.

x + 2, g ( x) = x3

x +1 2

x2 y2

=4 x2 + y2 + 1 Intercepts: If y = 0, then 0 = 4, which is impossible; if x = 0, then 0 = 4, which is impossible. Testing for symmetry gives: x 2 (– y ) 2 =4 x-axis: x 2 + (− y )2 + 1 x2 y2 x2 + y2 + 1 equation.

84

= 4 , which is the original


y-axis:

(– x)2 y 2 (– x) 2 + y 2 + 1

Chapter 2 Review

10

=4

y

x2 y2

origin:

= 4 , which is the x2 + y2 + 1 original equation. (− x)2 (− y )2 =4 ( − x) 2 + (− y ) 2 + 1 x2 y2

x 10

28. y = 3x – 7

= 4, which is the

x2 + y 2 + 1 original equation.

line y = x: (a, b) on graph, then and b 2 = then b2 =

4(a 2 + 1) a2 − 4

b2 a 2 b2 + a 2 + 1

a 2b2 a 2 + b2 + 1

Intercepts: If y = 0, then 0 = 3x – 7, or x =

If x = 0, then y = –7. Testing for symmetry gives: x-axis: –y = 3x – 7 y = –3x + 7, which is not the original equation. y-axis: y = 3(–x) – 7 y = –3x – 7, which is not the original equation. origin: –y = 3(–x) – 7 y = 3x + 7, which is not the original equation. line y = x: (a, b) on graph, then b = 3a − 7 and 1 a = (b + 7) ≠ 3b − 7 for all b, so 3 (b, a) is not on the graph. ⎛7 ⎞ Answer: (0, –7), ⎜ , 0 ⎟ ; no symmetry of the ⎝3 ⎠ given types

=4

. (b, a) on graph,

= 4 and

4(a 2 + 1)

, so (a, b) and (b, a) a2 − 4 are both on the graph. Answer: no intercepts; symmetry about x-axis, y-axis, origin, and y = x.

27. y = 9 – x 2 Intercepts: If y = 0, then 0 = 9 – x 2 = (3 + x)(3 – x ) , or x = ±3 If x = 0, then y = 9. Testing for symmetry gives:

x-axis:

10

y

– y = 9 – x2 y = –9 + x 2 , which is not the original equation.

y-axis:

7 3

x 10

y = 9 – (– x)2 y = 9 – x 2 , which is the original equation.

origin:

7 . 3

− y = 9 − (− x)

29. G (u ) = u + 4

2

If G(u) = 0, then 0 = u + 4 . 0 = u + 4, u = –4 If u = 0, then G(u) = 4 = 2 . Intercepts: (0, 2), (–4, 0) Domain: all real numbers u such that u ≥ –4 Range: all real numbers ≥ 0

2

y = −9 + x , which is not the original equation.

line y = x: (a, b) on graph, then b = 9 − a 2 and a = ± 9 − b ≠ 9 − b 2 for all b, so (b, a) is not on the graph. Answer: (0, 9), (±3, 0); symmetry about y-axis.

85


10


G(u)

32. h(u ) = −5u If h(u) = 0, then 0 = −5u , u = 0. If u = 0, h(u) = 0. Intercept: (0, 0) Domain: all reals ≤ 0 Range: all reals ≥ 0

u 10

h(u)

30.

8

f ( x) = x + 1

If f(x) = 0, then 0 = x + 1. x = –1 , which has no solution.

If x = 0, then f(x) = 1. Intercept: (0, 1) Domain: all real numbers Range: all real numbers ≥ 1 10

u –8

33. Domain: all real numbers. Range: all real numbers ≤ 2

f(x)

5

y

x 10

x 5

31. y = g (t ) =

2 t –4

If y = 0, then 0 =

34.

y

2 , which has no solution. t –4

f(x) = √x

2 1 If t = 0, then y = = . 4 2 ⎛ 1⎞ Intercept: ⎜ 0, ⎟ ⎝ 2⎠ Domain: all real numbers t such that t ≠ 4 Range: all real numbers > 0 10

8

x y = √x – 2 – 1

35.

10

8

y f(x) = x2

g(t) x 10 y = – 1 x2 + 2

t

2

10

36. For 2006, t = 5. Hence S = 150,000 + 3000(5) = $165,000. S is a function of t.

86



37. From the vertical-line test, the graphs that represent functions of x are (a) and (c). 38. a.

a.

b. (1.92, 0), (0, 7)

729

20

44.

b. 359.43 39.

(–∞,∞)

8

–2

8

–8

2

–20

a.

–8

b. all real numbers ≥ −9.03

–0.67; 0.34, 1.73 40.

−9.03

c.

90

−5, ±2.

45. k = 0, 2, 4

2

6

–3

–3

3

–30

–1.38, 4.68 41.

5

k = 1, 3

–2 2

2

–2

3

–3

–5 –2

–1.50, –0.88, –0.11, 1.09, 1.40 42.

a.

20

0, 2, 4

b. none Mathematical Snapshot Chapter 2

8

–8

1.

f (23, 000) = 1510 + 0.15(23, 000 − 15,100) = 2695 The tax on $23,000 is $2695.

2.

f (85, 000) = 8440 + 0.25(85, 000 − 61,300) = 14,365 The tax on $85,000 is $14,365.

3.

f (290, 000) = 42,170 + 0.33(290, 000 − 188, 450) = 75, 681.5 The tax on $290,000 is $75,681.50.

–20

(–∞,∞) 43.

20

8

–8

–20

87

Chapter 2: Functions and Graphs 4.


f (462, 000) = 91, 043 + 0.35(462, 000 − 336,550) = 134,950.5 The tax on $462,000 is $134,950.50.

5. Answers may vary. 6. There should be no jump in tax as one moves from one tax bracket to the next, since it would be unfair for two couples whose incomes differ by only $0.01 to pay substantially different taxes. 7. g ( x) = x − f ( x) if 0 ≤ x ≤ 15,100 ⎧ x − 0.10 x ⎪ x − [1510 + 0.15( x − 15,100)] if 15,100 < x ≤ 61,300 ⎪ if 61,300 < x ≤ 123, 700 ⎪ x − [8440 + 0.25( x − 61,300)] =⎨ − + − x x [24, 040 0.28( 123, 700)] if 123, 700 < x ≤ 188, 450 ⎪ ⎪ x − [42,170 + 0.33( x − 188, 450)] if 188, 450 < x ≤ 336,550 ⎪⎩ x − [91, 043 + 0.35( x − 336,550)] if x > 336,550 if 0 ≤ x ≤ 15,100 ⎧0.90 x ⎪0.85 x + 755 if 15,100 < x ≤ 61,300 ⎪ if 61,300 < x ≤ 123, 700 ⎪0.75 x + 6885 =⎨ if 123, 700 < x ≤ 188, 450 ⎪0.72 x + 10,596 + x 0.67 20, 018.50 if 188, 450 < x ≤ 336,550 ⎪ ⎪⎩0.65 x + 26, 749.50 if x > 336,550 8.

400,000

g(x)

200,000

0

200,000

x 400,000

88

Chapter 3 F − F1 = m ( C − C1 )


9 (C − 5) 5 9 F − 41 = C − 9 5 9 F = C + 32 5

1. Let x = the time (in years) and let y = the selling price. Then, In 1991: x1 = 1991 and y1 = 32, 000 In 1994: x2 = 1994 and y2 = 26, 000 The slope is y −y m= 2 1 x2 − x1

F − 41 =

4. To find the slope and y-intercept, let a = 1000, then write the equation in slope-intercept form. 1 y= (t + 1)a 24 1 y= (t + 1)1000 24 1000 1000 y= t+ 24 24 125 125 y= t+ 3 3 125 Thus the slope, m, is and the y-intercept, b, 3 125 is . 3

26, 000 − 32, 000 1994 − 1991 −6000 = 3 = –2000 The car depreciated $2000 per year. Price (in thousands of dollars)

=

y (price) 40 30 20 10 x (time) 1990 1991 1992 1993 1994 Year

2. An equation relating the growth in enrollment to the number of years can be found by using the point-slope form of an equation of a line. If S = the number of students enrolled, and T = the number of years, then the point-slope form can be written as S − S1 = m (T − T1 )

5.

Let m = 14, S1 = 50 , and T1 = 3 . S – 50 = 14(T – 3) S – 50 = 14T – 42 S = 14T + 8

9 C + 32 5 ⎛9 ⎞ 5( F ) = 5 ⎜ C + 32 ⎟ 5 ⎝ ⎠ 5 F = 9C + 160 0 = 9C − 5F + 160 Thus, 9C – 5F + 160 = 0 is a general linear form 9 of F = C + 32 . 5 F=

6.

3. A linear function relating Fahrenheit temperature to Celsius temperature can be found by using the point-slope form of an equation of a line. F −F 77 − 41 36 9 = = m= 2 1 = 25 − 5 20 5 C2 − C1

F 100

–100

100

C

–100

To convert Celsius to Fahrenheit, locate the Celsius temperature on the horizontal axis, move vertically to the line, then move horizontally to read the Fahrenheit temperature of the vertical axis. 89

Chapter 3: Lines, Parabolas, and Systems


7. Right angles are formed by perpendicular lines. The slopes of the sides of the triangle are: 0−0 0 ⎧ AB ⎨m = = =0 6−0 6 ⎩

9.

7−0 7 ⎧ BC ⎨m = = =7 7−6 1 ⎩

y − 7 = −5[ x − (−1)] y − 7 = −5( x + 1) y − 7 = −5 x − 5 5x + y − 2 = 0 10

y

7−0 7 ⎧ AC ⎨m = = =1 7−0 7 ⎩ Since none of the slopes are negative reciprocals of each other, there are no perpendicular lines. Therefore, the points do not define a right triangle.

x 5

10.

Problems 3.1 1. m =

10 − 1 9 = =3 7−4 3

2. m =

10 − 3 7 = = −1 −2 − 5 −7

3. m =

−3 − (−2) −1 1 = =− 8−6 2 2

4. m =

−4 − (−4) 0 = =0 3− 2 1

y 100

x 10

11.

5. The difference in the x-coordinates is 5 – 5 = 0, so the slope is undefined. 6. m =

0 − (−6) 6 = =2 3−0 3

1 y − 5 = − [ x − (−2)] 4 4( y − 5) = −( x + 2) 4 y − 20 = − x − 2 x + 4 y − 18 = 0 10

y 9 2

−2 − (−2) 0 7. m = = =0 4−5 −1

8. m =

y − 0 = 75( x − 0) y = 75 x 75 x − y = 0

x 10

0 − (−7) 7 = 9 −1 8

90


12.

Section 3.1 −8 − (−4) −4 = = −4 −2 − (−3) 1 y − (−4) = −4[ x − (−3)] y + 4 = −4 x − 12 4 x + y + 16 = 0

1 ⎡ ⎛ 5 ⎞⎤ ⎢x − − ⎥ 3 ⎣ ⎜⎝ 2 ⎟⎠ ⎦ 5⎤ ⎡ 6( y − 5) = 2 ⎢ x + ⎥ 2⎦ ⎣ 6 y − 30 = 2 x + 5 2 x − 6 y + 35 = 0

15. m =

y −5 =

y 20

y

10

35 6

x x 10

20

3−0 3 = 2−0 2 3 y − 0 = ( x − 0) 2 3 y= x 2 2 y = 3x 3x − 2 y = 0

16. m =

4 −1 3 13. m = = 1 − (−6) 7 3 ( x − 1) 7 7( y − 4) = 3( x − 1) 7 y − 28 = 3x − 3 3x − 7 y + 25 = 0 y−4=

y

10

10

y

x 10

x 10

17.

2 − (−4) 6 = = −6 5−6 −1 y − 2 = −6( x − 5) y − 2 = −6 x + 30 6 x + y − 32 = 0

14. m =

y = 2x + 4 2x – y + 4 = 0 10

y

4

y

–2

50

x 10

91

x 10

Chapter 3: Lines, Parabolas, and Systems 18.


y = 5x – 7 5x – y – 7 = 0 10

21. A horizontal line has the form y = b. Thus y = –3, or y + 3 = 0.

y

5

y

x 10

7 5

x 5

–7

19.

22. A vertical line has the form x = a. Thus x = –1, or x + 1 = 0.

1 x−3 2 ⎛ 1 ⎞ 2 y = 2 ⎜ − x − 3⎟ ⎝ 2 ⎠ 2 y = −x − 6 x + 2y + 6 = 0 y=−

5

y 5

x (–1, –1)

y

x –6

23. A vertical line has the form x = a. Thus x = 2, or x − 2 = 0.

3

5

–3

20.

y = 0x − 1 2 2 y = −1 2y +1 = 0

5

y

x

1 2

5 (2, –3)

y=−

5

24. A horizontal line has the form y = b. Thus y = 0.

y

5

y

5x

x (0, 0)

–1 2

92

5


Section 3.1

25. y = 4x – 6 has the form y = mx + b, where m = 4 and b = –6. 10

29. x = –5 is a vertical line. Thus the slope is undefined. There is no y-intercept.

y

x

x 10

–5

5

30. x − 9 = 5 y + 3 5 y = x − 12 1 12 y = x− 5 5 1 12 m= ,b= − 5 5

26. x − 2 = 6 or x = 8, is a vertical line. Thus the slope is undefined. There is no y-intercept. 5

y

5

y

x 8

5

y

x

27. 3x + 5 y − 9 = 0 5 y = −3 x + 9 3 9 y = − x+ 5 5 3 9 m=− ,b= 5 5 5

20 – 12 5

31. y = 3x y = 3x + 0 m = 3, b = 0

y

5

y

9 5

x x

5 5

28. y + 4 = 7 y=3 y = 0x + 3 m = 0, b = 3 5

32. y − 7 = 3( x − 4) y − 7 = 3 x − 12 y = 3x − 5 m = 3, b = –5

y

3

y

3 x

x

5

5

–5

93



33. y = 3 y = 0x + 3 m = 0, b = 3

39.

y

5 3

x 5

1 x + 8 is in slope-intercept form. 300 ⎛ 1 ⎞ 300 y = 300 ⎜ x + 8⎟ 300 ⎝ ⎠ 300 y = x + 2400 x − 300 y + 2400 = 0 (general form)

40. y =

34. 6 y − 24 = 0 y=4 y = 0x + 4 m = 0, b = 4 5

x 2y 3 − + = −4 2 3 4 ⎛ x 2y ⎞ ⎛ 19 ⎞ 12 ⎜ − + ⎟ = 12 ⎜ − ⎟ ⎝ 2 3 ⎠ ⎝ 4⎠ −6 x + 8 y = −57 6 x − 8 y − 57 = 0 (general form) −8 y = −6 x + 57 3 57 y = x− (slope-intercept form) 4 8

y

41. The lines y = 7x + 2 and y = 7x – 3 have the same slope, 7. Thus they are parallel.

x 5

42. The lines y = 4x + 3 and y = 5 + 4x (or y = 4x + 5) have the same slope, 4. Thus they are parallel. 43. The lines y = 5x + 2 and –5x + y – 3 = 0 (or y = 5x + 3) have the same slope, 5. Thus they are parallel.

35. 2x = 5 – 3y, or 2x + 3y – 5 = 0 (general form) 2 5 3y = –2x + 5, or y = − x + (slope-intercept 3 3 form)

44. The line y = x has slope m1 = 1 and the line y = −x has slope m2 = −1 . m1 = −

36. 3x + 2y = 6, or 3x + 2y – 6 = 0 (general form) 3 2y = –3x + 6, or y = − x + 3 (slope-intercept 2 form)

1 so the m2

lines are perpendicular. 1 5⎞ ⎛ 45. The line x + 3y + 5 = 0 ⎜ or y = − x − ⎟ has 3 3⎠ ⎝ 1 slope m1 = − and the line y = −3x has slope 3 1 m2 = −3. Since m1 ≠ m2 and m1 ≠ − , the m2 lines are neither parallel nor perpendicular.

37. 4x + 9y – 5 = 0 is a general form. 4 5 9y = –4x + 5, or y = − x + (slope-intercept 9 9 form) 38. 3( x − 4) − 7( y + 1) = 2 3x − 12 − 7 y − 7 = 2 3x − 7 y − 21 = 0 (general form)

1 ⎞ ⎛ 46. The line x + 3y = 0 ⎜ or y = − x ⎟ has slope 3 ⎠ ⎝ 1 m1 = − and the line x + 6y − 4 = 0 (or 3 1 1 2⎞ y = − x + ⎟ has slope m2 = − . Since 6 6 3⎠

3 −7y = −3x + 21, or y = x − 3 (slope-intercept 7 form)

94


Section 3.1 56. y = –4 is a horizontal line. The perpendicular line must be vertical and has an equation of the form x = a. Since that line passes through (1, 1), its equation is x = 1.

1 , the lines are neither m2 parallel nor perpendicular. m1 ≠ m2 and m1 ≠ −

47. The line y = 3 is horizontal and the line x = −

1 3

57. y = −3 is a horizontal line, so the perpendicular line must be vertical with equation of the form x = a. Since that line passes through (5, 2), its equation is x = 5.

is vertical, so the lines are perpendicular. 48. Both lines are vertical and thus parallel.

58. The line 3 y = −

49. The line 3x + y = 4 (or y = –3x + 4) has slope m1 = –3, and the line x – 3y + 1 = 0 1 1 1⎞ ⎛ ⎜ or y = 3 x + 3 ⎟ has slope m2 = 3 . Since ⎝ ⎠ m2 = −

2x 2x ⎞ ⎛ + 1 has + 3 ⎜ or y = − 5 15 ⎟⎠ ⎝

2 , so the slope of a line perpendicular 15 15 . An equation of the to it must have slope 2 15 desired line is y − (−5) = ( x − 4) or 2 15 y = x − 35. 2

slope −

1 , the lines are perpendicular. m1

50. The line x − 2 = 3 (or x = 5) is vertical and the line y = 2 is horizontal, so the lines are perpendicular.

2 59. The line 2x + 3y + 6 = 0 has slope − , so the 3 2 slope of a line parallel to it must also be − . An 3 equation of the desired line is 2 2 29 y − (−5) = − [ x − (−7)], or y = − x − . 3 3 3

x 1 51. The slope of y = − − 2 is − , so the slope of 4 4 1 a line parallel to it must also be − . An 4 1 equation of the desired line is y − 1 = − ( x − 1) 4 1 5 or y = − x + . 4 4

60. The y-axis is vertical. A parallel line is also vertical and has an equation of the form x = a. Since it passes through (–4, 10), its equation is x = –4.

52. x = –4 is a vertical line. A line parallel to x = –4 has the form x = a. Since the line must pass through (2, –8), its equation is x = 2.

61. (1, 2), (–3, 8) 8−2 6 3 = =− m= −3 − 1 −4 2

53. y = 2 is a horizontal line. A line parallel to it has the form y = b. Since the line must pass through (2, 1) its equation is y = 1.

3 Point-slope form: y − 2 = − ( x − 1) . When the 2 x-coordinate is 5, 3 y − 2 = − (5 − 1) 2 3 y − 2 = − (4) 2 y − 2 = −6 y = −4 Thus the point is (5, –4).

54. The slope of y = 3 + 2x is 2, so the slope of a line parallel to it must also be 2. An equation of the desired line is y − (−4) = 2(x − 3), or y = 2x − 10. 55. The slope of y = 3x − 5 is 3, so the slope of a line 1 perpendicular to it must have slope − . An 3 1 equation of the desired line is y − 4 = − ( x − 3), 3 1 or y = − x + 5. 3

95


ISM: Introductory Mathematical Analysis Using the point-slope form with m = 28,000 and ( x1 , y1 ) = (0, − 100, 000) gives

62. m = 3, b = 1 Slope-intercept form: y = 3x + 1. The point (−1, −2) lies on the line if its coordinates satisfy the equation. If x = −1 and y = −2, then −2 = 3(−1) + 1 or −2 = −2, which is true. Thus (−1, −2) lies on the line.

y − y1 = m ( x − x1 ) y − (−100, 000) = 28, 000( x − 0) y + 100, 000 = 28, 000 x y = 28, 000 x − 100, 000

Price per share (dollars)

63. Let x = the time (in years) and y = the price per share. Then, In 1988: x1 = 1988 and y1 = 37 In 1998: x2 = 1998 and y2 = 8 The slope is −29 8 − 37 = = −2.9 m= 10 1998 − 1988 The stock price dropped an average of $2.90 per year.

66. Solve the equation for t. L = 1.53t – 6.7 L + 6.7 = 1.53t ( L + 6.7) =t 1.53 0.65L + 4.38 = t The slope is approximately 0.65 and the y-intercept is approximately 4.38.

40

67. A general linear form of d = 184 + t is –t + d – 184 = 0.

30

68. a.

y (price)

20 10

x (time)

0

1988

1993 Year

1998

64. The number of home runs hit increased as a function of time (in months). The given points are ( x1 , y1 ) = (3, 14) and ( x2 , y2 ) = (5, 20) .

b. Using the points (0.5, 0.5) and (−1, −2.5) −2.5 − 0.5 −3 = = 2. gives a slope of m = −1 − 0.5 −1.5 An equation is y − 0.5 = 2(x − 0.5) or 1 y = 2x − . 2

y −y 20 − 14 6 = =3 m= 2 1 = 5−3 2 x2 − x1 Using the point-slope form with m = 3 and ( x1 , y1 ) = (3, 14) gives

These two paths are not perpendicular to each other because the slopes are not negative reciprocals of each other.

y − y1 = m ( x − x1 ) y − 14 = 3( x − 3) y − 14 = 3x − 9 y = 3x + 5

69. The slopes of the sides of the figure are: 4−0 4 ⎧ AB ⎨m = = = undefined (vertical) −0 0 0 ⎩

65. The owner’s profits increased as a function of time. Let x = the time (in years) and let y = the profit (in dollars). The given points are ( x1 , y1 ) = (0, − 100, 000) and

( x2 ,

Using the points (3.5, −1.5) and (0.5, 0.5) 2 −1.5 − 0.5 =− . gives a slope of m = 3.5 − 0.5 3 2 An equation is y − 0.5 = − ( x − 0.5) or 3 2 5 y = − x+ . 3 6

7−3 4 ⎧ CD ⎨m = = = undefined (vertical) −2 0 2 ⎩

y2 ) = (5, 40, 000) .

3−0 3 ⎧ AC ⎨m = = 2−0 2 ⎩

y −y 40, 000 − (−100, 000) 140, 000 = m= 2 1 = 5−0 5 x2 − x1 = 28,000

7−4 3 ⎧ BD ⎨m = = 2−0 2 ⎩

Since AB is parallel to CD and AC is parallel to BC , ABCD is a parallelogram. 96


Section 3.2

70. Let x = the distance traveled and let y = the altitude. The path of descent is a straight line with a slope of –1 and y-intercept of 3600. Therefore, using the slope-intercept form with m = –1 and b = 3600 gives y = mx + b y = (–1)x + 3600 y = –x + 3600

10

76.

10

–10

–10 10

4000

–15

–1000

–10

If the airport is located 3800 feet from where the plane begins its landing approach, the plane will crash 200 feet short of the airport.

The slope of the first line is 0.1875 = 0.625 , and the slope of the m1 = 0.3 0.32 second line is m2 = − = −1.6 . Since 0.2 1 m1 = − , the lines are perpendicular. m2

71. The line has slope 59.82 and passes through (6, 1128.50). Thus C – 1128.50 = 59.82(T – 6) or C = 59.82T + 769.58. 72. The line has slope 50,000 and passes through (5, 330,000). Thus R – 330,000 = 50,000(T – 5) or R = 50,000T + 80,000.

Principles in Practice 3.2 1. Let x = the number of skis that are produced and let y = the number of boots that are produced. Then, the equation 8x + 14y = 1000 describes all possible production levels of the two products.

10

73.

10

–10

2. The quantity and price are linearly related such that p = 575 when q = 1200, and p = 725 when q = 800. Thus ( q1 , p1 ) = (1200, 575) and

–10

( q2 ,

The graph of the equation y = −0.9x − 7.3 shows that when x = 0, y = 7.3. Thus, the y-intercept is 7.3.

p2 ) = (800, 725) . The slope is

725 − 575 3 =− . 800 − 1200 8 An equation of the line is m=

10

74.

15

4000

–500

p − p1 = m ( q − q1 ) 3 p − 575 = − (q − 1200) 8 3 p − 575 = − q + 450 8 3 p = − q + 1025 8

10

–10

–10

The lines are parallel, which is expected because they have the same slope, 1.5. 75. The slope is 7.1.

97



3. Answers may vary, but two possible points are (0, 60) and (2, 140). f(x) = 40x + 60 x

f(x)

0

60

2

140

1000

Problems 3.2 1. y = f(x) = –4x = –4x + 0 has the form f(x) = ax + b where a = –4 (the slope) and b = 0 (the vertical-axis intercept). 5

y

x

f(x)

5

500

0

2. y = f(x) = x + 1 has the form f(x) = ax + b where a = 1 (the slope) and b = 1 (the vertical-axis intercept).

x 20

5

4. If t = the age of the child, then f(t) = the height of the child at any age t. The height and age are linearly related such that f(8) = 50.6. Since f(t) is a linear function it has the form f(t) = at + b. Since the height changes by 2.3 inches per year, a = 2.3. Then, f(t) = at + b f(8) = 2.3(8) + b 50.6 = 18.4 + b 32.2 = b Thus, f(t) = 2.3t + 32.2 is a function that describes the height of the child at age t.

y

1

x 5

3. h(t) = 5t − 7 has the form h(t) = at + b with a = 5 (the slope) and b = −7 (the vertical-axis intercept). 10

h(t)

5. Let y = f(x) = a linear function that describes the value of the necklace after x years. The problem states that f(3) = 360 and f(7) = 640. Thus, ( x1 , y1 ) = (3, 360) and ( x2 , y2 ) = (7, 640) . The

t 10

y2 − y1 640 − 360 280 = = = 70 x2 − x1 7−3 4 Using the point-slope form with m = 70 and ( x1 , y1 ) = (3, 360) gives

slope is m =

4. f(s) = 3(5 − 2s) = 15 − 6s has the form f(s) = as + b where a = –6 (slope) and b = 15 (the vertical-axis intercept).

y − y1 = m ( x − x1 ) y – 360 = 70(x – 3) y = f(x) = 70x + 150

f(s) 16

s 5

98


Section 3.2

2−q 2 1 = − q has the form 7 7 7 1 h(q) = aq + b where a = − (the slope) and 7 2 b = (the vertical-axis intercept). 7

⎛2⎞ −7 = −2 ⎜ ⎟ + b ⎝5⎠ 4 31 b = −7 + = − 5 5 31 so f ( x) = −2 x − . 5

5. h(q) =

5

h(q)

11. 2 7

f ( x) = ax + b = −

2 2 ⎛ 2⎞ x + b . Since f ⎜ − ⎟ = − , 3 3 ⎝ 3⎠

we have 2 2⎛ 2⎞ − = − ⎜− ⎟+b 3 3⎝ 3⎠

q 5

2 4 10 − =− , 3 9 9 2 10 so f ( x) = − x − . 3 9 b=−

6. h(q) = 0.5q + 0.25 has the form h(q) = aq + b with a = 0.5 (the slope) and b = 0.25 (the vertical-axis intercept). 1

h(q)

0.25 –0.5

12. Let y = f(x). The points (1, 1) and (2, 2) lie on 2 −1 =1. the graph of f. m = 2 −1 Thus y – 1 = 1(x – 1) ⇒ y = x, so f(x) = x.

q 1

13. Let y = f(x). The points (–2, –1) and (–4, –3) lie −3 + 1 = 1 . Thus on the graph of f. m = −4 + 2 y + 1 = 1(x + 2), so y = x + 1 ⇒ f(x) = x + 1.

7. f(x) = ax + b = 4x + b. Since f(2) = 8, 8 = 4(2) + b, 8 = 8 + b, b = 0 ⇒ f(x) = 4x.

14. f(x) = ax + b = 0.01x + b. Since f(0.1) = 0.01, we have 0.01 = (0.01)(0.1) + b ⇒ b = 0.009 ⇒ f(x) = 0.01x + 0.009.

8. Let y = f(x). The points (0, 3) and (4, –5) lie on −5 − 3 = −2 . Thus the graph of f. m = 4−0 y – 3 = –2(x – 0), so y = –2x + 3 ⇒ f ( x) = –2x + 3.

15. The points (40, 12.75) and (25, 18.75) lie on the graph of the equation, which is a line. 18.75 − 12.75 2 = − . Hence an equation of m= 25 − 40 5 2 the line is p − 12.75 = − (q − 40) , which can be 5 2 written p = − q + 28.75. When q = 37, then 5 2 p = − (37) + 28.75 = $13.95. 5

9. Let y = f(x). The points (1, 2) and (–2, 8) lie on 8−2 = −2 . Thus the graph of f. m = −2 − 1 y – 2 = –2(x – 1), so y = –2x + 4 ⇒ f ( x) = –2x + 4. 10. f(x) = ax + b = −2x + b. ⎛2⎞ Since f ⎜ ⎟ = −7, we have ⎝5⎠

16. The line passes through (26,000, 12) and (10,000, 18), so 18 − 12 = −0.000375. Then m= 10, 000 − 26, 000 p – 18 = –0.000375(q – 10,000) or p = –0.000375q + 21.75. 99



17. The line passes through (3000, 940) and 740 − 940 = 0.25. Then (2200, 740), so m = 2200 − 3000 p – 740 = 0.25(q – 2200) or p = 0.25q + 190.

20

18. The points (50, 35) and (35, 30) lie on the graph of the equation, which is a line. 30 − 35 −5 1 = = . Hence an equation of m= 35 − 50 −15 3 the line is 1 p − 35 = (q − 50) 3 1 55 p = q+ 3 3

r

d 0

20

23. Each year the value decreases by 0.10(1800). After t years the total decrease is 0.10(1800)t. Thus v = 1800 – 0.10(1800)t v = –180t + 1800 The slope is –180. v

19. The line passing through (10, 40) and (20, 70) 70 − 40 = 3 , so an equation for the has slope 20 − 10 line is c – 40 = 3(q – 10) c = 3q + 10 If q = 35, then c = 3(35) + 10 = 105 + 10 = $115.

1800

0

20. The line passing through (100, 79) and (400, 88) 88 − 79 = 0.03, so an equation for has slope 400 − 100 the line is c – 79 = 0.03(x – 100) c = 0.03x + 76

t 10

24. The line has slope –120 and passes through (4, 340). Thus y – 340 = –120(x – 4) or y = f(x) = –120x + 820. 25. The line has slope 45,000 and passes through (5, 960,000). Thus y – 960,000 = 45,000(x – 5) or y = f(x) = 45,000x + 735,000.

21. If x = the number of kilowatt hours used in a month, then f ( x) = the total monthly charges for x kilowatt hours of electricity. If f ( x) is a linear function it has the form f ( x) = ax + b. The problem states that f(380) = 51.65. Since 12.5 cents are charged per kilowatt hour used, a = 0.125. f(x) = ax + b 51.65 = 0.125(380) + b 51.65 = 47.5 + b 4.15 = b Hence, f ( x) = 0.125x + 4.15 is a linear function that describes the total monthly charges for any number of kilowatt hours x.

245, 000 49, 000 = and 15 3 y-intercept 245,000. So 49, 000 y = f ( x) = x + 245, 000. 3

26. The line has slope

27. If x = the number of hours of service, then f(x) = the price of x hours of service. Let y = f(x). f(1) = 159 and f(3) = 287, so (1, 159) and (3, 287) lie on the graph of f which has slope 287 − 159 = 64. Using (1, 159), we get a= 3 −1 y − 159 = 64(x − 1) or y = 64x + 95, so f(x) = 64x + 95.

22. The number of curative units from d cubic centimeters of the drug is 210d, and the number of curative units from r minutes of radiation is 305r. Thus 210d + 305r = 2410 42d + 61r = 482 100


Section 3.2

Suppose r = respiratory rate, l = wool length, and (l, r) lies on the graph, which is a line. The points (2, 160) and (4, 125) are on the line, so its slope is 125 − 160 35 = − . Thus 4−2 2 35 r − 160 = − (l − 2) 2 35 r = − l + 195 2

b. If l = 1, then r = −

100 − 65 35 = 100 − 56 44 35 y − 100 = ( x − 100) 44 35 3500 y= x− + 100 44 44 35 225 y= x+ 44 11

m=

31. a.

35 225 x+ 44 11 35 225 x = 62 − 44 11 1828 x= ≈ 52.2 35 52.2 is the lowest passing score on original scale. 62 =

b.

35 (1) + 195 = 177.5 2

29. At $200/ton, x tons cost 200x, and at $2000/acre, y acres cost 2000y. Hence the required equation is 200x + 2000y = 20,000, which can be written as x + 10y = 100.

32. R = 38N + 397 is a linear equation. Slope = 38.

30. P = 4x + 6y where x, y ≥ 0. a. 240 = 4x + 6y 100

R

587

y

549 511 473

40

435 0

60

x 100

0

33. p = f(t) = at + b, f(5) = 0.32, a = slope = 0.059.

b. Since the equation can be written 2 2 y = − x + 40 , slope = − . 3 3 c.

N 5

a.

p = f(t) = 0.059t + b. Since f(5) = 0.32, 0.32 = 0.059(5) + b, 0.32 = 0.295 + b, so b = 0.025. Thus p = 0.059t + 0.025. b. When t = 9, then p = 0.059(9) + 0.025 = 0.556.

600 = 4x + 6y. Since the equation can be 2 written y = − x + 100, 3 2 slope = − . 3

34. w = f(d) = ad + b, f(0) = 21, 6.3 = 0.63. Thus a = slope = 10 w = f(d) = 0.63d + b. Since f(0) = 21, we have 20 = 0.63(0) + b, so b = 21. Hence w = 0.63d + 21. When d = 55, then w = 0.63(55) + 21 = 34.65 + 21 = 55.65 kg.

d. Solving P = 4x + 6y for y gives 2 P y = − x + . Thus any isoprofit line has 3 6 2 slope − , and lines with the same slope are 3 parallel. Hence isoprofit lines are parallel.

101


35. a.


( )

t −t 80 − 68 12 1 = = . m= 2 1 = c2 − c1 172 − 124 48 4

h(1) = −16 12 + 32(1) + 8 = 24 . Thus, the vertex

is (1, 24). Since c = 8, the y-intercept is (0, 8). To find the x-intercepts we set y = h(t) = 0.

1 1 t − 68 = (c − 124) , t − 68 = c − 31 , or 4 4 1 t = c + 37 . 4

0 = −16t 2 + 32t + 8 t=

b. Since c is the number of chirps per minute, 1 1 then c is the number of chirps in 4 4 minute or 15 seconds. Thus from part (a), to estimate temperature add 37 to the number of chirps in 15 seconds.

=

−32 ± 322 − 4(−16)(8) −b ± b 2 − 4ac = 2a 2(−16)

−32 ± 1536 −32 ± 16 6 6 = = 1± −32 −32 2

⎛ 6 ⎞ Thus, the x-intercepts are ⎜1 + , 0 ⎟ and ⎜ ⎟ 2 ⎝ ⎠ ⎛ 6 ⎞ , 0⎟ . ⎜⎜1 − ⎟ 2 ⎝ ⎠

Principles in Practice 3.3 1. In the quadratic function y = P ( x) = − x 2 + 2 x + 399 , a = –1, b = 2, c = 399. Since a < 0, the parabola opens downward. The x-coordinate of the vertex is b 2 − =− =1. 2a 2(−1) The y-coordinate of the vertex is

30

–5

( )

P (1) = − 12 + 2(1) + 399 = 400 . Thus, the vertex

–20

is (1, 400). Since c = 399, the y-intercept is (0, 399). To find the x-intercepts we set y = p(x) = 0. 0 = − x 2 + 2 x + 399

(

0 = − x 2 − 2 x − 399

3. If we express the revenue r as a function of the quantity produced q, we obtain r = pq r = (6 – 0.003q)q

)

r = 6q − 0.003q 2 We note that this is a quadratic function with a = –0.003, b = 6, and c = 0. Since a < 0, the graph of the function is a parabola that opens downward, and r is maximum at the vertex (q, r). b 6 =− = 1000 q=− 2a 2(−0.003)

0 = –(x + 19)(x – 21) Thus, the x-intercepts are (–19, 0) and (21, 0). y 400

100 –25

25

5

r = 6(1000) − 0.003(1000)2 = 3000 Thus, the maximum revenue that the manufacturer can receive is $3000, which occurs at a production level of 1000 units.

x

If the model is correct, this is not a good business, since it will lose money if more than 21 minivans are sold.

Problems 3.3

2

2. In the quadratic function h(t ) = −16t + 32t + 8 , a = –16, b = 32, and c = 8. Since a < 0, the parabola opens downward. The x -coordinate of b 32 =− = 1 . The the vertex is − 2a 2(−16) y-coordinate of the vertex is

1.

f ( x) = 5 x 2 has the form f ( x) = ax 2 + bx + c

where a = 5, b = 0, and c = 0 ⇒ quadratic.

102


1

2. g ( x) =

2 x2 − 4

Section 3.3

10. y = f ( x) = 8 x 2 + 4 x − 1 a = 8, b = 4, c = –1

cannot be put in the form

g ( x) = ax 2 + bx + c where

a.

a ≠ 0 ⇒ not quadratic.

2

g ( x) = ax 2 + bx + c where

3⎞ ⎛ 1 Vertex: ⎜ − , − ⎟ 2⎠ ⎝ 4

a ≠ 0 ⇒ not quadratic. 4. k (v) = 3v 2 (v 2 + 2) = 3v 4 + 6v 2 cannot be put in

b. a = 8 > 0, so the vertex corresponds to the lowest point.

the form k (v) = av 2 + bv + c where a ≠ 0 ⇒ not quadratic.

11. y = x 2 + x − 6

2

5. h(q) = (3 − q ) = 9 − 6q + q has form

a = 1, b = 1, c = −6 a. c = –6. Thus the y-intercept is –6.

2

h(q) = aq + bq + c where a = 1, b = −6, and

c = 9 ⇒ quadratic. 6.

b.

f (t ) = 2t (3 − t ) + 4t = −2t 2 + 10t has the form

x 2 + x − 6 = ( x − 2)( x + 3) = 0, so x = 2, −3.

x-intercepts: 2, −3

2

f (t ) = at + bt + c where a = –2, b = 10, and

c.

c = 0 ⇒ quadratic.

(

)

2

−

b 1 =− 2a 2 2

1 25 ⎛ 1⎞ ⎛ 1⎞ f ⎜− ⎟ = ⎜− ⎟ − −6 = − 2 4 ⎝ 2⎠ ⎝ 2⎠

s2 − 9 1 2 9 = s − has the form 7. f ( s ) = 2 2 2 1 2 f ( s ) = as + bs + c where a = , b = 0, and 2 9 c = – ⇒ quadratic. 2

8. g (t ) = t 2 − 1

b 4 1 =− =− 2a 2 ⋅8 4

3 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = 8⎜ − ⎟ + 4 ⎜ − ⎟ −1 = − 4 4 4 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3. g(x) = 7 – 6x cannot be put in the form

2

−

25 ⎞ ⎛ 1 Vertex: ⎜ − , − ⎟ 4 ⎠ ⎝ 2

12. y = f ( x) = 5 − x − 3x 2 a = –3, b = −1, c = 5 a. c = 5. Thus the y-intercept is 5.

= t 4 − 2t 2 + 1 cannot be put in

the form g (t ) = at 2 + bt + c where a ≠ 0 ⇒ not quadratic.

b.

Vertex occurs when x = −

−b ± b 2 − 4ac 2a

−(−1) ± (−1)2 − 4(−3)(5) 2(−3) 1 ± 61 = −6 −1 ± 61 = 6 −1 + 61 −1 − 61 x-intercepts: , 6 6

9. y = f ( x) = −4 x 2 + 8 x + 7 a = –4, b = 8, c = 7 a.

x= =

b 8 =− =1. 2a 2(−4)

When x = 1, then y = f (1) = −4(1)2 + 8(1) + 7 = 11 . Vertex: (1, 11)

b. a = –4 < 0, so the vertex corresponds to the highest point.

103


c.

−


b −1 1 =− =− 2a 2(−3) 6

15. y = g ( x) = −2 x 2 − 6 x a = –2, b = –6, c = 0 b −6 6 3 =− =− =− Vertex: − 2a 2(−2) 4 2

2

61 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = 5 − ⎜ − ⎟ − 3⎜ − ⎟ = 12 ⎝ 6⎠ ⎝ 6⎠ ⎝ 6⎠

2

9 ⎛ −3 ⎞ ⎛ −3 ⎞ ⎛ −3 ⎞ −9 f ⎜ ⎟ = −2 ⎜ ⎟ − 6 ⎜ ⎟ = +9 = 2 2 ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠

⎛ 1 61 ⎞ Vertex: ⎜ − , ⎟ ⎝ 6 12 ⎠

⎛ 3 9⎞ Vertex: ⎜ − , ⎟ ⎝ 2 2⎠ y-intercept: c = 0

2

13. y = f ( x) = x − 6 x + 5 a = 1, b = –6, c = 5 b −6 =− =3 Vertex: − 2a 2 ⋅1

x-intercepts: −2 x 2 − 6 x = −2 x( x + 3) = 0 , so x = 0, –3. 9 Range: all y ≤ 2

f (3) = 32 − 6(3) + 5 = −4 Vertex = (3, –4) y-intercept: c = 5

y

2

x-intercepts: x − 6 x + 5 = (x – 1)(x – 5) = 0, so x = 1, 5. Range: all y ≥ –4

9 2

–3

x

–3 2

y 5

16. y = f ( x) = x 2 − 4 a = 1, b = 0, c = –4 b 0 =− =0 Vertex: − 2a 2 ⋅1

x 1

5

(3, –4)

f (0) = 02 − 4 = −4 Vertex = (0, –4) y-intercept: c = –4

14. y = f ( x) = –4 x 2 a = –4, b = 0, c = 0 b 0 =− =0 Vertex: − 2a 2(−4)

x-intercepts: x 2 − 4 = ( x + 2)( x − 2) = 0 , so x = –2, 2. Range: all y ≥ 4

f (0) = −4(0) 2 = 0 Vertex = (0, 0) y-intercept: c = 0

5

x-intercepts: −4 x 2 = 0 , so x = 0. Range: all y ≤ 0 5

y

x –2

y

2 –4

x 5

104

5


Section 3.3

17. s = h(t ) = t 2 + 6t + 9 a = 1, b = 6, c = 9 b 6 =− = −3 Vertex: − 2a 2 ⋅1

19. y = f ( x) = −9 + 8 x − 2 x 2 a = –2, b = 8, c = –9 b 8 =− =2 Vertex: − 2a 2(−2)

h(−3) = (−3) 2 + 6(−3) + 9 = 0 Vertex = (–3, 0) s-intercept: c = 9

f (2) = −9 + 8(2) − 2(2)2 = −1 Vertex = (2, –1) y-intercept: c = –9 x-intercepts: Because the parabola opens downward (a < 0) and the vertex is below the x-axis, there is no x-intercept. Range: y ≤ –1

t-intercepts: t 2 + 6t + 9 = (t + 3)2 = 0 , so t = –3. Range: all s ≥ 0 10

s

2

y

x (2, –1)

t –3

–9

3

18. s = h(t ) = 2t 2 + 3t − 2 a = 2, b = 3, c = –2 b 3 3 =− =− Vertex: − 2a 2⋅2 4

20. y = H ( x) = 1 − x − x 2 a = –1, b = –1, c = 1 b −1 1 =− =− Vertex: − 2a 2(−1) 2

2

⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ h ⎜ − ⎟ = 2 ⎜ − ⎟ + 3⎜ − ⎟ − 2 ⎝ 4⎠ ⎝ 4⎠ ⎝ 4⎠ 9 9 25 = − −2 = − 8 4 8 25 ⎞ ⎛ 3 Vertex = ⎜ − , − ⎟ 8 ⎠ ⎝ 4 s-intercept: c = –2

2

5 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ f ⎜ − ⎟ = 1− ⎜ − ⎟ − ⎜ − ⎟ = 4 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎛ 1 5⎞ Vertex = ⎜ − , ⎟ ⎝ 2 4⎠ y-intercept: c = 1

x-intercepts: Solving 1 − x − x 2 = 0 by the quadratic formula gives

t-intercepts: 2t 2 + 3t − 2 = (2t − 1)(t + 2) = 0 , so 1 t = , –2. 2

x=

Range: all s ≥ − 5

25 8

=

y

2(−1)

−1 ± 5 2

Range: all y ≤ 1 2

–2

(– 34 , – 258)

−(−1) ± (−1) 2 − 4(−1)(1)

5

x 5

(– 12 , 54 )

–2

–1 – 5 2

105

y

5 4

–1 + 5 2

x 5

=

1± 5 −2



21. t = f ( s ) = s 2 − 8s + 14 a = 1, b = –8, c = 14 b −8 Vertex: − =− =4 2a 2 ⋅1

23.

f (4) = 42 − 8(4) + 14 = −2 Vertex = (4, –2) t-intercept: c = 14

2

808 ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎞ f ⎜ ⎟ = 49 ⎜ ⎟ − 10 ⎜ ⎟ + 17 = . 49 49 49 49 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

s-intercepts: Solving s 2 − 8s + 14 = 0 by the quadratic formula: s=

f ( x) = 49 x 2 − 10 x + 17 Since a = 49 > 0, the parabola opens upward and f(x) has a minimum value that occurs when 5 b −10 . The minimum value is x=− =− = 2a 2 ⋅ 49 49

24.

−(−8) ± (−8)2 − 4(1)(14) 2(1)

8± 8 8± 2 2 = = 4± 2 2 2 Range: all t ≥ –2 =

f ( x) = −3 x 2 − 18 x + 7 Since a = –3 < 0, the parabola opens downward and f(x) has a maximum value that occurs when b −18 =− = −3 x=− 2a 2(−3) The maximum value is f (−3) = −3(−3)2 − 18(−3) + 7 = 34.

t

25. 14

4–

2

4+

s

2

f (20) = 4(20) − 50 − 0.1(20)2 = −10 .

(4, –2)

26.

22. t = f ( s ) = s 2 + 6s + 11 a = 1, b = 6, c = 11 6 b Vertex: − =− = −3 2a 2 ⋅1 f (−3) = (−3)2 + 6(−3) + 11 = 2 Vertex: (–3, 2) t-intercept: c = 11 s-intercepts: Because the parabola opens upward (a > 0) and the vertex is above the s-axis, there is no s-intercept. Range: all t ≥ 2 16

f ( x) = 4 x − 50 − 0.1x 2 Since a = –0.1 < 0, the parabola opens downward and f(x) has a maximum value that 4 b occurs when x = − =− = 20 . The 2a 2(−0.1) maximum value is

f ( x) = x( x + 3) − 12 = x 2 + 3 x − 12 Because a = 1 > 0, the parabola opens upward and f(x) has a minimum value that occurs when b 3 3 =− = − . The minimum value is x=− 2a 2 ⋅1 2 2

57 ⎛ 3⎞ ⎛ 3⎞ ⎛ 3⎞ f ⎜ − ⎟ = ⎜ − ⎟ + 3 ⎜ − ⎟ − 12 = − 4 ⎝ 2⎠ ⎝ 2⎠ ⎝ 2⎠

27.

f ( x) = x 2 − 2 x + 4

a = 1, b = −2, c = 4 b −2 =− =1 v=− 2a 2(1)

t

The restricted function is g ( x) = x 2 − 2 x + 4,

11

x ≥ 1. From the quadratic formula applied to (–3, 2)

x 2 − 2 x + 4 − y = 0, we get s

x=

10

2 ± 4 − 4(1)(4 − y ) = 1 ± 1 − (4 − y ) 2(1)

So the inverse of g(x) is g −1 ( x) = 1 + x − 3, x ≥ 3. 106


8

Section 3.3

y

30. If we express the revenue r as a function of the quantity produced q, we obtain r = pq r = (0.85 − 0.00045q)q

g(x) g –1(x)

r = 0.85q − 0.00045q 2

This is a quadratic function with a = −0.00045, b = 0.85, and c = 0. Since a < 0, the graph of the function is a parabola that opens downward, and r is a maximum at the vertex (q, r). b 0.85 8500 q=− =− = ≈ 944 2a 2(−0.00045) 9

x 8

28.

f ( x) = − x 2 + 4 x − 3

a = −1, b = 4, c = −3 b 4 v=− =− =2 2a 2(−1)

r = 0.85(944) − 0.00045(944) 2 = 401.39 Thus, the maximum revenue that the manufacturer can receive is $401.39, which occurs at a production level of 944 units.

The restricted function is g ( x) = − x 2 + 4 x − 3, x ≥ 2. From the quadratic formula applied to − x 2 + 4 x − 3 − y = 0, we get

31. If we express the revenue r as a function of the quantity produced q, we obtain r = pq r = (2400 – 6q)q

−4 ± 16 − 4(−1)(−3 − y ) 2(−1) = 2 ± (−1) 4 + (−3 − y )

x=

r = 2400q − 6q 2 This is a quadratic function with a = –6, b = 2400, and c = 0. Since a < 0, the graph of the function is a parabola that opens downward, and r is maximum at the vertex (q, r). b 2400 q=− =− = 200 2a 2(−6)

So the inverse of g(x) is g −1 ( x) = 2 + 1 − x , x ≤ 1. 5

y

g –1(x) x

r = 2400(200) − 6(200)2 = 240, 000 Thus, the maximum revenue that the manufacturer can receive is $240,000, which occurs at a production level of 200 units.

5

g(x)

29. If we express the revenue r as a function of the quantity produced q, we obtain r = pq r = (200 − 5q)q

32.

r = 200q − 5q 2 This is a quadratic function with a = –5, b = 200, and c = 0. Since a < 0, the graph of the function is a parabola that opens downward, and r is maximum at the vertex (q, r). b 200 q=− =− = 20 2a 2(−5)

10 40 10 n(12 − n) = n − n 2 , where 9 3 9 10 0 ≤ n ≤ 12. Since a = − < 0 , f(n) has a 9 maximum value that occurs at the vertex. f ( n) =

40

−

b 3 =− =6 2a 2 − 10 9

( )

The maximum value of f(n) is 40 10 f (6) = (6) − (6) 2 = 80 − 40 = 40 , which 3 9 corresponds to 40,000 households.

r = 200(20) − 5(20)2 = 2000 Thus, the maximum revenue that the manufacturer can receive is $2000, which occurs at a production level of 20 units.

33. In the quadratic function P ( x ) = − x 2 + 18 x + 144, a = –1, b = 18, and c = 144. Since a < 0, the graph of the function is a parabola that opens downward. The x-coordinate of the vertex 107


is −

ISM: Introductory Mathematical Analysis f(P) is

18 b =− = 9 . The y-coordinate of the 2a 2(−1)

f (50) =

( )

vertex is P (9) = − 92 + 18(9) + 144 = 225 .

36. s = −4.9t 2 + 62.3t + 1.8 Since a = –4.9 < 0, s has a maximum value that occurs at the vertex where b 62.3 62.3 89 t=− =− = = ≈ 6.36 sec. 2a 2(−4.9) 9.8 14

Thus, the vertex is (9, 225). Since c = 144, the y-intercept is (0, 144). To find the x-intercepts, let y = P(x) = 0. 0 = − x 2 + 18 x + 144

(

0 = − x 2 − 18 x − 144

−1 (50)2 + 2(50) + 20 = 70 grams. 50

)

When t =

0 = –(x – 24)(x + 6) Thus, the x-intercepts are (24, 0) and (–6, 0).

89 , then 14 2

⎛ 89 ⎞ ⎛ 89 ⎞ s = −4.9 ⎜ ⎟ + 62.3 ⎜ ⎟ + 1.8 ⎝ 14 ⎠ ⎝ 14 ⎠ = 199.825 meters.

P(x) 400

37. h(t ) = −16t 2 + 85t + 22 –20

Since a = −16 < 0, h(t) has a maximum value that occurs at the vertex where b 85 t=− =− ≈ 2.7 sec. When t = 2.7, 2a 2(−16) then

x

30

34. If k = 2, then y = kx 2

h(t ) = −16(2.7) 2 + 85(2.7) + 22 = 134.86 feet.

y = 2 x2 This is a quadratic equation with a = 2, b = 0 and c = 0. Since a > 0, the graph of the function is a parabola that opens upward. The x-coordinate of b 0 =− =0. the vertex is − 2a 2(2) The y-coordinate is

38. h(t ) = −16t 2 + 16t + 4 Since a = –16 < 0, h(t) has a maximum value that occurs at the vertex where b 16 1 1 t=− =− = sec. When t = , 2a 2(−16) 2 2 2

⎛1⎞ ⎛1⎞ then, h(t ) = −16 ⎜ ⎟ + 16 ⎜ ⎟ + 4 = 8 feet. ⎝2⎠ ⎝2⎠

y = 2(0) 2 = 0 Thus, the vertex is (0, 0). 8

y

39. In the quadratic function h(t ) = −16t 2 + 80t + 16 , a = –16, b = 80, and c = 16. Since a < 0, the graph of the function is a parabola that opens downward. The x-coordinate of the vertex is b 80 5 − =− = . 2a 2(−16) 2 The y-coordinate of the vertex is

x 5

2

⎛5⎞ ⎛5⎞ ⎛5⎞ h ⎜ ⎟ = −16 ⎜ ⎟ + 80 ⎜ ⎟ + 16 = 116 ⎝2⎠ ⎝2⎠ ⎝2⎠

1 35. f ( P ) = − P 2 + 2 P + 20 , where 0 ≤ P ≤ 100. 50 1 Because a = − < 0 , f(P) has a maximum 50 value that occurs at the vertex. 2 b − =− = 50 . The maximum value of 2a 2 − 1

(

50

⎛5 ⎞ Thus, the vertex is ⎜ , 116 ⎟ . Since c = 16, the ⎝2 ⎠ y-intercept is (0, 16). To find the x-intercepts, we let y = h(t) = 0.

)

0 = −16t 2 + 80t + 16 108


t= =

Section 3.4 y = 78 − x = 78 − 39 = 39. Thus, two numbers whose sum is 78 and whose product is a maximum are 39 and 39.

−b ± b 2 − 4ac 2a −80 ± 802 − 4(−16)(16)

43. (1.11, 2.88)

2(−16) −80 ± 7424 5 ± 29 = = −32 2 ⎛ 5 + 29 ⎞ Thus, the x-intercepts are ⎜ , 0 ⎟ and ⎜ ⎟ 2 ⎝ ⎠

44. –1.61, 3.73 45. a.

none

b. one

⎛ 5 − 29 ⎞ , 0⎟ . ⎜⎜ ⎟ 2 ⎝ ⎠

c.

two

46. 14.18

h(t)

47. 4.89

160


–10

10

1. Let x = the number invested at 9% and let y = the amount invested at 8%. Then, the problem states ⎧ x + y = 200, 000, ⎨ ⎩0.09 x + 0.08 y = 17, 200. We eliminate x by multiplying the first equation by –0.09 and then adding ⎧−0.09 x − 0.09 y = −18, 000, ⎨ ⎩0.09 x + 0.08 y = 17, 200. −0.01 y = −800, y = 80, 000. Therefore, ⎧ x = 120, 000, ⎨ ⎩ y = 80, 000. Thus, $120,000 is invested at 9% and $80,000 is invested at 8%.

x

40. A = x(11 − x) = 11x − x 2 , so A is a quadratic function of x where a = –1 < 0. A has maximum value at the vertex where b 11 11 x=− =− = . 2a 2(−1) 2 41. Since the total length of fencing is 500, the side opposite the highway has length 500 – 2x. The area A is given by A = x(500 − 2 x) = 500 x − 2 x 2 , which is quadratic with a = –2 < 0. Thus A is 500 maximum when x = − = 125. Then the 2(−2) side opposite the highway is 500 – 2x = 500 – 2(125) = 250. Thus the dimensions are 125 ft by 250 ft.

2. Let A = the number of deer of species A, and let B = the number of deer of species B. Then, the number of pounds of food pellets that will be consumed is 4A + 2B = 4000. The number of pounds of hay that will be consumed is 5A + 7B = 9500. Then, we have ⎧4 A + 2 B = 4000, ⎨ ⎩5 A + 7 B = 9500. If we solve the first equation for B, we obtain ⎧ B = 2000 − 2 A ⎨5 A + 7 B = 9500. ⎩ Substituting 2000 – 2A for B in the second equation gives 5A + 7(2000 – 2A) = 9500 A = 500

Highway x

x 500 – 2x

42. Let x, y be two numbers whose sum is 78. Thus x + y = 78 and y = 78 − x. Their product is then p( x) = x(78 − x) = 78 x − x 2 . Since a = −1 < 0, p(x) has a maximum value that occurs at the b 78 vertex where x = − =− = 39 and 2a 2(−1) 109



Thus ⎧ B = 2000 − 2 A, ⎨ ⎩ A = 500. and ⎧ B = 1000, ⎨ ⎩ A = 500. The food will support 500 of species A and 1000 of species B.

1 ⎧ ⎪A = 6 , ⎪ 1 ⎪ ⎨C = , 2 ⎪ 1 ⎪ ⎪B = 3 . ⎩

Thus, the final mixture will consist of

3. Let A = the number of fish of species A, and let B = the number of fish of species B. Then, the number of milligrams of the first supplement that will be consumed is 15A + 20B = 100,000. The number of milligrams of the second supplement that will be consumed is 30A + 40B = 200,000. ⎧15 A + 20 B = 100, 000, ⎨ ⎩30 A + 40 B = 200, 000.

A,

1 lb of 6

1 1 lb of B, and lb of C. 3 2

Problems 3.4 ⎧ x + 4 y = 3, 1. ⎨ ⎩3 x − 2 y = −5.

(1) (2)

From Eq. (1), x = 3 – 4y. Substituting in Eq. (2) gives 3(3 – 4y) – 2y = –5 9 – 12y – 2y = –5 –14y = –14, or y = 1 ⇒ x = 3 – 4y = 3 – 4(1) = –1. Thus x = –1, y = 1.

1 We multiply the second equation by − and 2 then add. ⎧15 A + 20 B = 100, 000, ⎨ ⎩−15 A − 20 B = −100, 000, 0=0 Thus, there are infinitely many solutions of the 20, 000 4 − r , B = r, where form A = 3 3 0 ≤ r ≤ 5000.

⎧4 x + 2 y = 9, (1) 2. ⎨ ⎩5 y − 4 x = 5. (2)

Rewriting the system gives ⎧4 x + 2 y = 9, ⎨ ⎩−4 x + 5 y = 5. Adding the equations gives 7y = 14 y=2 From Eq. (1) we have 4x + 2(2) = 9 4x = 5 5 x= 4 5 Thus x = , y = 2. 4

4. Let A = the amount of type A used, let B = the amount of type B used, and let C = the amount of type C used. If the final blend will sell for $8.50 per pound, then 12A + 9B + 7C = 8.50, and A + B + C = 1. Furthermore, since the amount of type B is to be twice the amount of type A, B = 2A. Thus, the system of equations is ⎧12 A + 9 B + 7C = 8.50, ⎪ ⎨ A + B + C = 1, ⎪ B = 2 A. ⎩ Simplifying gives ⎧30 A + 7C = 8.50, ⎪ ⎨3 A + C = 1, ⎪ B = 2 A. ⎩

⎧3x − 4 y = 13, 3. ⎨ ⎩2 x + 3 y = 3.

(1) (2)

Multiplying Eq. (1) by 3 and Eq. (2) by 4 gives ⎧9 x − 12 y = 39, ⎨ ⎩8 x + 12 y = 12.

110


Section 3.4

Adding gives 17x = 51 x=3 From Eq. (2) we have 2(3) + 3y = 3 3y = –3 y = –1 Thus x = 3, y = –1. ⎧2 x − y = 1, 4. ⎨ ⎩− x + 2 y = 7.

⎧3x + 5 y = 7, 8. ⎨ ⎩5 x + 9 y = 7.

Multiplying Eq. (1) by 5 and Eq. (2) by –3 gives ⎧15 x + 25 y = 35, ⎨ ⎩−15 x − 27 y = −21. Adding gives –2y = 14, or y = –7. From Eq. (2) we have 5x + 9(–7) = 7 5x = 70 x = 14 Thus x = 14, y = –7.

(1) (2)

From Eq. (1), y = 2x – 1. Substituting in Eq. (2) gives –x + 2(2x – 1) = 7 3x = 9 x = 3 ⇒ y = 2x – 1 = 2(3) – 1 = 5. Thus x = 3, y = 5.

⎧4 x − 3 y − 2 = 3 x − 7 y, 9. ⎨ ⎩ x + 5 y − 2 = y + 4.

Simplifying, we have ⎧ x + 4 y = 2, ⎨ ⎩ x + 4 y = 6. Subtracting the second equation from the first gives 0 = –4, which is never true. Thus there is no solution.

⎧u + v = 5 5. ⎨ ⎩u − v = 7 From the first equation, v = 5 − u. Substituting in the second equation gives u − (5 − u ) = 7 2u − 5 = 7 2u = 12 or u = 6 so v = 5 − u = 5 − 6 = −1. Thus, u = 6, v = −1.

⎧2 p + q = 16, 6. ⎨ ⎩3 p + 3q = 33.

⎧5 x + 7 y + 2 = 9 y − 4 x + 6, ⎪ 10. ⎨ 21 4 11 3 2 5 ⎪ 2 x− 3 y− 4 = 2 x+ 3 y+ 4. ⎩

(1) (2)

By simplifying, we have ⎧9 x − 2 y = 4, ⎨ ⎩9 x − 2 y = 4. Both equations represent the same line, so we have infinitely many solutions. Let y = r. Then 2 4 9 x − 2r = 4 ⇒ x = r + . Thus a parametric 9 9 2 4 solution is x = r + , y = r, where r is any real 9 9 number.

From Eq. (1), q = 16 − 2p. Substituting in Eq. (2) gives 3 p + 3(16 − 2 p ) = 33 −3 p = −15 p = 5 ⇒ q = 16 − 2p = 16 − 10 = 6. Thus, p = 5, q = 6. ⎧ x − 2 y = –7, 7. ⎨ ⎩5 x + 3 y = –9.

(1) (2)

(1) (2)

1 ⎧2 ⎪⎪ 3 x + 2 y = 2, 11. ⎨ ⎪ 3 x + 5 y = − 11 . ⎪⎩ 8 6 2

From Eq. (1), x = 2y – 7. Substituting in Eq. (2) gives 5(2y – 7) + 3y = –9 13y = 26 y = 2 ⇒ x = 2y – 7 = 2(2) – 7 = –3. Thus x = –3, y = 2.

Clearing fractions gives the system ⎧4 x + 3 y = 12, ⎨ ⎩9 x + 20 y = −132. Multiplying the first equation by 9 and the second equation by –4 gives 111


ISM: Introductory Mathematical Analysis Multiplying the first equation by 5 and the second equation by –3 gives ⎧15 x + 50 z = 20, ⎨ ⎩−15 x − 18 z = −12.

⎧36 x + 27 y = 108, ⎨ ⎩−36 x − 80 y = 528. Adding gives –53y = 636 y = –12 From 4x + 3y = 12, we have 4x + 3(–12) = 12 4x = 48 ⇒ x = 12. Thus x = 12, y = –12.

Adding gives 32z = 8, or z = 3x + 10z = 4, we have ⎛1⎞ 3x + 10 ⎜ ⎟ = 4 ⎝4⎠

⎧⎪ 1 z − 1 w = 1 4 6 12. ⎨ 12 1w= 1 z + ⎪⎩ 2 4 6 Multiplying both equations by 12 gives ⎧6 z − 3w = 2 ⎨ 6 z + 3w = 2 ⎩

3 2 1 x= 2 From 2x + y + 6z = 3, we have ⎛1⎞ ⎛1⎞ 2⎜ ⎟ + y + 6⎜ ⎟ = 3 2 ⎝ ⎠ ⎝4⎠ 3x =

1 Adding gives 12z = 4 and so z = . 3 ⎛1⎞ From the first equation we have 6 ⎜ ⎟ − 3w = 2, ⎝3⎠ 1 from which w = 0. Thus z = , w = 0. 3 ⎧5 p + 11q = 7, 13. ⎨ ⎩10 p + 22q = 33.

y=

1 1 1 , y= , z= . 2 2 4

⎧ x + y + z = −1, ⎪ 16. ⎨3x + y + z = 1, ⎪4 x − 2 y + 2 z = 0. ⎩

(1) (2)

(1) (2) (3)

Subtracting Eq. (2) from Eq. (1) gives –2x = –2, or x = 1. Substituting x = 1 in Eqs. (2) and (3) and simplifying gives ⎧ y + z = −2, ⎨ ⎩−2 y + 2 z = −4. Multiplying the first equation by 2 gives ⎧2 y + 2 z = −4, ⎨ ⎩−2 y + 2 z = −4. By adding, we have 4z = –8 z = –2 From y + z = –2, we have y + (–2) = –2 y=0 Thus x = 1, y = 0, z = –2.

(1) (2)

Multiplying Eq. (1) by 2 gives ⎧10 x − 6 y = 4, ⎨ ⎩−10 x + 6 y = 4. Adding gives 0 = 8, which is never true, so the system has no solution. ⎧2 x + y + 6 z = 3, ⎪ 15. ⎨ x − y + 4 z = 1, ⎪3x + 2 y − 2 z = 2. ⎩

1 2

Therefore x =

Multiplying Eq. (1) by –2 gives ⎧−10 p − 22q = −14, ⎨ ⎩ 10 p + 22q = 33. Adding gives 0 = 19, which is never true, so the system has no solution. ⎧5 x − 3 y = 2, 14. ⎨ ⎩−10 x + 6 y = 4.

1 . From 4

(1) (2) (3)

⎧ x + 4 y + 3 z = 10 ⎪ 17. ⎨4 x + 2 y − 2 z = −2 ⎪⎩ 3x − y + z = 11 From the third equation, y = 3x + z − 11. Substituting in the first two equations gives ⎧ x + 4(3 x + z − 11) + 3 z = 10 ⎨4 x + 2(3 x + z − 11) − 2 z = −2 ⎩

Adding Eq. (1) and (2), and adding 2 times Eq. (2) to Eq. (3) gives ⎧3 x + 10 z = 4, ⎨ ⎩5 x + 6 z = 4. 112


Section 3.4

or ⎧13x + 7 z = 54 ⎨10 x = 20 ⎩ From the last equation we have x = 2. Thus 13(2) + 7z = 54, and 7z = 28, hence z = 4. Substitute these two values to solve for y: y = 3(2) + 4 − 11 = −1 Therefore, x = 2, y = −1, z = 4.

(1) ⎧ x − y + 2 z = 0, ⎪ 21. ⎨2 x + y − z = 0 (2) ⎪ x + 2 y − 3z = 0 (3) ⎩ Adding Eq. (1) to Eq. (3) gives ⎧ x − y + 2 z = 0, ⎪ ⎨2 x + y − z = 0 ⎪2 x + y − z = 0 ⎩ We can ignore the third equation because the second equation can be used to reduce it to 0 = 0. We have ⎧ x − y + 2 z = 0, ⎨ ⎩2 x + y − z = 0. Adding the first equation to the second gives 3x + z = 0 1 x=− z 3 Substituting in the first equation we have 1 − z − y + 2z = 0 3 5 y= z 3 Letting z = r gives the parametric solution 1 5 x = − r , y = r , z = r, where r is any real 3 3 number.

⎧ x + y + z = 18 (1) ⎪ 18. ⎨ x − y − z = 12 (2) ⎪⎩3x + y + 4 z = 4 (3) Adding Eq. (2) to both Eq. (1) and Eq. (3) gives 2 x = 30 ⎧ ⎨4 x + 3 z = 16 ⎩ From the first equation, x = 15. Substituting in the second equation gives 4(15) + 3 z = 16 3 z = −44 44 z=− 3 From x + y + z = 18 44 = 18 15 + y − 3 53 y= 3 53 44 Thus, x = 15, y = , z = − . 3 3

(1) ⎧ x − 2 y − z = 0, ⎪ 22. ⎨2 x − 4 y − 2 z = 0 (2) ⎪− x + 2 y + z = 0 (3) ⎩ Adding Eq. (1) to Eq. (3) gives ⎧ x − 2 y − z = 0, ⎪ ⎨2 x − 4 y − 2 z = 0 ⎪0 = 0 ⎩ We can ignore the third equation, so we have ⎧ x − 2 y − z = 0, ⎨ ⎩2 x − 4 y − 2 z = 0. Multiplying the first equation by –2 gives ⎧−2 x + 4 y + 2 z = 0, ⎨ ⎩2 x − 4 y − 2 z = 0. Adding the first equation to the second, we have ⎧−2 x + 4 y + 2 z = 0, ⎨ ⎩0 = 0. From the first equation, x = 2y + z. Setting y = r and z = s gives the parametric solution x = 2r + s, y = r, z = s, where r and s are any real numbers.

(1) ⎧ x − 2 z = 1, 19. ⎨ (2) ⎩ y + z = 3. From Eq. (1), x = 1 + 2z; from Eq. (2), y = 3 – z. Setting z = r gives the parametric solution x = 1 + 2r, y = 3 – r, z = r, where r is any real number. ⎧2 y + 3 z = 1, 20. ⎨ ⎩3x − 4 z = 0.

From Eq. (1), y =

(1) (2) 1 3 − z ; from Eq. (2), 2 2

4 z . Setting z = r gives the parametric 3 4 1 3 solution x = r , y = − r , z = r, where r is 3 2 2 any real number. x=

113

Chapter 3: Lines, Parabolas, and Systems ⎧2 x + 2 y − z = 3, 23. ⎨ ⎩4 x + 4 y − 2 z = 6.

ISM: Introductory Mathematical Analysis ⎧0.03 x + 0.11 y = 1.8, ⎨ ⎩ y = 20 − x. By substituting 20 – x for y in the first equation, and then simplifying, we obtain ⎧ x = 5, ⎨ ⎩ y = 15. Thus, the final mixture should contain 5 lb of 3% nitrogen fertilizer, and 15 lb of 11% nitrogen fertilizer.

(1) (2)

Multiplying Eq. (2) by −

1 gives 2

⎧2 x + 2 y − z = 3, ⎨ ⎩−2 x − 2 y + z = −3. Adding the first equation to the second equation gives ⎧2 x + 2 y − z = 3, ⎨ ⎩0 = 0. Solving the first equation for x, we have 3 1 x = − y + z . Letting y = r and z = s gives the 2 2 3 1 parametric solution x = − r + s , y = r, z = s, 2 2 where r and s are any real numbers.

27. Let C = the number of pounds of cotton, let P = the number of pounds of polyester, and let N = the number of pounds of nylon. If the final blend will cost $3.25 per pound to make, then 4C + 3P + 2N = 3.25. Furthermore, if we use the same amount of nylon as polyester to prepare, say, 1 pound of fabric, then N = P and C + P + N = 1. Thus, the system of equations is ⎧4C + 3P + 2 N = 3.25, ⎪ ⎨C + P + N = 1, ⎪ N = P. ⎩ Simplifying gives ⎧4C + 5 N = 3.25, ⎪ ⎨C + 2 N = 1, ⎪ N = P. ⎩

⎧5 x + y + z = 17 24. ⎨ ⎩4 x + y + z = 14 Subtracting the second equation from the first gives x = 3. From the first equation we have y + z = 17 − 5x = 17 − 5(3) = 2 Letting z = r we have the parametric solution x = 3, y = 2 − r, z = r, where r is any real number.

⎧ N = 0.25, ⎪ ⎨C = 0.5, ⎪ P = 0.25. ⎩ Thus, each pound of the final fabric will contain 0.25 lb each of nylon and polyester, and 0.5 lb of cotton.

25. Let x = number of gallons of 20% solution and y = number of gallons of 35% solution. Then (1) ⎧ x + y = 800, ⎨ (2) ⎩0.20 x + 0.35 y = 0.25(800). From Eq. (1), y = 800 – x. Substituting in Eq. (2) gives 0.20x + 0.35(800 – x) = 0.25(800) –0.15x + 280 = 200 –0.15x = −80 1600 x= ≈ 533.3 3 1600 800 y = 800 − x = 800 − = ≈ 266.7. Thus 3 3 533.3 gal of 20% solution and 266.7 gal of 35% solution must be mixed.

28. Let F = federal tax and S = state tax. Now solve the system ⎧ F = 0.25(312, 000 − S ), ⎨ S = 0.10(312, 000 − F ), ⎩ which is equivalent to ⎧ 4 F + S = 312, 000 ⎨ F + 10 S = 312, 000, ⎩ and has solution ⎧ F = 72, 000, ⎨ S = 24, 000. ⎩ Federal tax is $72,000 and state tax is $24,000.

26. Let x = the number of pounds of 3% nitrogen fertilizer, and let y = the number of pounds of 11% nitrogen fertilizer. Then ⎧0.03 x + 0.11 y = 0.09(20), ⎨ ⎩ x + y = 20.

29. Let p = speed of airplane in still air and w = wind speed. Now convert the time into minutes and solve the system

114


Section 3.4 (1) ⎧ x = 1.20 y, ⎨ + = 250 x 350 y 130, 000. (2) ⎩ Substituting 1.20y for x in Eq. (2) gives 250(1.20y) + 350y = 130,000 300y + 350y = 130,000 650y = 130,000 y = 200 Thus x = 1.20y = 1.20(200) = 240. Therefore 240 units of early American and 200 units of Contemporary must be sold.

900 ⎧ ⎪⎪ p + w = 175 ⎨ ⎪ p − w = 900 , ⎪⎩ 206 900 900 36 450 + = + Thus 2 p = 175 206 7 103 3429 miles per minute p= 721 279 miles per minute w= 721 Multiplying by 60 to get miles per hour we have p ≈ 285 and w ≈ 23.2 Plane speed in still air is about 285 mph and wind speed is about 23.2 mph.

32. Let x = number of favorable comments, y = number of unfavorable comments, and z = number of no comments. Then (1) ⎧ x + y + z = 250, ⎪ (2) ⎨ x = 1.625 y, ⎪ z = 0.16(250). (3) ⎩ From Eq. (3), z = 40. Substituting for x and z in Eq. (1), we obtain (1.625y) + y + (40) = 250 2.625y = 210 y = 80 Thus x = 1.625y = 1.625(80) = 130. Therefore 130 liked, 80 disliked, and 40 had no comment.

30. Let r = speed of raft in still water and c = speed of current. Then rate of raft downstream is r + c, and rate upstream is r – c. Since (rate)(time) = distance, we have ⎧ ⎛1⎞ ⎪(r + c) ⎜ ⎟ = 10, ⎪ ⎝2⎠ ⎨ ⎪(r − c) ⎛ 3 ⎞ = 10, ⎜4⎟ ⎪⎩ ⎝ ⎠ or, more simply, ⎧r + c = 20, ⎪ 40 ⎨ ⎪r − c = 3 . ⎩ Adding the equations gives 100 2r = 3 50 r= 3 10 Since r + c = 20, we have c = . Thus the 3 2 speed of the raft in still water is 16 mi/h; 3 1 speed of the current is 3 mi/h. 3

33. Let x = number of calculators produced at Exton, and y = number of calculators produced at Whyton. The total cost of Exton is 7.50x + 7000, and the total cost at Whyton is 6.00y + 8800. Thus 7.50x + 7000 = 6.00y + 8800. Also, x + y = 1500. This gives the system (1) ⎧ x + y = 1500, ⎨ (2) ⎩7.50 x + 7000 = 6.00 y + 8800. From Eq. (1), y = 1500 – x. Substituting in Eq. (2) gives 7.50x + 7000 = 6.00(1500 – x) + 8800 7.50x + 7000 = 9000 – 6x + 8800 13.5x = 10,800 x = 800 Thus y = 1500 – x = 1500 – 800 = 700. Therefore 800 calculators must be made at the Exton plant and 700 calculators at the Whyton plant.

31. Let x = number of early American units and y = number of Contemporary units. The fact that 20% more of early American styles are sold than Contemporary styles means that x = y + 0.20y x = 1.20y An analysis of profit gives 250x + 350y = 130,000. Thus we have the system

34. Let x, y, and z be the amounts of 2.20, 2.30, and 2.60 dollars/lb coffee, respectively. Then (1) ⎧ x + y + z = 100, ⎪ (2) ⎨2.20 x + 2.30 y + 2.60 z = 2.40(100), ⎪ y = z. (3) ⎩ From Eq. (3), y = z. Substituting for y in Eqs. (1) and (2) gives

115


ISM: Introductory Mathematical Analysis 20 x + 17.6 x = 3600 37.6 x = 3600 x ≈ 95.74 y = x ≈ 95.74 Thus, 95 boxes will be loose-filled and 8(95) = 760 clam-shells will be used, for a total of 190 boxes.

⎧ x + z + z = 100, ⎨ ⎩2.20 x + 2.30 z + 2.60 z = 240. or, by simplifying, ⎧ x + 2 z = 100, ⎨ ⎩2.20 x + 4.90 z = 240. From the first equation, x = 100 – 2z. Substituting in the second equation gives 2.20(100 – 2z) + 4.90z = 240 0.50z = 20 z = 40 From x = 100 – 2z, x = 100 – 2(40) = 20. From y = z, y = 40. Thus, 20, 40, and 40 lb of $2.20, $2.30, and $2.60 per lb coffee must be used, respectively.

38. Let p1 and p2 be the amounts of the two investments, respectively. Then the total amount invested was p1 + p2 , and from the statement of the problem we can write 3 ( p1 + p2 ) + 600 = p1 . The return on the 10 second investment was 1120 – 384 = 736. Since the percentage return on each was the same, and interest , we can write since rate = amt. invested 384 736 = . This can also be written as p1 p2

35. Let x = rate on first $100,000 and y = rate on sales over $100,000. Then (1) ⎧100, 000 x + 75, 000 y = 8500, ⎨ (2) ⎩100, 000 x + 180, 000 y = 14,800. Subtracting Eq. (1) from Eq. (2) gives 105,000y = 6300 y = 0.06 Substituting in Eq. (1) gives 100,000x + 75,000(0.06) = 8500 100,000x + 4500 = 8500, 100,000x = 4000, or x = 0.04. Thus the rate is 4% on the first $100,000 and 6% on the remainder.

p1 p = 2 . Hence we have the system 384 736 ⎧3 ⎪⎪10 ( p1 + p2 ) + 600 = p1 , ⎨ ⎪ p1 = p2 . ⎪⎩ 384 736 Simplifying, we have 3 ⎧ 7 ⎪⎪− 10 p1 + 10 p2 = −600, ⎨ ⎪ p = 12 p . ⎪⎩ 1 23 2 12 Substituting p1 = p2 in first equation gives 23 7 ⎛ 12 ⎞ 3 − ⎜ p2 + p2 = −600 10 ⎝ 23 ⎟⎠ 10

36. A system that describes the situation is ⎧T = L + 25, 000, 000 ⎨T = L + 0.30 L ⎩ We can rewrite this as ⎧T = L + 25, 000, 000 ⎨T = 1.30 L ⎩ Thus T = 1.3L and we can substitute this in the first equation: 1.3L = L + 25, 000, 000. Solving for L 0.3L = 25, 000, 000 L = 83,333,333 T = 1.3L = 1.3(83,333,333) = 108,333,333 thus T = $108,333,333 and L = $83,333,333.

3 p2 = −600 46 p2 = 9200

−

12 12 p2 = (9200) = 4800 . The total 23 23 amount invested was p1 + p2 = 4800 + 9200 = $14, 000 .

37. Let x = number of loose-filled boxes and y = number of boxes of clam-shells that will be filled. Then 8y clam-shells will be used. This will take 20x + 2.2(8y) pounds of peaches. (1) ⎧x = y ⎨20 x + 17.6 y = 3600 (2) ⎩ Substitute x = y in Eq. (2).

Thus p1 =

39. Let c = number of chairs company makes, r = number of rockers, and l = number of chaise lounges. Wood used: (1)c + (1)r + (1)l = 400 Plastic used: (1)c + (1)r + (2)l = 600 116


Section 3.4

Aluminum used: (2)c + (3)r + (5)l = 1500 Thus we have the system (1) ⎧c + r + l = 400, ⎪ (2) ⎨c + r + 2l = 600, ⎪2c + 3r + 5l = 1500. (3) ⎩ Subtracting Eq. (1) from Eq. (2) gives l = 200. Adding –2 times Eq. (1) to Eq. (3) gives r + 3l = 700, from which r + 3(200) = 700, r = 100 From Eq. (1) we have c + 100 + 200 = 400, or c = 100. Thus 100 chairs, 100 rockers and 200 chaise lounges should be made.

second gives: 5 x = 25 x=5 So y = 2x = 10 z = 70 − 3x = 70 − 15 = 55 The company should hire 5 skilled workers, 10 semiskilled workers, and 55 shipping clerks. 42. Method 1. Let a = number of minutes that pump for tank A operates, and b = number of minutes that pump for tank B operates. Then b = a + 5. 25a gallons are pumped from tank A and 35b from tank B. (1) ⎧b = a + 5, ⎨ (2) ⎩25a + 35b = 10, 000. Since b = a + 5, substituting in Eq. (2) gives 25a + 35(a + 5) = 10,000 60a = 9825 a = 163.75 b = a + 5, b = 163.75 + 5 = 168.75. Thus 25(163.75) = 4093.75 gallons are pumped from A, and 35(168.75) = 5906.25 gallons are pumped from B. Method 2. Let a = number of gallons from A, and let b = number of gallons from B. Then a + b = 10,000. The number of minutes the a . For the pump on B, pump on A operates is 25 b it is . Thus 35 b ⎧a (1) ⎪ +5 = 35 ⎨ 25 ⎪a + b = 10, 000. (2) ⎩

40. Let x, y, and z, be the amounts originally invested at 7%, 8%, and 9%, respectively. Then (1) ⎧ x + y + z = 35, 000, ⎪ (2) ⎨0.07 x + 0.08 y + 0.09 z = 2830, ⎪0.07 x + 0.08 y + 0.10 z = 2960. (3) ⎩ Subtracting Eq. (2) from Eq. (3) gives 0.01z = 130 z = 13,000 Subtracting 0.07 times Eq. (1) from Eq. (2) gives 0.01y + 0.02z = 380. Letting z = 13,000, we have 0.01y + 0.02(13,000) = 380 0.01y = 120 y = 12,000 From Eq. (1), x + 12,000 + 13,000 = 35,000 x = 10,000 The investments are $10,000 at 7%, $12,000 at 8%, $13,000 at 9% (later 10%). 41. Let x = number of skilled workers employed, y = number of semiskilled workers employed, z = number of shipping clerks employed. Then we have the system (1) ⎧number of workers: x + y + z = 70, ⎪ wages: 16 x + 9.5 y + 10 z = 725 (2) ⎨ ⎪semiskilled: y = 2x (3) ⎩ From the last equation, y = 2x so substitute into the first two equations: x + 2 x + z = 70 ⎧ ⎨16 x + 9.5(2 x) + 10 z = 725 ⎩ or ⎧ 3x + z = 70 ⎨35 x + 10 z = 725 ⎩ Adding −10 times the first equation to the

117

From Eq. (2), a = 10,000 – b. Substituting in Eq. (1) gives 10, 000 − b b +5 = 25 35 b b 400 − + 5 = 25 35 12b 405 = 175 5906.25 = b Thus a = 10,000 – b = 10,000 – 5906.25 = 4093.75. 45. x = 3, y = 2 46. x = 1.33, y = 0.67 47. x = 8.3, y = 14.0



Problems 3.5

4. From Eq. (2), y = x – 14. Substituting in Eq. (1) gives

In the following solutions, any reference to Eq. (1) or Eq. (2) refers to the first or second equation, respectively, in the given system.

( x − 14)2 − x 2 = 28 –28x + 196 = 28 –28x = –168 x=6 If x = 6, then y = x – 14 = 6 – 14 = –8. The only solution is x = 6, y = –8.

1. From Eq. (2), y = 3 − 2x. Substituting in Eq. (1) gives 3 − 2x = x2 − 9 0 = x 2 + 2 x − 12

5. Substituting y = x 2 into x = y 2 gives x = x 4 ,

−b ± b 2 − 4ac x= 2a

x4 − x = 0

(

)

x x3 − 1 = 0

−2 ± 22 − 4(1)(−12) 2(1) −2 ± 52 = 2 = −1 ± 13 =

Thus x = 0, 1. From y = x 2 , if x = 0, then y = 02 = 0 ; x = 1, then y = 12 = 1 . There are two solutions: x = 0, y = 0; x = 1, y = 1.

From y = 3 − 2x, if x = −1 + 13, then

⎧ 2 6. ⎨ p − q + 1 = 0 ⎩5q − 3 p − 2 = 0

y = 5 − 2 13; if x = −1 − 13, then y = 5 + 2 13. There are two solutions: x = −1 + 13, y = 5 − 2 13;

From the first equation q = p 2 + 1. Substituting into the second equation gives 5( p 2 + 1) − 3 p − 2 = 0

x = −1 − 13, y = 5 + 2 13.

5 p2 − 3 p + 3 = 0

2. From Eq. (2), y = x. Substituting in Eq. (1) gives

p=

x = x3 x − x3 = 0

(

=

)

−b ± b 2 − 4ac 2a 3 ± (−3) 2 − 4(5)(3)

2(5) 3 ± −51 = 10 Since −51 is not a real number, there are no real solutions.

x 1 − x2 = 0

x(1 + x)(1 – x) = 0. Thus x = 0, ±1. From y = x, if x = 0, then y = 0; if x = 1, then y = 1; if x = –1, then y = –1. There are three solutions: x = 0, y = 0; x = 1, y = 1; x = –1, y = –1. 3. From Eq. (2), q = p – 1. Substituting in Eq. (1) gives

7. Substituting y = x 2 − 2 x in Eq. (1) gives x2 − 2 x = 4 x − x2 + 8

p 2 = 5 − ( p − 1)

2 x2 − 6 x − 8 = 0

p2 + p − 6 = 0 (p + 3)(p – 2) = 0 Thus p = –3, 2. From q = p – 1, if p = –3, we have q = –3 – 1 = –4; if p = 2, then q = 2 – 1 = 1. There are two solutions: p = –3, q = –4; p = 2, q = 1.

x2 − 3x − 4 = 0 (x – 4)(x + 1) = 0

Thus x = 4, –1. From y = x 2 − 2 x , if x = 4, then we have y = 42 − 2(4) = 8 ; if x = –1, then y = (−1)2 − 2(−1) = 3 . There are two solutions: x = 4, y = 8; x = –1, y = 3.

118


Section 3.5 12. From Eq. (2), y = 3x – 5. Substituting in Eq. (1) gives

8. From Eq. (1), y = x 2 + 4 x + 4. Substituting in Eq. (2) gives

x 2 + (3 x − 5) 2 − 2 x(3 x − 5) = 1

x2 + 4 x + 4 − x2 − 4 x + 3 = 0 7=0 Since this is never true, the system has no solution.

9. Substituting p = q in Eq. (2) gives Squaring both sides gives

4 x 2 − 20 x + 24 = 0

x2 − 5x + 6 = 0 (x – 3)(x – 2) = 0 Thus x = 3, 2. From y = 3x – 5, if x = 3, then y = 3(3) – 5 = 4; if x = 2, then y = 3(2) – 5 = 1. Thus there are two solutions: x = 3, y = 4; x = 2, y = 1.

q = q2 .

q = q4

13. From Eq. (1), y = x − 1. Substituting in Eq. (2) gives x −1 = 2 x + 2

q4 − q = 0

(

)

q q3 − 1 = 0

( x − 1) 2 = 4( x + 2)

Thus q = 0, 1. From p = q , if q = 0, then

x2 − 2 x + 1 = 4 x + 8

p = 0 = 0 ; if q = 1, then p = 1 = 1 . There are two solutions: p = 0, q = 0; p = 1, q = 1.

10. Substituting z =

x2 − 6 x − 7 = 0 ( x + 1)( x − 7) = 0

4 in Eq. (2) gives w

Thus x = −1 or 7. From y = x − 1, if x = −1, then y = −2; if x = 7, then y = 6. However, from Eq. (2), y ≥ 0. The only solution is x = 7, y = 6.

⎛4⎞ 3 ⎜ ⎟ = 2w + 2 ⎝ w⎠ 12 = 2w2 + 2 w

14. Substituting y =

2

w + w−6 = 0 (w + 3)(w – 2) = 0

x2 1 = +1 x −1 x −1

4 , if w = –3, then w 4 4 z = − ; if w = 2, then z = = 2 . There are two 3 2 4 solutions: w = –3, z = − ; w = 2, z = 2. 3

Thus w = –3, 2. From z =

1 = x 2 + ( x − 1) x2 + x − 2 = 0 (x + 2)(x – 1) = 0 Thus x = –2, 1. But x cannot equal 1 in either of the original equations (division by zero). From 1 1 1 , if x = –2, then y = y= = − . The x −1 −2 − 1 3 1 solution is x = –2, y = − . 3

11. Replacing x 2 by y 2 + 13 in Eq. (2) gives

(

1 in Eq. (1) gives x −1

)

y = y 2 + 13 − 15 2

y − y−2 = 0 (y – 2)(y + 1) = 0 Thus y = 2, –1. If y = 2, then

15. We can write the following system of equations. ⎧⎪ y = 0.01x 2 + 0.01x + 7, ⎨ ⎪⎩ y = 0.01x + 8.0. By substituting 0.01x + 8.0 for y in the first equation and simplifying, we obtain

x 2 = y 2 + 13 = 22 + 13 = 17 , so x = ± 17 .

If y = –1, then x 2 = y 2 + 13 = (−1)2 + 13 = 14 , so x = ± 14 . The system has four solutions: x = 17 , y = 2; x = − 17 , y = 2; x = 14 ,

0.01x + 8.0 = 0.01x 2 + 0.01x + 7 0 = 0.01x 2 − 1 0 = (0.1x + 1)(0.1x – 1) x = −10 or x = 10 If x = –10 then y = 7.9, and if x = 10 then y = 8.1.

y = –1; x = − 14 , y = –1.

119



The rope touches the streamer twice, 10 feet away from center on each side at (–10, 7.9) and (10, 8.1).

20

16. We can write the following system of equations. ⎪⎧ y = 0.06 x 2 + 0.012 x + 8, ⎨ ⎪⎩ y = 0.912 x + 5. By substituting 0.912x + 5 for y in the first equation and then simplifying, we obtain

(100, 7) q 0

0.912 x + 5 = 0.06 x 2 + 0.012 x + 8

(

0 = 0.06 x 2 − 15 x + 50

200

2. Equating p-values gives 1 1 q+4= − q+9 1500 2000 7 q=5 6000 30, 000 5 = 4285 ≈ 4285.71 q= 7 7 5 When q = 4285 , then 7 1 1 ⎛ 5⎞ 6 p= q+4= ⎜ 4285 ⎟ + 4 = 6 ≈ 6.86 1500 1500 ⎝ 7⎠ 7 5 6⎞ ⎛ The equilibrium point is ⎜ 4285 , 6 ⎟ . 7 7⎠ ⎝

2

0 = 0.06 x − 0.9 x + 3

p

)

0 = 0.06(x – 10)(x – 5) x = 10 or x = 5 If x = 10 then y = 14.12, and if x = 5 then y = 9.56. The two holes are located at (10, 14.12) and (5, 9.56). 17. The system has 3 solutions. 18. x = 2, y = 4 19. x = –1.3, y = 5.1

p 10

20. x = −1.9, y = −3.6; x = −0.3, y = 1.2; x = 2.1, y = 8.3

6 6⎞ ⎛ ⎜ 4285 , 6 ⎟ 7 7⎠ ⎝

21. x = 1.76

5

22. x = 2.81 23. x = –1.46

5000

Problems 3.6

q 10,000

⎧35q − 2 p + 250 = 0, 3. ⎨ ⎩65q + p − 537.5 = 0.

1. Equating p-values gives 4 6 + 13 q+3= − 100 100 10 q = 10 100 q = 100

(1) (2)

Multiplying Eq. (2) by 2 and adding equations gives 165q – 825 = 0 q=5 From Eq. (2), 65(5) + p – 537.5 = 0 p = 212.50 Thus the equilibrium point is (5, 212.50).

4 (100) + 3 = 7 100 Thus, the equilibrium point is (100, 7). p=

120


Section 3.6

(1) ⎧246 p − 3.25q − 2460 = 0, 4. ⎨ 410 p + 3 q − 14, 452.5 = 0. (2) ⎩ Multiplying Eq. (1) by 3 and Eq. (2) by 3.25 gives ⎧738 p − 9.75q − 7380 = 0, ⎨ ⎩1332.5 p + 9.75q − 46,970.625 = 0. Adding gives 2070.5p – 54,350.625 = 0 54,350.625 p= = 26.25 2070.5 From Eq. (2) in original system, 14, 452.5 − 410 p 14, 452.5 − 410(26.25) q= = 3 3 14, 452.5 − 10, 762.5 3690 = = = 1230 3 3 The equilibrium point is (1230, 26.25).

8. Equating p-values gives q 2240 +6 = 4 q+2 (q + 24)(q + 2) = 2240(4)

5. Equating p-values:

9. Letting yTR = yTC gives 4q = 2q + 5000, or q = 2500 units.

2q + 20 = 200 − 2q

q 2 + 26q + 48 = 8960 q 2 + 26q − 8912 = 0 q= =

−b ± b 2 − 4ac 2a −26 ± (26) 2 − 4(1)(−8912)

2(1) q ≈ 82.29 or −108.29 q ≥ 0 so choose q ≈ 82.29. 82.29 + 6 ≈ 26.57. Then p ≈ 4 The equilibrium point is (82.29, 26.57).

2

TR

p

2q 2 + 2q − 180 = 0

TC

15,000

q 2 + q − 90 = 0 (q + 10)(q – 9) = 0 Thus q = –10, 9. Since q ≥ 0, choose q = 9. Then p = 2q + 20 = 2(9) + 20 = 38. The equilibrium point is (9, 38).

(2500, 10,000)

q

6. Equating p-values gives

5000

(q + 10)2 = 388 − 16q − q 2

10. Letting yTR = yTC gives

2q 2 + 36q − 288 = 0

14q =

q 2 + 18q − 144 = 0 (q + 24) (q – 6) = 0 Thus q = –24, 6. Since q ≥ 0, choose q = 6. Then

40 q + 1200 3

2 q = 1200 3 q = 1800 units

p = (q + 10)2 = (6 + 10)2 = 162 = 256 . The equilibrium point is (6, 256).

30,000

y

(1800, 25,200)

7. Equating p-values gives 20 − q = q + 10 . Squaring both sides gives

1800 units

400 − 40q + q 2 = q + 10

TC

q 2 − 41q + 390 = 0 (q – 26)(q – 15) = 0 Thus q = 26, 15. If q = 26, then p = 20 – q = 20 – 26 = –6. But p cannot be negative. If q = 15, then p = 20 – q = 20 – 15 = 5. The equilibrium point is (15, 5).

TR 0

q 1000

2000

11. Letting yTR = yTC gives 0.05q = 0.85q + 600 –0.80q = 600 q = –750, which is negative. Thus one cannot break even at any level of production. 121



12. Letting yTR = yTC gives 0.25q = 0.16q + 360 0.09q = 360 q = 4000 units

3 q + 9 + 0.27 200 3 p= q + 9.27 200 This equation can be written –3q + 200p – 1854 = 0, and the new system to solve is ⎧−3q + 200 p − 1854 = 0, ⎨ ⎩3q + 100 p − 1800 = 0. Adding gives 3654 300 p − 3654 = 0 ⇒ p = = $12.18 . 300 p=

900 = 1.1q + 37.3 q+3 90(q + 3) − 900 = (1.1q + 37.3)(q + 3)

13. Letting yTR = yTC gives 90 −

90q + 270 − 900 = 1.1q 2 + 40.6q + 111.9 1.1q 2 − 49.4q + 741.9 = 0

q= =

−b ± b 2 − 4ac 2a 49.4 ± (−49.4) 2 − 4(1.1)(741.9)

16. a.

2(1.1)

49.4 ± −824 2.2 There are no real solutions, therefore one cannot break even at any level of production.

Letting yTR = yTC gives 7q = 6q + 800, or q = 800 units. 6000

=

3000

14. Letting yTR = yTC gives

p

(800, 5600)

TC TR

2

0.1q + 9q = 3q + 400 2

0

0.1q + 6q − 400 = 0 q 2 + 60q − 4000 = 0 (q + 100)(q − 40) = 0 Thus q = –100, 40. Since q ≥ 0, choose q = 40 units. ⎧3q − 200 p + 1800 = 0, 15. ⎨ ⎩3q + 100 p − 1800 = 0.

a.

(1) (2)

17. Since profit = total revenue – total cost, then 4600 = 8.35q – (2116 + 7.20q). Solving gives 4600 = 1.15q – 2116 1.15q = 6716 6716 q= = 5840 units 1.15 For a loss (negative profit) of $1150, we solve −1150 = 8.35q – (2116 + 7.20q). Thus –1150 = 1.15q – 2116 1.15q = 966 q = 840 units To break even, we have yTR = yTC , or 8.35q = 2116 + 7.20q 1.15q = 2116 q = 1840 units

p

S

10

D 0

500

q 1000

b. The new total cost equation is yTC = 1.05(6q + 800) yTC = 6.3q + 840 Letting yTR = yTC gives 7q = 6.3q + 840 0.7q = 840 q = 1200 units

Subtracting Eq. (2) from Eq. (1) gives –300p + 3600 = 0 p = $12 20

500

q 1000

b. Before the tax, the supply equation is 3q – 200p + 1800 = 0 –200p = –3q – 1800 3 p= q+9 200 After the tax, the supply equation is 122


Section 3.6

18. For the supply equation we fit the points (0, 1) and (13,500, 4.50) to a straight line. We have

2500 ⎛ 1250 ⎞ q+⎜ ⎟ 3 ⎝ 3 ⎠ 1300 1,562,500 q2 + q+ =0 3 9 Using the quadratic formula, 400q = q 2 +

7

4.50 − 1 3.5 7 , = = 2 = 13,500 − 0 13,500 13,500 27, 000 so the line is 7 p −1 = (q − 0) 27, 000 27,000(p – 1) = 7q 7q – 27,000p + 27,000 = 0 For the demand equation, we fit the points (0, 20) and (13,500, 4.50) to a straight line. We have m=

2

2

1300 ⎛ 1300 ⎞ ⎛ 1,562,500 ⎞ ± ⎜ ⎟ − 4(1) ⎜ ⎟ 3 9 ⎝ 3 ⎠ ⎝ ⎠ q= , 2 which is not real. Thus total cost always exceeds total revenue; there is no break-even point. −

22. p =

31

4.50 − 20 15.5 m= =− =− 2 13,500 − 0 13,500 13,500 31 , so the line is =− 27, 000 31 p − 20 = − (q − 0) 27, 000 27,000(p – 20) = –31q 31q + 27,000p – 540,000 = 0

19. Let q = break-even quantity. Since total revenue is 5q, we have 5q = 200,000, which yields q = 40,000. Let c be the variable cost per unit. Then at the break even point, Tot. Rev. = Tot. Cost = Variable Cost + Fixed Cost. Thus 200,000 = 40,000c + 40,000 160,000 = 40,000c c = $4.

1000 q

a.

4=

1000 1000 = 250 units gives q = q 4

b.

2=

1000 1000 = 500 units gives q = q 2

c.

0.50 =

1000 1000 = 2000 units gives q = 0.50 q

⎛ 1000 ⎞ The revenue is qp = q ⎜ ⎟ = 1000 , so ⎝ q ⎠ revenue of $1000 is received regardless of price.

23. After the subsidy the supply equation is ⎡ 8 ⎤ p=⎢ q + 50 ⎥ − 1.50 ⎣ 100 ⎦ 8 q + 48.50 100 The system to consider is 8 ⎧ ⎪⎪ p = 100 q + 48.50, ⎨ ⎪ p = − 7 q + 65. ⎪⎩ 100 Equating p-values gives 8 7 q + 48.50 = − q + 65 100 100 15 q = 16.5 100 q = 110 When q = 110, then 8 8 p= q + 48.50 = (110) + 48.50 100 100 = 8.8 + 48.50 = 57.30 . Thus the original equilibrium price decreases by $0.70. p=

20. Let q = number of pairs sold. Total Revenue = 2.63q Total Cost = 0.85q + 0.96q + 0.32q + 70,500 At the break-even point, Total Revenue = Total cost, or 2.63q = 0.85q + 0.96q + 0.32q + 70,500 Solving for q gives 2.63q = 2.13q + 70,500 or 0.5q = 70,500 q = 141,000 21. yTC = 3q + 1250 : yTR = 60 q . Letting yTR = yTC gives 60 q = 3q + 1250 1250 3 Squaring gives 20 q = q +

123

Chapter 3: Lines, Parabolas, and Systems 24. a.


Profit = Total Revenue – Total Cost = 280,000(2.00) – [110,000 + 280,000(1.75)] = 560,000 – 600,000 = –40,000. There is a net loss of $40,000.

1 ( x − 10) 2 1 y −4 = x−5 2 1 y = x − 1 , which is slope-intercept form. 2 Clearing fractions, we have ⎛1 ⎞ 2 y = 2 ⎜ x − 1⎟ ⎝2 ⎠ 2y = x – 2 x – 2y – 2 = 0, which is a general form.

5. y − 4 =

b. Let q = unit sales volume. Then 40,000 = 2.00q – [110,000 + 1.75q] 150,000 = 0.25q q = 600,000 units 25. Equating qA -values gives 7 − pA + pB = −3 + 4 pA − 2 pB 10 = 5 pA − 3 pB Equating qB -values gives 21 + pA − pB = −5 − 2 pA + 4 pB 26 = −3 pA + 5 pB Now we solve ⎧10 = 5 pA − 3 pB ⎨26 = −3 p + 5 p A B ⎩ Adding 3 times the first equation to 5 times the second equation gives 160 = 16 pB pB = 10 From 5 pA − 3 pB = 10, 5 pA − 3(10) = 10 or pA = 8. Thus pA = 8 and pB = 10.

6. Slope of a vertical line is undefined, so slopeintercept form does not exist. An equation of the vertical line is x = 3. General form: x – 3 = 0. 7. Slope of a horizontal line is 0. Thus y – 4 = 0[x – (–2)] y – 4 = 0, so slope-intercept form is y = 4. A general form is y – 4 = 0. 5 5 7⎞ ⎛ 8. –3y + 5x = 7 ⎜ or y = x − ⎟ has slope . 3 3 3 ⎝ ⎠

Thus the line perpendicular to it has slope −

3 and its equation is y − 2 = − ( x − 1) , or 5 3 13 y = − x + . A general form is 3x + 5y – 13 = 5 5 0.

26. $17.80; 2.6 thousand units 27. 2.4 and 11.3 Chapter 3 Review Problems 1. Solving

5 ⎛ ⎞ 9. The line 2y + 5x = 2 ⎜ or y = − x + 1⎟ has slope 2 ⎝ ⎠ 2 5 − , so the line perpendicular to it has slope . 5 2 Since the y-intercept is −3, the equation is 2 y = x − 3. A general form is 2x − 5y − 15 = 0. 5

k −5 = 4 gives k – 5 = 4, k = 9. 3− 2

2. The equation

3 5

4−4 = 0 is true for any real 5−k

number k ≠ 5. 3. (−2, 3) and (0, −1) lie on the line, so −1 − 3 m= = −2. Slope-intercept form: 0 − (−2)

8−2 6 = = 3, so an 1 − (−1) 2 equation of the line is y – 8 = 3(x – 1). If x = 3, then y – 8 = 3(3 − 1) y–8=6 y = 14 Thus (3, 13) does not lie on the line.

10. The line has slope

y = mx + b ⇒ y = –2x − 1. A general form: 2x + y + 1 = 0. 4. Slope of y = 3x – 4 is m = 3, so slope of parallel line is also m = 3. Thus y – (–1) = 3[x – (–1)] y + 1 = 3x + 3, Slope-intercept form: y = 3x + 2. General form: 3x – y + 2 = 0. 124


Chapter 3 Review

In Problems 11–16, m1 = slope of first line, and m2 = slope of second line.

5

1 1⎞ ⎛ 11. x + 4y + 2 = 0 ⎜ or y = − x − ⎟ has slope 4 2⎠ ⎝ 1 m1 = − and 8x – 2y – 2 = 0 (or y = 4x – 1) has 4 1 slope m2 = 4 . Since m1 = − , the lines are m2 perpendicular to each other. 12. y – 2 = 2(x – 1) (or y = 2x) has slope m1 = 2 , and

y

x 4 3

–2

5

18. x = –3y + 4 3y = –x + 4 1 4 y = − x+ 3 3 1 m=− 3

1 3⎞ ⎛ 2x + 4y – 3 = 0 ⎜ or y = − x + ⎟ has slope 2 4⎠ ⎝ 1 1 , the lines are m2 = − . Since m1 = − m2 2 perpendicular.

5

y 4 3

1 11 ⎞ ⎛ 13. x – 3 = 2(y + 4) ⎜ or y = x − ⎟ has slope 2 2⎠ ⎝

x 4

1 , and y = 4x + 2 has slope m2 = 4 . Since 2 1 m1 ≠ m2 and m1 ≠ − , the lines are neither m2 parallel nor perpendicular to each other. 2 4⎞ ⎛ 14. 2x + 7y − 4 = 0 ⎜ or y = − x + ⎟ has slope 7 7⎠ ⎝ 2 m1 = − , and 6x + 21y = 90 7 2 2 30 ⎞ ⎛ ⎜ or y = − x + ⎟ has slope m2 = − . Since 7 7 7 ⎝ ⎠ m1 = m2 , the lines are parallel. m1 =

19. 4 – 3y = 0 –3y = –4 4 y= 3 m=0 5

y

4 3

x 5

15. y = 3x + 5 has slope 3, and 6x − 2y = 7 7⎞ ⎛ ⎜ or y = 3x − 2 ⎟ has slope 3. Since m1 = m2 , the ⎝ ⎠ lines are parallel. 16. y = 7x has slope m1 = 7 , and y = 7 has slope

20. y = 2x m=2 5

1 , the m2 lines are neither parallel nor perpendicular. 17. 3x – 2y = 4 –2y = –3x + 4 3 y = x−2 2 3 m= 2

y

m2 = 0 . Since m1 ≠ m2 and m1 ≠ −

x 5

125



21. y = f(x) = 17 − 5x has the linear form f(x) = ax + b, where a = −5 and b = 17. Slope = −5; y-intercept (0, 17).

10

y (0, 9)

y 25

x

–3

x

3

5

5

24. y = f(x) = 3x – 7 has the linear form f(x) = ax + b, where a = 3, b = –7. Slope = 3; y-intercept (0, –7)

22. s = g (t ) = 5 − 3t + t 2 has the quadratic form

2

y

x

g (t ) = at 2 + bt + c , where a = 1, b = –3, c = 5.

Vertex: −

5

b −3 3 =− = 2a 2(1) 2 2

11 ⎛3⎞ ⎛3⎞ ⎛3⎞ g ⎜ ⎟ = 5 − 3⎜ ⎟ + ⎜ ⎟ = 2 2 2 4 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ 3 11 ⎞ ⇒ Vertex = ⎜ , ⎟ ⎝2 4 ⎠ s-intercept: c = 5 t-intercepts: Because the parabola opens upward (a > 0) and the vertex is above the t-axis, there is no t-intercept. s

–7

25. y = h(t ) = t 2 − 4t − 5 has the quadratic form h(t ) = at 2 + bt + c , where a = 1, b = –4, and c = –5. b −4 =− =2 Vertex: − 2a 2 ⋅1

8

h(2) = 22 − 4(2) − 5 = −9

⇒ Vertex = (2, –9) y-intercept: c = –5 t -intercepts: t 2 − 4t − 5 = (t − 5)(t + 1) = 0

t

⇒ t = 5, –1

6

y

23. y = f ( x) = 9 − x 2 has the quadratic form

–1

2

f ( x) = ax + bx + c , where a = –1, b = 0 and c = 9. b 0 =− =0 Vertex: − 2a 2(−1)

–5 –9

f (0) = 9 − 02 = 9

⇒ Vertex = (0, 9) y-intercept: c = 9 x-intercepts: 9 − x 2 = (3 − x)(3 + x) = 0 , so x = 3, –3.

126

2

5

t


Chapter 3 Review

(

26. y = k(t) = −3 − 3t has the linear form k(t) = at + b, where a = −3, b = −3. Slope = −3, y-intercept (0, −3) 5

)

29. y = F ( x) = − x 2 + 2 x + 3 = − x 2 − 2 x − 3 has the quadratic form F ( x ) = ax 2 + bx + c , where a = –1, b = –2, and c = –3 b −2 =− = −1 Vertex: − 2a 2(−1)

y

t

F (−1) = − ⎡(−1)2 + 2(−1) + 3⎤ = −2 ⎣ ⎦ ⇒ Vertex = (–1, –2) y-intercept: c = –3 x-intercepts: Because the parabola opens downward (a < 0) and the vertex is below the x-axis, there is no x-intercept.

5

27. p = g(t) = −7t has the linear form g(t) = at + b, where a = −7 and b = 0. Slope = −7; p-intercept (0, 0)

2

y

x

p

–1

10

–2 –3

5

t 5

x 1 − 2 = x − 2 has the linear form 3 3 1 f(x) = ax + b, where a = , b = –2. 3 1 Slope = ; y-intercept (0, –2) 3

30. y = f ( x) = 28. y = F ( x) = (2 x − 1) 2 = 4 x 2 − 4 x + 1 has the quadratic form F ( x ) = ax 2 + bx + c , where a = 4, b = –4, c = 1. b −4 1 =− = Vertex: − 2a 2⋅4 2

5

y

2

⎛1⎞ ⎡ ⎛1⎞ ⎤ F ⎜ ⎟ = ⎢ 2 ⎜ ⎟ − 1⎥ = 0 ⎝2⎠ ⎣ ⎝2⎠ ⎦

x

⎛1 ⎞ ⇒ Vertex = ⎜ , 0 ⎟ ⎝2 ⎠ y-intercept: c = 1

–2

x-intercepts: (2 x − 1)2 = 0 , so x = y

1 1 2

1 2

⎧2 x − y = 6, 31. ⎨ ⎩3x + 2 y = 5.

8

(1) (2)

From Eq. (1), y = 2x – 6. Substituting in Eq. (2) gives 3x + 2(2x – 6) = 5 7x – 12 = 5, 7x = 17 17 17 8 ⇒ y = 2x − 6 = 2 ⋅ − 6 = − . x= 7 7 7 17 8 Thus x = , y=− . 7 7

x 5

127

Chapter 3: Lines, Parabolas, and Systems ⎧8 x − 4 y = 7, 32. ⎨ ⎩ y = 2 x − 4.


(1)

1 1 ⎧1 ⎪⎪ 3 x − 4 y = 12 , 36. ⎨ ⎪ 4 x + 3y = 5 . ⎪⎩ 3 3

(2)

Replacing y by 2x – 4 in Eq. (1) gives 8x – 4(2x – 4) = 7 16 = 7, which is never true. There is no solution.

(2)

Multiplying Eq. (1) by –4 gives 1 ⎧ 4 ⎪⎪− 3 x + y = − 3 , ⎨ ⎪ 4 x + 3y = 5 . ⎪⎩ 3 3 4 1 Adding gives 4 y = ⇒ y = . From Eq. (2), 3 3 4 ⎛1⎞ 5 x + 3⎜ ⎟ = 3 ⎝3⎠ 3

⎧7 x + 5 y = 5 33. ⎨ ⎩6 x + 5 y = 3 Subtracting the second equation from the first equation gives x = 2. Then 7(2) + 5y = 5, or 9 9 5y = −9, so y = − . Thus x = 2, y = − . 5 5 ⎧2 x + 4 y = 8 (1) 34. ⎨ ⎩ 3x + 6 y = 12 (2) Multiplying Eq. (1) by 3 and Eq. (2) by −2 gives ⎧ 6 x + 12 y = 24 ⎨−6 x − 12 y = −24. ⎩ Adding gives 0 = 0. Thus, the equations are equivalent. From EQ. (1), x = −2y + 4. Letting y = r gives the parametric solution x = −2r + 4, y = r, where r is any real number.

3 ⎧1 ⎪⎪ 4 x − 2 y = −4, 35. ⎨ ⎪ 3 x + 1 y = 8. ⎪⎩ 4 2

(1)

4 2 x= 3 3 1 x= 2

Thus x =

1 1 , y= . 3 2

⎧3x − 2 y + z = −2, ⎪ 37. ⎨2 x + y + z = 1, ⎪ x + 3 y − z = 3. ⎩

(1) (2)

(1) (2) (3)

Subtracting Eq. (2) from Eq. (1) and adding Eq. (2) to Eq. (3) gives ⎧ x − 3 y = −3, ⎨ ⎩3 x + 4 y = 4. Multiplying the first equation by –3 gives ⎧−3x + 9 y = 9, ⎨ ⎩3 x + 4 y = 4. Adding the first equation to the second gives 13y = 13 y=1 From the equation x – 3y = –3, we get x – 3(1) = –3 x=0 From 3x – 2y + z = –2, we get 3(0) – 2(1) + z = –2 z=0 Thus x = 0, y = 1, z = 0.

Multiplying Eq. (2) by 3 gives 3 ⎧1 ⎪⎪ 4 x − 2 y = −4, ⎨ ⎪ 9 x + 3 y = 24. ⎪⎩ 4 2 Adding the first equation to the second gives 5 x = 20 2 x=8 From Eq. (1), 1 3 (8) − y = −4 4 2 3 − y = −6 2 y=4 Thus x = 8, y = 4.

128


Chapter 3 Review

⎧2 x + 3 y + x = 9 ⎪ 3 38. ⎨ 5x+2 y ⎪⎩ y + 4 = 7 simplifies to ⎧7 x + 3 y = 27 (1) ⎨5 x + 6 y = 28 (2) ⎩

x=

−5 − 65 −21 − 5 65 , y= . 4 8 (1) (2)

From Eq. (2), y = x + 7. Substituting in Eq. (1) we have 18 x+7 = x+4 (x + 7)(x + 4) = 18 x 2 + 11x + 28 = 18 x 2 + 11x + 10 = 0 (x + 1)(x + 10) = 0 Thus x = –1, –10. From y = x + 7, if x = –1, then y = –1 + 7 = 6; if x = –10, then y = –10 + 7 = –3. Thus the two solutions are x = –1, y = 6, and x = –10, y = –3. (1) ⎧ x + 2 z = −2, 41. ⎨ (2) ⎩ x + y + z = 5. From Eq. (1) we have x = –2 – 2z. Substituting in Eq. (2) gives –2 – 2z + y + z = 5, so y = 7 + z. Letting z = r gives the parametric solution x = –2 – 2r, y = 7 + r, z = r, where r is any real number.

(1) (2)

From Eq. (2), y = 3 − x 2 . Substituting in Eq. (1) gives x 2 − (3 − x 2 ) + 5 x = 2

⎧ x + y + z = 0, ⎪ 42. ⎨ x − y + z = 0, ⎪ x + z = 0. ⎩

2

2 x + 5x − 5 = 0 x=

−5 + 65 −21 + 5 65 , y= , and 4 8

18 ⎧ , ⎪y = 40. ⎨ x+4 ⎪ x − y + 7 = 0. ⎩

Multiplying Eq. (1) by −2 gives ⎧−14 x − 6 y = −54 ⎨ 5 x + 6 y = 28 ⎩ Adding the equations gives −9 x = −26 26 x= 9 Multiplying Eq. (1) by −5 and Eq. (2) by 7 gives ⎧−35 x − 15 y = −135 ⎨ 35 x + 42 y = 196 ⎩ Adding the equations gives 27 y = 61 61 y= 27 26 61 , y= . Thus, x = 9 27 ⎧⎪ x 2 − y + 5 x = 2, 39. ⎨ 2 ⎪⎩ x + y = 3.

x=

2

−b ± b − 4ac 2a

(1) (2) (3)

Subtracting Eq. (3) from both Eqs. (1) and (2) gives ⎧ y = 0, ⎪ ⎨− y = 0, ⎪ x + z = 0. ⎩ The first two equations state that y = 0, and the third implies that x = –z. Letting z = r gives the parametric solution x = –r, y = 0, z = r, where r is any real number.

−5 ± 52 − 4(2)(−5) = 2(2) −5 ± 65 = 4 −5 + 65 , then 4 −21 + 5 65 −5 − 65 y= ; if x = , then 8 4

Since y = 3 − x 2 , if x =

−21 − 5 65 . 8 Thus, the two solutions are y=

129

Chapter 3: Lines, Parabolas, and Systems ⎧ x − y − z = 0, 43. ⎨ ⎩2 x − 2 y + 3 z = 0.


(1)

47. Slope is

(2)

f(1) = 5, 4 5 = − (1) + b 3 19 b= 3

Multiplying Eq. (1) by –2 gives ⎧−2 x + 2 y + 2 z = 0, ⎨ ⎩2 x − 2 y + 3 z = 0. Adding the first equation to the second gives ⎧−2 x + 2 y + 2 z = 0, ⎨ ⎩5 z = 0. From the second equation, z = 0. Substituting in Eq. (1) gives x – y – 0 = 0, so x = y. Letting y = r gives the parametric solution x = r, y = r, z = 0, where r is any real number. ⎧2 x − 5 y + 6 z = 1, 44. ⎨ ⎩4 x − 10 y + 12 z = 2.

Thus f ( x) = −

4 19 x+ . 3 3

5−8 −3 = = −1 . Thus 2 − (−1) 3 f(x) = ax + b = –x + b. Since f(2) = 5, 5 = –2 + b b=7 Thus f(x) = –x + 7.

48. The slope of f is

(1) (2)

49. r = pq = (200 − 2q )q = 200q − 2q 2 , which is a quadratic function with a = –2, b = 200, c = 0. Since a < 0, r has a maximum value when b 200 q=− =− = 50 units. If q = 50, then 2a −4 r = [200 – 2(50)](50) = $5000.

Multiplying Eq. (1) by –2 gives ⎧−4 x + 10 y − 12 z = −2, ⎨ ⎩4 x − 10 y + 12 z = 2. Adding the first equation to the second gives ⎧−4 x + 10 y − 12 z = −2, ⎨ ⎩0 = 0. Solving the first equation for x, we have 1 5 x = + y − 3 z . Letting y = r and z = s gives 2 2 1 5 the parametric solution x = + r − 3s , y = r, 2 2 z = s, where r and s are any real numbers.

50. Let p1 and p2 be the prices (in dollars) of the two items, respectively, before the tax. At the time the difference in prices is p1 − p2 = 3.5. After the tax, the prices are 1.05 p1 and 1.05 p2 , so their difference is 1.05 p1 − 1.05 p2 , or 4.1. This gives the system p1 − p2 = 3.5 ⎧ ⎨1.05 p − 1.05 p = 4.1 1 2 ⎩

45. a = 1 when b = 2; a = 5 when b = 3, so a −a 5 −1 4 m= 2 1 = = = 4. b2 − b1 3 − 2 1 Thus an equation relating a and b is a − 1 = 4(b − 2) a − 1 = 4b − 8 a − 4b = −7 When b = 5, then a = 4b − 7 = 4(5) − 7 = 13. 46. a.

−4 4 ⇒ f ( x) = ax + b = − x + b . Since 3 3

Adding −1.05 times the first equation to the second equation gives 0 = 0.425, which indicates that the system does not have a solution. Thus this scenario is not possible. ⎧120 p − q − 240 = 0, 51. ⎨ ⎩100 p + q − 1200 = 0.

r = 206 when T = 36; r = 122 when T = 30. r −r 122 − 206 −84 = = 14 Thus m = 2 1 = T2 − T1 30 − 36 −6 r − 206 = 14(T − 36) r = 14T − 298

Adding gives 220p − 1440 = 0, or 1440 p= ≈ 6.55. 220

b. If T = 27, then r = 14T − 298 = 14(27) − 298 = 80.

130



R = aL + b. If L = 0, then R = 1310. Thus we have 1310 = 0 · L + b, or b = 1310. So R = aL + 1310. Since R = 1460 when L = 2, 1460 = a(2) + 1310 150 = 2a a = 75 Thus R = 75L + 1310.

58. x = 3.02, y = 0.14 59. x = 0.75, y = 1.43 60. x = 2.68 Mathematical Snapshot Chapter 3 1. P1 (6000) = 39.99 + 0.45(6000 − 450) = 2537.49 P6 (6000) = 199.99

b. If L = 1, then R = 75(1) + 1310 = 1385 milliseconds. c.

He loses $2537.49 − $199.99 = $2337.50 by using P1.

Since R = 75L + 1310, the slope is 75. The slope gives the change in R for each 1-unit increase in L. Thus the time necessary to travel from one level to the next level is 75 milliseconds.

2. The graph shows that P2 and P3 intersect when the second branch of P2 crosses the first branch of P3 . Thus 59.99 + 0.40(t − 900) = 79.99 t = 950 P2 is best for usage between 494.44 and 950 minutes.

53. yTR = 16q ; yTC = 8q + 10, 000 . Letting yTR = yTC gives 16q = 8q + 10,000 8q = 10,000 q = 1250 If q = 1250, then yTR = 16(1250) = 20, 000 . Thus the break-even point is (1250, 20,000) or 1250 units, $20,000.

3. The graph shows that P3 and P4 intersect when the second branch of P3 crosses the first branch of P4 Thus 79.99 + 0.35(t − 1350) = 99.99 t ≈ 1407.14 P3 is best for usage between 950 and 1407.14 minutes.

54. C = aF + b. The points (32, 0) and (212, 100) lie on the graph of the function. Thus its slope is 100 − 0 100 5 5 = = , so C = F + b . Since 212 − 32 180 9 9 5 C = 0 when F = 32, 0 = (32) + b , so 9 160 5 160 . Thus C = F − or b=− 9 9 9 5 C = ( F − 32) . When 9 5 5 F = 50, then C = (50 − 32) = (18) = 10 . 9 9

4. The graph shows that P4 and P5 intersect when the second branch of P4 crosses the first branch of P5 Thus 99.99 + 0.25(t − 2000) = 149.99 t = 2200 P4 is best for usage between 1407.14 and 2200 minutes. 5. The graph shows that P5 and P6 intersect when the second branch of P5 crosses the first branch of P6 Thus 149.99 + 0.25(t − 4000) = 199.99 t = 4200 P5 is best for usage between 2200 and 4200 minutes.

55. Equating L-values gives 0.0042 0.0378 = 0.0005 + 0.0183 − p p 0.042 0.0178 = p 0.0178 p = 0.042 p ≈ 2.36 The equilibrium pollution level is about 2.36 tons per square kilometer.

6. P6 is best for usage of greater than 4200 minutes.

56. x = 12, y = –4

7. No; answers may vary.

57. x = 7.29, y = −0.78 131

Chapter 4 Principles in Practice 4.1

second year is (1 − r )2 = (1 − 0.15)2 = 0.72 . This pattern will continue as shown in the table.

1. The shapes of the graphs are the same. The value of A scales the value of any point by A.

Year

Multiplicative Decrease

Expression

0

1

0.850

1

0.85

0.851

2

0.72

0.852

3

0.61

0.853

2. If P = the amount of money invested and r = the annual rate at which P increases, then after 1 year, the investment has grown from P to P + Pr = P(1 + r). Since r = 0.10, the factor by which P increases for the first year is 1 + r = 1 + 0.1 = 1.1. Similarly, during the second year the investment grows from P(1 + r) to P (1 + r ) + r[ P (1 + r )] = P (1 + r )2 . Again, since r = 0.10, the multiplicative increase for the second year is (1 + 0.10)2 = (1.1) 2 = 1.21. This pattern will continue as shown in the table. Year

Thus, the depreciation is exponential with a base of 1 – r = 1 – 0.15 = 0.85. If we graph the multiplicative decrease as a function of years, we obtain the following.

Multiplicative Increase

Expression

0

1

1.10

2

1

1.1

1.11

1

2

1.21

1.12

3

1.33

1.13

1.46

4

4

y

1 2 3 4 5

4. Let t = the time at which George’s sister began saving, then since George is 3 years behind, t – 3 = the time when George began saving.

1.1

Therefore, if y = 1.08t represents the multiplicative increase in George’s sister’s

Thus, the growth of the initial investment is exponential with a base of 1 + r = 1 + 0.1 = 1.1. If we graph the multiplicative increase as a function of years we obtain the following.

account y = 1.08t −3 represents the multiplicative increase in George’s account. A graph showing the projected increase in George’s money will have the same shape as the graph of the projected increase in his sister’s account, but will be shifted 3 units to the right.

y 2 1

1 2 3 4 5

x years

5. S = P(1 + r ) n

x years

S = 2000(1 + 0.13)5 = 2000(1.13)5 ≈ 3684.87 The value of the investment after 5 years will be $3684.87. The interest earned over the first 5 years is 3684.87 – 2000 = $1684.87.

3. If V = the value of the car and r = the annual rate at which V depreciates, then after 1 year the value of the car is V – rV = V(1 – r). Since r = 0.15, the factor by which V decreases for the first year is 1 – r = 1 – 0.15 = 0.85. Similarly, after the second year the value of the car is

6. Let N(t) = the number of employees at time t, where t is in years. Then, N (4) = 5(1 + 1.2) 4 = 5(2.2) 4 = 117.128 Thus, there will be 117 employees at the end of 4 years.

V (1 − r ) − r[V (1 − r )] = V (1 − r ) 2 . Again, since r = 0.15, the multiplicative decrease for the 132


Section 4.1 3.

0.06t

⎛1⎞ 7. P = e−0.06t = ⎜ ⎟ ⎝e⎠ 1 Since 0 < < 1 , the graph is that of an e exponential function falling from left to right.

x

y

0

1

2

0.89

4

0.79

6

0.70

8

0.62

10

0.55

8

y

x 5

4.

y 60

x 10

P

5.

y

8

1

10

20

t years

x 5

Problems 4.1 1.

8

6.

y

8

x

x

5

5

2.

8

y

7.

y

y 9

x

x

5

5

133

Chapter 4: Exponential and Logarithmic Functions 8.

8


y

14. y = 0.4 x has base b = 0.4 and 0 < b < 1, so its graph falls from left to right. Thus the graph is A. 15. For 2015 we have t = 20, so 20

P = 125, 000(1.11) 20 = 125, 000(1.11)1 = 138, 750 .

x 5

9.

8

16. a.

y

For 1999, t = 1 and P = 1,527, 000(1.015)1 = 1,549,905

b. For 2000, t = 2 and P = 1,527, 000(1.015) 2 ≈ 1,573,154 x 5

10.

8

1 1⎛1⎞ 17. With c = , P = 1 − ⎜ ⎟ 2⎝2⎠ 2

y

n −1

1

1 1 ⎛1⎞ n = 1: P = 1 − ⎜ ⎟ = 1 − = 2 2 2 ⎝ ⎠ 2

1 3 ⎛1⎞ n = 2: P = 1 − ⎜ ⎟ = 1 − = 4 4 ⎝2⎠ 3

x

1 7 ⎛1⎞ n = 3: P = 1 − ⎜ ⎟ = 1 − = 8 8 ⎝2⎠

5

11.

8

( )

y

18. y = 23 x = 23 19. a.

5

20. a. 8

= 8 x . Thus y = 8 x .

4000(1.06)7 ≈ $6014.52

b. 6014.52 – 4000 = $2014.52

x

12.

x

y

5000(1.05)20 ≈ $13, 266.49

b. 13,266.49 – 5000 = $8266.49 21. a.

700(1.035)30 ≈ $1964.76

b. 1964.76 – 700 = $1264.76 x 10

22. a.

4000(1.0375)24 ≈ $9677.75

b. 9677.75 – 4000 = $5677.75

13. For the curves, the bases involved are 0.4, 2, and 5. For base 5, the curve rises from left to right, and in the first quadrant it rises faster than the

23. a.

curve for base 2. Thus the graph of y = 5 x is B.

⎛ 0.0875 ⎞ 3000 ⎜1 + 4 ⎟⎠ ⎝

64

≈ 11,983.37

b. 11,983.37 − 3000 = $8983.37 134

n

⎛1⎞ = 1− ⎜ ⎟ . ⎝2⎠


24. a.

⎛ 0.07 ⎞ 2000 ⎜1 + ⎟ 4 ⎠ ⎝

Section 4.1

48

≈ $4599.20

b. 4599.20 − 2000 = $2599.20 25. a.

Hours

Bacteria

Expression

0

100,000

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

1

90,000

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

81,000

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

2

2

72,900

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

3

3

65,610

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

4

4

1

5000(1.0075)30 ≈ $6256.36

b. 6256.36 – 5000 = $1256.36 10

26. a.

⎛ 0.11 ⎞ 500 ⎜ 1 + 2 ⎟⎠ ⎝

≈ $854.07

b. 854.07 – 500 = $354.07 27. a.

⎛ 0.0625 ⎞ 8000 ⎜ 1 + 365 ⎟⎠ ⎝

0

3(365)

≈ $9649.69

b. 9649.69 – 8000 = $1649.69 28. a. b.

10

900(1.0225)

30. a.

≈ $1124.28

900(1.045)5 ≈ $1121.56

⎛ 0.04 ⎞ 29. 6500 ⎜ 1 + ⎟ 4 ⎠ ⎝

t

⎛ 9 ⎞ after t hours is given by N (t ) = 100, 000 ⎜ ⎟ . ⎝ 10 ⎠

24

≈ $8253.28

33. Let P = the amount of plastic recycled and let r = the rate at which P increases each year. Then after the first year, the amount of plastic recycled, increases from P to P + rP = P(1 + r), since r = 0.3, the factor by which P increases for the first year, is 1 + r = 1 + 0.3 = 1.3. Similarly, during the second year, the amount of plastic recycled increases from P(1 + r) to

P = 5000(1.03)t

N = 400(1.05)t

b. When t = 1, then N = 400(1.05)1 = 420. c.

t

Thus, in general, the number of bacteria present

b. When t = 3, then P = 5000(1.03)3 ≈ 5464. 31. a.

⎛ 9 ⎞ 100, 000 ⎜ ⎟ ⎝ 10 ⎠

t

P(1 + r) + r[ P (1 + r )] = P (1 + r )2 . Again, since r = 0.3, the multiplicative increase for the second

When t = 4, then N = 400(1.05)4 ≈ 486.

year is (1 + r ) 2 = (1 + 0.3)2 = (1.3) 2 = 1.69 . This pattern will continue as shown in the table.

32. If N = N(t) = the number of bacteria present at any time t, where t is in hours, and if r = the rate at which the bacteria are reduced, then, after the first hour, the number of bacteria remaining is N – rN = N(1 – r) = 100,000(1 – 0.1) = 100,000(0.9) = 90,000. Similarly, after the second hour, the number of bacteria remaining is N(1 – r) – r[N(1 – r)] = N (1 − r )

2

= 100, 000(1 − 0.1) 2 = 100, 000(0.9)2 = 81, 000 This pattern will continue as shown in the table. 135

Year

Multiplicative Increase

Expression

0

1

1.30

1

1.3

1.31

2

1.69

1.32

3

2.20

1.33

Chapter 4: Exponential and Logarithmic Functions


Thus, the increase in recycling is exponential with a base = 1 + r = 1 + 0.3 = 1.3. If we graph the multiplicative increase as function of years, we obtaining the following.

42.

5

y

x 5

y 3 2 1 1 2 3 4 5

x years

43. For x = 3, P =

From the graph it appears that recycling will triple after about 4 years.

44. f(0) ≈ 0.399; f(–1) = f(1) ≈ 0.242

( )

34. Population of city A after 5 years:

45. e kt = e k

5

70, 000(1.04) . Population of city B after 5 years: 60, 000(1.05)5 . Difference in populations:

46.

70, 000(1.05)5 − 60, 000(1.05)5 ≈ 8589 .

t

= bt , where b = e k

x

1 ⎛1⎞ = ⎜ ⎟ = b x , where b = x e ⎝e⎠ e 1

47. a.

35. P = 350, 000(1 − 0.015)t = 350, 000(0.985)t , where P is the population after t years.

When t = 0, N = 12e −0.031(0) = 12 ⋅1 = 12.

b. When t = 10, N = 12e −0.031(10) = 12e−0.31 = 8.8.

When t = 3, P = 350, 000(0.985)3 ≈ 334, 485.

c.

When t = 44, N = 12e −0.031(44) = 12e−1.364 ≈ 3.1.

36. E = 14, 000(1 − 0.03)t = 14, 000(0.97)t , where E is the enrollment after t years. When t = 12,

1 of the 4 initial amount remains. Because 1 ⎛ 1 ⎞⎛ 1 ⎞ = , 44 hours corresponds to 2 4 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠ half-lives. Thus the half-life is approximately 22 hours.

d. After 44 hours, approximately

E = 14, 000(0.97)12 ≈ 9714.

37. 4.4817 38. 29.9641 39. 0.4966 40. 0.5134 41.

e−3 33 ≈ 0.2240 3!

48. N = 75e −0.045(10) ≈ 48 y

2

49. After one half-life,

x 5

1 gram remains. After two 2 2

1 1 ⎛1⎞ 1 ⋅ = ⎜ ⎟ = gram remains. 2 2 ⎝2⎠ 4 Continuing in this manner, after n half-lives,

half-lives, n

4

1 ⎛1⎞ ⎛1⎞ ⎜ 2 ⎟ gram remains. Because 16 = ⎜ 2 ⎟ , after ⎝ ⎠ ⎝ ⎠ 1 4 half-lives, gram remains. This corresponds 16 to 4 · 8 = 32 years. 136


50.

51.

f ( x) =

e −0.5 (0.5) x x!

f (2) =

e−0.5 (0.5) 2 ≈ 0.0758 2!

f ( x) =

e −4 4 x x!

f (2) =

e −4 42 ≈ 0.1465 2!

Section 4.2 58. a.

When p = 10, then q = 10, 000(0.95123)10 ≈ 6065 .

b. Using a graphics calculator, 0.95123 = e − x when x ≈ 0.05. Thus, 0.95123 ≈ e −0.05 .

(

)

p q = 10, 000(0.95123) p ≈ 10, 000 e−0.05 .

= 10, 000e−0.05 p q = 10, 000e −0.05(10) ≈ 6065 .

c.

5

52.

59. The first integer t for which the graph of P = 2500(1.043)t lies on or above the horizontal line P = 5000 is 17.

2

–2

Principles in Practice 4.2 –5

The intersection point is (0, 1). 53.

1. If 16 = 2t is the exponential form then t = log 2 16 is the logarithmic form, where t represents the number of times the bacteria have doubled.

8

–5

⎛ I ⎞ 2. If 8.3 = log10 ⎜ ⎟ is the logarithmic form, then ⎝ I0 ⎠ I = 108.3 is the exponential form. I0

5 –2

If f ( x) = 2 x , then

3. Let R = the amount of material recycled every year. If the amount being recycled increases by 50% every year, then the amount recycled at the end of y years is

y = 2a ⋅ 2 x = 2 x + a = f ( x + a). Thus, the graph

of y = 2a ⋅ 2 x is the graph of y = 2 x shifted a units to the left. 54. 0.71

R (1 + r ) y = R (1 + 0.5) y = R (1.5) y Thus, the multiplicative increase in recycling at the end of

55. 3.17

y years is (1.5) y . If we let

56. The first integer t for which the graph of

x = the multiplicative increase, then x = (1.5) y and, in logarithmic form, log1.5 x = y .

P = 1000(1.07)t lies on or above the horizontal line P = 3000 is 17.

⎛4⎞ 57. 300 ⎜ ⎟ ⎝3⎠

y 6

4.1

≈ 976

y = log1.5x

3

4.2

⎛4⎞ 300 ⎜ ⎟ ≈ 1004 ⎝3⎠ 4.2 minutes

5

137

10

x multiplicative increase



4. Let V = the value of the boat. If the value depreciates by 20% every year, then at the end of y years the value of the boat is

4. log8 4 =

V (1 − r ) y = V (1.02) y = V (0.8) y . Thus, the multiplicative decrease in value at the end of y

2 3

5. ln 20.0855 = 3

years is (0.8) y . If we let

6. ln 1.4 = 0.33647

x = the multiplicative decrease, then x = (0.8) y and, in logarithmic form, log 0.8 x = y

7. e1.09861 = 3 8. 100.6990 = 5

y

9. 8 4

5

y

y = log0.8x

x 1

x multiplicative decrease

5. The equation t (r ) =

5

ln 4 can be rewritten as r

10.

ln 4 . When this equation is graphed we find t (r ) that the annual rate r needed to quadruple the investment in 10 years is approximately 13.9%. Alternatively, we can solve for r by setting t(r) = 10. ln(4) r= t (r ) r=

r=

5

y

5x

11.

ln(4) ≈ 0.139 or ≈ 13.9% 10

6. Since m = e rt , then ln m = rt. ln m = rt ln m =r t Let m = 3 and t = 12. ln 3 =r 12 0.092 = r Thus, to triple your investment in 12 years, invest at an annual percentage rate of 9.2%.

5

y

x 5

12.

5

y

5x

Problems 4.2 1. log 10,000 = 4 2. (12)2 = 144 3. 26 = 64 138

ISM: Introductory Mathematical Analysis 13.

5

Section 4.2 23. Because 10−2 = 0.01, log 0.01 = −2

y

1 24. Because 21/ 3 = 3 2, log 2 3 2 = . 3

x 8

25. Because 50 = 1, log5 1 = 0

14.

5

26. Because 5−2 =

y

1 1 27. Because 2−3 = , log 2 = −3 8 8

x 5

15.

5

1 1 = −2 , log5 25 25

1 28. Because 41/ 5 = 5 4, log 4 5 4 = . 5

29. 34 = x x = 81

y

30. 28 = x x = 256

x 5

31. 53 = x x = 125 16.

5

32. 40 = x x=1

y

33. 10−1 = x 1 x= 10

x 5

34. e1 = x x=e

17. Because 62 = 36 , log 6 36 = 2

35. e −3 = x

18. Because 26 = 64, log 2 64 = 6.

36. x 2 = 25 Since x > 0, we choose x = 5.

19. Because 33 = 27, log3 27 = 3

37. x3 = 8 x=2

1 20. Because 161/ 2 = 4, log16 4 = 2

21. Because 71 = 7, log 7 7 = 1

38. x1/ 2 = 3 x=9

22. Because 104 = 10, 000, log10, 000 = 4

39. x −1 = x=6 139

1 6



40. y = x1 x=y

49. e3 x = 2 3x = ln 2 ln 2 x= 3

41. 3−3 = x 1 x= 27

50. 0.1e0.1x = 0.5 e0.1x = 5 0.1x = ln 5 x = 10 ln 5

1

42. x = 2 x − 3 x=3 43. x 2 = 6 − x

51. e2 x −5 + 1 = 4

2

e 2 x −5 = 3 2x – 5 = ln 3 5 + ln 3 x= 2

x + x−6 = 0 (x + 3)(x – 2) = 0 The roots of this equation are –3 and 2. But since x > 0, we choose x = 2.

44. log8 64 = x − 1

52. 6e2 x − 1 =

8 x −1 = 64 x–1=2 x=3

6e 2 x =

45. 2 + log 2 4 = 3x − 1 2 + 2 = 3x – 1 5 = 3x 5 x= 3

e2 x =

1 2

3 2

1 4

1 4 1 1 x = ln 2 4

2 x = ln

46. 3−2 = x + 2 1 = x+2 9 17 x=− 9

53. 1.60944 54. 1.45161 55. 2.00013 56. 2.30058

47. x 2 = 2 x + 8

57. If V = the value of the antique. If the value appreciates by 10% every year, then at the end of y years the value of the antique is

x2 − 2 x − 8 = 0 (x – 4)(x + 2) = 0 The roots of this equation are 4 and –2. But since x > 0, we choose x = 4.

V (1 + r ) y = V (1 + 0.10) y = V (1.10) y . Thus, the multiplicative increase in value at the end of

48. x 2 = 6 + 4 x − x 2

y years is (1.10) y . If we let

2

2x − 4x − 6 = 0

x = the multiplicative increase, then x = (1.10) y , and, in logarithm form, log1.10 x = y .

x2 − 2x − 3 = 0 ( x − 3)( x + 1) = 0 The roots of the equation are 3 and −1. But since x > 0, we choose x = 3.

140


Section 4.2

Years

y 8 7 6 5 4 3 2 1

63. T =

ln 2 ≈ 36.1 minutes 0.01920

64. T =

ln 2 ≈ 21.7 years 0.03194

x 0

1 2 Multiplicative increase

65. From log y x = 3 , y 3 = x ; from log z x = 2 , 3

z 2 = x . Thus z 2 = y 3 or z = y 2 .

58. c = 3(6) ln 6 + 12 ≈ 44.25

66. x + 3e2 y − 8 = 0

1980 ⎤ ⎡ 59. p = log ⎢10 + = log[10 + 990] = log1000 2 ⎥⎦ ⎣ =3 ⎛ ⎞ E 60. 1.5M = log ⎜ 11 ⎟ ⎝ 2.5 × 10 ⎠ E 101.5M = 2.5 × 1011

(

)(

E = 2.5 × 1011 101.5M

3e2 y = 8 − x 8− x e2 y = 3 ⎡8 − x ⎤ ln[e2 y ] = ln ⎢ ⎥ ⎣ 3 ⎦ ⎡8 − x ⎤ 2 y = ln ⎢ ⎥ ⎣ 3 ⎦ 1 ⎡8 − x ⎤ y = ln ⎢ 2 ⎣ 3 ⎥⎦

)

E = 2.5 × 1011+1.5 M

61. a.

( ) = 2N 1

If t = k, then N = N 0 2

67.

b. From part (a), N = 2 N 0 when t = k. Thus k is the time it takes for the population to double. c.

3

0

–1

4

–2

t

N1 = N 0 2 k

a.

t N1 = 2k N0

b. [−0.37, ∞)

N t = log 2 1 k N0 t = k log 2

68.

–1

x22 2

x1 = e

4 –1

x2 u0 − 2 = A ln ( x1 ) 2

ln ( x1 ) =

4

N1 N0

62. u0 = A ln ( x1 ) +

u0 −

(0, 1)

(1, 0)

x22 2

A

( )

u0 − x 2 2 2 A

141


ISM: Introductory Mathematical Analysis R1 − R2 = log(900, 000) − log 9000

69. For y = e x , if y = 3, then 3 = e x or x = ln 3. 10

= log

–10

=2 Thus, the two earthquakes differ by 2 on the Richter scale.

10

2. The magnitude (Richter Scale) of an earthquake ⎛ I ⎞ is given by R = log ⎜ ⎟ where I is the intensity ⎝ I0 ⎠

–10

From the graph of y = e x , when y = 3, then x = ln 3 ≈ 1.10.

of the earthquake and I 0 is the intensity of a I = how I0 many times greater the earthquake is than a zeroI = 10, 000 , then level earthquake. Thus, if I0

70. For y = ln x, when y = 2, then 2 = ln x or x = e2 .

zero-level reference earthquake.

5

0

900, 000 = log 100 = log102 = 2 log 10 9000

10

R = log 10,000 = log104 = 4 log 10 = 4 The earthquake measures 4 on the Richter scale. –5

From the graph of y = ln x, when y = 2, then

Problems 4.3

2

x = e ≈ 7.39 .

71.

1. log 30 = log(2 ⋅ 3 ⋅ 5) = log 2 + log 3 + log 5 = a+b+c

4

2. log16 = log 24 = 4 log 2 = 4a 0

5

3. log

2 = log 2 − log 3 = a − b 3

4. log

5 = log 5 − log 2 = c − a 2

–1

1.41, 3.06 Principles in Practice 4.3

8 = log 8 − log 3 = log 23 − log 3 3 = 3 log 2 – log 3 = 3a – b

1. The magnitude (Richter Scale) of an earthquake ⎛ I ⎞ is given by R = log ⎜ ⎟ where I is the intensity ⎝ I0 ⎠

5. log

of the earthquake and I 0 is the intensity of a

6. log

I = how I0 many times greater the earthquake is than a zeroI = 900, 000, level earthquake. Thus, when I0

7. log 36 = log(2 ⋅ 3)2 = 2 log(2 ⋅ 3) = 2(log 2 + log 3) = 2(a + b)


R1 = log(900, 000)

When

I = 9000 I0

R2 = log(9000) 142

6 2⋅3 = log 25 52 = log 2 + log 3 − 2 log 5 = a + b − 2c


Section 4.3

8. log 0.00003 = log(3 ⋅10−5 )

24. ln[ x( x + 1)]3 = 3ln[ x( x + 1)] = 3[ln x + ln( x + 1)]

= log 3 + log10−5 = log 3 − 5log10 = log 3 − 5log(2 ⋅ 5) = log 3 − 5(log 2 + log 5) = b − 5(a + c) = −5a + b − 5c

9. log 2 3 =

log10 3 log 3 b = = log10 2 log 2 a

10. log3 5 =

log10 5 log 5 c = = log10 3 log 3 b

11. log 7 7

48

4

x +1 ⎛ x +1 ⎞ 25. ln ⎜ = 4[ln( x + 1) − ln( x + 2)] ⎟ = 4 ln x + 2 x +2 ⎝ ⎠

26. ln x( x + 1)( x + 2) = ln[ x( x + 1)( x + 2)]1/ 2 1 = [ln x( x + 1)( x + 2)] 2 1 = [ln x + ln( x + 1) ln( x + 2)] 2 27. ln

= ln x − [ln( x + 1) + ln( x + 2)] = ln x − ln( x + 1) − ln( x + 2)

= 48

( )

12. log5 5 5

5

5

15 15 ⎛ 3⎞ = log5 ⎜ 5 2 ⎟ = log5 5 2 = 2 ⎝ ⎠

28. ln

14. 10log 3.4 = 10log10 3.4 = 3.4

e

= − ln e2 = − log e e2 = −2

18. log3 81 = log3 34 = 4 19. log

30. ln

1 1 + ln e3 = log10 + log e e3 = −1 + 3 = 2 10 10

21. ln ⎡ x( x + 1)2 ⎤ = ln x + ln( x + 1) 2 ⎣ ⎦ = ln x + 2 ln( x + 1)

2

= ln

x5 1

( x + 2)( x + 1) 5 2 1 ⎡ ⎤ = ln x 5 − ln ⎢ ( x + 2)( x + 1) 5 ⎥ ⎣ ⎦ 1 2 ⎡ ⎤ = ln x − ⎢ ln( x + 2) + ln( x + 1) 5 ⎥ 5 ⎣ ⎦ 2 1 = ln x − ln( x + 2) − ln( x + 1) 5 5

1 x 1 = ln x 2 − ln( x + 1) = ln x − ln( x + 1) x +1 2

x2

= ln x 2 − ln( x + 1)3 ( x + 1)3 = 2 ln x − 3ln( x + 1)

23. ln

x = ln x − [ln( x + 1) + ln( x + 2)] ( x + 1)( x + 2) = ln x − ln( x + 1) − ln( x + 2)

1 ⎡ ⎤ ⎡ 1 2 ⎤ ⎛ x2 ⎞ 5 ⎥ x 1 ⎢ 31. ln ⎢ 5 ⎥ = ln ⎢ x + 2 ⎜⎜ x + 1 ⎟⎟ ⎥ ⎢ x + 2 x +1 ⎥ ⎝ ⎠ ⎥ ⎢⎣ ⎣ ⎦ ⎦

20. eln π = elog e π = π

22. ln

1

= ln x 2 − ln ⎡( x + 1)2 ( x + 2)3 ⎤ ⎣ ⎦ ( x + 1)2 ( x + 2)3 1 = ln x − ⎡ln( x + 1) 2 + ln( x + 2)3 ⎤ ⎣ ⎦ 2 1 = ln x − [2 ln( x + 1) + 3ln( x + 2)] 2 1 = ln x − 2 ln( x + 1) − 3ln( x + 2) 2

16. ln e = log e e = 1

2

x

29. ln

15. ln e5.01 = log e e5.01 = 5.01

1

x 2 ( x + 1) = ln ⎡ x 2 ( x + 1) ⎤ − ln( x + 2) ⎣ ⎦ x+2

= ln x 2 + ln( x + 1) − ln( x + 2) = 2 ln x + ln( x + 1) − ln( x + 2)

13. log 0.0000001 = log10−7 = −7

17. ln

x = ln x − ln[( x + 1)( x + 2)] ( x + 1)( x + 2)

143


32. ln 3

x3 ( x + 2)2


⎡ ⎤ 42. log 2 ⎢ ln ⎛⎜ 5 + e2 + 5 ⎞⎟ + ln ⎛⎜ 5 + e2 − 5 ⎞⎟ ⎥ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ ⎡ ⎛ ⎤ ⎞ ⎛ ⎞ 2 2 = log 2 ⎢ ln ⎜ 5 + e + 5 ⎟ ⎜ 5 + e − 5 ⎟ ⎥ ⎠⎝ ⎠⎦ ⎣ ⎝ 2 = log 2 [ln(5 + e − 5)]

1 x3 ( x + 2) 2 = ln 3 ( x + 1)3 1 = ln[ x3 ( x + 2)2 ] − ln( x + 1)3 3 1 = [ln x3 + ln( x + 2)2 − ln( x + 1)3 ] 3 1 = [3ln x + 2 ln( x + 2) − 3ln( x + 1)] 3 2 = ln x + ln( x + 2) − ln( x + 1) 3

( x + 1)3

{

}

= log 2 [ln e2 ] = log 2 (2) =1

43. log 6 54 − log 6 9 = log 6

33. log (6 · 4) = log 24

44. log3 3 + log 2 3 2 − log5 4 5

⎛ 10 ⎞ 34. log3 ⎜ ⎟ = log3 2 ⎝ 5⎠

35. log 2

= log3 31/ 2 + log 2 21/ 3 − log5 51/ 4 1 1 1 = + − 2 3 4 7 = 12

2x x +1

36. log x 2 − log x − 2 = log

x2

45. eln(2 x ) = 5 2x = 5 5 x= 2

x−2

37. 5log 2 10 + 2 log 2 13 = log 2 105 + log 2 132 = log 2 (105 ⋅132 )

46. 4log 4 ( x ) + log 4 (2) = 3

38. 5(log x 2 + log y 3 − log z 2 ) ⎛ x2 y3 ⎞ = 5log ⎜ ⎟ ⎜ z2 ⎟ ⎝ ⎠ ⎡ ⎛ 2 3 ⎞5 ⎤ x y = log ⎢⎜ ⎟ ⎥ ⎢⎜ z 2 ⎟ ⎥ ⎠ ⎦⎥ ⎣⎢⎝

4log 4 (2 x ) = 3 2x = 3 3 x= 2 2

47. 10log x = 4 x2 = 4 x=±2

39. log100 + log(1.05)10 = log ⎡100(1.05)10 ⎤ ⎣ ⎦

48. e3ln x = 8

( )

3

215 68 1 1 8 3 40. log 215 + log 6 − log169 = log 2 2 1693

(

= log

41. e

)

eln x = 8 x3 = 8 x=2

215(6)8

49. From the change of base formula with b = 2, m = 2x + 1, and a = e, we have log e (2 x + 1) ln(2 x + 1) log 2 (2 x + 1) = = log e 2 ln 2

1693

4 ln 3−3ln 4

=e

54 = log 6 6 = 1 9

ln 34 − ln 43

4 ln ⎛⎜ 33 ⎞⎟ 4 ⎠ ⎝ =e

=

34 3

4

=

81 64

144


Section 4.3 b. Given M1 = log ( A1 ) + 3 , let

50. From the change of base formula with b = 3, m = x 2 + 2 x + 2 and a = e,

M = log (10 A1 ) + 3

log e ( x 2 + 2 x + 2) log3 ( x 2 + 2 x + 2) = log e 3 =

M = log10 + log ( A1 ) + 3

M = 1 + ⎡⎣ log ( A1 ) + 3⎤⎦

ln( x 2 + 2 x + 2) ln 3

M = 1 + M1

51. From the change of base formula with b = 3,

57. y = log 6 x =

m = x 2 + 1 , and a = e, we have

(

)

log3 x 2 + 1 =

(

) = ln ( x + 1) .

log e x 2 + 1

2

log e 3

ln 3

ln x ln 6

2

0

10

52. From the change of base formula with b = 5, m = 9 − x 2 , and a = e, we have

(

log5 9 − x

2

)=

(

log e 9 − x 2 log3 5

) = ln (9 − x )

–2

2

ln 5

58. y = log 4 ( x + 2) =

53. eln z = 7e y

ln( x + 2) ln 4

4

z = 7e y z = ey 7 z y = ln 7

–3

10

–4

54. y = ab x so

ln x . ln10 ln x Thus the graphs of y = log x and y = are ln10 identical.

59. By the change of base formula, log x =

x

log y = log(ab ) = log a + log b x = log a + x log b. This is a linear expression because it is in the form Ax + B, where A = log b and B = log a.

60.

4

55. C = B + E ⎛ E⎞ C = B ⎜1 + ⎟ B⎠ ⎝ ⎡ ⎛ E ⎞⎤ ln C = ln ⎢ B ⎜ 1 + ⎟ ⎥ B ⎠⎦ ⎣ ⎝

0 –1

⎛ E⎞ ln C = ln B + ln ⎜ 1 + ⎟ B⎠ ⎝

y = ln(4x) = ln 4 + ln x. If f(x) = ln x, then y = ln(4x) = f(x) + ln 4. Thus the graph of y = ln(4x) is the graph of y = ln x shifted ln 4 units upward.

56. M = log(A) + 3 a.

5

M = log(10) + 3 = 1 + 3 = 4

145



8

61.

–2

3. The magnitude (Richter Scale) of an earthquake ⎛ I ⎞ is given by R = log ⎜ ⎟ where I is the intensity ⎝ I0 ⎠ of the earthquake and I 0 is the intensity of a

8

I = how I0 many times greater the earthquake is than a zerolevel earthquake. R1 = log(675, 000)


–2

ln(6x) = ln(3 ⋅ 2x) = ln 3 + ln(2x). If f(x) = ln(2x), then y = ln(6x) = f(x) + ln 3. Thus, the graph of y = ln(6x) is the graph of y = ln(2x) shifted ln 3 units upward.

⎛ I ⎞ R2 = log ⎜ ⎟ ⎝ I0 ⎠ Since R1 − 4 = R2

Principles in Practice 4.4 1. Let x = the number and let y = the unknown exponent. Then

⎛ I ⎞ log(675, 000) − 4 = log ⎜ ⎟ ⎝ I0 ⎠ ⎛ I ⎞ log 6.75 × 105 − 4 = log ⎜ ⎟ ⎝ I0 ⎠ ⎛ I ⎞ log 6.75 + 5log10 − 4 = log ⎜ ⎟ ⎝ I0 ⎠ ⎛ I ⎞ 1.829 = log ⎜ ⎟ ⎝ I0 ⎠ I 101.829 = I0

x ⋅ 32 y = x ⋅ 4(3 y −9) 32 y = 4(3 y −9)

(

log 32 y = log 4(3 y −9) y log 32 = (3y – 9) log 4 y log 32 = 3y log 4 – 9 log 4 y(log 32 – 3 log 4) = –9 log 4 −9 log 4 −18log 2 −18log 2 y= = = − log 2 log 1 log 323 2

4

y = 18 Thus, Greg used 32 to the power of 18. 2. Let S = 450. ⎛4⎞ S = 800 ⎜ ⎟ ⎝3⎠

450 ⎛ 4 ⎞ = 800 ⎜⎝ 3 ⎟⎠ log

I I0 Thus, the other earthquake is 67.5 times as intense as a zero-level earthquake. 67.5 =

−0.1d

⎛4⎞ 450 = 800 ⎜ ⎟ ⎝3⎠

−0.1d

Problems 4.4

−0.1d

1. log(3x + 2) = log(2 x + 5) 3x + 2 = 2 x + 5 x=3

450 ⎛4⎞ = −0.1d log ⎜ ⎟ 800 ⎝3⎠

450 log 800

( )

−0.1log 43

)

2. log x − log 5 = log 7 log x = log 5 + log 7 log x = log 35 x = 35

=d

20 = d Thus, he should start the new campaign 20 days after the last one ends.

146


Section 4.4

3. log 7 – log(x – 1) = log 4 7 log = log 4 x −1 7 =4 x −1 7 = 4x – 4 4x = 11 11 x= = 2.75 4 4. log 2 x + log 2 23 = log 2 log 2 (8 x) = log 2 8x =

8. (e3 x − 2 )3 = e3 e3(3 x − 2) = e3 3(3 x − 2) = 3 3x − 2 = 1 3x = 3 x =1

9. (81) 4 x = 9 (34 ) 4 x = 32 316 x = 32 16 x = 2 2 1 x= = = 0.125 16 8

2 x

2 x

2 x

10. (27) 2 x +1 = 3−1

(3 ) 3

8x2 = 2 1 x2 = 4 1 x = = 0.5 since x > 0 2

(

5. ln(− x) = ln x 2 − 6

= 3−1

36 x +3 = 3−1 6x + 3 = –1 6x = –4 2 x = − ≈ −0.667 3

)

11. e 2 x = 9

− x = x2 − 6

(e x ) 2 = 32

x2 + x − 6 = 0 (x + 3)(x – 2) = 0 x = –3 or x = 2 However, x = –3 is the only value that satisfies the original equation. x = −3

ex = 3 x = ln 3 ≈ 1.099

12. e4 x =

6. ln(4 – x) + ln 2 = 2 ln x ln[(4 − x)2] = ln x

2 x+1

3 4

4 x = ln

2

(4 − x)2 = x 2

x=

x2 + 2 x − 8 = 0 (x + 4)(x – 2) = 0 x = –4 or x = 2 However, x = 2 is the only value that satisfies the original equation. x=2

ln

3 4

( 34 ) ≈ −0.072 4

13. 2e5 x + 2 = 17 17 e5 x + 2 = 2 ⎛ 17 ⎞ 5 x + 2 = ln ⎜ ⎟ ⎝ 2 ⎠ ⎛ 17 ⎞ 5 x = ln ⎜ ⎟ − 2 ⎝ 2 ⎠ 1 ⎡ ⎛ 17 ⎞ ⎤ x = ⎢ ln ⎜ ⎟ − 2 ⎥ ≈ 0.028 5⎣ ⎝ 2 ⎠ ⎦

7. e 2 x e5 x = e14 e7 x = e14 7x = 14 x=2

147



14. 5e2 x −1 − 2 = 23

19. 2 x = 5

5e2 x −1 = 25

ln 2 x = ln 5 x ln 2 = ln 5 ln 5 x= ≈ 2.322 ln 2

e2 x −1 = 5 2 x − 1 = ln 5 1 + ln 5 x= ≈ 1.305 2

ln(7 2 x +3 ) = ln 9 (2 x + 3) ln 7 = ln 9 ln 9 2x + 3 = ln 7 ln 9 2x = −3 ln 7 1 ⎛ ln 9 ⎞ − 3 ⎟ ≈ −0.935 x= ⎜ 2 ⎝ ln 7 ⎠

4

15. 10 x = 6 4 = log 6 x 4 x= ≈ 5.140 log 6 16.

4(10)0.2 x =3 5 15 (10)0.2 x = 4 15 0.2 x = log 4 x=

17.

( ) ≈ 2.870

=7 5 7

102 x =

2 x = log

x=

ln 73 x − 2 = ln 5 (3x − 2) ln 7 = ln 5 ln 5 3x − 2 = ln 7 ln 5 +2 3x = ln 7 ln 5 + 2 x = ln 7 ≈ 0.942 3

0.2

102 x

log

73 x − 2 = 5

21.

log 15 4

5

72 x +3 = 9

20.

x

22. 4 2 = 20 x

5 7

ln 4 2 = ln 20 x ln 4 = ln 20 2 x ln 20 = 2 ln 4 2 ln 20 x= ≈ 4.322 ln 4

( 75 ) ≈ −0.073

2

18. 2(10) x + (10) x +1 = 4 2(10) x + 10(10) x = 4 12(10) x = 4

23. 2

1 3 1 x = log ≈ −0.477 3

(10) x =

− 23x

ln 2 −

=

− 23x

4 5 = ln

4 5

2x 4 ln 2 = ln 3 5

( )

4 2 x ln 5 − = 3 ln 2

x=− 148

3ln

( 54 ) ≈ 0.483

2 ln 2


(

Section 4.4

)

29. log 4 (9 x − 4) = 2

24. 5 3x − 6 = 10

42 = 9 x − 4

3x − 6 = 2

9 x = 42 + 4

3x = 8

x=

ln 3x = ln 8 x ln 3 = ln 8 ln 8 x= ≈ 1.893 ln 3

30. log 4 (2 x + 4) − 3 = log 4 3 log 4 (2 x + 4) − log 4 3 = 3 2x + 4 =3 3 2x + 4 43 = 3

25. (4)53− x − 7 = 2 53− x =

9 4

(3 − x) ln 5 = ln 3− x = x = 3−

26.

log 4

9 4

ln 53− x = ln

ln

( 94 )

2 x + 4 = 3 ⋅ 43 9 4

x=

3 ⋅ 43 − 4 188 = = 94 2 2

31. log(3x – 1) – log(x – 3) = 2 3x − 1 log =2 x−3 3x − 1 102 = x −3 100(x – 3) = 3x – 1 97x = 299 299 x= ≈ 3.082 97

ln 5 ln

42 + 4 20 = ≈ 2.222 9 9

( 94 ) ≈ 2.496

ln 5

7

= 13 3x 7 = 3x 13 ⎛7⎞ ln ⎜ ⎟ = ln(3x ) ⎝ 13 ⎠ ⎛7⎞ ln ⎜ ⎟ = x ln 3 ⎝ 13 ⎠ 7 ln 13 x= ≈ −0.563 ln 3

32. log( x − 3) + log( x − 5) = 1 log[( x − 3)( x − 5)] = 1 x 2 − 8 x + 15 = 10 x2 − 8x + 5 = 0

( )

x=

27. log(x – 3) = 3

8 ± (−8)2 − 4(1)(5) 2(1)

= 4 ± 11

However, x = 4 + 11 ≈ 7.317 is the only value that satisfies the original equation. x ≈ 7.317

103 = x − 3 x = 103 + 3 = 1003

33.

28. log 2 ( x + 1) = 4 24 = x + 1 x = 24 − 1 = 15

log 2 (5 x + 1) = 4 − log 2 (3x − 2) log 2 (5 x + 1) + log 2 (3x − 2) = 4 log[(5 x + 1)(3 x − 2)] = 4 (5 x + 1)(3x − 2) = 24 15 x 2 − 7 x − 2 = 16 15 x 2 − 7 x − 18 = 0 x ≈ 1.353 or x ≈ −0.887 However, x ≈ 1.353 is the only value that satisfies the original equation. x ≈ 1.353

149



34. log( x + 2)2 = 2 2 log(x + 2) = 2 log(x + 2) = 1 101 = x + 2 x=8

⎛2⎞ 35. log 2 ⎜ ⎟ = 3 + log 2 x ⎝x⎠ ⎛2⎞ log 2 ⎜ ⎟ − log 2 x = 3 ⎝x⎠ log 2 log 2 23 =

2 x

x 2

=3

x2 2

=3

x2 1 x2 = 4 1 x=± 2

However, x = x=

36.

1 is the only value that satisfies the original equation. 2

1 = 0.5 2

ln( x − 2) = ln(2 x − 1) + 3 ln( x − 2) − ln(2 x − 1) = 3 ⎛ x−2 ⎞ ln ⎜ ⎟=3 ⎝ 2x −1 ⎠ x−2 = e3 2x −1 e3 (2 x − 1) = x − 2 2e3 x − e3 = x − 2 x(2e3 − 1) = −2 + e3 x=

−2 + e3

≈ 0.462 2 e3 − 1 However, this value does not satisfy the original equation. The equation has no solution.

37. log S = log 12.4 + 0.26 log A log S = log12.4 + log A0.26 log S = log ⎡12.4 A0.26 ⎤ ⎣ ⎦ S = 12.4 A0.26

150


Section 4.4

38. log T = 1.7 + 0.2068log P − 0.1334(log P )2 log T = log 50 + 0.2068 log P – 0.1334(log P)(log P) log T = log 50 + 0.2068 log P + [–0.1334 log P] log P log T = log 50 + log P 0.2068 + log P[ −0.1334 log P ]

(

log T = log ⎡ (50) P 0.2068 ⎣⎢

)( P

−0.1334 log P

)⎤⎦⎥

T = 50 P 0.2068−(0.1334 log P ) (logb x)2 = (logb x)(logb x) = logb ( x logb x )

39. a.

When t = 0, Q = 100e −0.035(0) = 100e0 = 100 ⋅1 = 100 .

b. If Q = 20, then 20 = 100e −0.035t . Solving for t gives 20 = e −0.035t 100 1 = e−0.035t 5 1 ln = −0.035t 5 –ln 5 = –0.035t ln 5 t= ≈ 46 0.035 −

N

40. 100 = 225e 225 N 225 9 e 225 = = 100 4 N 9 = ln 225 4 9 N = 225ln ≈ 182 4 41. If P = 1,500,000, then 1,500, 000 = 1, 000, 000(1.02)t . Solving for t gives 1,500, 000 = (1.02)t 1, 000, 000 1.5 = (1.02)t ln1.5 = ln(1.02)t ln 1.5 = t ln 1.02 ln1.5 t= ≈ 20.5 ln1.02

151


42. If F(0) = 0, then 0 =

ISM: Introductory Mathematical Analysis Thus

q − pe−C ( p + q ) . Thus q ⎡1 + eC ( p + q ) ⎤ ⎣ ⎦

0.8t =

q − pe−C ( p + q ) = 0 − pe

−C ( p + q )

e −C ( p + q ) =

⎛ 3 − log q ⎞ t log(0.8) = log ⎜ ⎟. ⎝ log 2 ⎠

= −q q p

t=

q −C ( p + q ) = ln p

3− log q log 2

)

log(0.8) x x

log y = log A + a x log b log y − log A = a x log b log y − log A ax = log b ⎛ log y − log A ⎞ log a x = log ⎜ ⎟ log b ⎝ ⎠ ⎛ log y − log A ⎞ x log a = log ⎜ ⎟ log b ⎝ ⎠

2 p = 80 − q log 2 p = log(80 − q )

p log 2 = log(80 − q) log(80 − q) log 2

log 20 When q = 60, then p = ≈ 4.32 . log 2

x=

log

(

log y − log A log b

)

log a The previous solution was the special case y = q, 1 A = 1000, b = , a = 0.8, and x = t. 2

44. The investment doubles when A = 2P. Thus 2 P = P (1.105)t , or 2 = (1.105)t . Solving for t gives

(

ln 2 = ln(1.105)t ln 2 = t ln1.105 ln 2 ≈7 t= ln1.105

46. q = 500 1 − e−0.2t a.

) (

)

If t = 1, then q = 500 1 − e−0.2 ≈ 91 .

(

0.8t

⎛1⎞ log q = log1000 + log ⎜ ⎟ ⎝2⎠ 1 log q = 3 + 0.8t log 2

(

log y = log A + log b a

43. q = 80 − 2 p

⎛1⎞ 45. q = 1000 ⎜ ⎟ ⎝2⎠

log

y = Ab a

1 q C=− ln . p+q p

p=

log(q) − 3 3 − log q = − log 2 log 2

)

b. If t = 10, then q = 500 1 − e−2 ≈ 432 . c.

0.8t

We solve the equation

(

400 = 500 1 − e−0.2t 4 = 1 − e−0.2t 5 1 e−0.2t = 5 1 −0.2t = ln = − ln 5 5 ln 5 ≈8 t= 0.2

log q = 3 + 0.8t (− log 2) log(q) − 3 = 0.8t (− log 2)

152

)


Chapter 4 Review

47. log 2 x = 5 − log 2 ( x + 4) is equivalent to 0 = 5 − log 2 ( x + 4) − log 2 x , or

y=

ln( x + 4) ln x . Thus the solutions of the 0 = 5− − ln 2 ln 2 original equation are the zeros of the function ln( x + 4) ln x . y = 5− − ln 2 ln 2

ln

( 4 x3+5 ) . ln 2

8

–2

5

8 –2

0


10

1. log3 243 = 5 2. 54 = 625

–5

From the graph of this function, the only zero is x = 4. Thus 4 is the only solution of the original equation. 48.

1

3. 814 = 3 4. log 100,000 = 5

20

5. ln 54.598 = 4 6. 91 = 9

0

7. Because 53 = 125 , log5 125 = 3 0

2

8. Because 42 = 16 , log 4 16 = 2

1.20 49.

9. Because 3−4 =

10

1 1 = −4 , log3 81 81

3

1 1 ⎛1⎞ 10. Because ⎜ ⎟ = , log 1 = 3. 4 64 64 ⎝4⎠

10

–10

⎛1⎞ 11. Because ⎜ ⎟ ⎝3⎠

–10

−2

= 32 = 9 , log 1 9 = −2

3.33 1

12. Because 4 2 = 2 , log 4 2 =

y

50. (3)2 − 4 x = 5 (3)2 y = 4 x + 5 4x + 5 2y = 3 ⎛ 4x + 5 ⎞ y ln 2 = ln ⎜ ⎟ ⎝ 3 ⎠ ⎛ 4x + 5 ⎞ y ln 2 = ln ⎜ ⎟ ⎝ 3 ⎠ ln 4 x3+5 y= ln 2 The graph of the original equation is the graph of

(

13. 5 x = 625 x=4 1 = −4 81 1 x −4 = 81 1 1 = 4 81 x

14. log x

)

x 4 = 81 x=3

153

3

1 2



15. 2−5 = x 1 1 = x= 5 32 2

25.

1 log 2 x + 2 log 2 x 2 − 3log 2 ( x + 1) − 4 log 2 ( x + 2) 2

1 = e −1 e x = –1

= log 2

0

e = 2x + 3 1 = 2x + 3 2 x = −2 x = −1

( x + 1)3 ( x + 2) 4

= log x 4 + log y 2 − 3log zw = log x 4 + log y 2 − log( zw)3 = log x 4 y 2 − log z 3 w3 = log

19. log 8000 = log(2 ⋅10)3 = 3log(2 ⋅10) = 3(log 2 + log10) = 3(a + 1)

2

x2

26. 4 log x + 2 log y − 3(log z + log w)

18. Because eln( x + 4) = x + 4 , x+4=7 x=3

= log

− ⎡ log 2 ( x + 1)3 + log 2 ( x + 2) 4 ⎤ ⎣ ⎦

9

17. ln(2 x + 3) = 0

9

2

⎛ 1 ⎞ = log 2 ⎜ x 2 x 4 ⎟ − log 2 ⎡ ( x + 1)3 ( x + 2)4 ⎤ ⎣ ⎦ ⎝ ⎠

16. e x =

20. log

( )

1

= log 2 x 2 + log 2 x 2

32 1

27. ln

x4 y 2 z 3 w3

x3 y 2

= ln x3 y 2 − ln z −5

z −5

= ln x3 + ln y 2 − ln z −5 = 3ln x + 2 ln y + 5ln z

1

= log 32 − log 2 2

22

1 1 a = 2 log 3 − log 2 = 2b − a = 2b − 2 2 2

21. 3log 7 − 2 log 5 = log 73 − log 52 = log

28. ln 73

=

52

= ln( x5 y 2 z )

23. 2 ln x + ln y − 3ln z = ln x 2 + ln y − ln z 3 x2 y

1 ln x − 2(ln y + ln z ) 2

4

z3

(

(

= 4 ln x + ln y 3 − ln z 2 9⎤

= log 6 2 − ⎡ log 6 4 + log 6 3 ⎣ ⎦

)

( yz )2

⎡ xy 3 ⎤ xy 3 30. ln ⎢ = 4 ln = 4 ln xy 3 − ln z 2 ⎥ 2 2 z ⎢⎣ z ⎥⎦

24. log 6 2 − log 6 4 − 9 log 6 3

(

1

= ln x − ln( yz )2 = ln x 2 − 2 ln( yz )

1 1 29. ln 3 xyz = ln( xyz ) 3 = ln( xyz ) 3 1 = (ln x + ln y + ln z ) 3

22. 5ln x + 2 ln y + ln z = ln x5 + ln y 2 + ln z

= ln x 2 y − ln z 3 = ln

x

= log 6 2 − log 6 4 ⋅ 39 = log 6

)

)

= 4(ln x + 3ln y − 2 ln z ) 2 9

4⋅3

= log 6

1 39,366

⎡1 31. ln ⎢ ⎣⎢ x

() y⎤ ⎥ = ln y z

1/ 2

1

⎛ y ⎞2 = ln ⎜ ⎟ − ln x ⎝z⎠

z ⎥⎦ x 1 y 1 = ln − ln x = (ln y − ln z ) − ln x 2 z 2

154


Chapter 4 Review 42.

⎡⎛ x ⎞ 2 x 3 ⎤ x5 ⎛ ⎞ ⎢ 32. ln ⎜ ⎟ ⎜ ⎟ ⎥ = ln = ln x5 − ln y 2 z 3 2 3 ⎢⎝ y ⎠ ⎝ z ⎠ ⎥ y z ⎣ ⎦

(

5

y

y = 3x y = log 3 x

)

x

= ln x5 − ln y 2 + ln z 3 = 5ln x − 2 ln y − 3ln z

33. log3 ( x + 5) =

5

log e ( x + 5) ln( x + 5) = log e 3 ln 3

43.

log10 (7 x3 + 5) 34. log 2 (7 x + 5) = log10 2

9

3

=

35. log5 19 =

log(7 x3 + 5) log 2

x

log 2 19 4.2479 = ≈ 1.8295 log 2 5 2.3219

5

44.

ln 5 ≈ 1.1610 36. log 4 5 = ln 4

(

)

= 2y +

1 x 2

37. ln 16 3 = ln 42 + ln 3 = 2 ln 4 +

38. log

y

5

x

1 ln 3 2

5

45. log(5 x + 1) = log(4 x + 6) 5x + 1 = 4 x + 6 x=5

x3 3 x + 1 5 2

x +2 5

= log x3 3 x + 1 − log x 2 + 2

46. log 3x + log 3 = 2 log 9 x = 2

5

= log x3 + log 3 x + 1 − log x 2 + 2 1 1 = 3log x + log( x + 1) − log( x 2 + 2) 3 5

9 x = 102 9 x = 100 100 x= 9

39. 10log x + log10 x + log10 = x + x + 1 = 2 x + 1

( )

40. log102 + log(1000) − 5 = log102 + log 103 − 5

47. 34 x = 9 x +1

( )

=2+3–5=0 41. In exponential form, y = e x

34 x = 32 2

y

+2

x +1

34 x = 32( x +1) 4x = 2(x + 1) 4x = 2x + 2 2x = 2 x=1

.

155


48. 43− x =


3x

1 16

54. 10 2 = 5 3x = log 5 2 2 x = log 5 ≈ 0.466 3

43− x = 4−2 3 – x = –2 x=5

49. log x + log(10x) = 3 log x + log 10 + log x = 3 2 log(x) + 1 = 3 2 log(x) = 2 log x = 1

(

10 x + 4 − 3 = 3 10 x + 4 = 6 x + 4 = log 6 x = log(6) – 4 ≈ –3.222

x = 101 = 10

50.

log 2 ( x + 4) = log 2 ( x − 2) + 3 ⎛ x+4⎞ log 2 ⎜ ⎟=3 ⎝ x−2⎠ x+4 = 23 = 8 x−2 x + 4 = 8( x − 2) = 8 x − 16 20 = 7 x 20 x= 7

56. 7e3 x −1 − 2 = 1 7e3 x −1 = 3 3 e3 x −1 = 7 3x − 1 = ln 3x = ln

51. ln(log x 3) = 2 log x 3 = e

2

x=

2

xe = 3 2 2 ( x e ) −e = 3−e

2

52. log 2 x + log 4 x = 3 log 2 x =3 log 2 4

log 2 x +

log 2 x =3 2

3 +1 7

ln 73 + 1 3

≈ 0.051

ln 4 x +3 = ln 7 (x + 3)ln 4 = ln 7 ln 7 x+3= ln 4 ln 7 x= − 3 ≈ −1.596 ln 4

2 2 2 x e ⋅−e = 3− e

log 2 x +

3 7

57. 4 x +3 = 7

2

x1 = 3−e 1 x= 2 3e

)

55. 3 10 x + 4 − 3 = 9

58.

3 log 2 x = 3 2 log 2 x = 2

35 / x = 2 5 ln 3 = ln 2 x 5ln 3 x= ≈ 7.925 ln 2

59. Quarterly rate =

x = 22 x=4

6

a.

53. e3 x = 14 3x = ln 14 ln14 x= ≈ 0.880 3

0.06 = 0.015 4

1 yr = 26 quarters 2 2600(1.015) 26 ≈ $3829.04

b. 3829.04 – 2600 = $1229.04

156


60. Monthly rate = 5 yr = 60 mo. ⎛ 0.11 ⎞ 4000 ⎜1 + 12 ⎟⎠ ⎝

Chapter 4 Review

0.11 12

3

66. Because

60

for

≈ $6915.66

62. a.

a. b.

N = 600(1.05)t

t − 40

If t = 20, R = 10e 5 = 10e

t − 40

,

− 20 40

1 =e 2

t − 40

= 10e

t 1 = ln = − ln 2 40 2 t = 40 ln 2 ≈ 28. 68. Let d = depth in centimeters.

When t = 5, N = 600(1.05)5 ≈ 766.

d

63. a.

(0.9) 20 = 0.0017

t

P = 6000[1 + (−0.005)] or P = 6000(0.995)

d

ln(0.9) 20 = ln 0.0017

t

d ln 0.9 = ln 0.0017 20 20 ln 0.0017 d= ≈ 1210 cm ln 0.9

b. When t = 10, then P = 6000(0.995)10 ≈ 5707.

64. If t = 2, R = 200, 000e −0.4 ≈ 134, 064

69. Tt − Te = (Tt − Te )o e − at

If t = 3, R = 200, 000e −0.6 ≈ 109, 762

e − at =

65. N = 10e −0.41t a.

When t = 0, then N = 10e0 = 10 ⋅1 = 10 mg

−at = ln

b. When t = 2, then N = 10e−0.82 ≈ 4.4 mg c.

When t = 10, then N = 10e −4.1 ≈ 0.2 mg

d.

ln 2 ≈ 1.7 0.41

e.

Tt − Te (Tt − Te )o Tt − Te (Tt − Te )o

T −T 1 a = − ln t e t (Tt − Te )o

1 (Tt − Te )o a = ln t Tt − Te

If N = 1, then 1 = 10e −0.41t . Solving for t gives 1 = e−0.41t 10 1 −0.41t = ln = − ln10 10 ln10 t= ≈ 5.6 0.41

157

− 12

. Thus

−

b. When t = 1, N = 600(1.05)1 = 630. c.

1 of the initial amount to be present. 8

67. R = 10e

⎛ 1 ⎞ 61. 12 ⎜ 1 % ⎟ = 14% ⎝ 6 ⎠

1 ⎛1⎞ , it will take 3 · 10 = 30 days = 8 ⎜⎝ 2 ⎟⎠

≈6.



70. For double-declining balance depreciation, the

73.

n

10

2⎞ ⎛ equation is V = C ⎜ 1 − ⎟ . ⎝ N⎠ 2 ⎞ ⎛ 700 = 1800 ⎜ 1 − ⎟ ⎝ 48 ⎠ n 700 ⎛ 46 ⎞ =⎜ ⎟ 1800 ⎝ 48 ⎠ 7 ⎛ 23 ⎞ =⎜ ⎟ 18 ⎝ 24 ⎠

10

–10

n

–10

2.53

n

20

74. n

⎛7⎞ ⎛ 23 ⎞ ln ⎜ ⎟ = ln ⎜ ⎟ ⎝ 18 ⎠ ⎝ 24 ⎠ ⎛7⎞ ⎛ 23 ⎞ ln ⎜ ⎟ = n ln ⎜ ⎟ ⎝8⎠ ⎝ 24 ⎠ 7 ln 18 n= ≈ 22 23 ln 24

( ) ( )

–5

0.37

The value drops below $700 at about 22 months.

(

1

71.

)

2

75. y = log 2 x + 1 =

0

10

5

(

)

ln x 2 + 1 ln 2

8

–2

–2

–10

(−∞, 0.37] 72.

5

0

76. (6)5 y + x = 2

10

2− x 6 2− x ln 5 y = ln 6 2− x y ln 5 = ln 6

5y = –10

10

–10 10

y= –10

ln 2−6 x ln 5 1

10

–10

3

–10

(–1.96, –3.17), (2.93, 1.60) –3

158


Mathematical Snapshot Chapter 4 2. From the text, the half-life H is given by ln 2 ln 2 or, equivalently, k = . If H = I, H= k H ln 2 then k = . Thus I − d ⋅ lnI 2 ⋅ I ⎞ ⎛ P ⎜1 − e ⎟ P 1 − e− dkI ⎠ T= = ⎝ ln 2 kI ⋅ I e −1 e I −1

8

–5

5

(

–2

y=

3x 3x = = 3x − 2 . 9 32 x

If f ( x) = 3 , then we have y = 3

x −2

−d ⎞ ⎛ P ⎜ 1 − ⎡ eln 2 ⎤ ⎟ P 1 − 2− d ⎣ ⎦ ⎠ = = ⎝ ln 2 2 −1 e −1 ⎛ 1 ⎞ = P 1 − 2− d = ⎜ 1 − ⎟P . ⎝ 2d ⎠

(

= f ( x − 2) .

x

(

3 is the graph of y = 3x 9 shifted 2 units to the right.

Thus the graph of y =

1. T = a.

(

( T (e

T e

kI kI

1− e

b.

)

e kI − 1

a.

) ( ) − 1) T (e = P or P =

−1 = P 1− e

− dkI

− dkI

(

kI

1− e

)

)

(

)

b.

)

(

) ⎤⎥

(

(

)

4.

⎥ ⎥⎦

e − 32

(

ln 2 ⋅4 8

−1

⎞ 3 ⎟⎟ 100 ⎜⎛1 − 2− 2 ⎟⎞ ⎠= ⎝ ⎠ ≈ 156 1 2 2 −1

)

)

⎞ ⎟ . Thus ⎠

5

) ⎤⎥

kI ⎡ 1 ⎢ P − T e −1 d = − ln kI ⎢ P ⎣⎢

⎡ 1 ⎢ P d = ln ⎢ kI kI ⎢⎣ P − T e − 1

−1

R = P 1 − e− dkI . From part (a),

(

⎡ P − T ekI − 1 −dkI = ln ⎢ ⎢ P ⎢⎣

)

−3 ⎛ P 1 − e− dkI = 100 ⎜ 1 − 2 2 ⎝ − 32 ⎞ ⎛ R = 100 ⎜ 1 − 2 ⎟ ≈ 65. ⎝ ⎠

P − T e kI − 1 P

e

kI

ln 2 ln 2 = H 8

−3⋅ ln 2 ⋅4 ⎞ ⎛ 100 ⎜1 − e 8 ⎟ ⎝ ⎠ =

⎡ eln 2 ⎤ 2 − 1 ⎣ ⎦

Pe− dkI = P − T e kI − 1 e − dkI =

(

P 1− e

1

− dkI

T ekI − 1 = P − Pe − dkI

(

T=

− dkI

⎛ 100 ⎜ 1 − ⎡eln 2 ⎤ ⎜ ⎣ ⎦ ⎝ =

−1

)

)

3. P = 100, I = 4, d = 3, H = 8, k =

Mathematical Snapshot Chapter 4 P 1 − e − dkI

)

⎥ ⎦⎥

0

⎤ ⎥ ⎥ ⎥⎦

0

10

As d changes, some of the coefficients need to change from P to Y1 or vice versa.

159


By graphing re as a function of r, we find that, when the nominal rate r = 0.077208 or 7.7208%, the effective rate re = 0.08 or 8%.

1. Let P = 518 and let n = 3(365) = 1095. S = P(1 + r ) n

0.1

1095

r ⎞ ⎛ S = 518 ⎜ 1 + ⎟ ⎝ 365 ⎠ By graphing S as a function of the nominal rate r, we find that when r = 0.049, S = 600. Thus, at the nominal rate of 4.9% compounded daily, the initial amount of $518 will grow to $600 after 3 years.

0

1000

0.1

0

4. The respective effective rates of interest are n

⎛ r⎞ found using the formula re = ⎜1 + ⎟ − 1 . ⎝ n⎠ Let n = 12 when r = 0.11: 12

0

⎛ 0.11 ⎞ re = ⎜1 + − 1 ≈ 0.1157 . Hence, when the 12 ⎟⎠ ⎝ nominal rate r = 11% is compounded monthly, the effective rate is re = 11.57% . When

0.2

400

2. Let P = 520 and let r = 0.052. S = P(1 + r ) n

4

⎛ 0.052 ⎞ S = 520 ⎜1 + 365 ⎟⎠ ⎝

⎛ 0.1125 ⎞ r = 0.1125: re = ⎜1 + − 1 ≈ 0.1173 . 4 ⎟⎠ ⎝ Hence in the second case when the nominal rate r = 11.25% is compounded quarterly, the effective rate is re = 11.73% . This is the better effective rate of interest. To find the better investment, compare the compound amounts, S at the end of n years. With P = 10,000 and re = 0.1157 ,

n

n

⎛ 365.052 ⎞ S = 520 ⎜ ⎟ ⎝ 365 ⎠ By graphing S as a function of n, we find that when n = 2571, S = 750. Thus, it will take 2571 ≈ 7.044 years, or 7 years and 16 days for 365 $520 to grow to $750 at the nominal rate of 5.2% compounded daily.

S1 = P (1 + r )n = 10, 000(1 + 0.1157)n , and, in the second case, when P = 9700 and re = 0.1173

800

S2 = P (1 + r )n = 9700(1 + 0.1173)n . S1 (20) = 10, 000(1.1157)20 ≈ 89,319.99

0

S2 (20) = 9700(1.1173)20 ≈ 89,159.52 The $10,000 investment is slightly better over 20 years.

3000

500

Problems 5.1

3. Let n = 12.

1. a.

n

⎛ r⎞ re = ⎜1 + ⎟ − 1 ⎝ n⎠

b. 11,105.58 – 6000 = $5105.58

12

r ⎞ ⎛ re = ⎜1 + ⎟ ⎝ 12 ⎠

6000(1.08)8 ≈ $11,105.58

−1 160


Section 5.1

750(1.07) = $802.50 c.

b. 802.5 – 750 = $52.50

⎛ 0.07 ⎞ (i) 1000 ⎜1 + 52 ⎟⎠ ⎝ ⎛ 0.07 ⎞ (ii) ⎜1 + 52 ⎟⎠ ⎝

3. (1.015) 2 − 1 ≈ 0.030225 or 3.023%

52(5)

− 1000 ≈ $418.73

52

− 1 ≈ 0.07246 or 7.246%

4

⎛ 0.05 ⎞ 4 4. ⎜1 + ⎟ − 1 = (1.0125) − 1 ≈ 0.05095 or 4 ⎠ ⎝ 5.095% ⎛ 0.04 ⎞ 5. ⎜1 + 365 ⎟⎠ ⎝

365

⎛ 0.06 ⎞ 6. ⎜1 + 365 ⎟⎠ ⎝

365

7. a.

⎛ 0.07 ⎞ d. (i) 1000 ⎜ 1 + 365 ⎟⎠ ⎝ ⎛ 0.07 ⎞ (ii) ⎜1 + 365 ⎟⎠ ⎝

− 1 ≈ 0.04081 or 4.081%

365(5)

− 1000 ≈ $419.02

365

− 1 ≈ 0.07250 or 7.250%

9. Let re be the effective rate. Then 2000 (1 + re ) = 2950 5

− 1 ≈ 0.06183 or 6.183%

(1 + re )5 =

A nominal rate compounded yearly is the same as the effective rate, so the effective rate is 10%.

1 + re = 5

2

b.

⎛ 0.10 ⎞ ⎜1 + 2 ⎟ − 1 = 0.1025 or 10.25% ⎝ ⎠

c.

⎛ 0.10 ⎞ ⎜1 + 4 ⎟ − 1 ≈ 0.10381 or 10.381% ⎝ ⎠

2950 2000

2950 2000

2950 −1 2000 re ≈ 0.0808 or 8.08%. re = 5

4

10. Let r be the monthly rate. Then (1 + r )84 = 1835 1835 (1 + r )84 = 1000 1835 1 + r = 84 1000 1835 −1 r = 84 1000 r = 0.0072529 This gives a nominal rate of approximately 12(0.0072529) = 0.0870 ≈ 8.70% compounded monthly.

12

d.

⎛ 0.10 ⎞ ⎜1 + 12 ⎟ ⎝ ⎠

e.

⎛ 0.10 ⎞ ⎜1 + 365 ⎟ ⎝ ⎠

8. a.

− 1 ≈ 0.10471 or 10.471% 365

− 1 ≈ 0.10516 or 10.516%

⎛ 0.07 ⎞ (i) 1000 ⎜ 1 + 4 ⎟⎠ ⎝

4(5)

− 1000 ≈ $414.78

4

⎛ 0.07 ⎞ (ii) ⎜1 + − 1 ≈ 0.07186 or 7.186% 4 ⎟⎠ ⎝

11. From Example 6, the number of years, n, is ln 2 ≈ 8.0 years. given by n = ln(1.09)

12(5)

⎛ 0.07 ⎞ b. (i) 1000 ⎜1 + 12 ⎟⎠ ⎝

− 1000 ≈ $417.63

12. From Example 6, the number of years, n, is ln 2 given by n = ≈ 14.2 years. ln(1.05)

12

⎛ 0.07 ⎞ (ii) ⎜1 + 12 ⎟⎠ ⎝

− 1 ≈ 0.07229 or 7.229%

13. 6000(1.08)7 ≈ $10, 282.95

161

Chapter 5: Mathematics of Finance

ISM: Introductory Mathematical Analysis 22. Let r be the required nominal rate.

14. 3P = P (1 + r ) n

12

r ⎞ ⎛ ⎜1 + 12 ⎟ ⎝ ⎠

3 = (1 + r )n ln 3 = n ln(1 + r) ln 3 n= ln(1 + r ) 10

15. 21,500(1.06)

17. a. b.

12

r ⎞ ⎛ ⎜1 + 12 ⎟ ⎝ ⎠ 1+

≈ $38,503.23

⎛ 0.02 ⎞ 16. 21,500 ⎜ 1 + 4 ⎟⎠ ⎝

− 1 = 0.045

40

≈ $26, 247.08

= 1.045

r 12 = 1.045 12 r 12 = 1.045 − 1 12 r = 12 ⎡12 1.045 − 1⎤ ≈ 0.0441 ⎣ ⎦

or 4.41%.

(0.015)(12) = 0.18 or 18% (1.015)12 − 1 ≈ 0.1956 or 19.56%

18. 2 P = P (1.01)n 2 = (1.01)n ln 2 = n ln(1.01) ln 2 n= ≈ 70 months ln(1.01)

⎛ 0.0475 ⎞ ⎜1 + 360 ⎟ ⎝ ⎠

365

23. a.

⎛ 0.0475 ⎞ ⎜1 + 365 ⎟ ⎝ ⎠

365

b.

− 1 ≈ 0.0493 or 4.93%

− 1 ≈ 0.0486 or 4.86%

24. Let r be the nominal rate. 8

⎛ r⎞ 801.06 = 700 ⎜1 + ⎟ ⎝ 4⎠

19. The compound amount after the first four years

1+

is 2000(1.06)4 . After the next four years the compound amount is ⎡ 2000(1.06) 4 ⎤ (1.03)8 ≈ $3198.54 . ⎣ ⎦

r 8 801.06 = 4 700

⎛ 801.06 ⎞ r = 4⎜ 8 − 1⎟ ≈ 0.0680 or 6.80% ⎜ ⎟ 700 ⎝ ⎠

20. 700 = 500(1.02) n

25. Let re = effective rate. 300, 000 = 100, 000 (1 + re )

1.4 = (1.02) ln(1.4) = n ln(1.02) ln(1.4) n= ≈ 17 quarters or 4 years, 3 months ln(1.02) n

10

(1 + re )10 = 3 1 + re = 10 3 re = 10 3 − 1 ≈ 0.1161 or 11.61%.

21. 7.8% compounded semiannually is equivalent to an effective rate of (1.039)2 − 1 = 0.079521 or 7.9521%. Thus 8% compounded annually, which is the effective rate, is the better rate.

162


Section 5.2

26. Let P = average price of such a good, n = number of days. ⎛ 0.0725 ⎞ 2 P = P ⎜1 + 365 ⎟⎠ ⎝ ⎛ 0.0725 ⎞ 2 = ⎜1 + 365 ⎟⎠ ⎝

⎛ 0.0875 ⎞ 8. 500 ⎜ 1 + 4 ⎟⎠ ⎝

n

⎛ 0.135 ⎞ 10. 1250 ⎜1 + 52 ⎟⎠ ⎝

⎛ 0.071 ⎞ 12. 12, 000 ⎜ 1 + 2 ⎟⎠ ⎝

⎛ r⎞ ⎜1 + 2 ⎟ ⎝ ⎠ 1+

28

=

= 1000

−12

≈ $11,381.89

−2

≈ $11,191.31

15. Let x be the payment 2 years from now. The equation of value at year 2 is

⎡ 50 ⎤ r = 2 ⎢ 28 − 1⎥ ≈ 0.0629 or 6.29% ⎣⎢ 21 ⎥⎦

x = 600(1.04) −2 + 800(1.04)−4

x ≈ $1238.58

28. 1000(1 − 0.01)20 = 1000(0.99) 20 ≈ $817.91 Problems 5.2 1. 6000(1.05) −20 ≈ $2261.34 2. 3500(1.06)−8 ≈ $2195.94 3. 4000(1.035)−24 ≈ $1751.83 4. 1740(1.015) −24 = $1217.21

⎛ 0.10 ⎞ 7. 8000 ⎜ 1 + 12 ⎟⎠ ⎝

≈ $1021.13

14. 550(1.025) −16 + 550(1.025)−20 ≈ $706.14

r 28 50 = 2 21

−22

≈ $5821.55

⎛ 0.10 ⎞ 6. 6000 ⎜ 1 + 2 ⎟⎠ ⎝

≈ $6838.95

13. 27, 000(1.03) −22 ≈ $14, 091.10

1000 50 = 420 21

⎛ 0.08 ⎞ 5. 9000 ⎜1 + 4 ⎟⎠ ⎝

−4(365)

−78

⎛ 0.053 ⎞ 11. 12, 000 ⎜ 1 + 12 ⎟⎠ ⎝

27. Let r = the required nominal rate.

28

≈ $385.65

⎛ 0.095 ⎞ 9. 10, 000 ⎜1 + 365 ⎟⎠ ⎝

n

⎛ 0.0725 ⎞ ln 2 = n ln ⎜1 + 365 ⎟⎠ ⎝ ln 2 n= ≈ 3489.98 days ⎛ 0.0725 ⎞ ln ⎜ 1 + 365 ⎟⎠ ⎝ or ≈ 9.56 years

⎛ r⎞ 420 ⎜ 1 + ⎟ ⎝ 2⎠

−12

−13

≈ $3181.93

−60

≈ $4862.31 163



16. Let x be the payment at the end of 5 years. The equation of value at year 5 is ⎛ 0.08 ⎞ 3000 ⎜ 1 + 12 ⎟⎠ ⎝

60

+ x = 7000 ⎛ 0.08 ⎞ x = 7000 − 3000 ⎜ 1 + 12 ⎟⎠ ⎝ x ≈ $2530.46

60

17. Let x be the payment at the end of 6 years. The equation of value at year 6 is 2000(1.025) 4 + 4000(1.025) 2 + x = 5000(1.025) + 5000(1.025) −4 x = 5000(1.025) + 5000(1.025)−4 − 2000(1.025) 4 − 4000(1.025) 2 x ≈ $3244.63.

18. Let x be the amount of each of the equal payments. The equation of value at year 3 is 1500(1.07)3 + x(1.07) 2 + x(1.07) + x = 3500(1.07) −1 + 5000(1.07) −3 x[(1.07)2 + 1.07 + 1] = 3500(1.07)−1 + 5000(1.07)−3 − 1500(1.07)3 x=

3500(1.07)−1 + 5000(1.07)−3 − 1500(1.07)3 (1.07)2 + 2.07

x ≈ $1715.44

19. a.

NPV = 8000(1.025) −6 + 10, 000(1.025) −8 + 14, 000(1.025)−12 − 25, 000 ≈ $515.62

b. Since NPV > 0, the investment is profitable. 20. a.

NPV = 8000(1.03) −6 + 10, 000(1.03) −8 + 14, 000(1.03) −12 − 25, 000 ≈ −$586.72

b. Since NPV < 0, the investment is not profitable. 21. We consider the value of each investment at the end of eight years. The savings account has a value of 10, 000(1.03)16 ≈ $16, 047.06. The business investment has a value of $16,000. Thus the better choice is the savings account.

22. The payments due B are 1000(1.07)5 at year 5 and 2000(1.04)14 at year 7. Let x be the payment at the end of 6 years. The equation of value at year 6 is x = 1000(1.07)5 (1.015)4 + 2000(1.04)14 (1.015) −4 x ≈ $4751.73 ⎛ 0.075 ⎞ 23. 1000 ⎜1 + 4 ⎟⎠ ⎝

−80

⎛ 0.058 ⎞ 24. 6500 ⎜ 1 + 360 ⎟⎠ ⎝

≈ $226.25 −1460

≈ $5137.67

164


Section 5.3

25. Let r be the nominal discount rate, compounded quarterly. Then ⎛ r⎞ 4700 = 10, 000 ⎜ 1 + ⎟ ⎝ 4⎠ 10, 000 4700 = 32 1 + 4r

(

⎛ r⎞ ⎜1 + 4 ⎟ ⎝ ⎠ 1+

b.

−32

14. With option (a), after 18 months they have 50, 000(1 + 0.0125)6 ≈ $53,869.16 with option (b), they have

)

32

=

P = 59, 081e − (0.045)(25) ≈ $19,181

50, 000e(0.045)(1.5) ≈ $53, 491.51 .

15. Effective rate = e r − 1 . Thus 0.05 = e r − 1 ,

10, 000 100 = 4700 47

e r = 1.05 , r = ln 1.05 ≈ 0.0488. Answer: 4.88%

r 32 100 = 4 47

16. If r is the annual rate compounded continuously, then at the end of 1 year the compound amount

⎡ 100 ⎤ − 1⎥ ≈ 0.0955 or 9.55% r = 4 ⎢32 ⎢⎣ 47 ⎥⎦

of a principal of P dollars is Per (1) = Per . This amount must equal the compound amount of P dollars at a nominal rate of 6% compounded

Problems 5.3

semiannually, which is P (1.03)2 . Thus

1. S = 4000e0.0625(6) ≈ $5819.97 5819.97 − 4000 = $1819.97

Per = P (1.03) 2 e r = (1.03)2 r = ln(1.03) 2 r = 2 ln 1.03 ≈ 0.0591 Answer: 5.91%

2. S = 4000e0.09(6) ≈ $6864.03 6864.03 – 4000 = $2864.03 3. P = 2500e−0.0675(8) ≈ $1456.87

17. 3P = Pe0.07t

4. P = 2500e−0.08(8) ≈ $1318.23

3 = e0.07t 0.07t = ln 3 ln 3 ≈ 16 t= 0.07 Answer: 16 years

5. e0.04 − 1 ≈ 0.0408 Answer: 4.08% 6. e0.08 − 1 ≈ 0.0833 Answer: 8.33%

18.

4 P = Per (30) 4 = e30r 30r = ln 4 ln 4 r= ≈ 0.046 30 Answer: 5%

7. e0.03 − 1 ≈ 0.0305 Answer: 3.05% 8. e0.11 − 1 = 0.1163 Answer: 11.63%

19. The accumulated amounts under each option are: 9. S = 100e0.045(2) ≈ $109.42 10. S = 1000e0.03(8) ≈ $1271.25 11. P = 1, 000, 000e −0.05(5) ≈ $778,800.78 12. P = 50, 000e −0.06(30) ≈ $8264.94 13. a.

25, 000(1 + 0.035) 25 = $59, 081 165

a.

1000e(0.035)(2) ≈ $1072.51

b.

1020(1.0175) 4 ≈ $1093.30

c.

500e(0.035)(2) + 500(1.0175) 4 ≈ 536.25 + 535.93 = $1072.18

Chapter 5: Mathematics of Finance 20. a.

ISM: Introductory Mathematical Analysis 4. The amount of profit earned in the first two years is the sum of the monthly profits. Let a = 2000, r = 1.1, and n = 24.

On Nov. 1, 2006 the accumulated amount is 10, 000e ≈ $14,918.25 . On Nov. 1, 2011 the accumulated amount is (0.04)(10)

14,918.25(1.05)5 ≈ $19, 039.89 .

b.

21. a.

s=

(

2000 1 − (1.1) 24

) ≈ 176,994.65

1 − 1.1 Thus, the company earned $176,994.65 in the first two years.

10, 000(1.045)15 ≈ $19,352.82 , which is $312.93 more than the amount in part (a). 9000(1.0125) 4 ≈ $9458.51

5. Let R = 500 and let n = 72. Then, the present value A of the annuity is given by ⎛ 1 − (1 + r ) − n ⎞ ⎛ 1 − (1 + r ) −72 ⎞ A = R⎜ ⎟ = 500 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ r r ⎝ ⎠ ⎝ ⎠ By graphing A as a function of r, we find that when r ≈ 0.005167, A = 30,000. Thus, if the present value of the annuity is $30,000, the monthly interest rate is 0.5167%, and the nominal rate is 12(0.005167) = 0.062 or 6.2%.

b. After one year the accumulated amount of the investment is 10, 000e0.055 ≈ $10,565.41 . The payoff for the loan (including interest) is 1000 + 1000(0.08) = $1080. The net return is 10,565.41 – 1080 = $9485.41. Thus, this strategy is better by 9485.41 – 9458.51 = $26.90.

50,000

Principles in Practice 5.4 1. Let a = 64 and let r =

3 . Then, the next five 4 2

⎛3⎞ ⎛3⎞ heights of the ball are 64 ⎜ ⎟ , 64 ⎜ ⎟ , 4 ⎝ ⎠ ⎝4⎠ 3

4

0 10,000

5

⎛3⎞ ⎛3⎞ ⎛3⎞ 64 ⎜ ⎟ , 64 ⎜ ⎟ , 64 ⎜ ⎟ , or 48 ft, 36 ft, ⎝4⎠ ⎝4⎠ ⎝4⎠ 3 1 27 ft, 20 ft, and 15 ft. 16 4

6. Since the man pays $2000 for 6 years and $3500 for 8 years, we can consider the payments to be an annuity of $3500 for 14 years minus an annuity of $1500 for 6 years so that the first 24 payments are $2000 each. Thus, the present value is 3500a56 0.015 − 1500a24 0.015

2. Let a = 500 and let r = 1.5. Then, the number of bacteria at the end of each minute for the first six minutes is 500(1.5), 500(1.5)2 , 500(1.5)3 , 4

5

≈ 3500(37.705879) − 1500(20.030405) = 101,924.97 Thus, the present value of the payments is $101,925. Since the man made an initial down payment of $20,000, list price was 101,925 + 20,000 = $121,925.

6

500(1.5) , 500(1.5) , 500(1.5) , or 750, 1125, 1688, 2531, 3797, 5695.

3. The total vertical distance traveled in the air after n bounces is equal to 2 times the sum of heights. 2 If a = 6 and r = , then when the ball hits the 3 ground for the twelfth time, n = 12 and the distance traveled in the air is 12 ⎞ ⎤ ⎡ ⎛ 2 ⎡ a 1− rn ⎤ ⎢ 6 ⎜1 − 3 ⎟ ⎥ ⎠⎥ ⎥ = 2⎢ ⎝ 2s = 2 ⎢ ⎢ 1− r ⎥ ⎢ ⎥ 1 − 23 ⎣⎢ ⎦⎥ ⎢ ⎥ ⎣ ⎦ ≈ 35.72 meters

(

)

0.05

0.048 = 0.012, and n = 24. 4 ⎛ 1 − (1 + r ) − n ⎞ A = R⎜ ⎟ ⎜ ⎟ r ⎝ ⎠ ⎛ 1 − (1 + 0.012) −24 ⎞ ⎛ 1 − (1.012)−24 ⎞ A = R⎜ ⎟ = R⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 0.012 0.012 ⎝ ⎠ ⎝ ⎠ By Graphing A as a function of R, we find that when R = 723.03, A = 15,000. Thus the monthly payment is $723.03 if the present value of the

7. Let r =

( )

166


Section 5.4

annuity is $15,000.

Problems 5.4

20,000

1. 64 ⎛1⎞ 64 ⎜ ⎟ = 32 ⎝2⎠ 2

0

0

⎛1⎞ 64 ⎜ ⎟ = 16 ⎝2⎠

1000

3

⎛1⎞ 64 ⎜ ⎟ = 8 ⎝2⎠

8. Find the annuity due. The man makes an initial payment of $1200 followed by an ordinary annuity of $1200 for 11 months. Thus, let 0.068 . The present R = 1200, n = 11, and r = 12 value of the annuity due is ⎛ ⎞ 1200 ⎜1 + a 0.068 ⎟ ≈ 1200(1 + 10.635005) 11 ⎜ ⎟ 12 ⎠ ⎝ ≈ 13,962.01 Thus, he should pay $13,962.01.

4

⎛1⎞ 64 ⎜ ⎟ = 4 ⎝2⎠

2. 2 2(–3) = –6 2(−3) 2 = 18 2(−3)3 = −54

3. 100 100(1.02) = 102

9. Let R = 2000 and let r = 0.057. Then, the value of the IRA at the end of 15 years, when n = 15, is given by ⎛ (1 + r )n − 1 ⎞ S = R⎜ ⎟ ⎜ ⎟ r ⎝ ⎠ ⎛ (1 + 0.057)15 − 1 ⎞ S = 2000 ⎜ ⎟ ≈ 45,502.06 ⎜ ⎟ 0.057 ⎝ ⎠ Thus, at the end of 15 years the IRA will be worth $45,502.06.

100(1.02) 2 = 104.04

4. 81 ⎛1⎞ 81⎜ ⎟ = 27 ⎝3⎠ 2

⎛1⎞ 81⎜ ⎟ = 9 ⎝3⎠ 3

⎛1⎞ 81⎜ ⎟ = 3 ⎝3⎠

10. Let R = 2000 and let r = 0.057. Since the deposits are made at the beginning of each year, the value of the IRA at the end of 15 years is given by ⎛ (1 + r )n +1 − 1 ⎞ S = R⎜ ⎟−R. ⎜ ⎟ r ⎝ ⎠ Let n = 15. ⎛ (1 + 0.057)16 − 1 ⎞ S = 2000 ⎜ ⎟ − 2000 ≈ 48, 095.67 ⎜ ⎟ 0.057 ⎝ ⎠ Thus, the IRA is worth $48,095.67 at the end of 15 years.

5. s =

4 7

( )

5 ⎡ 4 ⎤ ⎢1 − 7 ⎥ ⎣ ⎦ 1 − 74

=

4 ⎡ 15,783 ⎤ 7 ⎣ 16,807 ⎦ 3 7

=

21, 044 16,807

()

7⎤ ⎡ 1 ⎢1 − 15 ⎥ ⎦= 6. s = ⎣ 1 − 15

167

78,124 78,125 4 5

=

19,531 15, 625



⎛ 23. 1200 ⎜ s13 ⎝

1 ⎡1 − (0.1)6 ⎤ ⎦ = 1.11111 7. s = ⎣ 1 − 0.1

8. Observe that (1.1) −1 =

( )

10 ⎡1 − 10 11 ⎢ 11

⎣

1 − 10 11

9. a35

0.04

10. a15

0.07

6⎤

25. 175a

6⎤ ⎥ ⎡ ⎦ = 10 ⎢1 − ⎛ 10 ⎞ ⎥ ≈ 4.355 ⎜ ⎟ ⎢⎣ ⎝ 11 ⎠ ⎥⎦

32

⎞ − 1⎟ ≈ 1200(21.495297 − 1) ⎠ ≈ $24,594.36

0.025

⎞ − 1⎟ ≈ 600(46.000271 − 1) ⎠ ≈ $27, 000.16

0.04 12

− 25a 0.04 8 12

≈ 175(30.304595) − 25(7.881321) ≈ $5106.27

≈ 18.664613

26. 1500 + 1500a5

0.0075

12. s11

0.0125

≈ 11.713937

0.06

14. 1000a8

27. R =

≈ 8.213180

0.0075

0.05

≈ 1500 + 1500(4.889440) ≈ $8834.16

≈ 9.107914

11. s8

13. 600a6

⎛ 24. 600 ⎜ s31 ⎝

10 . 11

and r =

s=

1 10 10 = . Thus a = 1.1 11 11

0.08

5000 5000 ≈ ≈ $458.40 a12 0.015 10.907505

28. 3000 + 250a12

0.04

≈ 3000 + 250(9.385074) ≈ $5346.27

≈ 600(4.917324) ≈ $2950.39

29. a.

≈ 1000(6.463213) ≈ 6463.21

⎛ ⎞ 24 ⎜ 50s48 0.005 ⎟ (1.005) ⎝ ⎠ ≈ 50(54.097832)(1.005)24

15. 2000a18 16. 1500a15

0.02

≈ 2000(14.992031) ≈ $29,984.06

0.0075

17. 800 + 800a11

≈ $3048.85 b. 3048.85 – 48(50) = $648.85

≈ 1500(14.136995) ≈ $21, 205.49

0.035

30. Let R be the yearly payment. 275, 000 = R + Ra9 0.035

≈ 800 + 800(9.001551)

⎛ ⎞ 275, 000 = R ⎜ 1 + a9 0.035 ⎟ ⎝ ⎠ 275,000 ≈ R(8.607687), R ≈ $31,948.19

≈ $8001.24

18. 150 + 150a

59

19. 2000s36

0.07 12

0.0125

≈ 150 + 150(49.796588) ≈ $7619.49

31. R =

≈ 2000(45.115505) ≈ $90, 231.01

20. 600s16

0.02

≈ 600(18.639285) ≈ $11,183.57

21. 5000s20

0.07

22. 2000s20

0.03

≈ 5000(40.995492) ≈ $204,977.46 ≈ 2000(26.870374) ≈ 53, 740.75 168

48, 000 48, 000 ≈ ≈ $3474.12 s10 0.07 13.816448


Section 5.4

32. Let x be the purchase price. In the same manner as in Example 12, [50, 000 − 0.08x] s10 0.06 = x 50, 000 − 0.08 x = 50, 000 = 0.08 x +

⎛ x = (1.08)6 ⎜ 5000 − 1000a5 ⎝

≈ (1.08)6 [5000 − 1000(3.992710)] ≈ $1598.44

x s10

35. s60

0.06

=

0.017

x s10

36. a9

0.06

⎛ ⎞ 1 ⎟ 50, 000 = x ⎜ 0.08 + ⎜⎜ s10 0.06 ⎟⎟ ⎝ ⎠ 50, 000 50, 000 x= ≈ ≈ $320,800 . 1 0.08 + s 1 0.08 + 13.180795

=

0.052

37. 750a480

25, 000 . After s10 0.06

six years the value of the fund is 25, 000 s . s10 0.06 6 0.06

39. R =

This accumulates to ⎡ ⎤ ⎢ 25, 000 s ⎥ (1.07) 4 . 6 0.06 ⎥ ⎢s ⎢⎣ 10 0.06 ⎥⎦ Let x be the amount of the new payment. ⎡ ⎤ 25, 000 4⎥ ⎢ xs4 0.07 = 25, 000 − s6 0.06 (1.07) ⎢s ⎥ ⎢⎣ 10 0.06 ⎥⎦

40. R =

42. a.

b.

0.01375

3000(0.01375) (1.01375)20 − 1

≈ $131.34

( 0.1 ) )

25, 000 12 25, 000 = ≈ $531.18 a60 0.1 1 − 1 + 0.1 −60 12 12

(

0.10

$650(12)(15) = $117,000

650a

180

(

0.055 12

⎡ 0.055 ⎢ 1 − 1 + 12 = 650 ⎢ 0.055 ⎢ 12 ⎣ ≈ $79,551.24

)

−180

⎤ ⎥ ⎥ ⎥ ⎦

43. For the first situation, the compound amount is ⎡ ⎛ ⎞⎤ 30 ⎢ 2000 ⎜ s11 0.07 − 1⎟ ⎥ (1.07) ⎝ ⎠⎦ ⎣

34. Let x be the final payment.

0.08

=

⎡ 1 − (1.10) −19 ⎤ = 200, 000 + 200, 000 ⎢ ⎥ 0.10 ⎢⎣ ⎥⎦ ≈ $1,872,984.02

25,000 25, 000 − ⎡ 13.180795 (6.975319)(1.07)4 ⎤ ⎣ ⎦ x≈ 4.439943 x ≈ $1725

5000 − 1000a5

3000 s20

⎡ 1 − (1.0135) −480 ⎤ = 750 ⎢ ⎥ 0.0135 ⎢⎣ ⎥⎦ ≈ 55, 466.57

41. 200, 000 + 200, 000a19

⎡ ⎤ 25, 000 − ⎢ s25,000 s6 0.06 (1.07)4 ⎥ ⎣ 10 0.06 ⎦ x= s4 0.07

0.08

1 − (1.052) −9 ≈ 7.04494 0.052

⎡ (1.01)120 − 1 ⎤ = 1000 ⎢ ⎥ 0.01 0.01 ⎢⎣ ⎥⎦ ≈ 230, 038.69

38. 1000 s120 33. The original annual payment is

(1.017)60 − 1 ≈ 102.91305 0.017

0.0135

10 0.06

5000 = 1000a5

⎞ ⎠

0.08 ⎟

+ x(1.08) −6

⎡ (1.07)11 − 1 ⎤ = 2000 ⎢ − 1⎥ (1.07)30 ⎢⎣ 0.07 ⎥⎦ ≈ $225,073, so the net earnings are 225,073 – 20,000 = $205,073.

= x(1.08)−6

Thus

169



For the second situation, the compound amount is ⎡ (1.07)31 − 1 ⎤ ⎛ ⎞ 2000 ⎜ s31 0.07 − 1⎟ = 2000 ⎢ − 1⎥ ⎝ ⎠ ⎢⎣ 0.07 ⎥⎦ ≈ $202,146, so the net earnings are 202,146 – 60,000 = $142,146. 44. 100

1 − e−0.05(20) ≈ $1264 0.05

45. 40, 000

1 − e−0.04(5) ≈ $181, 269.25 0.04

Problems 5.5 1. R =

8000 8000 ≈ ≈ $273.42 a36 0.14 29.258904 12

2. A = 50a36 3. R =

0.01

≈ 50(30.107505) ≈ $1505.38

8000 8000 ≈ ≈ $236.19 a36 0.04 33.870766 12

Finance charge = 36(236.19) – 8000 = $502.84 4. a.

R=

500 a12

≈

0.0125

500 ≈ $45.13 11.079312

b. 12(45.13) – 500 = $41.56 5. a.

R=

7500 7500 ≈ ≈ $221.43 a36 0.04 33.870766 12

b. 7500 c. 6. a.

0.04 = $25 12

221.43 – 25 = $196.43 R=

35, 000 35, 000 ≈ ≈ $851.17 a48 0.078 41.119856 12

0.078 = $227.50 12

b.

35, 000

c.

851.17 − 227.50 = $623.67

170


7. R =

Section 5.5

5000 5000 ≈ ≈ $1476.14 a4 0.07 3.387211

The interest for the first period is (0.07)(5000) = $350, so the principal repaid at the end of that period is 1476.14 – 350 = $1126.14. The principal outstanding at the beginning of period 2 is 5000 – 1126.14 = $3873.86. The interest for period 2 is (0.07)(3873.86) = $271.17, so the principal repaid at the end of that period is 1476.14 – 271.17 = $1204.97. The principal outstanding at beginning of period 3 is 3873.86 – 1204.97 = $2668.89. Continuing in this manner, we construct the following amortization schedule.

Period

Prin. Outs. at Beginning

Int. for Period

Prin. Repaid at End

1

5000.00

350.00

1476.14

1126.14

2

3873.86

271.17

1476.14

1204.97

3

2668.89

186.82

1476.14

1289.32

4

1379.57

96.57

1476.14

1379.57

904.56

5904.56

5000.00

Total

8. R =

Pmt. at End

9000 a8

0.0475

≈

9000 ≈ $1378.46 6.529036

The interest for the first period is (0.0475)(9000) = $427.50, so the principal repaid at the end of that period is 1378.46 − 427.50 = $950.96. The principal outstanding at the beginning of period 2 is 9000 − 950.96 = $8049.04. The interest for period 2 is (0.0475)(8049.04) = $382.33, so the principal repaid at the end of that period is 1378.46 − 382.33 = $996.13. The principal outstanding at beginning of period 3 is 8049.04 − 996.13 = $7052.91. Continuing in this manner, we construct the following amortization schedule. Note the adjustment in the final payment. Period


Int. for Period

Pmt. at End

Prin. Repaid at End

1

9000.00

427.50

1378.46

950.96

2

8049.04

382.33

1378.46

996.13

3

7052.91

335.01

1378.46

1043.45

4

6009.46

285.45

1378.46

1093.01

5

4916.45

233.53

1378.46

1144.93

6

3771.52

179.15

1378.46

1199.31

7

2572.21

122.18

1378.46

1256.28

8

1315.93

62.51

1378.44

1315.93

2027.66

11,027.66

9000.00

Total

171


9. R =


900 900 ≈ ≈ $193.72 a5 0.025 4.645828

The interest for period 1 is (0.025)(900) = $22.50, so the principal repaid at the end of that period is 193.72 – 22.50 = $171.22. The principal outstanding at the beginning of period 2 is 900 – 171.22 = $728.78. The interest for that period is (0.025)(728.78) = $18.22, so the principal repaid at the end of that period is 193.72 – 18.22 = $175.50. The principal outstanding at the beginning of period 3 is 728.78 – 175.50 = $553.28. Continuing in this manner, we obtain the following amortization schedule. Note the adjustment in the final payment. Period


Int. for Period

Prin. Repaid at End

1

900.00

22.50

193.72

171.22

2

728.78

18.22

193.72

175.50

3

553.28

13.83

193.72

179.89

4

313.39

9.33

193.72

184.39

5

189.00

4.73

193.73

189.00

68.61

968.61

900.00

Total 10. R =

Pmt. at End

10, 000 10, 000 ≈ ≈ $2045.22 a5 0.0075 4.889440

The interest for period 1 is (0.0075)(10,000) = $75, so the principal repaid at the end of that period is 2045.22 – 75 = $1970.22. The principal outstanding at the beginning of period 2 is 10,000 – 1970.22 = $8029.78. The interest for period 2 is (0.0075)(8029.78) = $60.22, so the principal repaid at the end of that period is 2045.22 – 60.22 = $1985. The principal outstanding at the beginning of period 3 is 8029.78 – 1985 = $6044.78. Continuing in this manner, we construct the following amortization schedule. Note the adjustment in the final payment. Period


Int. for Period

Pmt. at End

1

10,000.00

75.00

2045.22

1970.22

2

8029.78

60.22

2045.22

1985.00

3

6044.78

45.34

2045.22

1999.88

4

4044.90

30.34

2045.22

2014.88

5

2030.02

15.23

2045.25

2030.02

226.13

10,226.13

10,000.00

Total 11. From Eq. (1), 100 ⎤ ln ⎡⎢ 100−1000(0.02) ⎣ ⎦⎥ ≈ 11.268 . n= ln(1.02) Thus the number of full payments is 11. 12. a.

Prin. Repaid at End

2000 2000 ≈ ≈ $52.67 a48 0.01 37.973959 172

ISM: Introductory Mathematical Analysis b.

52.67 a13

0.01

Section 5.5

≈ 52.67(12.133740)

b. 48(228.88) – 8500 = $2486.24

≈ $639.08

c.

100 ⎤ ln ⎡⎢ 100− 2000(0.015) ⎥⎦ ⎣ ≈ 23.956. Thus the 17. n = ln1.015 number of full payments is 23.

(639.08)(0.01) ≈ $6.39

d. 52.67 – 6.39 = $46.28 e.

48(52.67) – 2000 = $528.16

18. R =

18, 000 13. Each of the original payments is . a15 0.035

9500 a60

0.0077

⎡ ⎤ 0.0077 = 9500 ⎢ ⎥ −60 ⎣⎢ 1 − (1.0077) ⎦⎥ ≈ $198.31

After two years the value of the remaining ⎡ ⎤ 18, 000 ⎥ a . Thus the new payments is ⎢ ⎢a ⎥ 11 0.035 15 0.035 ⎣⎢ ⎦⎥ semi-annual payment is 18, 000a11 0.035 1 ⋅ a15 0.035 a11 0.04

19. Present value of mortgage payments is −360 ⎤ ⎡ 0.076 ⎢ 1 − 1 + 12 ⎥ 600a360 0.076 = 600 ⎢ ⎥ 0.076 12 ⎢ ⎥ 12 ⎣ ⎦ ≈ $84,976.84 This amount is 75% of the purchase price x. 0.75x = 84,976.84 x = $113,302.45 ≈ $113,302

18, 000(9.001551) 1 ⋅ 11.517411 8.760477 ≈ $1606.

20. For the 15-year mortgage, the monthly payment is ⎡ ⎤ 240, 000 0.005 ⎥ = 240, 000 ⎢ ⎢ 1 − (1 + 0.005 )−180 ⎥ a180 0.005 ⎣ ⎦

(

=

14. R =

15. a.

2000 2000(0.014) = ≈ $49.49 a60 0.014 1 − (1.014) −60

Monthly interest rate is Monthly payment is

≈ $2025.26 The finance charge is 180(2025.26) – 240,000 = $124,546.80 For the 25-year mortgage, the monthly payment is ⎡ ⎤ 240, 000 0.005 = 240, 000 ⎢ ⎥ − 300 a300 0.005 ⎣⎢ 1 − (1 + 0.005) ⎦⎥

0.092 . 12

⎡ 0.092 245, 000 ⎢ 12 = 245, 000 ⎢ a300 0.092 0.092 ⎢ 1 − 1 + 12 12 ⎣ ≈ $2089.69

(

)

⎤ ⎥ −300 ⎥ ⎥ ⎦

≈ $1546.32 The finance charge is 300(1546.32) − 240,000 = 223,896.00 Thus the savings is 223,896.00 − 124,546.80 = $99,349.20

⎛ 0.092 ⎞ b. 245,000 ⎜ ⎟ = $1878.33 ⎝ 12 ⎠

c.

21.

2089.69 – 1878.33 = $211.36

d. 300(2089.69) – 245,000 = $381,907 16. a.

Monthly interest rate is

)

25, 000 a60

0.0125

−

25, 000 a60 0.01

⎡ ⎤ 1 1 ⎢ ⎥ = 25, 000 − ⎢a ⎥ a 60 0.01 ⎥ ⎢⎣ 60 0.0125 ⎦ ⎡ ⎤ 0.0125 0.01 = 25, 000 ⎢ − ⎥ −60 1 − (1.01) −60 ⎥⎦ ⎢⎣ 1 − (1.0125) ≈ $38.64

0.132 = 0.011 . 12

Monthly payment is ⎡ ⎤ 8500 0.011 = 8500 ⎢ ⎥ − 48 a48 0.011 ⎢⎣ 1 − (1.011) ⎥⎦ ≈ $228.88

173



22. The government’s payment is ( y − x)a60 0.0925

7. a.

A = 200a13

0.04

≈ $1997.13

12

⎡ 5000 5000 =⎢ − ⎢ a 0.0925 a 0.04 60 12 ⎣⎢ 60 12 ⎡ a 0.0925 60 12 = 5000 ⎢1 − ⎢ a60 0.04 ⎢⎣ 12

⎤ ⎥a ⎥ 60 ⎦⎥

b.

0.0925 12

S = 200s13

0.04

≈ 200(16.626838) ≈ $3325.37

⎤ ⎥ ⎥ ⎥⎦

8. 150 s14

0.04

− 150 = 150(18.291911) − 150 ≈ 2593.79

⎡ 1−(1+ 0.0925 )−60 12 ⎢ 0.0925 ⎢ 12 = 5000 ⎢1 − −60 1−(1+ 0.04 12 ) ⎢ 0.04 12 ⎣⎢

(

⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥

9. 200 s13

10. 250a20

)

−60 ⎡ ⎤ 0.0925 0.04 ⎥ ⎢ 1 − 1 + 12 = 5000 ⎢1 − ⋅ −60 0.0925 ⎥ 1 − 1 + 0.04 ⎢ ⎥ 12 ⎣ ⎦ ≈ $589.89

(

≈ 200(9.985648)

11.

)

− 200 ≈ 200(13.532926) − 200

0.025

≈ $2506.59 ≈ 250(15.589162) ≈ $3897.29

5000 5000 ≈ ≈ $886.98 s5 0.06 5.637093

12. a.


0.08 12

7000 7000 ≈ ≈ $206.67 a36 0.04 33.870766 12

5

2 ⎛2⎞ 1. s = 3 + 2 + 2 ⋅ + " + 3 ⎜ ⎟ 3 ⎝3⎠ 6⎤ ⎡ 3 ⎢1 − 23 ⎥ 3 ⎡ 665 ⎤ ⎦ = ⎣ 729 ⎦ = 665 = ⎣ 2 1 81 1− 3 3

b. 36(206.67) – 7000 = $440.12

( )

13. Let x be the first payment. The equation of value now is x + 2 x(1.07) −3 = 500(1.05) −3 + 500(1.03) −8 x ⎡1 + 2(1.07) −3 ⎤ = 500(1.05)−3 + 500(1.03) −8 ⎣ ⎦

12

⎛ 0.05 ⎞ 2. ⎜1 + ⎟ 12 ⎠ ⎝

− 1 ≈ 0.0512 or 5.12%

x=

1 + 2(1.07)−3 x ≈ $314.00

3. 8.2% compounded semiannually corresponds to an effective rate of (1.041)2 − 1 = 0.083681 or 8.37%. Thus the better choice is 8.5% compounded annually.

⎡ 0.01375 ⎤ = 3500 ⎢ ⎥ −3 a3 0.01375 14. ⎣⎢ 1 − (1.01375) ⎦⎥ ≈ $1198.90 The interest for the first period is (0.01375)(3500) = $48.13, so the principal repaid at the end of that period is 1198.90 − 48.13 = $1150.77. The principal outstanding at the beginning of period 2 is 3500 − 1150.77 = $2349.23. The interest for that period is (0.01375)(2349.23) = $32.30. The principal repaid at the end of that period is 1198.90 − 32.30 = $1166.60. The principal outstanding at the beginning of period 3 is 2349.23 − 1166.60 = $1182.63. Continuing, we R=

4. NPV = 3400(1.035) −4 + 3500(1.035) −8 − 7000 ≈ −$1379.16 5. Let x be the payment at the end of 2 years. The equation of value at the end of year 2 is 1000(1.04) 4 + x = 1200(1.04) −4 + 1000(1.04) −8

x = 1200(1.04)−4 + 1000(1.04)−8 − 1000(1.04)4 ≈ $586.60

6. 250a48

0.005

500(1.05)−3 + 500(1.03) −8

≈ 250(42.580318) ≈ $10, 645.08

174

3500


Chapter 5 Review

obtain the following amortization schedule. Note the adjustment in the final payment. Period

Int. for Period


Prin. Repaid at End

1

3500.00

48.13

1198.90

1150.77

2

2349.23

32.30

1198.90

1166.60

3

1182.63

16.26

1198.89

1182.63

96.69

3596.69

3500.00

Total

15. R =

Pmt. at End

15, 000 15, 000 ≈ ≈ $3067.84 a5 0.0075 4.889440

The interest for period 1 is (0.0075)(15,000) = $112.50, so the principal repaid at the end of that period is 3067.84 – 112.50 = $2955.34. The principal outstanding at beginning of period 2 is 15,000 – 2955.34 = $12,044.66. The interest for period 2 is 0.0075(12,044.66) = $90.33, so the principal repaid at the end of that period is 3067.84 – 90.33 = $2977.51. Principal outstanding at the beginning of period 3 is 12,044.66 – 2977.51 = $9067.15. Continuing, we obtain the following amortization schedule. Note the adjustment in the final payment. Period

Int. for Period


Pmt. at End

Prin. Repaid at End

1

15,000

112.50

3067.84

2955.34

2

12,044.66

90.33

3067.84

2977.51

3

9067.15

68.00

3067.84

2999.84

4

6067.31

45.50

3067.84

3022.34

5

3044.97

22.84

3067.81

3044.97

339.17

15,339.17

15,000.00

Total

16. 540a84

(

0.10 12

)

−84 ⎤ ⎡ 0.10 ⎢ 1 − 1 + 12 ⎥ = 540 ⎢ ⎥ ≈ $32,527.80 0.10 ⎢ ⎥ 12 ⎣ ⎦

17. The monthly payment is ⎡ ⎤ 0.055 11, 000 ⎢ ⎥ 12 = 11, 000 ⎢ ⎥ ≈ $255.82 48 − a48 0.055 0.055 − + 1 1 ⎢ ⎥ 12 12 ⎣ ⎦ The finance charge is 48(255.82) – 11,000 = $1279.36

(

)

175




0.085 = 0.0425, thus R = 0.0425(25,000) = 1062.50. 2 1 − (1.0825) −25 P = 25, 000(1.0825)−25 + 1062.50 ⋅ 1.0825 − 1 ≈ $26,102.13

2.

0.065 = 0.0325 , thus R = 0.0325(10,000) = 325. 2

On a graphics calculator, let Y1 = 10,389 and Y2 = 10,000(1 + x)^ – 7 + 325(1 – (1 + x)^ – 7)/

(

)

(1 + x) − 1 .

The curves intersect at 0.0590. The yield is 5.9%. 3. The normal yield curve assumes a stable economic climate. By contrast, if investors are expecting a drop in interest rates, and with it a drop in yields from future investments, they will gladly give up liquidity for long-term investment at current, more favorable, interest rates. T-bills, which force the investor to find a new investment in a short time, are correspondingly less attractive, and so prices drop and yields rise.

176

Chapter 6 Principles in Practice 6.1 1. There are 3 rows, one for each source. There are two columns, one for each raw material. Thus, the size of the matrix is 3 × 2. Alternatively, she could use a 2 × 3 matrix. 2. The first column consists of 1’s each representing the 1 hour needed for each phase of project 1. The second column consists of 2’s for each phase of project 2 and so on. In general the nth column will consist of 2n ’s, each representing the 2n hours needed for each phase of project n. The time-analysis matrix is as follows. ⎡1 2 4 8 16 ⎤ ⎢1 2 4 8 16 ⎥ ⎢ ⎥ ⎢⎣1 2 4 8 16 ⎥⎦ Problems 6.1 1. a.

The size is the number of rows by the columns. Thus A is 2 × 3, B is 3 × 3, C is 3 × 2, D is 2 × 2, E is 4 × 4, F is 1 × 2, G is 3 × 1, H is 3 × 3, and J is 1 × 1.

b. A square matrix has the same number of rows as columns. Thus the square matrices are B, D, E, H, and J. c.

An upper triangular matrix is a square matrix where all entries below the main diagonal are zeros. Thus H and J are upper triangular. A lower triangular matrix is a square matrix where all entries above the main diagonal are zeros. Thus D and J are lower triangular.

d. A row vector (or row matrix) has only one row. Thus F and J are row vectors. e.

A column vector (or column matrix) has only one column. Thus G and J are column vectors.

2. A has 4 rows and 4 columns. Thus the order of A is 4. 3. a21 is the entry in the 2nd row and 1st column, namely 6. 4. a14 is the entry in the 1st row and 4th column, namely 6. 5. a32 is the entry in the 3rd row and 2nd column, namely 4. 6. a34 is the entry in the 3rd row and 4th column, namely 0. 7. a44 is the entry in the 4th row and 4th column, namely 0. 8. a55 is the entry in the 5th row and 5th column. But A has only 4 rows and 4 columns. Thus a55 does not exist. 9. The main diagonal entries are the entries on the diagonal extending from the upper left corner to the lower right corner. Thus the main diagonal entries are 7, 2, 1, 0. 10. ⎡ 2 ⎢0 ⎢ ⎢0 ⎢ ⎣⎢ 0

3 4 5⎤ 4 5 6 ⎥⎥ 0 6 7⎥ ⎥ 0 0 8 ⎦⎥ 177

Chapter 6: Matrix Algebra


11. ⎡ −2 ⋅1 + 3 ⋅1 −2 ⋅1 + 3 ⋅ 2 −2 ⋅1 + 3 ⋅ 3 −2 ⋅1 + 3 ⋅ 4 −2 ⋅ 1 + 3 ⋅ 5 ⎤ ⎡ 1 4 7 10 13⎤ ⎢ −2 ⋅ 2 + 3 ⋅1 −2 ⋅ 2 + 3 ⋅ 2 −2 ⋅ 2 + 3 ⋅ 3 −2 ⋅ 2 + 3 ⋅ 4 −2 ⋅ 2 + 3 ⋅ 5⎥ = ⎢ −1 2 5 8 11⎥ ⎢ ⎥ ⎢ ⎥ ⎣ −2 ⋅ 3 + 3 ⋅1 −2 ⋅ 3 + 3 ⋅ 2 −2 ⋅ 3 + 3 ⋅ 3 −2 ⋅ 3 + 3 ⋅ 4 −2 ⋅ 3 + 3 ⋅ 5 ⎦ ⎣ −3 0 3 6 9 ⎦

( (

1+1 2 ⎡ 1 + 12 12. ⎢ (−1) ⎢ 2 +1 2 2 + 12 ⎢⎣ (−1)

) )

( (

) )

(−1)1+ 2 12 + 22 ⎤ ⎡ 2 −5⎤ ⎥ =⎢ 2+ 2 2 2 ⎥ ⎣ −5 8⎥⎦ (−1) 2 +2 ⎥ ⎦

13. 12 · 10 = 120, so A has 120 entries. For a33 , i = 3 = j, so a33 = 1. Since 5 ≠ 2, a52 = 0. For a10, 10 , i = 10 = j, so a10, 10 = 1. Since 12 ≠ 10, a12, 10 = 0. 14. The main diagonal is the diagonal extending from the upper left corner to the lower right corner. a.

1, 0, –5, 2

b. x, y, z 15. A zero matrix is a matrix in which all entries are zeros.

a.

⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢⎣ 0

0 0 0⎤ 0 0 0 ⎥⎥ 0 0 0⎥ ⎥ 0 0 0 ⎥⎦

b.

⎡0 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0⎤ 0 ⎥⎥ 0⎥ ⎥ 0⎥ 0⎥ ⎥ 0 ⎥⎦

16. If A is 7 × 9, then A T is 9 × 7. T

⎡ 6 −3⎤ ⎡ 6 2⎤ =⎢ 17. A T = ⎢ ⎥ ⎥ ⎣2 4⎦ ⎣ −3 4 ⎦

18. A T = [ 2 4 6 8]

T

⎡ 2⎤ ⎢ 4⎥ =⎢ ⎥ ⎢6⎥ ⎢ ⎥ ⎣⎢ 8 ⎦⎥

3 −4 ⎤ T ⎡1 ⎡ 1 3 7 3⎤ ⎢3 2 5⎥⎥ 19. A T = ⎢⎢ 3 2 −2 0 ⎥⎥ = ⎢ ⎢ 7 −2 0 ⎥ ⎢ ⎥ ⎣⎢ −4 5 0 1⎦⎥ 1⎥⎦ ⎢⎣ 3 0 178


Section 6.1 b. From F, the entry in row 2 (deluxe) and column 3 (blue) is 3. Thus in February, 3 blue deluxe models were sold.

T

⎡ 2 −1 0 ⎤ ⎡ 2 −1 0 ⎤ ⎢ ⎥ T 20. A = ⎢ −1 5 1⎥ = ⎢⎢ −1 5 1⎥⎥ ⎢⎣ 0 1 3⎥⎦ ⎢⎣ 0 1 3⎥⎦

21. a.

c.

A and C are diagonal matrices.

b. All are them are triangular matrices. T

⎡ 2 −1 0 ⎤ ⎡ 2 −1 0 ⎤ ⎢ ⎥ 22. A = −1 5 1 = ⎢ −1 5 1⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 0 1 3⎦ ⎣ 0 1 3⎦

d. In both January and February, the deluxe blue models (row 2, column 3) sold the same number of units (3).

T

Since A T = A, the matrix of Problem 20 is symmetric.

e.

In January a total of 0 + 1 + 3 + 5 = 9 deluxe models were sold. In February a total of 2 + 3 + 3 + 2 = 10 deluxe models were sold. Thus more deluxe models were sold in February.

f.

In January a total of 2 + 0 + 2 = 4 red widgets were sold, while in February a total of 0 + 2 + 4 = 6 red widgets were sold. Thus more red widgets were sold in February.

g.

Adding all entries in matrix J yields that a total of 38 widgets were sold in January.

⎡ 1 7⎤ T ⎡ 1 0 −1⎤ ⎢ 0 0⎥ 23. A T = ⎢ = ⎢ ⎥ ⎣7 0 9 ⎥⎦ ⎣ −1 9 ⎦ T

⎡ 1 7⎤ ⎡ 1 0 −1⎤ ⎢ (A ) = 0 0⎥ = ⎢ =A ⎢ ⎥ ⎣ 7 0 9 ⎥⎦ ⎣ −1 9 ⎦ T T

24. Equating corresponding entries gives 2x = 4, y = 6, z = 0, and 3w = 7. Thus x = 2, y = 6, z = 0, 7 w= . 3

30. The sums of the entries in the columns are 680, 710, 1510, and 6690. The sum of the entries in the rows are 680, 710, 1510, and 6690. The amount an industry consumes is equal to the amount of its output. Industry B has to increase output by (0.20)(90) = 18 units and industry C has to increase output by (0.20)(120) = 24 units. All other producers have to increase it by (0.20)(420) = 84 units.

25. Equating corresponding entries gives 6 = 6, 2 = 2, x = 6, 7 = 7, 3y = 2, and 2z = 7. Thus 2 7 x = 6, y = , z = . 3 2 26. Equating entries in the 3rd row and 3rd column gives 7 = 8, which is never true, so there is no solution.

31. By equating entries we find that x must satisfy x 2 + 2000 x = 2001 and x 2 = − x . The second equation implies that x < 0. From the

27. Equating corresponding entries gives 2x = y, 7 = 7, 7 = 7, and 2y = y. Now 2y = y yields y = 0. Thus from 2x = y we get 2x = 0, so x = 0. The solution is x = 0, y = 0. 28. [125 275 ⎡ 0.95⎤ ⎢1.03 ⎥ ⎢ ⎥ ⎣1.25 ⎦ 29. a.

The entries in row 1 (regular) and column 4 (purple) give the number of purple regular models sold. For J the entry is 2 and for F the entry is 4. Thus more purple regular models were sold in February.

first equation, x 2 + 2000 x − 2001 = 0 , (x + 2001)(x – 1) = 0, so x = –2001.

400]

⎡ 3 −2 ⎤ 32. ⎢ 1⎥⎥ ⎢ −4 ⎢⎣ 5 6 ⎥⎦ ⎡3 33. ⎢ ⎢1 ⎢4 ⎢ ⎣⎢ 2

From J, the entry in row 3 (super-duper) and column 2 (white) is 7. Thus in January, 7 white super-duper models were sold.

179

1 7 3 6

1⎤ 4 ⎥⎥ 1⎥ ⎥ 2 ⎦⎥



Principles in Practice 6.2 ⎡120 80 ⎤ ⎡110 140 ⎤ 1. T = J + F = ⎢ ⎥+⎢ ⎥ ⎣105 130 ⎦ ⎣ 85 125⎦ ⎡120 + 110 80 + 140 ⎤ ⎡ 230 220 ⎤ =⎢ ⎥=⎢ ⎥ ⎣ 105 + 85 130 + 125⎦ ⎣190 255⎦ ⎡ x1 ⎤ ⎡ 40 ⎤ ⎡ 248⎤ ⎢ ⎥ ⎢ ⎥ 2. 0.8 ⎢ x2 ⎥ − ⎢ 30 ⎥ = 2 ⎢⎢319 ⎥⎥ ⎢⎣ x3 ⎥⎦ ⎢⎣ 60 ⎥⎦ ⎢⎣532 ⎥⎦ ⎡ 0.8 x1 ⎤ ⎡ 40 ⎤ ⎡ 496 ⎤ ⎢ 0.8 x ⎥ − ⎢ 30 ⎥ = ⎢ 638⎥ 2⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢⎣ 0.8 x3 ⎥⎦ ⎢⎣ 60 ⎥⎦ ⎢⎣1064 ⎥⎦ ⎡ 0.8 x1 − 40 ⎤ ⎡ 496 ⎤ ⎢ 0.8 x − 30 ⎥ = ⎢ 638⎥ 2 ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0.8 x3 − 60 ⎥⎦ ⎢⎣1064 ⎥⎦ Solve 0.8 x1 − 40 = 496 to get x1 = 670 . Solve 0.8 x2 − 30 = 638 to get x2 = 835 . Solve 0.8 x3 − 60 = 1064 to get x3 = 1405 .

Problems 6.2 −3 + 4 ⎤ ⎡ 4 −3 1⎤ 2 + 2 0 + (−3) ⎡ 2 0 −3⎤ ⎡ 2 −3 4 ⎤ ⎡ 4+6 0 + 5⎥⎥ = ⎢⎢ −2 10 5⎥⎥ 1. ⎢⎢ −1 4 0 ⎥⎥ + ⎢⎢ −1 6 5⎥⎥ = ⎢⎢ −1 + (−1) ⎢⎣ 1 −6 5⎥⎦ ⎢⎣ 9 11 −2 ⎥⎦ ⎢⎣ 1 + 9 −6 + 11 5 + (−2) ⎥⎦ ⎢⎣ 10 5 3⎥⎦ 2 + 7 + 2 −7 + (−4) + 7 ⎤ ⎡11 −4 ⎤ ⎡ 2 −7 ⎤ ⎡ 7 −4 ⎤ ⎡ 2 7 ⎤ ⎡ 2. ⎢ = ⎥ + ⎢ −2 ⎥ + ⎢ 7 2 ⎥ = ⎢ −6 + (−2) + 7 6 4 1 4 + 1 + 2 ⎥⎦ ⎢⎣ −1 7 ⎥⎦ − ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎡ 1 4 ⎤ ⎡ 6 −1⎤ ⎡ 1 − 6 4 − (−1) ⎤ ⎡ −5 5⎤ 7 − 2⎥⎥ = ⎢⎢ −9 5⎥⎥ 3. ⎢⎢ −2 7 ⎥⎥ − ⎢⎢ 7 2 ⎥⎥ = ⎢⎢ −2 − 7 ⎢⎣ 6 9 ⎥⎦ ⎢⎣ 1 0 ⎥⎦ ⎢⎣ 6 − 1 9 − 0⎥⎦ ⎢⎣ 5 9⎥⎦ ⎡1 6⎤ ⎢ 2 ⋅ 4 ⎡ 4 −2 1⎢ 4. 2 10 −12 ⎥⎥ = ⎢ 12 ⋅ 2 ⎢ 2⎢ ⎢⎣ 0 0 8⎥⎦ ⎢ 1 ⋅ 0 ⎣2

1 (−2) 2 1 ⋅10 2 1 ⋅0 2

⎤ ⎥ ⎡ 2 −1 3⎤ 1 ( −12) ⎥ = ⎢ 1 5 −6 ⎥⎥ 2 ⎥ ⎢ 0 4 ⎥⎦ 1 ⋅ 8 ⎥ ⎢⎣ 0 2 ⎦ 1 ⋅6 2

5. 2[2 −1 3] + 4[−2 0 1] − 0[2 3 1] = [4 −2 6] + [−8 0 4] − [0 0 0] = [4 − 8 − 0 −2 + 0 − 0 6 + 4 − 0] = [−4 −2 10] 6.

[7

7 ] is a matrix and 66 is a number, so the sum is not defined.

180


⎡1 2 ⎤ 7. ⎢ ⎥ has size 2 × 2, and ⎣3 4 ⎦

Section 6.2

⎡7 ⎤ ⎢ 2 ⎥ has size 2 × 1. Thus the sum is not defined. ⎣ ⎦

⎡ 2 −1⎤ ⎡ 0 0 ⎤ ⎡ 2 −1⎤ ⎡ 0 0 ⎤ ⎡ 2 −1⎤ 8. ⎢ + 3⎢ ⎥ ⎥=⎢ ⎥+⎢ ⎥=⎢ ⎥ ⎣7 4⎦ ⎣0 0⎦ ⎣7 4 ⎦ ⎣ 0 0⎦ ⎣7 4⎦ 1⎤ ⎡ −6 ⋅ 2 −6(−6) −6 ⋅ 7 −6 ⋅ 1⎤ ⎡ −12 36 −42 −6 ⎤ ⎡ 2 −6 7 9. −6 ⎢ = =⎢ ⎥ 1 6 −2 ⎦ ⎣ −6 ⋅ 7 −6 ⋅1 –6 ⋅ 6 −6(−2) ⎥⎦ ⎢⎣ −42 −6 −36 12 ⎥⎦ ⎣7 ⎡ 1 −1⎤ ⎡ −6 9 ⎤ ⎡ 1 −1⎤ ⎡ −18 ⎢2 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − 3 ⎢ 2 6⎥ = ⎢ 2 0⎥ − ⎢ 6 10. ⎢ ⎢ 3 −6 ⎥ ⎢ 1 −2 ⎥ ⎢ 3 −6 ⎥ ⎢ 3 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣⎢ 4 9 ⎦⎥ ⎣⎢ 4 5⎥⎦ ⎣⎢ 4 9 ⎦⎥ ⎣⎢ 12

27 ⎤ ⎡ 19 −28⎤ 18⎥⎥ ⎢⎢ −4 −18⎥⎥ = −6 ⎥ ⎢ 0 0⎥ ⎥ ⎢ ⎥ 15⎦⎥ ⎣⎢ −8 −6 ⎦⎥

⎡ 1 −5 0 ⎤ ⎡10 0 30 ⎤ ⎡ 1 −5 0 ⎤ ⎡ 2 0 6 ⎤ ⎡ 3 −5 6 ⎤ 1⎢ ⎢ ⎥ 11. ⎢ −2 7 0 ⎥ + ⎢ 0 5 0 ⎥⎥ = ⎢⎢ −2 7 0 ⎥⎥ + ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ −2 8 0 ⎥⎥ 5 ⎣⎢ 4 6 10 ⎦⎥ ⎣⎢ 5 20 25⎥⎦ ⎢⎣ 4 6 10 ⎦⎥ ⎣⎢ 1 4 5 ⎦⎥ ⎣⎢ 5 10 15⎦⎥ ⎡ 1 0 0 ⎤ ⎛ ⎡ 1 2 0 ⎤ ⎡ 4 −2 2 ⎤ ⎞ ⎡ 3 12. 3 ⎢ 0 1 0 ⎥ − 3 ⎜ ⎢0 −2 1⎥ − ⎢ −3 21 −9 ⎥ ⎟ = ⎢ 0 ⎢ ⎥ ⎜⎜ ⎢ ⎥ ⎢ ⎥⎟ ⎢ 1 0 ⎦ ⎠⎟ ⎣ 0 ⎣ 0 0 1⎦ ⎝ ⎣0 0 1⎦ ⎣ 0 ⎡3 = ⎢0 ⎢ ⎣0 = ⎡ 12 ⎢ −9 ⎢ ⎣ 0

0 0⎤ ⎡ −3 3 0⎥ − 3 ⎢ 3 ⎥ ⎢ 0 3⎦ ⎣ 0 0 0 ⎤ ⎡ −9 3 0⎥ − ⎢ 9 ⎥ ⎢ 0 3⎦ ⎣ 0 6⎤ −12 72 −30 ⎥ ⎥ 3 0⎦

4 −2 ⎤ −23 10 ⎥ ⎥ −1 1⎦ 12 −6 ⎤ −69 30 ⎥ ⎥ −3 3⎦

⎡ −6 −5⎤ ⎡ −6 −5⎤ ⎡ −1(−6) −1(−5) ⎤ ⎡ 6 5⎤ = (−1) ⎢ 13. −B = − ⎢ ⎥ ⎥=⎢ ⎥=⎢ ⎥ ⎣ 2 −3⎦ ⎣ 2 −3⎦ ⎣ −1(2) −1(−3) ⎦ ⎣ −2 3⎦ 1 − (−5) ⎤ ⎡ 2 − (−6) ⎡8 6 ⎤ ⎡ −8 −6 ⎤ 14. −( A − B) = − ⎢ ⎥ = − ⎢1 0 ⎥ = ⎢ −1 0 ⎥ 3 − 2 − 3 − ( − 3) ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡0 0⎤ ⎡ 2 ⋅ 0 2 ⋅ 0⎤ ⎡0 0⎤ 15. 2O = 2 ⎢ ⎥=⎢ ⎥=⎢ ⎥=O ⎣0 0⎦ ⎣ 2 ⋅ 0 2 ⋅ 0⎦ ⎣0 0⎦ ⎡ 2 − (−6) + (−2) 1 − (−5) + (−1) ⎤ ⎡ 6 5⎤ 16. A − B + C = ⎢ ⎥=⎢ ⎥ ⎣ 3 − 2 + (−3) −3 − (−3) + 3⎦ ⎣ −2 3⎦ ⎧⎪ ⎡ 4 2 ⎤ ⎡ −18 −15⎤ ⎫⎪ 1⎤ ⎡ −6 −5⎤ ⎪⎫ ⎡ 22 17 ⎤ ⎡ 66 51⎤ ⎪⎧ ⎡ 2 17. 3(2A – 3B) = 3 ⎨2 ⎢ ⎥ − 3 ⎢ 2 −3⎥ ⎬ = 3 ⎨ ⎢ 6 −6 ⎥ − ⎢ 6 −9 ⎥ ⎬ = 3 ⎢ 0 3⎥ = ⎢ 0 9⎥ − 3 3 ⎪⎩ ⎣ ⎦ ⎣ ⎦ ⎪⎭ ⎦ ⎣ ⎦ ⎭⎪ ⎣ ⎦ ⎣ ⎦ ⎩⎪ ⎣ ⎡ −4 −4 ⎤ ⎡ 0 0 ⎤ 18. 0(A + B) = 0 ⎢ ⎥=⎢ ⎥=O ⎣ 5 −6 ⎦ ⎣ 0 0 ⎦ 181



19. 3(A – C) is a 2 × 2 matrix and 6 is a number. Therefore 3(A – C) + 6 is not defined. 1⎤ ⎡ –2 + (−6) −1 + (−5) ⎤ ⎡ 2 + (−8) 1 + (−6) ⎤ ⎡ −6 −5⎤ ⎡2 = = 20. A + (C + B) = ⎢ ⎥ + ⎢ −3 + 2 − 3 3 3 + (−3) ⎥⎦ ⎢⎣3 + (−1) −3 + 0 ⎥⎦ ⎢⎣ 2 −3⎥⎦ ⎣ ⎦ ⎣ 1⎤ ⎡ −6 −5⎤ ⎡2 ⎡ −2 −1⎤ − 3⎢ + 2⎢ 21. 2B − 3A + 2C = 2 ⎢ ⎥ ⎥ ⎥ ⎣ 2 −3⎦ ⎣ 3 −3⎦ ⎣ −3 3⎦ 3⎤ ⎡ −4 −2 ⎤ ⎡ −12 −10 ⎤ ⎡ 6 =⎢ −⎢ ⎥ ⎥+⎢ ⎥ ⎣ 4 −6 ⎦ ⎣ 9 −9 ⎦ ⎣ −6 6 ⎦ ⎡ −18 −13⎤ ⎡ −4 −2 ⎤ ⎡ −22 −15⎤ =⎢ + = 3⎥⎦ ⎢⎣ −6 6 ⎥⎦ ⎢⎣ −11 9 ⎥⎦ ⎣ −5 ⎡ −6 −3⎤ ⎡ −12 −10 ⎤ ⎡ 6 7 ⎤ 22. 3C − 2B = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣ −9 9 ⎦ ⎣ 4 −6 ⎦ ⎣ −13 15⎦

23.

⎧⎪ ⎡ −6 −5⎤ 1⎤ ⎡ −2 −1⎤ ⎫⎪ 1 1 ⎡2 A − 2(B + 2C) = ⎢ ⎥ − 2 ⎨ ⎢ 2 −3⎥ + 2 ⎢ −3 3⎥ ⎬ − 3 3 2 2⎣ ⎦ ⎦ ⎣ ⎦ ⎪⎭ ⎩⎪ ⎣ 1 ⎡1 ⎤ ⎪⎧ ⎡ −6 −5⎤ ⎡ −4 −2 ⎤ ⎪⎫ 2⎥ =⎢ − 2 ⎨⎢ ⎥+⎢ ⎥⎬ 3 3 ⎢ ⎥ ⎪⎩ ⎣ 2 −3⎦ ⎣ −6 6 ⎦ ⎪⎭ ⎣2 − 2⎦ 29 ⎤ 1⎤ 1⎤ ⎡ 21 ⎡1 ⎡ −10 −7 ⎤ ⎡⎢ 1 2⎥ 2 ⎥ ⎡ −20 −14 ⎤ 2 ⎥ ⎢ =⎢ −2⎢ = − = ⎥ ⎢3 ⎢ −8 ⎥ ⎢ 19 3 15 ⎢3 − 3⎥ − 4 3 ⎥ 6 ⎣ ⎦ ⎣2 − 2⎦ ⎣ ⎦ ⎣ 2 − 2 ⎥⎦ 2⎦ ⎣2

24.

1⎤ 1⎤ ⎡ 41 61 ⎤ ⎡1 ⎡ −8 −6 ⎤ ⎡⎢ 1 1 2⎥ 2 ⎥ ⎡ 40 30 ⎤ ⎢ 2⎥ −5⎢ = + = A − 5(B + C) = ⎢ ⎥ ⎢3 ⎢ 5 0 ⎥ ⎢ 13 3 1 0 ⎢3 − 3⎥ − ⎥ 2 ⎣ ⎦ ⎣2 − 2⎦ ⎣ ⎦ ⎣ 2 − 32 ⎥⎦ 2⎦ ⎣2

⎡ −4 −4 ⎤ ⎡ −12 −12 ⎤ 25. 3( A + B) = 3 ⎢ ⎥=⎢ ⎥ ⎣ 5 −6 ⎦ ⎣ 15 −18⎦ 3⎤ ⎡ −18 −15⎤ ⎡ −12 −12 ⎤ ⎡6 3A + 3B = ⎢ ⎥+⎢ ⎥=⎢ ⎥ ⎣ 9 −9 ⎦ ⎣ 6 −9 ⎦ ⎣ 15 −18⎦ Thus 3(A + B) = 3A + 3B. 5⎤ ⎡10 26. (2 + 3) A = 5A = ⎢ ⎥ ⎣15 −15⎦ 3⎤ ⎡10 5⎤ ⎡ 4 2⎤ ⎡6 2A + 3A = ⎢ ⎥ + ⎢9 −9 ⎥ = ⎢15 −15⎥ − 6 6 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Thus (2 + 3)A = 2A + 3A. ⎡ 2k 27. k1 ( k2 A ) = k1 ⎢ 2 ⎣ 3k2 ⎡ 2k k

( k1k2 ) A = ⎢ 3k1k2

k2 ⎤ ⎡ 2k1k2 = −3k2 ⎥⎦ ⎢⎣ 3k1k2

k1k2 ⎤ −3k1k2 ⎥⎦

k1k2 ⎤ −3k1k2 ⎥⎦

⎣ 1 2 Thus k1 ( k2 A ) = ( k1k2 ) A.

182


Section 6.2

⎛ ⎡2 1⎤ ⎡ −12 −10 ⎤ ⎡ −2 −1⎤ ⎞ ⎡ 12 10 ⎤ ⎡ 12k 10k ⎤ −⎢ +⎢ 28. k ( A − 2B + C) = k ⎜⎜ ⎢ ⎟⎟ = k ⎢ ⎥ ⎥ ⎥ ⎥=⎢ ⎥ ⎣ −4 6 ⎦ ⎣ − 4 k 6 k ⎦ ⎝ ⎣ 3 −3⎦ ⎣ 4 −6 ⎦ ⎣ −3 3⎦ ⎠ k ⎤ ⎡ −12k −10k ⎤ ⎡ −2k − k ⎤ ⎡ 12k 10k ⎤ ⎡ 2k kA − 2kB + kC = ⎢ + = ⎥−⎢ −6k ⎥⎦ ⎢⎣ −3k 3k ⎥⎦ ⎢⎣ −4k 6k ⎥⎦ ⎣ 3k −3k ⎦ ⎣ 4k Thus k(A − 2B + C) = kA − 2kB + kC. ⎡ 1 2 ⎤ ⎡ 1 1⎤ ⎡ 3 6 ⎤ ⎡ 1 1⎤ ⎡ 4 7 ⎤ 29. 3A + D = 3 ⎢⎢ 0 −1⎥⎥ + ⎢⎢ 2 0 ⎥⎥ = ⎢⎢ 0 −3⎥⎥ + ⎢⎢ 2 0⎥⎥ = ⎢⎢ 2 −3⎥⎥ ⎢⎣ 7 0 ⎥⎦ ⎢⎣ −1 2 ⎥⎦ ⎢⎣ 21 0 ⎥⎦ ⎢⎣ −1 2⎥⎦ ⎢⎣ 20 2 ⎥⎦ T

T

T

⎡ 0 3⎤ ⎡ 0 3⎤ ⎪⎧ ⎡ 1 3⎤ ⎡1 0 ⎤ ⎪⎫ 30. (B − C) = ⎨ ⎢ ⎥ − ⎢1 2 ⎥ ⎬ = ⎢ 3 −3⎥ = ⎢ 3 −3⎥ − 4 1 ⎦ ⎣ ⎦ ⎪⎭ ⎣ ⎦ ⎣ ⎦ ⎩⎪ ⎣ T

⎡ 1 4⎤ ⎡ 1 1 ⎤ ⎡ 2 8 ⎤ ⎡ 3 3 ⎤ ⎡ −1 5 ⎤ 31. 2BT − 3CT = 2 ⎢ − 3⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎥ ⎥ =⎢ ⎣3 −1⎦ ⎣ 0 2 ⎦ ⎣ 6 −2 ⎦ ⎣ 0 6 ⎦ ⎣ 6 −8⎦ ⎡ 1 3⎤ ⎡ 1 4 ⎤ ⎡ 2 6⎤ ⎡ 1 4 ⎤ ⎡ 3 10 ⎤ 32. 2B + BT = 2 ⎢ ⎥ ⎥+⎢ ⎥=⎢ ⎥+⎢ ⎥ =⎢ ⎣ 4 −1⎦ ⎣3 −1⎦ ⎣ 8 −2⎦ ⎣3 −1⎦ ⎣11 −3⎦ T

⎡1 0 ⎤ ⎡1 2 −1⎤ −⎢ 33. C − D = ⎢ is impossible because CT and D are not of the same size. ⎥ ⎥ ⎣1 2 ⎦ ⎣1 0 2 ⎦ T

34.

(

D − 2AT

)

T

T

⎧⎪ ⎡1 2 −1⎤ ⎡ 1 0 7 ⎤ ⎫⎪ = ⎨⎢ −2⎢ ⎥ ⎥⎬ ⎣ 2 −1 0 ⎦ ⎪⎭ ⎩⎪ ⎣1 0 2 ⎦ T

⎧⎪ ⎡1 2 −1⎤ ⎡ 2 0 14 ⎤ ⎫⎪ ⎡ −1 2 −15⎤ = ⎨⎢ ⎥ − ⎢ 4 −2 0 ⎥ ⎬ = ⎢ −3 2 1 0 2 2 ⎥⎦ ⎦ ⎣ ⎦ ⎪⎭ ⎣ ⎩⎪ ⎣ ⎡ −1 −3⎤ = ⎢⎢ 2 2 ⎥⎥ ⎢⎣ −15 2 ⎥⎦

T

⎡ 3⎤ ⎡ −4 ⎤ ⎡2⎤ 35. x ⎢ ⎥ − y ⎢ ⎥ = 3 ⎢ ⎥ ⎣2⎦ ⎣ 7⎦ ⎣4⎦ ⎡ 3x ⎤ ⎡ −4 y ⎤ ⎡ 6⎤ ⎡ 3 x + 4 y ⎤ ⎡ 6 ⎤ ⎢ 2 x ⎥ − ⎢ 7 y ⎥ = ⎢12⎥ = ⎢ 2 x − 7 y ⎥ = ⎢12 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Equating corresponding entries gives ⎧3x + 4 y = 6 ⎨2 x − 7 y = 12 ⎩

Multiply the first equation by 2 and the second equation by −3 to get ⎧ 6 x + 8 y = 12 ⎨−6 x + 21y = −36 ⎩ Now add the two equations to get

183



29 y = −24 24 y=− 29 Therefore ⎛ 24 ⎞ 270 3x = 6 − 4 ⎜ − ⎟ = ⎝ 29 ⎠ 29 90 x= 29 90 24 The solution is x = , y=− . 29 29

10 ⎡2⎤ ⎡ −1⎤ ⎡ 0⎤ ⎡ ⎤ ⎥ 6 40. x ⎢⎢ 0 ⎥⎥ + 2 ⎢⎢ 0 ⎥⎥ + y ⎢⎢ 2 ⎥⎥ = ⎢⎢ ⎥ ⎢⎣ 2 ⎥⎦ ⎢⎣ 6 ⎥⎦ ⎢⎣ −5⎥⎦ ⎢⎣ 2 x + 12 − 5 y ⎥⎦ 10 ⎡ 2x − 2 ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ = 2 6 y ⎢ ⎥ ⎢ ⎥ ⎢⎣ 2 x + 12 − 5 y ⎥⎦ ⎢⎣ 2 x + 12 − 5 y ⎥⎦

2x − 2 = 10, 2x = 12, or x = 6. 2y = 6 or y = 3. 2x + 12 − 5y = 2x + 12 − 5y, which is true for all values of x and y. Thus x = 6, y = 3.

⎡ 2 x − 4 y ⎤ ⎡16 ⎤ 36. ⎢ ⎥=⎢ ⎥ ⎣ 5 x + 7 y ⎦ ⎣ − 3⎦ ⎡ 2 x ⎤ ⎡ −4 y ⎤ ⎡16 ⎤ ⎢ 5 x ⎥ + ⎢ 7 y ⎥ = ⎢ −3⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ 30 50 ⎤ ⎡ 15 25 ⎤ 41. X + Y = ⎢⎢800 720 ⎥⎥ + ⎢⎢960 800 ⎥⎥ ⎢⎣ 25 30 ⎥⎦ ⎢⎣ 10 5 ⎥⎦ 50 + 25 ⎤ ⎡ 45 75⎤ ⎡ 30 + 15 ⎢ ⎥ ⎢ = ⎢800 + 960 720 + 800 ⎥ = ⎢1760 1520 ⎥⎥ ⎢⎣ 25 + 10 30 + 5 ⎥⎦ ⎢⎣ 35 35⎥⎦

⎡2⎤ ⎡ −4 ⎤ ⎡ 16 ⎤ x⎢ ⎥+ y⎢ ⎥ = ⎢ ⎥ 5 ⎣ ⎦ ⎣ 7 ⎦ ⎣ −3⎦ ⎡ x⎤ ⎡ −2 ⎤ ⎡ 6⎤ 37. 3 ⎢ ⎥ − 3 ⎢ ⎥ = 4 ⎢ ⎥ ⎣ y⎦ ⎣ 4⎦ ⎣ −2 ⎦ ⎡ 3x + 6 ⎤ ⎡ 24 ⎤ ⎢3 y − 12 ⎥ = ⎢ −8⎥ ⎣ ⎦ ⎣ ⎦ 3x + 6 = 24, 3x = 18, or x = 6. 4 3y – 12 = –8, 3y = 4, or y = . 3 4 Thus x = 6, y = . 3

⎡ 380 330 220 ⎤ ⎡ 400 350 150 ⎤ 42. 2B − A = 2 ⎢ ⎥−⎢ ⎥ ⎣ 460 320 750 ⎦ ⎣ 450 280 850 ⎦ ⎡ 2 ⋅ 380 2 ⋅ 330 2 ⋅ 220 ⎤ ⎡ 400 350 150 ⎤ =⎢ ⎥−⎢ ⎥ ⎣ 2 ⋅ 460 2 ⋅ 320 2 ⋅ 750 ⎦ ⎣ 450 280 850 ⎦ ⎡760 =⎢ ⎣920 ⎡360 =⎢ ⎣ 470

⎡ x⎤ ⎡ 7⎤ ⎡− x ⎤ 38. 3 ⎢ ⎥ − 4 ⎢ ⎥ = ⎢ ⎥ ⎣2⎦ ⎣− y ⎦ ⎣2 y ⎦ ⎡3x − 28⎤ ⎡ − x ⎤ ⎢ 6 + 4 y ⎥ = ⎢2 y ⎥ ⎣ ⎦ ⎣ ⎦ 3x – 28 = –x, 4x = 28, or x = 7. 6 + 4y = 2y, 2y = –6, or y = –3. Thus x = 7, y = –3.

660 440 ⎤ ⎡ 400 350 150 ⎤ − 640 1500 ⎥⎦ ⎢⎣ 450 280 850 ⎥⎦ 310 290 ⎤ 360 650 ⎥⎦

43. P + 0.1P = [ p1 p2 p3 ] + [0.1 p1 0.1 p2 0.1 p3 ] = [1.1 p1 1.1 p2 1.1 p3 ] = 1.1P Thus P must be multiplied by 1.1. 44. ( A − B)T = [ A + (−1)B]T [definition of subtraction] = A T + [(−1)B]T [transpose of a sum] = A T + (−1)BT [transpose of a scalar multiple]

⎡ 2⎤ ⎡ x ⎤ ⎡ −10⎤ ⎢ ⎥ 39. ⎢ 4 ⎥ + 2 ⎢⎢ y ⎥⎥ = ⎢⎢ −24⎥⎥ ⎣⎢ 6 ⎦⎥ ⎣⎢ 4 z ⎦⎥ ⎣⎢ 14 ⎦⎥

= A T − BT [definition of subtraction] ⎡15 −4 26 ⎤ 45. ⎢ ⎥ ⎣ 4 7 30 ⎦

⎡ 2 + 2 x ⎤ ⎡ −10 ⎤ ⎢ 4 + 2 y ⎥ = ⎢ −24 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 6 + 8 z ⎥⎦ ⎢⎣ 14 ⎥⎦ 2 + 2x = –10, 2x = –12, or x = –6. 4 + 2y = –24, 2y = –28, or y = –14. 6 + 8z = 14, 8z = 8, or z = 1. Thus x = –6, y = –14, z = 1.

⎡ −16 −11 −24 ⎤ 46. ⎢ ⎥ ⎣ −16 −3 −36 ⎦

184


Section 6.3 5. c31 = 0(0) + 4(−2) + 3(3) = 1

⎡ −10 22 12 ⎤ 47. ⎢ ⎥ ⎣ 24 36 −44 ⎦

6. c12 = 1(−2) + 3(4) + (−2)(1) = 8


7. A is 2 × 3 and E is 3 × 2, so AE is 2 × 2; 2 · 2 = 4 entries.

1. Represent the value of each book by [ 28 22 16] and the number of each book by

8. D is 4 × 3 and E is 3 × 2, so DE is 4 × 2; 4 · 2 = 8 entries.

⎡100 ⎤ ⎢ 70 ⎥ . ⎢ ⎥ ⎢⎣ 90 ⎥⎦

9. E is 3 × 2 and C is 2 × 5, so EC is 3 × 5; 3 · 5 = 15 entries.

The total value is given by the following matrix product. ⎡100 ⎤ [ 28 22 16] ⎢⎢ 70 ⎥⎥ = [2800 + 1540 + 1440] ⎢⎣ 90 ⎥⎦ = [5780] The total value is $5780.

10. D is 4 × 3 and B is 3 × 1, so DB is 4 × 1; 4 · 1 = 4 entries. 11. F is 2 × 3 and B is 3 × 1, so FB is 2 × 1; 2 · 1 = 2 entries. 12. B is 3 × 1 and C is 2 × 5. Because the number of columns of B does not equal the number of rows of C, BC is not defined.

2. The total cost is given by the matrix product PQ. ⎡ 250 ⎤ PQ = [ 26.25 34.75 28.50] ⎢⎢ 325 ⎥⎥ ⎢⎣175 ⎥⎦

13. E is 3 × 2, ET is 2 × 3, and B is 3 × 1, so EET B is 3 × 1; 3 ⋅ 1 = 3 entries.

= [ 6562.5 + 11, 293.75 + 4987.5] = [22,843.75] The total cost is $22,843.75.

14. A is 2 × 3 and E is 3 × 2, so AE is 2 × 2. Thus E(AE) is 3 × 2; 3 · 2 = 6 entries. 15. E is 3 × 2. F is 2 × 3 and B is 3 × 1, so FB is 2 × 1. Thus E(FB) is 3 × 1; 3 · 1 = 3 entries.

3. First, write the equations with the variable terms on the left-hand side. 8 8 ⎧ ⎪⎪ y + 5 x = 5 ⎨ ⎪y + 1 x = 5 ⎪⎩ 3 3 ⎡8⎤ ⎡1 8 ⎤ ⎡ y⎤ 5 5⎥ Let A = ⎢ , X = ⎢ ⎥ , and B = ⎢ ⎥ . 5⎥ ⎢ ⎢1 1 ⎥ x ⎣ ⎦ ⎣ 3⎦ ⎣3⎦ Then the pair of lines is equivalent to the matrix ⎡1 8 ⎤ ⎡ y ⎤ ⎡ 8 ⎤ 5 5⎥ equation AX = B or ⎢ = ⎢ ⎥. ⎢1 1 ⎥ ⎢⎣ x ⎥⎦ ⎢ 5 ⎥ ⎣ 3⎦ ⎣3⎦

16. Both F and A are 2 × 3, so F + A is 2 × 3. Because B is 3 × 1, (F + A)B is 2 × 1; 2 · 1 = 2 entries. 17. An identity matrix is a square matrix (in this case 4 × 4) with 1's on the main diagonal and all other entries 0's. ⎡1 0 0 0 ⎤ ⎢0 1 0 0⎥ ⎥ I4 = ⎢ ⎢0 0 1 0⎥ ⎢ ⎥ ⎢⎣ 0 0 0 1 ⎥⎦

Problems 6.3

⎡1 ⎢0 ⎢ ⎢0 18. I 6 = ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

1. c11 = 1(0) + 3(−2) + (−2)(3) = −12 2. c23 = −2(3) + 1(−2) + (−1)(−1) = −7 3. c32 = 0(−2) + 4(4) + 3(1) = 19 4. c33 = 0(3) + 4(−2) + 3(−1) = −11 185

0 0 0 0 0⎤ 1 0 0 0 0 ⎥⎥ 0 1 0 0 0⎥ ⎥ 0 0 1 0 0⎥ 0 0 0 1 0⎥ ⎥ 0 0 0 0 1⎥⎦



⎡ 2 −4 ⎤ ⎡ 4 0 ⎤ ⎡ 2(4) + (−4)(−1) 2(0) + (−4)(3) ⎤ ⎡12 −12 ⎤ 19. ⎢ = ⎥⎢ ⎥=⎢ 3(0) + 2(3) ⎥⎦ ⎢⎣10 6 ⎥⎦ ⎣ 3 2 ⎦ ⎣ −1 3⎦ ⎣3(4) + 2(−1) ⎡ −1 1⎤ ⎡ −1(1) + 1(3) −1(−2) + 1(4) ⎤ ⎡ 2 6 ⎤ ⎡ 1 −2 ⎤ ⎢ ⎢ ⎥ 20. ⎢ 0 4 ⎥ ⎢ = 0(1) + 4(3) 0(−2) + 4(4) ⎥⎥ = ⎢⎢12 16 ⎥⎥ 3 4 ⎥⎦ ⎢ ⎣ ⎢⎣ 2(1) + 1(3) 2(−2) + 1(4) ⎥⎦ ⎢⎣ 5 0 ⎥⎦ ⎣⎢ 2 1⎦⎥ ⎡1 ⎤ ⎡ 2 0 3⎤ ⎢ ⎥ ⎡ 2(1) + 0(4) + 3(7) ⎤ ⎡ 23⎤ 21. ⎢ ⎥ ⎢4⎥ = ⎢ ⎥=⎢ ⎥ ⎣ −1 4 5⎦ ⎢7 ⎥ ⎣ −1(1) + 4(4) + 5(7) ⎦ ⎣50 ⎦ ⎣ ⎦ ⎡0⎤ ⎢1 ⎥ 22. [1 0 6 2] ⎢ ⎥ = [1(0) + 0(1) + 6(2) + 2(3)] = [18] ⎢2⎥ ⎢ ⎥ ⎣⎢ 3 ⎦⎥ ⎡ 1 4 −1⎤ ⎡ 2 1 23. ⎢⎢ 0 0 2 ⎥⎥ ⎢⎢ 0 −1 ⎣⎢ −2 1 1⎥⎦ ⎢⎣ 1 1 ⎡1(2) + 4(0) + (−1)1 = ⎢⎢ 0(2) + 0(0) + 2(1) ⎢⎣ −2(2) + 1(0) + 1(1)

0⎤ 1⎥⎥ 2 ⎦⎥ 1(1) + 4(−1) + (−1)(1) 1(0) + 4(1) + (−1)(2) ⎤ ⎡ 1 −4 2 ⎤ 0(1) + 0(−1) + 2(1) 0(0) + 0(1) + 2(2) ⎥⎥ = ⎢⎢ 2 2 4 ⎥⎥ −2(1) + 1(−1) + 1(1) −2(0) + 1(1) + 1(2) ⎥⎦ ⎢⎣ −3 −2 3⎥⎦

⎡ 4 2 −2 ⎤ ⎡ 3 1 1 24. ⎢⎢ 3 10 0 ⎥⎥ ⎢⎢ 0 0 0 ⎣⎢ 1 0 2 ⎦⎥ ⎣⎢ 0 1 0 ⎡ 4(3) + 2(0) + (−2)(0) ⎢ = ⎢ 3(3) + 10(0) + 0(0) ⎢⎣ 1(3) + 0(0) + 2(0) ⎡12 2 4 −2 ⎤ = ⎢⎢ 9 3 3 0 ⎥⎥ ⎢⎣ 3 3 1 2 ⎥⎦

0⎤ 0 ⎥⎥ 1⎦⎥

4(1) + 2(0) + (−2)(1) 4(1) + 2(0) + (−2)(0) 4(0) + 2(0) + (−2)(1) ⎤ 3(1) + 10(0) + 0(1) 3(1) + 10(0) + 0(0) 3(0) + 10(0) + 0(1) ⎥⎥ 1(1) + 0(0) + 2(0) 1(0) + 0(0) + 2(1) ⎥⎦ 1(1) + 0(0) + 2(1)

⎡ 1 5 −2 −1⎤ 1⎥ 25. [1 − 2 5] ⎢ 0 0 2 ⎢ ⎥ 1 −3⎦ ⎣ −1 0 = [1 + 0 − 5 5 + 0 + 0 −2 − 4 + 5 −1 − 2 − 15] = [−4 5 −1 −18]

26. The first matrix is 1 × 2 and the second is 3 × 2, so the product is not defined.

186


Section 6.3

6 −4 6⎤ ⎡ 2⎤ ⎡ 2(2) 2(3) 2(−2) 2(3) ⎤ ⎡ 4 ⎢ 3⎥ ⎢ 3(2) ⎥ ⎢ 6 9 6 9 ⎥⎥ 3(3) 3( − 2) 3(3) − ⎥ =⎢ 27. ⎢ ⎥ [ 2 3 −2 3] = ⎢ ⎢ −4 ⎥ ⎢ −4(2) −4(3) −4(−2) −4(3) ⎥ ⎢ −8 −12 8 −12 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 3 −2 3⎥⎦ 1(3) 1(−2) 1(3) ⎥⎦ ⎢⎣ 2 ⎢⎣ 1⎥⎦ ⎢⎣ 1(2) ⎡ 0 1⎤ ⎪⎧ ⎡1 0 28. ⎢ ⎥ ⎨⎢ ⎣ 2 3⎦ ⎩⎪ ⎣1 1 ⎡ 0(1) + 1(1) =⎢ ⎣ 2(1) + 3(1)

1 ⎤ ⎡ 0 1 0 ⎤ ⎪⎫ ⎡ 0 1⎤ ⎡1 1 1⎤ + ⎬= 0 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎭⎪ ⎢⎣ 2 3⎥⎦ ⎢⎣1 1 1⎥⎦ 0(1) + 1(1) 0(1) + 1(1) ⎤ ⎡1 1 1 ⎤ = 2(1) + 3(1) 2(1) + 3(1) ⎥⎦ ⎢⎣5 5 5⎥⎦

⎡1 ⎡ −1 0 2 ⎤ ⎪⎫ ⎢ ⎪⎧ ⎡ −2 0 2 ⎤ 29. 3 ⎨ ⎢ ⎥ + 2 ⎢ 1 1 −2 ⎥ ⎬ ⎢3 ⎪⎩ ⎣ 3 −1 1⎦ ⎣ ⎦ ⎪⎭ ⎢5 ⎣ ⎡1 ⎪⎧ ⎡ −2 0 2 ⎤ ⎡ −2 0 4 ⎤ ⎪⎫ ⎢ = 3 ⎨⎢ + ⎥ ⎢ ⎥ ⎬ ⎢3 ⎩⎪ ⎣ 3 −1 1⎦ ⎣ 2 2 −4 ⎦ ⎭⎪ ⎢5 ⎣ 1 2 ⎡ ⎤ ⎪⎧ ⎡ −4 0 6 ⎤ ⎪⎫ ⎢ ⎥ = ⎡ −12 0 = 3 ⎨⎢ 3 4 ⎥⎬ ⎢ ⎥ ⎢ ⎩⎪ ⎣ 5 1 −3⎦ ⎭⎪ ⎢5 6 ⎥ ⎣ 15 3 ⎣ ⎦

2⎤ 4 ⎥⎥ 6 ⎥⎦ 2⎤ 4 ⎥⎥ 6 ⎥⎦

⎡1 2 ⎤ 18⎤ ⎢ 3 4 ⎥⎥ −9 ⎥⎦ ⎢ ⎢⎣5 6 ⎥⎦ ⎡ −12(1) + 0(3) + 18(5) −12(2) + 0(4) + 18(6) ⎤ ⎡ 78 84 ⎤ =⎢ ⎥=⎢ ⎥ ⎣15(1) + 3(3) + (−9)(5) 15(2) + 3(4) + (−9)(6) ⎦ ⎣ −21 −12 ⎦

⎡ 1 −1⎤ ⎡ −1 0 −1 0 0 ⎤ 30. ⎢ ⎣ 0 3⎥⎦ ⎢⎣ 2 1 2 1 1⎥⎦ ⎡1(−1) + (−1)(2) 1(0) + (−1)(1) 1(−1) + (−1)(2) 1(0) + (−1)(1) 1(0) + (−1)(1) ⎤ =⎢ 0(0) + 3(1) 0(−1) + 3(2) 0(0) + 3(1) 0(0) + 3(1) ⎥⎦ ⎣ 0(−1) + 3(2) = ⎡ −3 1 −3 −1 −1⎤ ⎢⎣ 6 3 6 3 3 ⎥⎦ ⎧ ⎡ 1 −2 ⎤ ⎫ 1⎤ ⎢ ⎡1 2 ⎤ ⎪ ⎡ 2 0 ⎪ ⎡1 2⎤ ⎪⎧ ⎡ 2 + 0 + 3 −4 + 0 + 0 ⎤ ⎪⎫ 1⎥⎥ ⎬ = ⎢ 31. ⎢ ⎥ ⎨ ⎢ 1 0 −2 ⎥ ⎢ 2 ⎥ ⎨⎢ ⎥⎬ 3 4 ⎣ ⎦ ⎪⎣ ⎦ ⎢ 3 0 ⎥ ⎪ ⎣3 4⎦ ⎩⎪ ⎣1 + 0 − 6 −2 + 0 + 0⎦ ⎭⎪ ⎣ ⎦⎭ ⎩ ⎡1 2 ⎤ ⎡ 5 −4 ⎤ ⎡ 5 − 10 −4 − 4 ⎤ ⎡ −5 −8⎤ =⎢ ⎥⎢ ⎥=⎢ ⎥=⎢ ⎥ ⎣3 4 ⎦ ⎣ −5 −2 ⎦ ⎣15 − 20 −12 − 8⎦ ⎣ −5 −20⎦ ⎧⎪ ⎡ 1 0 ⎤ ⎡ −2 4⎤ ⎫⎪ ⎡ 3 6 ⎤ ⎧⎪ ⎡ −2 4 ⎤ ⎫⎪ ⎡ 1 2⎤ 32. 3 ⎢ − 4 ⎨⎢ − 4 ⎨I ⎢ ⎬=⎢ ⎥ ⎥ ⎢ ⎥ ⎥ ⎥⎬ ⎪⎩ ⎣0 1⎦ ⎣ 6 1⎦ ⎪⎭ ⎣ −3 12 ⎦ ⎪⎩ ⎣ 6 1⎦ ⎪⎭ ⎣ −1 4 ⎦ ⎧⎪ ⎡ −2 4 ⎤ ⎫⎪ ⎡ 3 6 ⎤ ⎡ −8 16 ⎤ ⎡ 11 −10 ⎤ ⎡ 3 6⎤ =⎢ − 4 ⎨⎢ ⎥ ⎥⎬ = ⎢ ⎥−⎢ ⎥=⎢ 8⎥⎦ ⎣ −3 12 ⎦ ⎩⎪ ⎣ 6 1⎦ ⎭⎪ ⎣ −3 12 ⎦ ⎣ 24 4 ⎦ ⎣ −27

187



⎡0 0 1 ⎤ ⎡ x ⎤ ⎡0 ⋅ x + 0 ⋅ y + 1⋅ z ⎤ ⎡ z ⎤ 33. ⎢⎢ 0 1 0 ⎥⎥ ⎢⎢ y ⎥⎥ = ⎢⎢ 0 ⋅ x + 1 ⋅ y + 0 ⋅ z ⎥⎥ = ⎢⎢ y ⎥⎥ ⎢⎣1 0 0 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣1 ⋅ x + 0 ⋅ y + 0 ⋅ z ⎥⎦ ⎢⎣ x ⎥⎦ a ⎤ ⎡ x ⎤ ⎡a x + a x ⎤ ⎡a 34. ⎢ 11 12 ⎥ ⎢ 1 ⎥ = ⎢ 11 1 12 2 ⎥ ⎣ a21 a22 ⎦ ⎣ x2 ⎦ ⎣ a21 x1 + a22 x2 ⎦

⎡ x1 ⎤ ⎡ 2 1 3 ⎤ ⎢ ⎥ ⎡ 2 x1 + x2 + 3x3 ⎤ 35. ⎢ ⎥ ⎥ ⎢ x2 ⎥ = ⎢ ⎣ 4 9 7 ⎦ ⎢ x ⎥ ⎣ 4 x1 + 9 x2 + 7 x3 ⎦ ⎣ 3⎦ ⎡ 2 −3⎤ ⎡ 2 x1 − 3 x2 ⎤ ⎡ x1 ⎤ ⎢ ⎢ ⎥ ⎥ 1⎥ ⎢ ⎥ = ⎢ x2 36. ⎢ 0 ⎥ x2 ⎦ ⎣ ⎢⎣ 2 ⎢⎣ 2 x1 + x2 ⎥⎦ 1⎥⎦ ⎡1 0 0 ⎤ ⎡3 0 0⎤ 1 1 1⎢ ⎢ ⎥ 37. D − EI = D − E = ⎢0 1 1 ⎥ − ⎢ 0 6 0 ⎥⎥ 3 3 3 ⎢⎣ 0 0 3 ⎦⎥ ⎣⎢1 2 1 ⎦⎥ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ = ⎢⎢0 1 1 ⎥⎥ − ⎢⎢0 2 0 ⎥⎥ ⎢⎣1 2 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ ⎡0 0 0 ⎤ = ⎢⎢0 −1 1⎥⎥ ⎢⎣ 1 2 0 ⎥⎦ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ ⎡1 + 0 + 0 0 + 0 + 0 0 + 0 + 0⎤ ⎡ 1 0 0 ⎤ 38. DD = ⎢⎢ 0 1 1 ⎥⎥ ⎢⎢ 0 1 1 ⎥⎥ = ⎢⎢ 0 + 0 + 1 0 + 1 + 2 0 + 1 + 1 ⎥⎥ = ⎢⎢ 1 3 2⎥⎥ ⎢⎣1 2 1 ⎥⎦ ⎢⎣1 2 1 ⎥⎦ ⎢⎣1 + 0 + 1 0 + 2 + 2 0 + 2 + 1⎥⎦ ⎢⎣ 2 4 3⎥⎦ ⎡ −1 1⎤ 3 0⎤ ⎢ ⎡ 1 −2 ⎤ ⎡ −2 ⎥ 39. 3A − 2BC = 3 ⎢ ⎥ − 2 ⎢ 1 −4 1⎥ ⎢ 0 3⎥ 0 3 ⎣ ⎦ ⎣ ⎦ ⎢ 2 4⎥ ⎣ ⎦ ⎡ 3 −6 ⎤ ⎡ 2 + 0 + 0 −2 + 9 + 0 ⎤ =⎢ −2⎢ ⎥ ⎥ ⎣0 9 ⎦ ⎣ −1 + 0 + 2 1 − 12 + 4 ⎦ ⎡ 3 −6 ⎤ ⎡ 4 14 ⎤ ⎡ −1 −20 ⎤ =⎢ ⎥−⎢ ⎥=⎢ 23⎥⎦ ⎣0 9 ⎦ ⎣ 2 −14 ⎦ ⎣ −2 ⎡4 0 0⎤ 3 0⎤ ⎢ ⎡ −2 ⎡ −8 + 0 + 0 0 + 21 + 0 0 + 3 + 0 ⎤ 0 7 1 ⎥⎥ = ⎢ 40. B(D + E) = ⎢ ⎥ ⎥ ⎢ ⎣ 1 −4 1⎦ ⎢ 1 2 4 ⎥ ⎣ 4 + 0 + 1 0 − 28 + 2 0 − 4 + 4 ⎦ ⎣ ⎦ 21 3⎤ ⎡ −8 =⎢ 5 − 26 0 ⎥⎦ ⎣ 188


Section 6.3

⎡ 1 0 0⎤ ⎢3 ⎥ ⎡3 0 0⎤ 2 2 41. 3I − FE = 3I − ⎢ 0 16 0 ⎥ ⎢⎢ 0 6 0 ⎥⎥ ⎥ 3 3⎢ ⎢ 0 0 1 ⎥ ⎢⎣ 0 0 3 ⎥⎦ 3⎦ ⎣ 1 ⎡ ⋅3+ 0 + 0 0 + 0 + 0 0 + 0 + 0 ⎤ ⎢3 ⎥ 2⎢ 1 ⋅6 + 0 ⎥ 0+0+0 0 0 0 = 3I − + + 6 ⎥ 3⎢ ⎢ 0 + 0 + 0 0 + 0 + 0 0 + 0 + 1 ⋅ 3⎥ 3 ⎦ ⎣ ⎡ 2 0 0⎤ ⎡ 7 ⎡1 0 0 ⎤ ⎡ 3 0 0 ⎤ ⎢ 3 ⎥ ⎢3 2⎢ ⎥ ⎢ ⎥ 2 ⎢ = 3I − ⎢ 0 1 0 ⎥ = ⎢ 0 3 0 ⎥ − 0 3 0 ⎥ = ⎢ 0 ⎢ ⎥ ⎢ 3 ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 3 ⎥⎦ ⎢ 0 0 23 ⎥ ⎢⎢ 0 ⎣ ⎦ ⎣ ⎡1 ⎢3 42. FE(D − I ) = ⎢ 0 ⎢ ⎢⎣ 0 ⎡1 = ⎢0 ⎢ ⎣0 = ⎡0 ⎢0 ⎢ ⎣1

0 0⎤ 3 ⎥⎡ 1 0 ⎢0 ⎥ 6 ⎥⎢ 0 13 ⎥⎦ ⎣ 0 0 0⎤ ⎡0 1 0⎥ ⎢0 ⎥⎢ 0 1⎦ ⎣ 1 0 0⎤ 0 1⎥ ⎥ 2 0⎦

0 7 3

0

0⎤ ⎥ 0⎥ ⎥ 7⎥ 3 ⎥⎦

0 0⎤ ⎛ ⎡ 1 0 0⎤ ⎡ 1 0 0⎤ ⎞ 6 0 ⎥ ⎜ ⎢0 1 1⎥ − ⎢ 0 1 0 ⎥ ⎟ ⎥⎜⎢ ⎥ ⎢ ⎥⎟ 0 3⎦ ⎜⎝ ⎣ 1 2 1⎦ ⎣ 0 0 1⎦ ⎟⎠ 0 0⎤ 0 1⎥ ⎥ 2 0⎦

⎧ ⎡1 0 0 ⎤ ⎡ −1 1⎤ ⎫ ⎡ −1 + 0 + 0 ⎪⎢ ⎪ ⎥ ⎢ ⎥ 43. (DC) A = ⎨ ⎢0 1 1 ⎥ ⎢ 0 3⎥ ⎬ A = ⎢⎢ 0 + 0 + 2 ⎪ ⎢1 2 1 ⎥ ⎢ 2 4 ⎥ ⎪ ⎢⎣ −1 + 0 + 2 ⎦⎣ ⎦⎭ ⎩⎣ 2 + 3 ⎤ ⎡ −1 ⎡ −1 1⎤ ⎡ −1 + 0 ⎡ 1 −2 ⎤ ⎢ ⎢ ⎥ = ⎢ 2 7⎥ ⎢ = 2 + 0 −4 + 21⎥⎥ = ⎢⎢ 2 0 3⎥⎦ ⎢ ⎣ ⎢⎣ 1 11⎥⎦ ⎢⎣ 1 + 0 −2 + 33⎥⎦ ⎢⎣ 1

1 + 0 + 0⎤ 0 + 3 + 4 ⎥⎥ A 1 + 6 + 4 ⎥⎦ 5⎤ 17 ⎥⎥ 31⎥⎦

⎧ ⎡ –1 1⎤ ⎫ 3 0⎤ ⎢ ⎡ 2 + 0 + 0 −2 + 9 + 0 ⎤ ⎡ 1 − 2 ⎤ ⎡ 2 7 ⎤ ⎪ ⎡ −2 ⎪ 44. A(BC) = A ⎨ ⎢ 0 3⎥⎥ ⎬ = A ⎢ ⎥ ⎥=⎢ ⎢ 3⎥⎦ ⎢⎣ 1 −7 ⎥⎦ ⎣ −1 + 0 + 2 1 − 12 + 4 ⎦ ⎣ 0 ⎪ ⎣ 1 −4 1⎦ ⎢ 2 4 ⎥ ⎪ ⎣ ⎦⎭ ⎩ ⎡ 2 − 2 7 + 14 ⎤ ⎡ 0 21⎤ =⎢ ⎥=⎢ ⎥ ⎣ 0 + 3 0 − 21⎦ ⎣ 3 −21⎦

45. Impossible: A is not a square matrix, so A 2 is not defined. ⎡ 1 0⎤ ⎡ 1 −1 0 ⎤ ⎡ 1 −1 0 ⎤ ⎢ ⎢ ⎥ ⎥ 46. A A = ⎢ −1 1⎥ ⎢ ⎥ = ⎢ −1 2 1⎥ 0 1 1 ⎦ ⎢ 0 1 1⎥ ⎢⎣ 0 1⎥⎦ ⎣ ⎣ ⎦ T

189



2

⎡ 0 0 −1⎤ ⎡ 0 0 −1⎤ ⎡ 0 0 −1⎤ ⎢ ⎥ 3 2 47. B = B B = ⎢ 2 −1 0 ⎥ B = ⎢⎢ 2 −1 0 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ B ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎡ 0 0 −2 ⎤ ⎡ 0 0 −1⎤ ⎡ 0 0 −4 ⎤ = ⎢⎢ −2 1 −2 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ = ⎢⎢ 2 −1 −2 ⎥⎥ ⎢⎣ 0 0 4 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 8⎥⎦

( )

⎡ 0 2 0⎤ ⎡ 0 2 0⎤ ⎢ ⎥⎢ ⎥ 48. A(B ) C = A ⎢ 0 −1 0 ⎥ ⎢ 0 −1 0 ⎥ C ⎢⎣ −1 0 2 ⎥⎦ ⎢⎣ −1 0 2 ⎥⎦ ⎡ 0 −2 0 ⎤ = A ⎢⎢ 0 1 0 ⎥⎥ C ⎢⎣ −2 −2 4 ⎥⎦ ⎡ 0 −2 0 ⎤ ⎡ 1 −1 0 ⎤ ⎢ 0 1 0 ⎥⎥ C =⎢ ⎥ ⎢ ⎣ 0 1 1⎦ ⎢ −2 −2 4 ⎥ ⎣ ⎦ T 2

⎡ 1 0⎤ ⎡ 0 −3 0 ⎤ ⎢ ⎥ =⎢ ⎥ ⎢ 2 −1⎥ ⎣ −2 −1 4 ⎦ ⎢ 0 1⎥ ⎣ ⎦ ⎡ −6 3⎤ =⎢ ⎥ ⎣ −4 5 ⎦ ⎛ ⎡ 1 0 0⎤ ⎡ 1 0⎤ ⎞ ⎡ 1 −1 0 ⎤ ⎢ 0 1 0 ⎥ ⎢ 2 −1⎥ ⎟ 49. ( AIC)T = ⎜ ⎢ ⎜⎜ ⎣0 1 1⎥⎦ ⎢ ⎥⎢ ⎥ ⎟⎟ ⎣ 0 0 1⎦ ⎣ 0 1⎦ ⎠ ⎝ ⎛ ⎡1 = ⎜⎢ ⎜⎜ ⎣0 ⎝ ⎡ −1 =⎢ ⎣ 2 ⎡ −1 =⎢ ⎣ 1

50. A

T

( 2C ) T

⎡ 1 0⎤ ⎞ −1 0 ⎤ ⎢ 2 −1⎥ ⎟ 1 1⎥⎦ ⎢ ⎥ ⎟⎟ ⎣ 0 1⎦ ⎠ T 1⎤ 0 ⎥⎦ 2⎤ 0 ⎥⎦

T

⎡ 1 0⎤ ⎡ 2 4 0⎤ ⎡2 4 0⎤ ⎢ ⎢ ⎥ = ⎢ −1 1⎥ ⎢ = −2 −6 2 ⎥⎥ 0 −2 2 ⎥⎦ ⎢ ⎣ ⎢⎣ 0 1⎥⎦ ⎢⎣ 0 −2 2 ⎥⎦ T

51.

( BA ) T

T

T

T

⎧ ⎡ 0 0 −1⎤ ⎡ 1 0 ⎤ ⎫ ⎡ 0 −1⎤ ⎡ 0 3 0⎤ ⎪⎢ ⎪ ⎥ ⎢ ⎥ ⎢ = ⎨ ⎢ 2 −1 0 ⎥ ⎢ −1 1⎥ ⎬ = ⎢ 3 −1⎥⎥ = ⎢ ⎥ ⎣ −1 −1 2 ⎦ ⎪ ⎢ 0 0 2 ⎥ ⎢ 0 1⎥ ⎪ ⎢ ⎥ 0 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭

190


T

Section 6.3

T

⎧ ⎡ 0 0 −1⎤ ⎫ ⎡ 0 0 −2 ⎤ ⎡ 0 4 0⎤ ⎪ ⎥ ⎢ ⎥ ⎢ ⎥ T ⎪ ⎢ (2 ) 2 2 1 0 4 2 0 52. B =⎨ ⎢ − ⎥⎬ = ⎢ − ⎥ = ⎢ 0 −2 0 ⎥ ⎪ ⎢0 0 2⎥ ⎪ ⎢⎣ 0 0 4 ⎥⎦ ⎢⎣ −2 0 4 ⎥⎦ ⎦⎭ ⎩ ⎣ 2

⎡2 0 0⎤ ⎡2 2 2 2 53. (2I ) − 2I = (2I ) − 2I = ⎢⎢ 0 2 0 ⎥⎥ − ⎢⎢ 0 ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 ⎡2 0 0⎤ ⎡2 0 0⎤ ⎡2 0 0⎤ ⎡4 = ⎢⎢ 0 2 0 ⎥⎥ ⎢⎢ 0 2 0 ⎥⎥ − ⎢⎢ 0 2 0 ⎥⎥ = ⎢⎢ 0 ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0

0 0⎤ 2 0 ⎥⎥ 0 2 ⎥⎦ 0 0⎤ ⎡2 0 0⎤ ⎡2 0 0⎤ 4 0 ⎥⎥ − ⎢⎢ 0 2 0 ⎥⎥ = ⎢⎢ 0 2 0 ⎥⎥ 0 4 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦

⎡ 1 0 0⎤ 54. A T is 3 × 2, CT is 2 × 3, and B is 3 × 3, so A T CT B is 3 × 3 and ( A T CT B)0 = I = ⎢ 0 1 0 ⎥ . ⎢ ⎥ ⎣ 0 0 1⎦ ⎡ 1 −1 0 ⎤ 55. A(I – O) = A(I) = AI. Since I is 3 × 3 and A has three columns, AI = A. Thus A(I − O) = A = ⎢ ⎥. ⎣ 0 1 1⎦ ⎡0 0 0 ⎤ 56. I O = IO = O = ⎢⎢0 0 0 ⎥⎥ ⎢⎣0 0 0 ⎥⎦ T

⎡ 0 0 −1⎤ ⎡ 1 −1 0 ⎤ ⎢ ⎡ −2 1 −1⎤ T 2 −1 0 ⎥⎥ ( AB)T = ⎢ 57. ( AB)( AB) = ⎢ ⎥ ⎥ ( AB) ⎢ − 0 1 1 2 1 2 ⎣ ⎦ ⎢ 0 0 2⎥ ⎣ ⎦ ⎣ ⎦ ⎡ −2 2 ⎤ ⎡ −2 1 −1⎤ ⎢ ⎡ 6 −7 ⎤ =⎢ 1 −1⎥⎥ = ⎢ ⎥ ⎢ 9 ⎥⎦ ⎣ 2 −1 2 ⎦ ⎢ −1 2 ⎥ ⎣ −7 ⎣ ⎦ T

58. B 2 − 3B + 2I ⎡ 0 0 −1⎤ ⎡ 0 0 −1⎤ ⎡0 = ⎢⎢ 2 −1 0 ⎥⎥ ⎢⎢ 2 −1 0 ⎥⎥ − 3 ⎢⎢ 2 ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 ⎡ 0 0 −2 ⎤ ⎡ 0 0 −3⎤ ⎡ 2 = ⎢⎢ −2 1 −2 ⎥⎥ − ⎢⎢ 6 −3 0 ⎥⎥ + ⎢⎢ 0 ⎢⎣ 0 0 4 ⎥⎦ ⎢⎣ 0 0 6 ⎥⎦ ⎢⎣ 0

0 −1⎤ ⎡1 0 0⎤ −1 0 ⎥⎥ + 2 ⎢⎢ 0 1 0 ⎥⎥ ⎢⎣ 0 0 1 ⎥⎦ 0 2 ⎥⎦ 0 0⎤ 2 0 ⎥⎥ 0 2 ⎥⎦

1⎤ 1⎤ ⎡ 2 0 0 ⎤ ⎡ 2 0 ⎡ 0 0 ⎢ ⎢ ⎥ ⎢ ⎥ = ⎢ −8 4 −2 ⎥ + ⎢ 0 2 0 ⎥ = ⎢ −8 6 −2 ⎥⎥ ⎢⎣ 0 0 −2 ⎥⎦ ⎢⎣ 0 0 2 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦

59. AX = B 1⎤ ⎡3 A=⎢ ⎥ − 2 9 ⎣ ⎦ 191



⎡ x⎤ X=⎢ ⎥ ⎣ y⎦ ⎡6⎤ B=⎢ ⎥ ⎣5⎦ 1⎤ ⎡ x ⎤ ⎡6 ⎤ ⎡3 The system is represented by ⎢ ⎥⎢ ⎥ = ⎢ ⎥ . ⎣ 2 −9 ⎦ ⎣ y ⎦ ⎣5 ⎦

60. AX = B ⎡ 3 1 1⎤ A = ⎢⎢ 1 −1 1⎥⎥ ⎢⎣5 −1 2 ⎥⎦ ⎡ x⎤ X = ⎢⎢ y ⎥⎥ ⎢⎣ z ⎥⎦ ⎡ 2⎤ B = ⎢⎢ 4 ⎥⎥ ⎢⎣12 ⎥⎦ ⎡ 3 1 1⎤ ⎡ x ⎤ ⎡ 2 ⎤ The system is represented by ⎢ 1 −1 1⎥ ⎢ y ⎥ = ⎢ 4 ⎥ . ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣5 −1 2 ⎦ ⎣ z ⎦ ⎣12 ⎦

61. AX = B ⎡ 2 −1 3⎤ A = ⎢⎢ 5 −1 2 ⎥⎥ ⎢⎣ 3 −2 2 ⎥⎦ ⎡r ⎤ X = ⎢⎢ s ⎥⎥ ⎢⎣ t ⎥⎦ ⎡ 9⎤ B = ⎢⎢ 5⎥⎥ ⎢⎣11⎥⎦ ⎡ 2 −1 3⎤ ⎡ r ⎤ ⎡ 9 ⎤ The system is represented by ⎢⎢ 5 −1 2 ⎥⎥ ⎢⎢ s ⎥⎥ = ⎢⎢ 5⎥⎥ . ⎢⎣ 3 −2 2 ⎥⎦ ⎢⎣ t ⎥⎦ ⎢⎣11⎥⎦

62. “the/falcon/has/landed” converted to corresponding numbers and slashes is “20, 8, 5/ 6, 1, 12, 3, 15, 14/ 8, 1, 19/ 12, 1, 14, 4, 5, 4.” Taking the numbers two at a time as 2 × 1 matrices and multiplying them by E gives: ⎡ 1 3 ⎤ ⎡ 20 ⎤ ⎡ 1 ⋅ 20 + 3 ⋅ 8 ⎤ ⎡ 20 + 24⎤ ⎡ 44⎤ ⎢ 2 4 ⎥ ⎢ 8⎥ = ⎢ 2 ⋅ 20 + 4 ⋅ 8⎥ = ⎢ 40 + 32 ⎥ = ⎢ 72 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ 1 3 ⎤ ⎡5 ⎤ ⎡ 1 ⋅ 5 + 3 ⋅ 6 ⎤ ⎡ 5 + 18 ⎤ ⎡ 23⎤ ⎢ 2 4 ⎥ ⎢6 ⎥ = ⎢ 2 ⋅ 5 + 4 ⋅ 6 ⎥ = ⎢10 + 24⎥ = ⎢34 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

192


⎡1 ⎢2 ⎣ ⎡1 ⎢2 ⎣

3 ⎤ ⎡ 1⎤ ⎡ 1 ⋅1 + 3 ⋅12 ⎤ ⎡ 1 + 36 ⎤ ⎡37 ⎤ = = = 4 ⎥⎦ ⎢⎣12 ⎥⎦ ⎢⎣ 2 ⋅1 + 4 ⋅12 ⎥⎦ ⎢⎣ 2 + 48⎥⎦ ⎢⎣50 ⎥⎦

⎡1 ⎢2 ⎣ ⎡1 ⎢2 ⎣

3 ⎤ ⎡14 ⎤ ⎡ 1 ⋅14 + 3 ⋅ 8 ⎤ ⎡14 + 24 ⎤ ⎡ 38 ⎤ = = = 4 ⎥⎦ ⎢⎣ 8⎥⎦ ⎢⎣ 2 ⋅14 + 4 ⋅ 8⎥⎦ ⎢⎣ 28 + 32 ⎥⎦ ⎢⎣ 60 ⎥⎦

⎡1 ⎢2 ⎣ ⎡1 ⎢2 ⎣

3 ⎤ ⎡12 ⎤ ⎡ 1 ⋅12 + 3 ⋅1 ⎤ ⎡12 + 3 ⎤ ⎡15 ⎤ = = = 4 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢⎣ 2 ⋅12 + 4 ⋅1⎥⎦ ⎢⎣ 24 + 4⎥⎦ ⎢⎣ 28⎥⎦

Section 6.3

3 ⎤ ⎡ 3⎤ ⎡ 1 ⋅ 3 + 3 ⋅15 ⎤ ⎡ 3 + 45 ⎤ ⎡ 48⎤ = = = 4 ⎥⎦ ⎢⎣15⎥⎦ ⎢⎣ 2 ⋅ 3 + 4 ⋅15⎥⎦ ⎢⎣6 + 60 ⎥⎦ ⎢⎣ 66 ⎥⎦

3 ⎤ ⎡ 1⎤ ⎡ 1 ⋅1 + 3 ⋅19 ⎤ ⎡ 1 + 57 ⎤ ⎡ 58⎤ = = = 4 ⎥⎦ ⎢⎣19 ⎥⎦ ⎢⎣ 2 ⋅1 + 4 ⋅19 ⎥⎦ ⎢⎣ 2 + 76 ⎥⎦ ⎢⎣ 78⎥⎦

3 ⎤ ⎡14 ⎤ ⎡ 1 ⋅14 + 3 ⋅ 4 ⎤ ⎡14 + 12 ⎤ ⎡ 26 ⎤ = = = 4 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ 2 ⋅14 + 4 ⋅ 4 ⎥⎦ ⎢⎣ 28 + 16⎥⎦ ⎢⎣ 44 ⎥⎦

⎡ 1 3 ⎤ ⎡ 5⎤ ⎡ 1 ⋅ 5 + 3 ⋅ 4 ⎤ ⎡ 5 + 12⎤ ⎡17 ⎤ ⎢ 2 4 ⎥ ⎢ 4 ⎥ = ⎢ 2 ⋅ 5 + 4 ⋅ 4 ⎥ = ⎢10 + 16 ⎥ = ⎢ 26 ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ The encoded message is 44, 72, 23/ 34, 37, 50, 48, 66, 38/ 60, 58, 78/ 15, 28, 26, 44, 17, 26. ⎡ 55 ⎤ 63. [ 6 10 7 ] ⎢⎢150 ⎥⎥ = [6 ⋅ 55 + 10 ⋅150 + 7 ⋅ 35] ⎢⎣ 35 ⎥⎦ = [330 + 1500 + 245] = [2075] The value of the inventory is $2075. ⎡100 ⎤ ⎢150 ⎥ ⎥ = [240, 000] 64. [ 200 300 500 250] ⎢ ⎢ 200 ⎥ ⎢ ⎥ ⎢⎣300 ⎥⎦ The total cost of the stocks is $240,000.

65. Q = [5 ⎡5 R = ⎢⎢ 7 ⎢⎣ 6

2 4] 20 16 7 17 ⎤ 18 12 9 21⎥⎥ 25 8 5 13⎥⎦ ⎡ 2500 ⎤ ⎢ 1200 ⎥ ⎢ ⎥ C = ⎢ 800 ⎥ ⎢ ⎥ ⎢ 150 ⎥ ⎢⎣ 1500 ⎥⎦

193



⎡5 ⋅ 2500 + 20 ⋅1200 + 16 ⋅ 800 + 7 ⋅150 + 17 ⋅1500 ⎤ QRC = Q(RC) = Q ⎢⎢ 7 ⋅ 2500 + 18 ⋅1200 + 12 ⋅ 800 + 9 ⋅150 + 21 ⋅1500 ⎥⎥ ⎢⎣ 6 ⋅ 2500 + 25 ⋅1200 + 8 ⋅ 800 + 5 ⋅150 + 13 ⋅1500 ⎥⎦ ⎡75,850 ⎤ = [5 2 4] ⎢⎢ 81,550 ⎥⎥ ⎢⎣ 71, 650 ⎥⎦ = [5(75,850) + 2(81,550) + 4(71, 650)] = [828,950] The total cost of raw materials is $828,950.

66. a.

b.

⎡3500 50 ⎤ ⎢ ⎥ ⎡ 5 20 16 7 17 ⎤ ⎢1500 50 ⎥ RC = ⎢⎢ 7 18 12 9 21⎥⎥ ⎢1000 100 ⎥ ⎢ ⎥ ⎢⎣ 6 25 8 5 13 ⎥⎦ ⎢ 250 10 ⎥ ⎢3500 0 ⎥⎦ ⎣ ⎡ 17,500 + 30, 000 + 16, 000 + 1750 + 59,500 250 + 1000 + 1600 + 70 + 0 ⎤ = ⎢⎢ 24,500 + 27, 000 + 12, 000 + 2250 + 73,500 350 + 900 + 1200 + 90 + 0 ⎥⎥ ⎢⎣ 21, 000 + 37,500 + 8000 + 1250 + 45,500 300 + 1250 + 800 + 50 + 0 ⎥⎦ ⎡124, 750 2920 ⎤ = ⎢⎢139, 250 2540 ⎥⎥ ⎢⎣113, 250 2400 ⎥⎦ ⎡124, 750 2920 ⎤ QRC = Q(RC) = [5 7 12] ⎢⎢139, 250 2540 ⎥⎥ ⎢⎣113, 250 2400 ⎥⎦ = [ 623, 750 + 974, 750 + 1,359, 000 14, 600 + 17, 780 + 28,800] = [ 2,957,500 61,180]

c.

67. a.

⎡1⎤ QRCZ = (QRC)Z = [ 2,957,500 61,180] ⎢ ⎥ ⎣1⎦ = [2,957,500 + 61,180] = [3,018,680]

Amount spent on goods: ⎡10, 000 ⎤ coal industry: DC P = [ 0 1 4] ⎢⎢ 20, 000 ⎥⎥ = [180, 000] ⎢⎣ 40, 000 ⎥⎦ ⎡10, 000 ⎤ elec. industry: DE P = [ 20 0 8] ⎢⎢ 20, 000 ⎥⎥ = [520, 000] ⎢⎣ 40, 000 ⎥⎦ ⎡10, 000 ⎤ steel industry: DS P = [30 5 0] ⎢⎢ 20, 000 ⎥⎥ = [400, 000] ⎢⎣ 40, 000 ⎥⎦ The coal industry spends $180,000, the electric industry spends $520,000, and the steel industry spends 194


Section 6.3

$400,000. ⎡10, 000 ⎤ consumer 1: D1P = [3 2 5] ⎢⎢ 20, 000 ⎥⎥ = [270, 000] ⎢⎣ 40, 000 ⎥⎦ ⎡10, 000 ⎤ consumer 2: D2 P = [ 0 17 1] ⎢⎢ 20, 000 ⎥⎥ = [380, 000] ⎢⎣ 40, 000 ⎥⎦ ⎡10, 000 ⎤ consumer 3: D3 P = [ 4 6 12] ⎢⎢ 20, 000 ⎥⎥ = [640, 000] ⎢⎣ 40, 000 ⎥⎦

Consumer 1 pays $270,000, consumer 2 pays $380,000, and consumer 3 pays $640,000. b. From Example 3 of Sec. 6.2, the number of units sold of coal, electricity, and steel are 57, 31, and 30, respectively. Thus the profit for coal is 10,000(57) – 180,000 = $390,000, the profit for elec. is 20,000(31) – 520,000 = $100,000, and the profit for steel is 40,000(30) – 400,000 = $800,000. c.

From (a), the total amount of money that is paid out by all the industries and consumers is 180,000 + 520,000 + 400,000 + 270,000 + 380,000 + 640,000 = $2,390,000.

d. The proportion of the total amount in (c) paid out by the industries is 180, 000 + 520, 000 + 400, 000 110 . = 2,390, 000 239 The proportion of the total amount in (c) paid by consumers is 270, 000 + 380, 000 + 640, 000 129 . = 2,390, 000 239 68. (A + B)(A – B) = A(A – B) + B(A – B) [dist. prop.] = A 2 − AB + BA − B 2 [dist prop.] = A 2 − BA + BA − B 2 [AB = BA, given] = A2 − B2 ⎡ ⎡1 2 ⎤ ⎡ 2 −3⎤ ⎢1(2) + (2)(−1) 1(−3) + 2 = 69. ⎢ ⎢ ⎥ ⎥ 3 ⎣1 2 ⎦ ⎣⎢ −1 2 ⎥⎦ ⎢⎢ 1(2) + 2(−1) 1(−3) + 2 ⎣ ⎡a 0 0⎤ ⎡d 0 ⎢ ⎥ 70. Let D1 = ⎢ 0 b 0 ⎥ and D2 = ⎢⎢ 0 e ⎢⎣ 0 0 c ⎥⎦ ⎢⎣ 0 0

a.

⎡a D1D2 = ⎢⎢ 0 ⎢⎣ 0 ⎡d D2 D1 = ⎢⎢ 0 ⎢⎣ 0 Both D1D2

0 0⎤ ⎡d b 0 ⎥⎥ ⎢⎢ 0 0 c ⎥⎦ ⎢⎣ 0 0 0 ⎤ ⎡a e 0 ⎥⎥ ⎢⎢ 0 0 f ⎥⎦ ⎢⎣ 0 and D2 D1

( 32 )⎤⎥ = ⎡0 ⎢ ( 32 )⎦⎥⎥ ⎣0

0⎤ 0 ⎥⎥ . f ⎥⎦

0 ⎤ ⎡ ad 0 0 ⎤ 0 ⎥⎥ = ⎢⎢ 0 be 0 ⎥⎥ 0 f ⎥⎦ ⎢⎣ 0 0 cf ⎥⎦ 0 0 ⎤ ⎡ ad 0 0 ⎤ b 0 ⎥⎥ = ⎢⎢ 0 be 0 ⎥⎥ 0 c ⎥⎦ ⎢⎣ 0 0 cf ⎥⎦ are diagonal matrices. 0 e

195

0⎤ 0 ⎦⎥



b. From part (a), D1D2 = D2 D1 . Thus D1 and

⎡1 0 0 5 ⎤ 9 R +R − 13 3 1 > ⎢⎢ 0 1 0 2 ⎥⎥ 6 R +R − 13 3 2 ⎢⎣ 0 0 1 1 ⎥⎦ Thus there should be 5 blocks of A, 2 blocks of B, and 1 block of C suggested.

D2 commute. [In fact, all n × n diagonal matrices commute.] −9.8⎤ ⎡ 72.82 71. ⎢ ⎥ ⎣ 51.32 −36.32 ⎦

2. Let x be the number of tablets of X, y be the number of tablets of Y, and z be the number of tablets of Z. The system is 40x + 10y + 10z = 180 20x + 10y + 50z = 200 10x + 30y + 20z = 190 Reduce the augmented coefficient matrix of the system. ⎡ 40 10 10 180 ⎤ ⎢ 20 10 50 200 ⎥ ⎢ ⎥ ⎢⎣10 30 20 190 ⎥⎦

⎡ 23.994 −20.832 −12.648⎤ 72. ⎢ 7.44 −168.64 ⎥⎦ ⎣ 26.164 64.08⎤ ⎡ 15.606 73. ⎢ ⎥ ⎣ −739.428 373.056 ⎦ 54.06 ⎤ ⎡11.952 74. ⎢ ⎥ ⎣86.496 278.648⎦


⎡10 30 20 190 ⎤ R1 ↔ R3 ⎢ > ⎢ 20 10 50 200 ⎥⎥ ⎢⎣ 40 10 10 180 ⎥⎦ ⎡ 1 3 2 19 ⎤ 1 R 10 1 > ⎢⎢ 2 1 5 20 ⎥⎥ 1 R 10 2 ⎢⎣ 4 1 1 18 ⎥⎦ 1 R 3 10

1. The corresponding system is ⎧ 6 A + B + 3C = 35 ⎪ ⎨3 A + 2 B + 3C = 22 ⎪⎩ A + 5 B + 3C = 18 Reduce the augmented coefficient matrix of the system. ⎡ 6 1 3 35 ⎤ ⎢ 3 2 3 22 ⎥ ⎢ ⎥ ⎢⎣1 5 3 18 ⎥⎦

3 2 19 ⎤ ⎡1 −2R1 + R 2 ⎢ > 0 −5 1 −18⎥⎥ −4R1 + R3 ⎢ ⎢⎣ 0 −11 −7 −58⎥⎦ 3 2 19 ⎤ ⎡1 − 15 R 2 ⎢ 18 ⎥ 1 1 −5 > ⎢0 5⎥ ⎢ ⎥ 0 11 7 58 − − − ⎣ ⎦ 13 41 ⎤ ⎡1 0 5 5 ⎥ ⎢ −3R 2 + R1 ⎢ 18 1 ⎥ > 0 1 −5 5⎥ 11R 2 + R3 ⎢ ⎢ 0 0 − 46 − 92 ⎥ 5 5 ⎦⎥ ⎣⎢ 13 41 ⎡1 0 ⎤ 5 5 ⎥ 5 R ⎢ − 46 3 ⎥ > ⎢ 0 1 − 15 18 5⎥ ⎢ ⎢0 0 1 2⎥ ⎣⎢ ⎦⎥

⎡1 5 3 18 ⎤ R1 ↔ R3 ⎢ > ⎢ 3 2 3 22 ⎥⎥ ⎢⎣ 6 1 3 35 ⎥⎦ 5 3 18⎤ ⎡1 −3R1 + R 2 ⎢ > ⎢0 −13 −6 −32 ⎥⎥ −6R1 + R3 ⎢⎣0 −29 −15 −73⎥⎦ 5 3 18⎤ ⎡1 1 R − 13 ⎢ 2 6 32 ⎥ > ⎢0 1 13 13 ⎥ ⎢ ⎥ ⎣ 0 −29 −15 −73⎦ 9 74 ⎤ ⎡1 0 13 13 ⎥ ⎢ −5R 2 + R1 ⎢ 6 32 ⎥ > 0 1 13 13 ⎥ 29R 2 + R 3 ⎢ ⎢0 0 – 21 – 21 ⎥ ⎢⎣ 13 13 ⎥⎦ ⎡1 0 9 74 ⎤ 13 13 ⎥ 13 − 21 R3 ⎢ 6 32 ⎥ ⎢ > 0 1 13 13 ⎥ ⎢ ⎢0 0 1 1 ⎥ ⎣⎢ ⎦⎥

⎡1 0 0 3 ⎤ − 13 R + R1 ⎢ 5 3 > ⎢0 1 0 4 ⎥⎥ 1R +R 2 5 3 ⎢⎣0 0 1 2 ⎥⎦ She should take 3 tablets of X, 4 tablets of Y, and 2 tablets of Z.

196


Section 6.4

3. Let a, b, c, and d be the number of bags of foods A, B, C, and D, respectively. The corresponding system is ⎧ 5a + 5b + 10c + 5d = 10, 000 ⎪ ⎨10a + 5b + 30c + 10d = 20, 000 ⎪⎩5a + 15b + 10c + 25d = 20, 000

Problems 6.4 1. The first nonzero entry in row 2 is not to the right of the first nonzero entry in row 1, hence not reduced. 2. Reduced.

Reduce the augmented coefficient matrix of the system. ⎡ 5 5 10 5 10, 000 ⎤ ⎢10 5 30 10 20, 000 ⎥ ⎢ ⎥ ⎢⎣ 5 15 10 25 20, 000 ⎥⎦

3. Reduced. 4. In row 2, the first nonzero entry is in column 2, but not all other entries in column 2 are zeros, hence not reduced.

2000 ⎤ ⎡ 1 1 2 1 ⎢ > ⎢10 5 30 10 20, 000 ⎥⎥ ⎣⎢ 5 15 10 25 20, 000 ⎦⎥

1R 5 1

5. The first row consists entirely of zeros and is not below each row containing a nonzero entry, hence not reduced.

1 2000 ⎤ ⎡1 1 2 −10R1 + R 2 ⎢ > ⎢ 0 −5 10 0 0 ⎥⎥ –5R1 + R 3 ⎢⎣ 0 10 0 20 10, 000 ⎥⎦

6. The first nonzero entry of row 2 is to the left of the first nonzero entry of row 1, hence not reduced.

⎡1 1 > ⎢⎢ 0 1 ⎢⎣ 0 10 ⎡1 −R 2 + R1 > ⎢⎢ 0 −10R 2 + R 3 ⎢⎣ 0

3⎤ ⎡ 1 3⎤ −4R1 + R 2 ⎡ 1 > ⎢ 7. ⎢ ⎥ ⎥ − 4 0 0 12 ⎣ ⎦ ⎣ ⎦

− 15 R 2

2 1 2000 ⎤ −2 0 0 ⎥⎥ 0 20 10, 000 ⎥⎦

1 R − 12 2

⎡1 > ⎢ ⎣0 −3R 2 + R1 >

0 4 1 2000 ⎤ 1 −2 0 0 ⎥⎥ 0 20 20 10, 000 ⎥⎦

⎡ 0 0 4 0 2000 ⎤ R3 ⎢ > ⎢ 0 1 −2 0 0 ⎥⎥ 1 1 500 ⎥⎦ ⎣⎢ 0 0 0⎤ ⎡ 1 0 0 −3 −4R 3 + R1 ⎢ > 0 1 0 2 1000 ⎥⎥ 2R 3 + R 2 ⎢ ⎢⎣ 0 0 1 1 500 ⎥⎦ 1 20

3⎤ 1⎥⎦ ⎡1 0 ⎤ ⎢0 1 ⎥ ⎣ ⎦

⎡ 0 −3 0 2 ⎤ R1 ↔ R 2 ⎡ 1 5 0 2 ⎤ 8. ⎢ ⎯⎯⎯⎯⎯→ ⎢ ⎣ 1 5 0 2 ⎥⎦ ⎣ 0 −3 0 2 ⎥⎦ 1 − R2 ⎡ 1 5 0 2⎤ 3 → ⎯⎯⎯⎯ ⎢0 1 0 − 2 ⎥ 3⎦ ⎣ 16 ⎤ −5R 2 + R1 ⎡ 1 0 0 3⎥ ⎢ ⎯⎯⎯⎯⎯⎯ → ⎢⎣ 0 1 0 − 23 ⎥⎦

This reduced matrix corresponds to the system ⎧ a − 3d = 0 ⎪ ⎨b + 2d = 1000 ⎪⎩ c + d = 500

⎡2 4 6⎤ ⎡1 2 3⎤ R1 ↔ R3 ⎢ ⎢ ⎥ 9. ⎢ 1 2 3 ⎥ > ⎢ 1 2 3 ⎥⎥ ⎢⎣ 1 2 3 ⎥⎦ ⎢⎣ 2 4 6 ⎥⎦ ⎡1 2 3 ⎤ −R1 + R 2 ⎢ > 0 0 0 ⎥⎥ –2R1 + R 3 ⎢ ⎢⎣0 0 0 ⎥⎦

Letting d = r, we get the general solution of the system: a = 3r b = –2r + 1000 c = –r + 500 d=r Note that a, b, c, and d cannot be negative, given the context, hence 0 ≤ r ≤ 500. One specific solution is when r = 250, then a = 750, b = 500, c = 250, and d = 250.

197


3⎤ ⎡2 ⎢ 1 −6 ⎥ ⎥ R1 ↔ R 2> 10. ⎢ ⎢ 4 8⎥ ⎢ ⎥ ⎢⎣ 1 7 ⎥⎦ ⎡ 2 ⎢ 1 11. ⎢ ⎢ −1 ⎢ ⎣⎢ 0


⎡ 1 −6 ⎤ –2R1 + R 2 ⎢ 3⎥⎥ –4R1 + R 3 ⎢2 > ⎢ 4 8⎥ −R1 + R 4 ⎢ ⎥ ⎢⎣ 1 7 ⎥⎦ 2 3 1 1

2⎤ 1⎥⎥ 4⎥ ⎥ 0 ⎦⎥

⎡ 1 4 2 2⎤ ⎢ ⎥ 1 −2R1 + R 2 ⎢ 0 −8 −1 −3⎥ − 8 R 2 > > ⎢0 7 3 6⎥ R1 + R3 ⎢ ⎥ ⎣⎢ 0 2 1 0 ⎦⎥

⎡1 ⎢ ⎢0 ⎢ ⎢0 ⎢⎣ 0

0 4 3 2

3 2 1 1

1⎤ 2 ⎥⎥ R1 ↔ R 2 > 4⎥ ⎥ 0 ⎦⎥

⎡1 −4R 2 + R1 ⎢ ⎢ −7 R 2 + R3 ⎢ 0 > −2 R 2 + R 4 ⎢ 0 ⎢ ⎢0 ⎣⎢ ⎡1 ⎢ 1 − 8 R3 + R 2 ⎢0 > ⎢ − 34 R3 + R 4 ⎢ 0 ⎢ ⎢0 ⎢⎣ − 23 R3 + R1

⎡1 ⎢ ⎢0 − 17 R 33 4 > ⎢ ⎢0 ⎢ ⎢⎣ 0 ⎡0 0 ⎢2 0 12. ⎢ ⎢ 0 −1 ⎢ ⎣⎢ 0 4

0 1 0 0

3 2 1 8 17 8 3 4

1⎤ 2⎥ 3⎥ 8 8 ⎥ 17 27 ⎥ 8 ⎥ − 34 ⎥⎦⎥

4 0 3 2

⎡1 ⎢ R3 ⎢0 > ⎢ ⎢0 ⎢ ⎢0 ⎣⎢

32 ⎤ 32 0 0 − 17 R + R1 ⎥ 17 4 3 3⎥− R + R2 1 0 17 ⎥ 17 4 27 ⎥ − 27 R 4 + R 3 0 1 17 ⎥ 17 0 0 1⎥⎦

3⎤ 2⎥

⎡2 0 ⎢0 0 ⎢ ⎢ 0 −1 ⎢ ⎣⎢ 0 4

⎡ 1 −6⎤ 6R 2 + R1 ⎢ 1⎥⎥ −32R 2 + R 3 ⎢0 > ⎢ 0 32⎥ −13R 2 + R 4 ⎢ ⎥ ⎢⎣ 0 13⎥⎦

⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

4 2 2⎤ ⎥ 1 18 83 ⎥ ⎥ 7 3 6⎥ 2 1 0 ⎥⎦ 0 1

3 2 1 8

0

1

0

3 4

1⎤ 2⎥ 3⎥ 8⎥ 27 ⎥ 17 ⎥ − 34 ⎥⎦⎥

⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎢⎣ 0

0 1 0 0

0 0 1 0

0⎤ 0 ⎥⎥ 0⎥ ⎥ 1 ⎥⎦

⎡1 0 3 ⎤ 3⎤ 2⎥ ⎢ 2 ⎥⎥ 12 R1 ⎢ 0 0 2 ⎥ R 2 ↔ R3 > ⎢ > ⎥ 0⎥ 0 −1 0 ⎥ ⎢ ⎥ 1⎦⎥ ⎢⎣ 0 4 1⎥⎦

⎡1 ⎢ 1 0 ⎥ −4R 2 + R 4 ⎢ 0 > ⎢ ⎥ 0 2⎥ ⎢0 ⎢⎣ 0 4 1⎥⎦

0

−6 ⎤ 1 R 15⎥⎥ 15 2 > 32 ⎥ ⎥ 13⎥⎦

32 ⎤ 0 0 − 17 ⎥ 3⎥ 1 0 17 ⎥ 27 ⎥ 0 1 17 ⎥ 33 ⎥ 0 0 − 17 ⎥⎦

2⎤ 3⎥⎥ R1 ↔ R 2 > 0⎥ ⎥ 1⎦⎥

⎡1 ⎢ −R 2 ⎢0 > ⎢ ⎢0 ⎢⎣ 0

⎡ 1 ⎢ 2 ⎢ ⎢ −1 ⎢ ⎣⎢ 0

⎡1 ⎢ ⎢0 ⎢0 ⎢ ⎢⎣ 0

0

3⎤ 2⎥

1 0⎥ ⎥ 0 2⎥ 0 1⎥⎦

⎡1 ⎢ 1R 2 3> ⎢ 0 ⎢ ⎢0 ⎢⎣ 0

198

3⎤ 2⎥

⎡1 0 3 ⎤ 2⎥ ⎢ ⎢0 −1 0 ⎥ ⎢ ⎥ ⎢0 0 2⎥ ⎢⎣0 4 1⎥⎦

⎡1 + R1 ⎢ 0 1 0⎥ − > ⎢ ⎥ ⎢0 0 1⎥ −R 3 + R 4 ⎢ 0 1⎥⎦ ⎣⎢ 0 0

3R 2 3

0 1 0 0

0⎤ 0 ⎥⎥ 1⎥ ⎥ 0 ⎦⎥

0⎤ 1 ⎥⎥ 0⎥ ⎥ 0 ⎥⎦


Section 6.4

⎡ 2 −7 50 ⎤ ⎡ 1 3 10 ⎤ → 13. ⎢ ⎣ 1 3 10 ⎥⎦ ⎢⎣ 2 −7 50 ⎥⎦ 10 ⎤ ⎡1 3 3 10 ⎤ ⎡1 →⎢ → ⎢ 30 ⎥ ⎣ 0 −13 30 ⎦⎥ ⎣ 0 1 − 13 ⎦ ⎡ 1 0 220 ⎤ 13 ⎥ →⎢ 30 ⎢⎣ 0 1 − 13 ⎥⎦ 220 30 Thus x = and y = − . 13 13 ⎡ 1 −3 −11⎤ ⎡ 1 0 − 52 ⎤ ⎡ 1 −3 −11⎤ ⎡ 1 −3 −11⎤ ⎢ ⎥ → → 14. ⎢ ⎢ 53 ⎥ → ⎢ 53 ⎥ 0 1 15 9 ⎦⎥ ⎣⎢ 0 15 53⎦⎥ ⎣4 3 ⎣⎢ ⎦⎥ ⎣ 0 1 15 ⎦ 2 53 Thus x = − , y = . 5 15 4⎤ ⎡1 1 4 ⎤ ⎡1 1 ⎡1 1 4⎤ ⎡ 3 1 4⎤ ⎡ 3 1 3 3⎥ → ⎢ 3 3⎥ → ⎢ 3 15. ⎢ →⎢ →⎢ ⎥ ⎥ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0 0 ⎢⎣ 0 0 −14 ⎥⎦ ⎣12 4 2 ⎦ ⎣ 0 0 −14 ⎦ The last row indicates 0 = 1, which is never true, so there is no solution.

0⎤ ⎥ 1 ⎥⎦

⎡ 1 2 −3 0 ⎤ ⎡ 1 2 −3 0 ⎤ 16. ⎢ ⎥→⎢ ⎥ ⎣ −2 −4 6 1⎦ ⎣ 0 0 0 1⎦ The last row indicates that 0 = 1, which is never true. There is no solution. ⎡1 2 1 ⎡1 2 1 4 ⎤ ⎡ 1 2 1 4 ⎤ →⎢ →⎢ 17. ⎢ ⎥ ⎥ 1 ⎢⎣0 1 6 ⎣3 0 2 5 ⎦ ⎣ 0 −6 −1 −7 ⎦

⎡1 0 4⎤ ⎢ → ⎥ 7 ⎢0 1 ⎥ 6⎦ ⎣

2 3 1 6

5⎤ 3⎥ 7⎥ 6⎦

,

⎧x + 2 z = 5 ⎪ 3 3 which gives ⎨ . 1z= 7 + y ⎪⎩ 6 6 2 5 1 7 Thus, x = − r + , y = − r + , z = r, where r is any real number. 3 3 6 6 ⎡ 1 0 13 2 1⎤ ⎡1 3 2 1⎤ ⎡ 1 3 2 1⎤ ⎡ 1 3 2 →⎢ →⎢ 18. ⎢ ⎥→⎢ ⎥ ⎥ 3 9 − − 0 1 ⎢0 1 − 3 ⎣1 1 5 10 ⎦ ⎣ 0 −2 3 9 ⎦ ⎢⎣ 2 2 ⎥⎦ 2 ⎣

Thus x = −

29 ⎤ 2 ⎥ − 92 ⎥⎦

13 29 3 9 r + , y = r − , z = r, where r is any real number. 2 2 2 2

9⎤ ⎡1 0 ⎡ 1 −3 0 ⎤ 8⎥ ⎡ 1 −3 0 ⎤ ⎡ 1 −3 0 ⎤ ⎢ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 3 3⎥ ⎢ 19. ⎢ 2 2 3⎥ → ⎢ 0 8 3⎥ → ⎢0 1 8⎥ → 0 1 8⎥ ⎢ ⎢ ⎥ ⎢ 0 0 − 17 ⎥ ⎣⎢ 5 −1 1⎦⎥ ⎣⎢ 0 14 1⎦⎥ ⎣0 14 1⎦ ⎢⎣ 4 ⎥⎦

From the third row, 0 = −

17 , which is never true, so there is no solution. 4

199



4 9⎤ ⎡ 1 4 9⎤ ⎡ 1 20. ⎢3 −1 6 ⎥ → ⎢0 −13 −21⎥ ⎢ ⎥ ⎢ ⎥ ⎣ 1 −1 2 ⎦ ⎣0 −5 −7 ⎦ ⎡ 1 0 33 ⎤ ⎡ 1 4 9⎤ 13 ⎥ ⎢ 21 ⎥ → 0 1 21 → ⎢0 1 13 ⎢ ⎥ 13 ⎢ ⎥ ⎢ 14 ⎥ ⎢⎣ 0 −5 −7 ⎥⎦ ⎢⎣0 0 13 ⎥⎦ ⎡ 1 0 33 ⎤ 13 ⎥ ⎡ 1 0 0⎤ ⎢ 21 → ⎢ 0 1 13 ⎥ → ⎢ 0 1 0 ⎥ ⎢ ⎥ ⎢ 0 0 1⎥ ⎦ ⎢⎣ 0 0 1⎥⎦ ⎣ The last row indicates that 0 = 1, which is never true. There is no solution. ⎡ 1 −1 −3 −5⎤ ⎡ 1 −1 21. ⎢⎢ 2 −1 −4 −8⎥⎥ → ⎢⎢ 0 1 ⎢⎣ 1 1 −1 −1⎥⎦ ⎢⎣ 0 2 ⎡ 1 0 −1 −3⎤ ⎡ 1 0 → ⎢⎢ 0 1 2 2 ⎥⎥ → ⎢⎢ 0 1 ⎢⎣ 0 0 1 0 ⎥⎦ ⎢⎣ 0 0 Thus, x = –3, y = 2, z = 0.

−3 − 5 ⎤ ⎡ 1 0 − 1 − 3 ⎤ 2 2 ⎥⎥ → ⎢⎢ 0 1 2 2 ⎥⎥ 2 4 ⎥⎦ ⎢⎣ 0 0 −2 0 ⎥⎦ 0 −3⎤ 0 2 ⎥⎥ 1 0 ⎥⎦

7⎤ ⎡ 1 ⎡ 1 1 −1 7 ⎤ ⎡ 1 1 −1 ⎢ ⎥ ⎢ 22. ⎢ 2 −3 −2 4 ⎥ → ⎢ 0 −5 0 −10 ⎥⎥ → ⎢⎢ 0 ⎢⎣ 1 −1 −5 23⎥⎦ ⎢⎣ 0 −2 −4 16 ⎥⎦ ⎢⎣ 0 ⎡ 1 0 −1 5⎤ ⎡ 1 0 −1 5⎤ ⎡ 1 → ⎢⎢ 0 1 0 2 ⎥⎥ → ⎢⎢0 1 0 2 ⎥⎥ → ⎢⎢ 0 ⎢⎣ 0 0 −4 20 ⎥⎦ ⎢⎣0 0 1 −5⎥⎦ ⎢⎣ 0 Thus x = 0, y = 2, z = –5.

1 −1 7 ⎤ 1 0 2 ⎥⎥ −2 −4 16 ⎥⎦ 0 0 0⎤ 1 0 2 ⎥⎥ 0 1 −5⎥⎦

⎡ 2 0 −4 8⎤ ⎡ 1 0 −2 4 ⎤ ⎡ 1 0 ⎢ 1 −2 −2 14 ⎥ ⎢ 1 −2 −2 14 ⎥ ⎢ 0 −2 ⎥→⎢ ⎥→⎢ 23. ⎢ ⎢ 1 1 −2 −1⎥ ⎢ 1 1 −2 −1⎥ ⎢ 0 1 ⎢ ⎥ ⎢ ⎥ ⎢ 1 1 0 ⎦⎥ ⎣⎢3 1 1 0 ⎥⎦ ⎣⎢ 0 1 ⎣⎢ 3 4 ⎤ ⎡ 1 0 −2 4⎤ ⎡1 ⎡ 1 0 −2 ⎢0 ⎢0 ⎥ ⎢ ⎥ 1 0 −5⎥ ⎢ 0 1 0 −5⎥ →⎢ → →⎢ ⎢ 0 −2 0 10 ⎥ ⎢ 0 0 0 ⎢0 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 7 −12 ⎥⎦ ⎢⎣ 0 0 7 −7 ⎥⎦ ⎢⎣ 0 ⎢⎣ 0 Thus x = 2, y = –5, z = –1.

−2 4⎤ 0 10 ⎥⎥ 0 −5 ⎥ ⎥ 7 −12 ⎦⎥ 0 −2 4 ⎤ ⎡ 1 1 0 −5⎥⎥ ⎢⎢ 0 → 0 1 −1⎥ ⎢ 0 ⎥ ⎢ 0 0 0 ⎥⎦ ⎢⎣ 0

⎡ 1 0 3 −1⎤ ⎡ 1 0 3 −1⎤ ⎡ 1 0 3 −1⎤ ⎡ 1 ⎢ 3 2 11 1⎥ ⎢ 0 2 2 4 ⎥ ⎢ 0 1 1 2 ⎥⎥ ⎢⎢0 ⎥→⎢ ⎥→⎢ 24. ⎢ → ⎢1 1 4 1⎥ ⎢ 0 1 1 2⎥ ⎢0 1 1 2⎥ ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎣⎢ 2 −3 3 −8⎦⎥ ⎣⎢ 0 −3 −3 −6 ⎦⎥ ⎣⎢ 0 −3 −3 −6 ⎦⎥ ⎣⎢0 Thus x = –3r – 1, y = –r + 2, z = r, where r is any real number. 200

0 1 0 0

0 1 0 0

0 2⎤ 0 −5⎥⎥ 1 −1⎥ ⎥ 0 0 ⎥⎦

3 −1⎤ 1 2 ⎥⎥ 0 0⎥ ⎥ 0 0 ⎦⎥


Section 6.4

⎡1 −1 ⎢1 1 25. ⎢ ⎢1 1 ⎢⎣1 1 ⎡1 ⎢0 →⎢ ⎢0 ⎢⎣ 0 ⎡1 ⎢0 →⎢ ⎢0 ⎣⎢ 0 ⎡1 ⎢0 →⎢ ⎢0 ⎣⎢ 0

−1 −1 −1 0 ⎤ ⎡ 1 −1 −1 −1 −1 0 ⎤ −1 −1 −1 0 ⎥ ⎢ 0 2 0 0 0 0 ⎥ ⎥→⎢ ⎥ 1 −1 −1 0 ⎥ ⎢ 0 2 2 0 0 0 ⎥ 1 1 −1 0 ⎥⎦ ⎢⎣ 0 2 2 2 0 0 ⎥⎦ −1 −1 −1 −1 0 ⎤ ⎡ 1 0 −1 −1 −1 0 ⎤ 1 0 0 0 0⎥ ⎢0 1 0 0 0 0⎥ ⎥→⎢ ⎥ 2 2 0 0 0⎥ ⎢0 0 2 0 0 0⎥ 2 2 2 0 0 ⎥⎦ ⎢⎣ 0 0 2 2 0 0 ⎥⎦ 0 −1 −1 −1 0 ⎤ ⎡ 1 0 0 −1 −1 0 ⎤ 1 0 0 0 0⎥ ⎢0 1 0 0 0 0⎥ ⎥→⎢ ⎥ 0 1 0 0 0⎥ ⎢0 0 1 0 0 0⎥ 0 2 2 0 0 ⎥⎦ ⎢⎣ 0 0 0 2 0 0 ⎥⎦ 0 0 −1 −1 0 ⎤ ⎡ 1 0 0 0 −1 0 ⎤ ⎢0 1 0 0 0 0 ⎥ 1 0 0 0 0⎥ ⎥→⎢ ⎥ 0 1 0 0 0⎥ ⎢0 0 1 0 0 0 ⎥ 0 0 1 0 0 ⎦⎥ ⎣⎢0 0 0 1 0 0 ⎦⎥ Thus, x1 = r , x2 = 0, x3 = 0, x4 = 0, and x5 = r , where r is any number.

⎡1 1 ⎢1 −1 26. ⎢ ⎢1 1 ⎢ ⎣⎢1 1 ⎡1 ⎢0 →⎢ ⎢0 ⎢ ⎣⎢ 0

1 −1 −1 −1 1 1 1 0 0 1 0 0

−1 1 −1 1 −1

0⎤ ⎡ 1 0 ⎥⎥ ⎢⎢ 0 → 0⎥ ⎢0 ⎥ ⎢ 0 ⎦⎥ ⎣⎢ 0 0⎤ ⎡1 ⎢0 ⎥ 0 0⎥ →⎢ ⎢0 0 0⎥ ⎢ ⎥ 1 0 ⎦⎥ ⎣⎢0

1 −1 0 ⎤ ⎡ 1 −2 2 0 ⎥⎥ ⎢⎢ 0 → −2 0 0 ⎥ ⎢ 0 ⎥ ⎢ −2 2 0 ⎦⎥ ⎣⎢ 0 0 0 0⎤ 1 0 0 0 ⎥⎥ 0 1 0 0⎥ ⎥ 0 0 1 0 ⎦⎥

1 −2 0 0 0

1 1 0 0

1 −1 1 −1 1 0 1 −1

0⎤ ⎡ 1 0 ⎥⎥ ⎢⎢ 0 → 0⎥ ⎢0 ⎥ ⎢ 0 ⎦⎥ ⎣⎢ 0

1 1 0 0

1 −1 0 ⎤ 1 −1 0 ⎥⎥ 1 0 0⎥ ⎥ 0 1 0 ⎦⎥

Thus x1 = 0, x2 = 0, x3 = 0, x4 = 0. 27. Let x = federal tax and y = state tax. Then x = 0.25(312,000 – y) and y = 0.10(312,000 – x). Equivalently, ⎧ x + 0.25 y = 78, 000 ⎨ ⎩0.10 x + y = 31, 200. 0.25 78, 000 ⎤ ⎡1 0.25 78, 000 ⎤ ⎡ 1 ⎡1 0.25 78, 000 ⎤ ⎡1 0 72, 000 ⎤ →⎢ . →⎢ → ⎢ 0.10 ⎥ ⎥ 1 31, 200 ⎦ ⎣ 0 0.975 23, 400 ⎦ 1 24, 000 ⎥⎦ ⎢⎣ 0 1 24, 000 ⎥⎦ ⎣0 ⎣ Thus x = 72,000 and y = 24,000, so the federal tax is $72,000 and the state tax is $24,000.

28. x = no. of units of A to be sold and y = no. of units of B to be sold. Then x = 1.25y and 8x + 11y = 42,000. Equivalently, ⎧ x − 1.25 y = 0, ⎨ ⎩8 x + 11y = 42, 000. 0 ⎤ ⎡1 −1.25 0 ⎤ 0 ⎤ ⎡1 0 2500 ⎤ ⎡1 −1.25 ⎡1 −1.25 . →⎢ → →⎢ ⎢8 ⎥ ⎥ 11 42, 000 ⎦ ⎣ 0 21 42, 000 ⎦ 1 2000 ⎥⎦ ⎢⎣0 1 2000 ⎥⎦ ⎣0 ⎣ Thus x = 2500 and y = 2000, so 2500 units of A and 2000 units of B must be sold.

29. Let x = number of units of A produced, y = number of units of B produced, and z = number of units of C produced. Then no. of units: x + y + z = 11,000 total cost: 4x + 5y + 7z + 17,000 = 80,000 201



total profit: x + 2y + 3z = 25,000 Equivalently, ⎧ x + y + z = 11, 000 ⎪ ⎨4 x + 5 y + 7 z = 63, 000 ⎪ x + 2 y + 3z = 25, 000 ⎩ ⎡ 1 1 1 11, 000 ⎤ ⎡1 1 1 11, 000 ⎤ ⎢ 4 5 7 63, 000 ⎥ → ⎢ 0 1 3 19, 000 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 2 3 25, 000 ⎥⎦ ⎢⎣ 0 1 2 14, 000 ⎥⎦ ⎡1 0 0 2000 ⎤ ⎡ 1 0 −2 −8, 000 ⎤ ⎡ 1 0 −2 −8, 000 ⎤ ⎢ ⎥ ⎢ ⎥ → ⎢ 0 1 3 19, 000 ⎥ → ⎢0 1 3 19, 000 ⎥ → ⎢⎢ 0 1 0 4000 ⎥⎥ ⎢⎣ 0 0 1 5000 ⎥⎦ ⎢⎣ 0 0 −1 −5, 000 ⎥⎦ ⎢⎣0 0 1 5, 000 ⎥⎦ Thus x = 2000, y = 4000, and z = 5000, so 2000 units of A, 4000 units of B and 5000 units of C should be produced.

30. Let x = number of desks to be produced at the East Coast plant and y = number of desks to be produced at the West Coast plant. Then x + y = 800 and 90x +20,000 = 95y + 18,000. Equivalently, ⎧ x + y = 800 ⎨ ⎩90 x − 95 y = −2000. 1 800 ⎤ ⎡ 1 1 800 ⎤ ⎡ 1 1 800 ⎤ ⎡ 1 0 400⎤ ⎡ 1 ⎢90 −95 −2000 ⎥ → ⎢ 0 −185 −74, 000 ⎥ → ⎢ 0 1 400 ⎥ → ⎢0 1 400⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ x = 400 and y = 400 Thus the production order is 400 units at the East Coast plant and 400 units at the West Coast plant.

31. Let x = number of brand X pills, y = number of brand Y pills, and z = number of brand Z pills. Considering the unit requirements gives the system ⎧2 x + 1y + 1z = 10 (vitamin A) ⎪ ⎨3x + 3 y + 0 z = 9 (vitamin D) ⎪5 x + 4 y + 1z = 19 (vitamin E) ⎩ 1 ⎡ 2 1 1 10 ⎤ ⎡⎢ 1 2 ⎢ ⎥ ⎢ 3 3 0 9 ⎥ → ⎢⎢ 3 3 ⎢⎣ 5 4 1 19 ⎥⎦ ⎢5 4 ⎣

5⎤ ⎡ 1 ⎥ ⎢ 0 9⎥ → ⎢0 ⎥ ⎢ 1 19 ⎥ ⎢ 0 ⎦ ⎣

1 2

1 2 3 2 3 2

1 2 − 32 − 32

5⎤ ⎥ −6 ⎥ ⎥ −6 ⎥⎦

1 ⎡1 1 5⎤ ⎡ 1 0 1 7 ⎤ 2 2 ⎢ ⎥ → ⎢ 0 1 −1 −4 ⎥ → ⎢⎢ 0 1 −1 −4 ⎥⎥ ⎢ ⎥ ⎢⎣ 0 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦ ⎧x = 7 − r ⎪ Thus ⎨ y = r − 4 where r = 4, 5, 6, 7. ⎪z = r ⎩ The only solutions for the problem are z = 4, x = 3, and y = 0; z = 5, x = 2, and y = 1; z = 6, x = 1, and y = 2; z = 7, x = 0, and y = 3. Their respective costs (in cents) are 15, 23, 31, and 39.

a.

The possible combinations are 3 of X, 4 of Z; 2 of X, 1 of Y, 5 of Z; 1 of X, 2 of Y, 6 of Z; 3 of Y, 7 of Z.

b. The combination 3 of X, 4 of Z costs 15 cents a day. 202

ISM: Introductory Mathematical Analysis c.

Section 6.4

The least expensive combination is 3 of X, 4 of Z; the most expensive is 3 of Y, 7 of Z.

32. Let x, y, and z be the numbers of units of A, B, and C, respectively. ⎧3x + 1 y + 2 z = 490 (machine I) ⎪ ⎨1x + 2 y + 1z = 310 (machine (II) ⎪2 x + 4 y + 1z = 560 (machine III) ⎩ ⎡ 3 1 2 490 ⎤ ⎡ 1 2 1 ⎢ 1 2 1 310 ⎥ → ⎢ 3 1 2 ⎢ ⎥ ⎢ ⎣ 2 4 1 560 ⎦ ⎣ 2 4 1 ⎡ 1 2 1 310 ⎤ ⎡ 1 → ⎢ 0 −5 −1 −440 ⎥ → ⎢ 0 ⎢ ⎥ ⎢ ⎣ 0 0 −1 −60 ⎦ ⎢⎣ 0 ⎡1 ⎢ → ⎢0 ⎢ 0 ⎣⎢ ⎡1 → ⎢0 ⎢ ⎣0

0 1

3 5 1 5

0 −1 0 0 1 0 0 1

310 ⎤ 490 ⎥ ⎥ 560 ⎦ 2 1 310 ⎤ 1 15 88⎥ ⎥ 0 −1 −60 ⎥⎦ 134 ⎤ ⎡ 1 0 53 134 ⎤ ⎥ ⎢ ⎥ 88⎥ → ⎢ 0 1 15 88⎥ ⎥ ⎢ ⎥ −60 ⎥ ⎢ 0 0 1 60 ⎥ ⎦ ⎣ ⎦ 98⎤ 76 ⎥ ⎥ 60 ⎦

x = 98, y = 76, z = 60 Thus, 98 units of A, 76 units of B, and 60 units of C should be produced. 33. a.

Let s, d, and g represent the number of units of S, D, and G, respectively. Then ⎧12s + 20d + 32 g = 220 (stock A) ⎪ ⎨16s + 12d + 28 g = 176 (stock B) ⎪8s + 28d + 36 g = 264 (stock C) ⎩ ⎡12 20 32 220 ⎤ ⎢16 12 28 176 ⎥ ⎢ ⎥ ⎢⎣ 8 28 36 264 ⎥⎦

( 14 ) R1> ( 14 ) R2 ( 18 ) R3

⎡3 5 ⎢ ⎢4 3 ⎢ 7 ⎢⎣ 1 2

8 55⎤ ⎥ 7 44 ⎥ 9 33⎥ ⎥⎦ 2

⎡ 1 7 9 33⎤ 2 2 ⎥ R1 ↔ R3 ⎢ > ⎢ 4 3 7 44 ⎥ ⎢ ⎥ ⎢⎣ 3 5 8 55⎥⎦ 7 9 ⎡1 33⎤ 2 2 ⎢ ⎥ −4R1 + R 2 > ⎢ 0 −11 −11 −88⎥ −3R1 + R3 ⎢ ⎥ 11 11 ⎢⎣ 0 − 2 − 2 −44 ⎥⎦

1 R − 11 2

7 9 ⎡1 33⎤ 2 2 ⎢ ⎥ > ⎢0 1 1 8⎥ ⎢ ⎥ 11 11 ⎢⎣ 0 − 2 − 2 −44 ⎥⎦

203



⎡1 0 1 5 ⎤ − 27 R 2 + R1 ⎢ > ⎢ 0 1 1 8 ⎥⎥ 11 R + R 3 2 2 ⎢⎣ 0 0 0 0 ⎥⎦ Thus s = 5 – r, d = 8 – r, and g = r, where r = 0, 1, 2, 3, 4, 5. The six possible combinations are given by

COMBINATION r

0

1

2

3

4

5

S

5

4

3

2

1

0

D

8

7

6

5

4

3

G

0

1

2

3

4

5

b. Computing the cost of each combination, we find that they are 4700, 4600, 4500, 4400, 4300, and 4200 dollars, respectively. Buying 3 units of Deluxe and 5 units of Gold Star (s = 0, d = 3, g = 5) minimizes the cost. Principles in Practice 6.5 1. Write the coefficients matrix and reduce. ⎡1 3 4 ⎤ ⎡5 3 4⎤ 1 5 5 ⎢ 6 8 7 ⎥ 5 R1> ⎢⎢ 6 8 7 ⎥⎥ −6R1 + R 2> ⎢ ⎥ ⎢ ⎥ −3R1 + R 3 ⎣⎢ 3 1 2 ⎦⎥ ⎢⎣ 3 1 2 ⎥⎦

3 ⎡1 5 ⎢ ⎢ 0 22 5 ⎢ ⎢0 − 4 5 ⎣

4⎤ 5⎥ 11 ⎥ 5⎥ − 52 ⎥ ⎦

3 4⎤ ⎡1 ⎡1 0 1 ⎤ 5 5⎥ 3 2⎥ ⎢ − 5 R 2 + R1 ⎢ 1 1 ⎢ ⎥ ⎢ 1 > 0 > 0 1 2⎥ 2⎥ 4 R + R ⎢ ⎢ ⎥ 2 3 ⎢0 − 4 − 2 ⎥ 5 ⎢0 0 0 ⎥ 5 5⎦ ⎣ ⎦ ⎣ The system has infinitely many solutions since there are two nonzero rows in the reduced coefficient matrix. 1 x+ z =0 2 1 y+ z =0 2 1 1 Let z = r, so x = − r and y = − r , where r is any real number. 2 2 5 R 22 2

Problems 6.5 −1 −9 −3⎤ ⎡ 1 1 −1 − 9 3 ⎤ ⎡ 1 1 − 1 − 9 − 3⎤ ⎡ 1 1 −1 −9 −3⎤ 2 15 12 ⎥⎥ → ⎢⎢ 0 1 4 33 18⎥⎥ → ⎢⎢ 0 1 4 33 18⎥⎥ → ⎢⎢0 1 4 33 18⎥⎥ ⎢⎣ 0 −1 4 23 14 ⎥⎦ ⎢⎣ 0 0 8 56 32 ⎥⎦ ⎢⎣0 0 1 7 4 ⎥⎦ 2 5 8⎥⎦ 1 0 −2 1⎤ ⎡ 1 0 0 −7 −1⎤ ⎥ 1 0 5 2 ⎥ → ⎢⎢0 1 0 5 2 ⎥⎥ ⎢⎣0 0 1 7 4 ⎥⎦ 0 1 7 4 ⎥⎦ Thus w = −1 + 7r, x = 2 − 5r, y = 4 − 7r, z = r (where r is any real number).

⎡1 1 1. ⎢⎢ 2 3 ⎢⎣ 2 1 ⎡1 → ⎢⎢ 0 ⎢⎣ 0

204


Section 6.5

1 10 15 −5⎤ ⎡2 ⎡ 1 −5 2 15 −10 ⎤ 2. ⎢ 1 −5 2 15 −10 ⎥ → ⎢ 2 1 10 15 −5⎥ ⎢ ⎥ ⎢ ⎥ 9⎦ 9⎦ ⎣ 1 1 6 12 ⎣ 1 1 6 12 15 −10 ⎤ ⎡ 1 −5 2 ⎡ 1 −5 2 15 −10 ⎤ ⎢ 6 15 ⎥ 1 11 − 15 → ⎢ 0 11 6 −15 15⎥ → 0 11 11 ⎥ ⎢ ⎢ ⎥ ⎣ 0 6 4 −3 19 ⎦ ⎣⎢0 6 4 −3 19 ⎦⎥ 90 − 35 ⎤ 90 − 35 ⎤ ⎡ 1 0 52 ⎡ 1 0 52 11 11 ⎥ 11 11 11 ⎥ 11 ⎢ ⎢ 15 6 − 15 15 → 0 1 6 − 15 → ⎢ 0 1 11 ⎥ ⎢ ⎥ 11 11 11 11 11 ⎢ ⎥ ⎢ 57 119 ⎥ 8 57 119 ⎢⎣ 0 0 11 ⎢⎣0 0 1 8 8 ⎥⎦ 11 11 ⎥⎦ 51 147 ⎡1 0 0 − − 2 ⎤ 2 ⎢ ⎥ 27 → ⎢ 0 1 0 − 21 − ⎥ 4 4 ⎢ ⎥ 57 119 ⎢⎣ 0 0 1 8 8 ⎥⎦ 57 119 51 147 21 27 , z = r (where r is any real number). , x= r− , y=− r+ Thus, w = r − 8 8 2 2 4 4 ⎡ 1 − 1 −1 − 1 − 2 ⎤ ⎡ 3 −1 −3 −1 −2 ⎤ 3 3 3⎥ ⎢ ⎢ 2 −2 −6 −6 −4 ⎥ ⎢ 2 2 6 6 4⎥ − − − − ⎥→ 3. ⎢ ⎢ ⎥ ⎢ 2 −1 −3 −2 −2 ⎥ ⎢ 2 −1 −3 −2 −2 ⎥ ⎢ ⎥ 1 3 7 2 ⎥⎦ ⎢⎣ 3 ⎢⎣ 3 1 3 7 2 ⎥⎦ ⎡ 1 − 1 −1 − 1 − 2 ⎤ ⎡ 1 − 1 −1 − 1 − 2 ⎤ 3 3 3⎥ ⎡ 1 0 0 1 0⎤ 3 3 3⎥ ⎢ ⎢ ⎢0 1 3 4 2⎥ 16 8 4 ⎢0 − −4 − 3 − 3 ⎥ ⎢0 1 3 4 2⎥ 3 ⎢ ⎥ ⎥→⎢ →⎢ → ⎥ 1 −1 − 4 − 2 ⎢ ⎥ 0 0 0 0 0 ⎢ 0 − 1 −1 − 4 − 2 ⎥ − 0 ⎢ 3 3 3⎥ ⎢ ⎥ 3 3 3⎥ ⎢ ⎢ ⎥ ⎣⎢ 0 0 0 0 0 ⎦⎥ 2 6 8 4 ⎦⎥ ⎢⎣ 0 2 6 8 4 ⎥⎦ ⎣⎢ 0 Thus, w = –s, x = –3r – 4s + 2, y = r, z = s (where r and s are any real numbers). 5 1⎤ 5 1⎤ 5 1⎤ ⎡1 1 0 ⎡1 1 0 ⎡1 1 0 ⎡1 ⎢ 1 0 1 2 1⎥ ⎢0 −1 1 −3 0 ⎥ ⎢0 ⎥ ⎢ 1 1 3 0 − ⎥→⎢ ⎥ →⎢ ⎥ → ⎢0 4. ⎢ ⎢ 1 −3 4 −7 1⎥ ⎢0 −4 4 −12 0 ⎥ ⎢ 0 −4 4 −12 0 ⎥ ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 −1 3 0 ⎥⎦ 1 −1 3 0 ⎥⎦ 1 −1 3 0 ⎥⎦ ⎢⎣ 0 ⎢⎣0 ⎢⎣ 0 ⎢⎣ 0 Thus, w = –r – 2s + 1, x = r – 3s, y = r, z = s (where r and s are any real numbers).

0 1 2 1⎤ 1 −1 3 0 ⎥⎥ 0 0 0 0⎥ ⎥ 0 0 0 0 ⎥⎦

2⎤ 2⎤ ⎡ 1 1 3 −1 2 ⎤ ⎡ 1 1 3 −1 ⎡ 1 1 3 −1 ⎡1 ⎢ 2 1 5 −2 0 ⎥ ⎢ 0 −1 −1 0 −4 ⎥ ⎢0 ⎥ ⎢0 1 1 0 4⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 5. ⎢ 2 −1 3 −2 −8⎥ → ⎢ 0 −3 −3 0 −12 ⎥ → ⎢ 0 −3 −3 0 −12 ⎥ → ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 3 2 8 −3 2 ⎥ ⎢ 0 −1 −1 0 −4 ⎥ ⎢ 0 −1 −1 0 −4 ⎥ ⎢0 ⎢⎣ 1 0 2 −1 −2 ⎥⎦ ⎢⎣ 0 −1 −1 0 −4 ⎥⎦ ⎢⎣ 0 −1 −1 0 −4 ⎥⎦ ⎢⎣ 0 Thus, w = –2r + s – 2, x = –r + 4, y = r, z = s (where r and s are any real numbers).

205

0 2 −1 −2 ⎤ 1 1 0 4 ⎥⎥ 0 0 0 0⎥ ⎥ 0 0 0 0⎥ 0 0 0 0 ⎥⎦



2 4⎤ 2 4⎤ ⎡ 1 1 1 2 4⎤ ⎡ 1 1 1 ⎡1 1 1 ⎡1 ⎢2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 1 2 2 7 ⎥ ⎢ 0 −1 0 −2 −1⎥ 1 0 2 1⎥ ⎢ ⎢0 ⎢ 6. ⎢ 1 2 1 4 5⎥ → ⎢ 0 1 0 2 1⎥ → ⎢ 0 1 0 2 1⎥ → ⎢0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ 3 −2 3 −4 7 ⎥ ⎢ 0 −5 0 −10 −5⎥ ⎢ 0 −5 0 −10 −5⎥ ⎢0 ⎢⎣ 4 −3 4 −6 9 ⎥⎦ ⎢⎣ 0 −7 0 −14 −7 ⎥⎦ ⎢⎣ 0 −7 0 −14 −7 ⎥⎦ ⎢⎣0 Thus, w = –r + 3, x = –2s + 1, y = r, z = s (where r and s are any real numbers).

0 1 0 0 0

1 0 3⎤ 0 2 1⎥⎥ 0 0 0⎥ ⎥ 0 0 0⎥ 0 0 0 ⎥⎦

⎡ 4 −3 5 −10 11 −8⎤ ⎡ 0 −5 −5 −10 5 −20 ⎤ 7. ⎢ →⎢ ⎥ 1 5 0 3 6⎦ 1 5 0 3 6 ⎥⎦ ⎣2 ⎣2 1 5 0 3 6⎤ ⎡2 ⎡2 1 5 0 3 6⎤ →⎢ →⎢ ⎥ ⎥ ⎣ 0 −5 −5 −10 5 −20 ⎦ ⎣ 0 1 1 2 −1 4 ⎦ ⎡ 2 0 4 −2 4 2 ⎤ ⎡ 1 0 2 −1 2 1⎤ →⎢ ⎥→⎢ ⎥ ⎣ 0 1 1 2 −1 4 ⎦ ⎣ 0 1 1 2 −1 4 ⎦

Thus, x1 = −2r + s − 2t + 1, x2 = − r − 2 s + t + 4, x3 = r , x4 = s, x5 = t (where r, s, and t are any real numbers). ⎡1 0 ⎢0 1 8. ⎢ ⎢ 2 −2 ⎢ ⎢⎣ 1 2

1⎤ ⎡ 1 0 0 ⎥⎥ ⎢⎢ 0 1 → 3 10 15 10 ⎥ ⎢ 0 −2 ⎥ ⎢ 3 −2 2 −2 ⎥⎦ ⎢⎣ 0 2 ⎡1 1 4 1⎤ ⎡1 0 3 ⎢ ⎢ 0 1 1 −2 ⎥ ⎢0 0 0⎥ ⎢ → →⎢ ⎢ 0 0 1 −4 −7 −8⎥ ⎢0 ⎢ ⎥ ⎢ ⎢⎣ 0 0 0 −7 −16 −19 ⎥⎦ ⎢0 ⎢⎣ 72 33 18 17 Thus x1 = − + r , x2 = − r , 7 7 7 7 3

1

4

3

1

4

1 −2

0

1 −2

0

−3

8

7

0 −3 −2 0 3 0 1 1 0 0

1 0

0 0 x3 =

1

12 7 32 7 15 7 16 7

20 15 − r, 7 7

1⎤ 3 ⎡1 0 ⎢0 1 1 0 ⎥⎥ →⎢ ⎢ 0 0 −1 8⎥ ⎥ ⎢ −3⎥⎦ ⎢⎣0 0 −2 12 − 7 ⎤ ⎡1 0 0 ⎥ ⎢ 38 ⎥ ⎢0 1 0 7 ⎥ →⎢ 20 ⎥ ⎢0 0 1 7 ⎥ ⎢ ⎢0 0 0 19 ⎥ 7 ⎦⎥ ⎣⎢ 19 16 x4 = − r , and 7 7

1⎤ 0 ⎥⎥ 4 7 8⎥ ⎥ 1 −2 −3⎥⎦ ⎤ − 72 0 − 33 7 7 ⎥ 17 18 ⎥ 0 7 7⎥ 15 20 ⎥ 0 7 7 ⎥ 16 19 ⎥ 1 7 7 ⎦⎥ 1

4

−2

0

x5 = r , where r is any real number.

9. The system is homogeneous with fewer equations than unknowns (2 < 3), so there are infinitely many solutions. 10. The system is homogeneous with fewer equations than unknowns (2 < 4), so there are infinitely many solutions. 5⎤ 5⎤ ⎡ 3 −4 ⎤ ⎡ 1 5⎤ ⎡1 ⎡1 ⎡1 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎥ → ⎢⎢ 0 1 ⎥⎥ = A 11. ⎢ 1 5⎥ → ⎢ 3 −4 ⎥ → ⎢ 0 −19 ⎥ → ⎢0 ⎢⎣ 4 −1⎥⎦ ⎢⎣ 4 −1⎥⎦ ⎢⎣ 0 −21⎥⎦ ⎢⎣0 −21⎥⎦ ⎢⎣ 0 0 ⎥⎦ A has k = 2 nonzero rows. Number of unknowns is n = 2. Thus k = n, so the system has the trivial solution only. 3 ⎡ 3 6⎤ ⎡1 3 6 ⎤ ⎡ 1 0 3⎤ 3 12 ⎤ ⎡ 1 2 6 ⎤ ⎢ 1 2 ⎡2 2 ⎥ ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ⎥ ⎥ → ⎢0 ⎥ → ⎢ 0 1 2 ⎥⎥ = A 12. ⎢ 3 −2 5⎥ → ⎢ 3 −2 5⎥ → ⎢ 0 − 13 1 2 13 − 2 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 1 14 ⎦⎥ ⎢ 4 1 14 ⎥ ⎢ 0 −5 −10 ⎥ ⎣⎢ 4 ⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎣ ⎦ ⎣ ⎦ A has k = 2 nonzero rows. Number of unknowns is n = 3. Thus k < n, so the system has infinitely many solutions.

206


Section 6.5

⎡1 1 1⎤ ⎡ 1 1 1⎤ ⎡ 1 1 1⎤ ⎡ 1 0 −1⎤ 1 2 ⎥⎥ → ⎢⎢ 0 1 2 ⎥⎥ = A 13. ⎢⎢1 0 −1⎥⎥ → ⎢⎢0 −1 −2 ⎥⎥ → ⎢⎢0 ⎢⎣1 −2 −5⎥⎦ ⎢⎣0 −3 −6 ⎥⎦ ⎢⎣0 −3 −6 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ A has k = 2 nonzero rows. Number of unknowns is n = 3. Thus k < n, so the system has infinitely many solutions. ⎡1 2 − 2 ⎤ ⎡1 2 − 2⎤ 3 3⎥ ⎡ 3 2 −2 ⎤ ⎡ 1 0 0⎤ ⎡ 1 0 0⎤ 3 3⎥ ⎢ ⎢ 14. ⎢⎢ 2 2 −2 ⎥⎥ → ⎢ 2 2 −2 ⎥ → ⎢ 0 23 − 23 ⎥ → ⎢⎢0 1 −1⎥⎥ → ⎢⎢ 0 1 0 ⎥⎥ = A ⎢ ⎥ ⎢ ⎥ 5⎦⎥ 5⎥ ⎢ 0 −4 ⎥ ⎣⎢ 0 0 1⎦⎥ 5 ⎣⎢ 0 −4 ⎣⎢ 0 0 1⎦⎥ ⎢⎣ 0 −4 ⎦ ⎣ ⎦ A has k = 3 nonzero rows. Number of unknowns is n = 3. Thus k = n, so the system has the trivial solution only. ⎡ 1 1⎤ ⎡ 1 1⎤ ⎡1 15. ⎢ →⎢ →⎢ ⎥ ⎥ ⎣3 −4 ⎦ ⎣ 0 −7 ⎦ ⎣0 The solution is x = 0, y = 0.

1⎤ ⎡1 0⎤ → 1⎥⎦ ⎢⎣ 0 1 ⎥⎦

⎡1 − 5 ⎤ ⎡ 2 −5⎤ ⎡ 2 −5⎤ 2⎥ 16. ⎢ → → ⎢ ⎥ ⎢ 0 0⎥ 8 20 − 0 ⎥⎦ ⎣ ⎦ ⎣ ⎦ ⎣⎢ 0 5 The solution is x = r , y = r. 2 6⎤ ⎡1 0 5⎥ ⎢ ⎢0 1 − 8 ⎥ 15 ⎦ ⎣

6 −2 ⎤ ⎡ 1 6 −2 ⎤ ⎡ 1 6 −2 ⎤ ⎡ 1 →⎢ 17. ⎢ ⎥ ⎥→⎢ 8⎥→ ⎣ 2 −3 4 ⎦ ⎣ 0 −15 8⎦ ⎣⎢ 0 1 − 15 ⎦⎥ 6 8 The solution is x = − r , y = r , z = r. 5 15 7⎤ ⎡ 1 7 ⎤ ⎡1 0 ⎤ ⎡ 4 7 ⎤ ⎡ 1 74 ⎤ ⎡⎢ 1 4⎥ 4⎥ → →⎢ →⎢ 18. ⎢ ⎥→ ⎥ ⎢ ⎥ ⎢⎣ 0 1⎥⎦ ⎣ 0 1 ⎦ ⎣ 2 3 ⎦ ⎢⎣ 2 3 ⎥⎦ ⎢⎣ 0 − 12 ⎥⎦ The solution is x = 0, y = 0.

1⎤ ⎡ 1 1⎤ ⎡ 1 1⎤ ⎡1 ⎡1 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎥ → ⎢⎢ 0 1 ⎥⎥ 19. ⎢ 3 −4 ⎥ → ⎢0 −7 ⎥ → ⎢ 0 ⎢⎣5 −8⎥⎦ ⎢⎣0 −13⎥⎦ ⎢⎣ 0 −13⎥⎦ ⎢⎣ 0 0⎥⎦ The solution is x = 0, y = 0. 3⎤ ⎡ 1 0 3⎤ ⎡ 2 −3 1⎤ ⎡ 0 −5 −1⎤ ⎡ 1 1 1⎤ ⎡1 0 ⎡ 1 0 1⎤ 1 −2 ⎥ → ⎢ 0 1 −2 ⎥ → ⎢ 0 1 − 2 ⎥ → ⎢ 0 1 − 2 ⎥ → ⎢ 0 1 0 ⎥ 20. ⎢ 1 2 −1⎥ → ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1⎦ ⎣ 1 1 1⎦ ⎣ 1 1 1⎦ ⎣ 0 −5 −1⎦ ⎣0 0 −11⎦ ⎣ 0 0 ⎣ 0 0 1⎦ The solution is x = 0, y = 0, z = 0. 1⎤ ⎡ 1 1 1⎤ ⎡ 1 ⎡1 1 ⎢ 0 −7 −14 ⎥ ⎢ 0 1 2 ⎥⎥ ⎢⎢ 0 ⎥→⎢ 21. ⎢ → ⎢ 0 −2 −4 ⎥ ⎢ 0 −2 −4 ⎥ ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0 −5 −10 ⎥⎦ ⎢⎣ 0 The solution is x = r, y = –2r, z = r.

0 −1⎤ 1 2 ⎥⎥ 0 0⎥ ⎥ 0 0 ⎥⎦

207



7⎤ 7 ⎤ ⎡1 ⎡ 1 1 7⎤ ⎡1 1 ⎡1 1 ⎢ 1 −1 −1⎥ ⎢ 0 −2 −8⎥ ⎢0 1 4 ⎥⎥ ⎢⎢ 0 ⎥→⎢ ⎥→⎢ 22. ⎢ → ⎢ 2 −3 −6 ⎥ ⎢ 0 −5 −20 ⎥ ⎢0 −5 −20 ⎥ ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎣ 3 1 13⎥⎦ ⎢⎣ 0 −2 −8⎥⎦ ⎢⎣0 −2 −8⎥⎦ ⎢⎣ 0 The solution is x = –3r, y = –4r, z = r. ⎡1 1 ⎢1 1 23. ⎢ ⎢2 1 ⎢ − 1 3 ⎣⎢

0 3⎤ 1 4 ⎥⎥ 0 0⎥ ⎥ 0 0 ⎥⎦

4⎤ 4⎤ 4⎤ ⎡1 1 1 ⎡1 1 1 ⎥ ⎢ ⎥ ⎢ 5⎥ 0 0 −1 1⎥ 0 −1 1 −4 ⎥⎥ →⎢ →⎢ ⎢0 −1 1 −4 ⎥ ⎢ 0 0 −1 3 4⎥ 1⎥ ⎥ ⎢ ⎥ ⎢ ⎥ 2 −9 ⎦⎥ ⎣⎢0 −4 1 −13⎦⎥ ⎣⎢ 0 −4 1 −13⎦⎥

1 0

4⎤ ⎡ 1 0 2 0⎤ ⎡ 1 ⎡1 1 1 ⎢0 1 −1 4 ⎥⎥ ⎢⎢ 0 1 −1 4 ⎥⎥ ⎢⎢ 0 →⎢ → → ⎢ 0 0 −1 1⎥ ⎢ 0 0 −1 1⎥ ⎢ 0 ⎢ ⎥ ⎢ ⎥ ⎢ ⎣⎢ 0 −4 1 −13⎦⎥ ⎣⎢ 0 0 −3 3⎦⎥ ⎣⎢ 0 The solution is w = –2r, x = –3r, y = r, z = r.

0⎤ ⎡1 ⎥ ⎢0 4⎥ →⎢ ⎢0 0 1 −1⎥ ⎥ ⎢ 0 −3 3⎦⎥ ⎣⎢ 0 0 2 1 −1

7⎤ ⎡ 1 1 2 7⎤ ⎡ 1 1 2 7⎤ ⎡1 1 2 ⎡1 ⎢ 1 −2 −1 1⎥ ⎢ 0 −3 −3 −6 ⎥ ⎢ 0 ⎥ ⎢0 1 1 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 24. → → →⎢ ⎢ 1 2 3 9⎥ ⎢0 ⎢0 1 1 2⎥ ⎢0 1 1 2⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎢⎣ 2 −3 −1 4 ⎥⎦ ⎢⎣ 0 −5 −5 −10 ⎥⎦ ⎢⎣ 0 −5 −5 −10 ⎥⎦ ⎢⎣ 0 The solution is w = –r – 5s, x = –r – 2s, y = r, z = s.

Principles in Practice 6.6 ⎡ 1 3 ⎤ ⎡ −2 1.5⎤ ⎡1 0 ⎤ 1. ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣ 2 4 ⎦ ⎣ 1 −0.5⎦ ⎣ 0 1 ⎦ Yes, they are inverses. ⎡ −2 1.5⎤ ⎡ 28⎤ ⎡13⎤ ⎡ M ⎤ 2. ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣ 1 −0.5⎦ ⎣ 46 ⎦ ⎣ 5 ⎦ ⎣ E ⎦ ⎡ −2 1.5⎤ ⎡ 65⎤ ⎡ 5 ⎤ ⎡ E ⎤ ⎢ 1 −0.5⎥ ⎢90 ⎥ = ⎢ 20 ⎥ = ⎢ T ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡ −2 ⎢ 1 ⎣ ⎡ −2 ⎢ 1 ⎣ ⎡ −2 ⎢ 1 ⎣ ⎡ −2 ⎢ 1 ⎣ ⎡ −2 ⎢ 1 ⎣

1.5⎤ ⎡ 61⎤ ⎡ 1 ⎤ ⎡ A ⎤ = = −0.5⎥⎦ ⎢⎣82 ⎥⎦ ⎢⎣ 20 ⎥⎦ ⎢⎣ T ⎥⎦ 1.5⎤ ⎡59 ⎤ ⎡14 ⎤ ⎡ N ⎤ = = −0.5⎥⎦ ⎢⎣88 ⎥⎦ ⎢⎣15⎥⎦ ⎢⎣ O ⎥⎦ 1.5⎤ ⎡57 ⎤ ⎡15 ⎤ ⎡ O ⎤ = = −0.5⎥⎦ ⎢⎣86 ⎥⎦ ⎢⎣14 ⎥⎦ ⎢⎣ N ⎥⎦ 1.5⎤ ⎡ 60 ⎤ ⎡ 6 ⎤ ⎡ F ⎤ = = −0.5⎥⎦ ⎢⎣84 ⎥⎦ ⎢⎣18⎥⎦ ⎢⎣ R ⎥⎦

1.5⎤ ⎡ 21⎤ ⎡ 9 ⎤ ⎡ I ⎤ = = −0.5⎥⎦ ⎢⎣34 ⎥⎦ ⎢⎣ 4 ⎥⎦ ⎢⎣ D ⎥⎦ ⎡ −2 1.5⎤ ⎡ 76 ⎤ ⎡ 1 ⎤ ⎡ A ⎤ ⎢ 1 −0.5⎥ ⎢102 ⎥ = ⎢ 25⎥ = ⎢ Y ⎥ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎣ ⎦ The message is “MEET AT NOON FRIDAY.” 208

2⎤ 3⎥⎥ 0 1 −1⎥ ⎥ 0 0 0 ⎦⎥

0 0 1 0

0 1 5⎤ 1 1 2 ⎥⎥ 0 0 0⎥ ⎥ 0 0 0 ⎥⎦


⎡3 1 2 1 3. ⎡⎣ E I ⎤⎦ = ⎢⎢ 2 2 2 0 ⎢⎣ 2 1 3 0 ⎡2 2 2 R1 ↔ R 2 ⎢ > ⎢3 1 2 ⎢⎣ 2 1 3

Section 6.6

0 0⎤ 1 0 ⎥⎥ 0 1 ⎥⎦ 0 1 0⎤ 1 0 0 ⎥⎥ 0 0 1 ⎥⎦

⎡ 1 1 1 1 0 0⎤ 2 2 ⎢ ⎥ > ⎢ 3 2 3 0 1 0⎥ ⎢ ⎥ 4 3 4 0 0 1⎥ ⎣⎢ ⎦

1R 2 1

⎡1 ⎢ −3R1 + R 2 ⎢ > 0 −4R1 + R3 ⎢ ⎢0 ⎣

⎡1 1 1 0 0⎤ ⎢ ⎥ > ⎢3 1 2 1 0 0⎥ ⎢ ⎥ ⎢⎣ 2 1 3 0 0 1 ⎥⎦ 1 2

1R 2 1

1 0⎤ ⎡1 1 1 0 2 ⎢ ⎥ R 2 ↔ R3 > ⎢ 0 −1 1 0 −1 1⎥ ⎢ ⎥ 3 ⎢⎣ 0 −2 −1 1 − 2 0 ⎥⎦

⎡1 0 2 0 − 1 1⎤ 2 ⎢ ⎥ −R 2 + R1 > ⎢ 0 1 −1 0 1 −1⎥ 2R 2 + R 3 ⎢ 1 −2 ⎥ ⎢⎣ 0 0 −3 1 ⎥⎦ 2 ⎡1 0 2 0 − 12 1⎤ ⎥ − 13 R3 ⎢ > ⎢0 1 −1 0 1 −1⎥ ⎢ 1 1 2⎥ ⎢⎣0 0 1 − 3 − 6 3 ⎥⎦

⎡ 2 ⎢ 3 E−1 = ⎢ − 13 ⎢ ⎢− 1 ⎣ 3 ⎡2 ⎡⎣ F I ⎤⎦ = ⎢⎢ 3 ⎢⎣ 4

− 16 5 6 − 16

1 2 3

− 16 5 6 1 −6

0

1 0

0 0⎤ ⎥ 1 0⎥ ⎥ −2 0 1⎥ ⎦

1 2 − 32

4. Let x be the number of shares of A, y be the number of shares of B, and z be the number of shares of C. We get the following equations from the given conditions. 50x + 20y + 80z = 500,000 x = 2z 0.13(50x) + 0.15(20y) + 0.10(80z) = 0.12(50x + 20y + 80z) Simplify the first equation. 5x + 2y + 8z = 50,000 Simplify the second equation. x – 2z = 0 Simplify the third equation. 6.5x + 3y + 8z = 6x + 2.4y + 9.6z 0.5x + 0.6y – 1.6z = 0 5x + 6y – 16z = 0 Thus, we solve the following system of equations. x – 2z = 0 5x + 6y – 16z = 0 5x + 2y + 8z = 50,000 ⎡ 1 0 −2 ⎤ The coefficient matrix is A = ⎢⎢5 6 −16 ⎥⎥ . ⎢⎣5 2 8⎥⎦

1 ⎡1 1 1 0 0⎤ 2 ⎢ ⎥ > ⎢0 1 −1 0 1 −1⎥ ⎢ ⎥ 3 0⎥ ⎢⎣ 0 −2 −1 1 − 2 ⎦

2 ⎡1 0 0 3 ⎢ −2R 3 + R1 ⎢ > 0 1 0 − 13 ⎢ R3 + R 2 ⎢0 0 1 − 1 3 ⎣

1

⎡ 1 1 1 1 0 0⎤ 2 2 ⎥ 2R 2 ⎢ > ⎢ 0 1 0 −3 2 0 ⎥ ⎢ ⎥ ⎢⎣ 0 1 0 −2 0 1⎥⎦ ⎡ 1 0 1 2 −1 0 ⎤ − 12 R 2 + R1 ⎢ > ⎢ 0 1 0 −3 2 0 ⎥⎥ −R 2 + R3 ⎢⎣ 0 0 0 1 −2 1⎥⎦ F does not reduce to I so F is not invertible.

1 0⎤ ⎡1 1 1 0 2 ⎢ ⎥ −3R1 + R 2 ⎢ > 0 −2 −1 1 − 32 0 ⎥ ⎥ −2R1 + R3 ⎢ ⎢0 −1 1 0 −1 1⎥ ⎣ ⎦

−R 2

1 2 1 2

− 13 ⎤ ⎥ − 13 ⎥ ⎥ 2⎥ 3⎦

− 13 ⎤ ⎥ − 13 ⎥ ⎥ 2⎥ 3⎦ 2 1 0 0⎤ 3 0 1 0 ⎥⎥ 4 0 0 1 ⎥⎦

⎡ 1 0 −2 1 0 0 ⎤ ⎡⎣ A I ⎤⎦ = ⎢⎢5 6 −16 0 1 0 ⎥⎥ ⎢⎣5 2 8 0 0 1⎥⎦ 1 0 0⎤ ⎡ 1 0 −2 −5R1 + R 2 ⎢ > ⎢ 0 6 −6 −5 1 0 ⎥⎥ 5 − R1 + R3 ⎢⎣ 0 2 18 −5 0 1⎥⎦ 209

Chapter 6: Matrix Algebra ⎡ 1 0 −2 ⎢ > ⎢ 0 1 −1 ⎢ ⎣ 0 2 18 ⎡1 0 ⎢ −2 R 2 + R 3 ⎢ > 0 1 ⎢ ⎢0 0 ⎢⎣ ⎡ 1 0 −2 1 R ⎢ 20 3 > ⎢ 0 1 −1 ⎢ ⎢0 0 1 ⎣ 1R 6 2


1 0 0⎤ 1 0⎥ ⎥ 6 ⎥ −5 0 1⎦

− 56

−2

1

−1

− 56

20 − 10 3 1

0

− 56 − 16

1 6 1 − 60

2 − 1 ⎡1 0 0 3 30 ⎢ 2R3 + R1 ⎢ 3 > 0 1 0 −1 20 R3 + R 2 ⎢ ⎢0 0 1 − 1 − 1 6 60 ⎣

⎡ 2 − 1 30 ⎢ 3 −1 ⎢ 3 A = −1 20 ⎢ ⎢− 1 − 1 60 ⎣ 6

0 0⎤ ⎥ 1 0⎥ 6 ⎥ − 13 1⎥⎥ ⎦ 0⎤ ⎥ 0⎥ ⎥ 1 ⎥ 20 ⎦ 1⎤ 10 ⎥ 1 ⎥ 20 ⎥ 1 ⎥ 20 ⎦

1⎤ 10 ⎥ 1 ⎥ 20 ⎥ 1 ⎥ 20 ⎦

1⎤ ⎡ 2 − 1 30 10 ⎥ ⎡ 0 ⎤ ⎡ 5000 ⎤ ⎡ x⎤ ⎢ 3 ⎢ ⎥ ⎢ 3 1 ⎥ ⎢ 0 ⎥ = ⎢ 2500 ⎥ ⎢ y ⎥ = ⎢ −1 ⎥ ⎢ ⎥ 20 20 ⎥ ⎢ ⎢⎣ z ⎥⎦ ⎢ 1 ⎢⎣50, 000 ⎥⎦ ⎢⎣ 2500 ⎥⎦ 1 1 ⎥ − − 60 20 ⎣ 6 ⎦ They should buy 5000 shares of Company A, 2500 shares of Company B, and 2500 shares of Company C.

Problems 6.6 ⎡ 6 1 1 0 ⎤ ⎡ 1 16 1. ⎢ ⎥→⎢ ⎣ 7 1 0 1 ⎦ ⎢⎣ 7 1

1 0⎤ ⎡ 1 6 ⎥→⎢ 1 ⎢ 0 1 ⎥⎦ ⎣ 0 − 6 1 6

1 6 7 −6

0⎤ ⎥ 1⎥ ⎦

0 −1 1⎤ ⎡ 1 0 −1 1⎤ ⎡1 →⎢ ⎥→ 1 0 − 6 − 76 1⎥ ⎢⎣ 0 1 7 −6 ⎥⎦ ⎣⎢ ⎦ ⎡ −1 1⎤ The inverse is ⎢ ⎥. ⎣ 7 −6 ⎦ 1 1 ⎡1 2 ⎡ 2 4 1 0 ⎤ ⎡⎢1 2 2 0 ⎤⎥ 2 ⎢ 2. ⎢ → → ⎣ 3 6 0 1⎥⎦ ⎢⎣1 2 0 13 ⎥⎦ ⎢⎣ 0 0 − 12 The given matrix is not invertible.

0⎤ ⎥ 1 ⎥ 3⎦

⎡ 2 2 1 0 ⎤ ⎡ 2 2 1 0 ⎤ ⎡ 1 1 12 0 ⎤ → →⎢ 3. ⎢ ⎥ ⎣ 2 2 0 1⎥⎦ ⎢⎣ 0 0 −1 1⎥⎦ ⎣ 0 0 −1 1⎦ The given matrix is not invertible. 210


3 ⎡1 4 8 ⎢ 4. ⎢0 − 1 6 ⎣

Section 6.6

1 0⎤ ⎡ 1 3 4 0⎤ ⎡ 1 0 4 9⎤ 2 ⎥→⎢ ⎥→⎢ ⎥ 0 1⎥ ⎢⎣ 0 1 0 −6 ⎥⎦ ⎣ 0 1 0 −6 ⎦ ⎦

⎡4 9⎤ The inverse is ⎢ ⎥. ⎣ 0 −6 ⎦ ⎡ 0 ⎡ 1 0 0 1 0 0⎤ ⎢ 1 0 0 1 5. ⎢⎢ 0 −3 0 0 1 0 ⎥⎥ → ⎢ 0 1 0 0 − 13 ⎢ ⎢⎣ 0 0 4 0 0 1⎥⎦ ⎢ 0 ⎣0 0 1 0 ⎡1 0 ⎢ 1 ⎢ The inverse is 0 − 3 ⎢ ⎢0 0 ⎣

0⎤ ⎥ 0⎥ ⎥ 1⎥ 4⎦

0⎤ ⎥ 0⎥ . ⎥ 1⎥ 4⎦

⎡ 2 0 8 1 0 0 ⎤ ⎡⎢ 1 0 4 6. ⎢⎢ −1 4 0 0 1 0 ⎥⎥ → ⎢ −1 4 0 ⎢⎣ 2 1 0 0 0 1⎥⎦ ⎢⎢ 2 1 0 ⎣

0 0⎤ ⎥ 0 1 0⎥ ⎥ 0 0 1⎥ ⎦

1 2

⎡ 1 0 4 1 0 0⎤ ⎡1 0 4 1 2 2 ⎢ ⎥ ⎢ 1 ⎢ ⎥ ⎢ → 0 4 4 2 1 0 → 0 1 1 18 ⎢ ⎥ ⎢ ⎢ 0 1 −8 −1 0 1⎥ ⎢0 1 −8 −1 ⎣ ⎦ ⎣ ⎡1 ⎢ → ⎢0 ⎢ ⎢0 ⎢⎣ ⎡1 ⎢ → ⎢0 ⎢ ⎢0 ⎣

0

4

1

1

0 −9 0 0 1 0 0

0 0⎤ ⎡ 1 0 4 ⎥ ⎢ 1 0⎥ → ⎢0 1 1 4 ⎥ ⎢ 1 − 4 1⎥⎥ ⎢⎣ 0 0 1 ⎦

1 2 1 8 9 −8

0 − 19 2 0 9

1 18

1 36

⎡0 − 1 9 ⎢ 2 ⎢ The inverse is 0 9 ⎢ 1 ⎢1 ⎣ 8 36

1 2 1 8 1 8

0 0⎤ ⎥ 1 0⎥ 4 ⎥ 0 1⎥ ⎦ 0 1 4 1 36

0⎤ ⎥ 0⎥ ⎥ − 19 ⎥ ⎦

4⎤ 9⎥ 1⎥ 9⎥ − 19 ⎥ ⎦ 4⎤ 9⎥ 1⎥ 9⎥ 1 − 9⎥ ⎦

.

211



⎡1 2 3 1 0 0 ⎤ ⎡1 2 3 1 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ 7. ⎢ 0 0 4 0 1 0 ⎥ → ⎢0 0 1 0 14 0 ⎥ ⎢ ⎥ ⎣⎢ 0 0 5 0 0 1 ⎥⎦ ⎣0 0 5 0 0 1 ⎦ ⎡ 1 2 0 1 − 3 0⎤ 4 ⎢ ⎥ 1 ⎢ ⎥ → 0 0 1 0 0 4 ⎢ ⎥ ⎢ 0 0 0 0 − 5 1⎥ 4 ⎣ ⎦

The given matrix is not invertible. ⎡1 0 0 ⎡2 0 0 1 0 0⎤ ⎢ 8. ⎢⎢ 0 0 0 0 1 0 ⎥⎥ → ⎢ 0 0 1 ⎢ ⎣⎢ 0 0 −4 0 0 1 ⎦⎥ ⎢ 0 0 0 ⎣ The given matrix is not invertible.

0⎤ ⎥ 0 0 − 14 ⎥ ⎥ 0 1 0⎥ ⎦

1 2

0

9. The matrix is not square, so it is not invertible. ⎡0 0 0⎤ ⎡0 0 0⎤ 10. For any 3 × 3 matrix B, B ⎢⎢ 0 0 0 ⎥⎥ = ⎢⎢ 0 0 0 ⎥⎥ ≠ I. ⎢⎣ 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ Thus the matrix is not invertible. ⎡1 2 3 1 11. ⎢ 0 1 2 0 ⎢ ⎣0 0 1 0 ⎡1 0 0 → ⎢0 1 0 ⎢ ⎣0 0 1

0 1 0 1 0 0

0 ⎤ ⎡ 1 0 −1 1 −2 0 ⎤ 0⎥ → ⎢0 1 2 0 1 0⎥ ⎥ ⎢ ⎥ 1⎦ ⎣ 0 0 1 0 0 1⎦ −2 1⎤ 1 −2 ⎥ ⎥ 0 1⎦

1⎤ ⎡ 1 −2 The inverse is ⎢ 0 1 −2 ⎥ . ⎢ ⎥ 1⎦ ⎣0 0 ⎡ 1 2 −1 1 0 0 ⎤ ⎡ 1 2 −1 1 0 0 ⎤ ⎢ ⎥ 1 4 0 1 0 ⎥⎥ 12. ⎢ 0 1 4 0 1 0 ⎥ → ⎢⎢0 ⎢⎣ 1 −1 2 0 0 1⎥⎦ ⎢⎣0 −3 3 −1 0 1⎥⎦ 2 ⎡1 0 0 1 −2 0 ⎤ 5 ⎡ 1 0 −9 1 −2 0 ⎤ ⎡ 1 0 −9 ⎢ ⎢ ⎥ 4 → ⎢⎢ 0 1 4 0 1 0 ⎥⎥ → ⎢ 0 1 4 0 1 0⎥ → ⎢0 1 0 15 ⎢ 1 1 1⎥ ⎢⎣ 0 0 15 −1 3 1⎥⎦ ⎢ 0 0 ⎢0 0 1 − 1 1 − 15 ⎢⎣ 5 15 ⎥⎦ 15 ⎣

⎡ 2 ⎢ 5 4 The inverse is ⎢ 15 ⎢ ⎢− 1 ⎣ 15

− 15 1 5 1 5

3⎤ 5⎥ 4 ⎥. − 15 ⎥ 1⎥ 15 ⎦

212

− 15 1 5 1 5

3⎤ 5⎥ 4⎥ − 15 ⎥ 1⎥ 15 ⎦


Section 6.6

⎡ 1 0 −2 ⎡ 7 0 −2 1 0 0 ⎤ 7 ⎢ 0 13. ⎢⎢ 0 1 0 0 1 0 ⎥⎥ → ⎢ 0 1 ⎢ ⎢⎣ −3 0 ⎥ 1 0 0 1⎦ 1 ⎢⎣ −3 0 ⎡1 0 − 2 7 ⎢ → ⎢0 1 0 ⎢ 1 ⎢⎣ 0 0 7 ⎡1 The inverse is ⎢⎢ 0 ⎣⎢ 3

1 7

0 3 7

0 0⎤ ⎡ 1 0 ⎥ ⎢ 1 0⎥ → ⎢0 1 ⎥ ⎢ 0 1⎥ ⎢ 0 0 ⎦ ⎣ 0 2⎤ 1 0 ⎥⎥ . 0 7 ⎦⎥

⎡ 2 3 −1 1 0 0 ⎤ ⎡ 1 2 14. ⎢⎢ 1 2 1 0 1 0 ⎥⎥ → ⎢⎢ 1 1 ⎢⎣ −1 −1 3 0 0 1⎥⎦ ⎢⎣ 2 3 1 0 1 0⎤ ⎡ 1 ⎡1 2 → ⎢⎢ 0 1 4 0 1 1⎥⎥ → ⎢⎢ 0 ⎢⎣ 0 −1 −3 1 −2 0 ⎥⎦ ⎢⎣ 0

0 0⎤ ⎥ 0 1 0⎥ ⎥ 0 0 1⎥ ⎦

1 7

1 0 2⎤ ⎡1 0 0 1 0 2 ⎤ ⎥ 0 1 0 ⎥ → ⎢⎢ 0 1 0 0 1 0 ⎥⎥ 3 0 1⎥ ⎣⎢ 0 0 1 3 0 7 ⎦⎥ ⎥⎦ 7

0 0 1 7

1 0

1

−3 0 0 −1 1 0 2 1 0 1 4 0 0

1 1

0⎤ ⎡ 1 2 −1⎥⎥ → ⎢⎢ 0 −1 0 ⎥⎦ ⎢⎣ 0 −1 1 0⎤ ⎡1 2 1 1⎥⎥ → ⎢⎢0 1 ⎢⎣0 0 −1 1⎥⎦

1 0

1

−4 0

−1

−3 1 − 2 0 −1 2 0 −4 1

5

1 −1

0⎤ −1⎥⎥ 0 ⎥⎦ −1⎤ ⎡ 1 0 0 7 −8 5⎤ −3⎥⎥ → ⎢⎢ 0 1 0 −4 5 −3⎥⎥ 1⎥⎦ ⎢⎣ 0 0 1 1 −1 1⎥⎦

⎡ 7 −8 5⎤ The inverse is ⎢⎢ −4 5 −3⎥⎥ . ⎢⎣ 1 −1 1⎥⎦ ⎡ 2 1 0 1 0 0 ⎤ ⎡ 1 −1 15. ⎢⎢ 4 −1 5 0 1 0 ⎥⎥ → ⎢⎢ 4 −1 ⎢⎣ 1 −1 2 0 0 1⎥⎦ ⎢⎣ 2 1 1⎤ ⎡ 1 ⎡ 1 −1 2 0 0 ⎢ ⎢ → ⎢ 0 3 −3 0 1 −4 ⎥⎥ → ⎢ 0 ⎢⎣ 0 3 −4 1 0 −2 ⎥⎦ ⎢ 0 ⎣

1⎤ 1 0 ⎥⎥ 0 1 0 0 ⎥⎦ −1 2 0 0 2 0 0 5 0

1 −1 0 3 −4

1 3

1 0

⎡1 0 1 0 1 − 1⎤ ⎡1 0 1 0 3 3⎥ ⎢ ⎢ → ⎢ 0 1 −1 0 13 − 43 ⎥ → ⎢0 1 −1 0 ⎢ ⎥ ⎢ ⎢ 0 0 −1 1 −1 ⎢0 0 1 −1 2⎥ ⎣ ⎦ ⎣

1 3 1 3

1

1⎤ ⎥ − 43 ⎥ ⎥ −2 ⎦ 5⎤ ⎡1 0 0 1 − 2 − 13 ⎤ 3 3⎥ ⎢ ⎥ 10 4 4 ⎢ ⎥ − 3 → 0 1 0 −1 − 3 ⎥. 3 ⎢ ⎥ ⎥ ⎢ 0 0 1 −1 −2 ⎥ 1 −2 ⎥ ⎢⎣ ⎦ ⎦⎥

5⎤ ⎡ 1 −2 3 3⎥ ⎢ 10 4 ⎢ The inverse is −1 − 3 ⎥. 3 ⎢ ⎥ ⎢ −1 1 −2 ⎥ ⎥⎦ ⎣⎢

213



⎡ −1 2 −3 1 0 0 ⎤ ⎡ 1 −2 3 1 0 0 1 0⎥ → ⎢2 1 0 16. ⎢ 2 ⎢ ⎥ ⎢ − − 4 2 5 0 0 1 4 2 5 ⎣ ⎦ ⎣ 1 − 2 ⎡ 3 −1 0 0 ⎤ ⎡ 1 −2 1 → ⎢ 0 5 −6 2 1 0 ⎥ → ⎢ 0 ⎢ ⎢ ⎥ 0 6 7 4 0 1 − ⎣ ⎦ ⎣⎢ 0 6 3 1 2 ⎡1 0 ⎡1 0 −5 0⎤ 5 5 ⎢ ⎥ ⎢ 6 2 1 → ⎢0 1 − 5 0 ⎥ → ⎢0 1 5 5 ⎢ ⎢ 8 − 6 1⎥ 1 0 0 ⎢⎣ 0 0 ⎥⎦ 5 5 5 ⎣⎢ ⎡ 1 0 0 −5 4 −3⎤ → ⎢ 0 1 0 10 −7 6 ⎥ ⎢ ⎥ ⎣ 0 0 1 8 −6 5 ⎦

−1 0 0 3 6 −5

0 0⎤ 1 0⎥ ⎥ 0 1⎦ −1 0 0 ⎤ 2 1 0⎥ 5 5 ⎥ −7 4 0 1⎦⎥

3 5 − 65

0⎤ ⎥ 0⎥ ⎥ 8 −6 5⎥ ⎦

− 15

1

2 5

2 5 1 5

⎡ −5 4 −3⎤ The inverse is ⎢ 10 −7 6 ⎥ . ⎢ ⎥ ⎣ 8 −6 5⎦ ⎡1 2 3 1 0 0 ⎤ ⎡ 1 2 3 1 0 0⎤ ⎢ ⎥ 17. ⎢1 3 5 0 1 0 ⎥ → ⎢⎢ 0 1 2 −1 1 0 ⎥⎥ ⎢⎣1 5 12 0 0 1 ⎥⎦ ⎢⎣ 0 3 9 −1 0 1⎥⎦ 1⎤ ⎡ 1 0 0 11 −3 ⎡ 1 0 −1 3 −2 0 ⎤ 3 3⎥ ⎡ 1 0 −1 3 −2 0 ⎤ ⎢ ⎢ ⎥ ⎢ ⎥ 7 2 → ⎢ 0 1 2 −1 1 0 ⎥ → ⎢0 1 2 −1 1 0 ⎥ → ⎢ 0 1 0 − 3 3 − 3⎥ ⎢ ⎥ ⎢ ⎥ 2 1 2 −1 1⎥ ⎢0 0 1 ⎣⎢ 0 0 3 2 −3 1⎥⎦ ⎢⎣0 0 1 3 −1 3 ⎥⎦ 3 3⎦ ⎣ 1⎤ ⎡ 11 −3 3⎥ ⎢ 3 The inverse is ⎢ − 73 3 − 23 ⎥ . ⎢ ⎥ 1⎥ ⎢ 2 −1 3⎦ ⎣ 3

⎡ 2 −1 3 1 0 0 ⎤ ⎡ 2 −1 18. ⎢⎢ 0 2 0 0 1 0 ⎥⎥ → ⎢⎢ 0 2 ⎢⎣ 2 1 1 0 0 1⎥⎦ ⎢⎣ 0 2 3 1 0 0⎤ ⎡1 − 1 ⎡1 2 2 2 ⎢ ⎥ ⎢ ⎢0 1 0 0 12 0 ⎥ → ⎢0 ⎢ ⎥ ⎢ ⎢0 ⎢0 2 −2 −1 0 1⎥ ⎣ ⎦ ⎣ ⎡1 0 3 2 ⎢ → ⎢0 1 0 ⎢ ⎢0 0 1 ⎣

1 2

0 1 2

1 4 1 2 1 2

1 0 0⎤ 1 0 ⎥⎥ −2 −1 0 1⎥⎦ 3

0

0

0⎤ ⎥ 1 0 0 0⎥ ⎥ 0 −2 −1 −1 1⎥ ⎦ 0

3 2

1 2

1 4 1 2

0 ⎤ ⎡ 1 0 0 − 14 − 12 ⎥ ⎢ 1 0⎥ → ⎢0 1 0 0 2 ⎥ ⎢ 1 1 1 ⎥ ⎢ 0 0 1 −2 2 2 ⎦ ⎣

⎡− 1 − 1 2 ⎢ 4 1 The inverse is ⎢ 0 2 ⎢ 1 ⎢ 1 2 ⎣ 2

3⎤ 4⎥

0⎥ ⎥ − 12 ⎥ ⎦

3⎤ 4⎥

0⎥ . ⎥ − 12 ⎥ ⎦ 214


Section 6.6

⎡x ⎤ ⎡1 2 ⎤ ⎡ 2 ⎤ ⎡10 ⎤ 19. X = ⎢ 1 ⎥ = A −1B = ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⇒ x1 = 10, x2 = 20 ⎣8 1 ⎦ ⎣ 4 ⎦ ⎣ 20⎦ ⎣ x2 ⎦ ⎡ 1 0 1 ⎤ ⎡10 ⎤ ⎡ 9 ⎤ ⎡ x1 ⎤ −1 20. X = ⎢ ⎥ = A B = ⎢⎢ 0 3 0 ⎥⎥ ⎢⎢ 2 ⎥⎥ = ⎢⎢ 6 ⎥⎥ ⇒ x1 = 9, x2 = 6, x3 = 16 ⎣ x2 ⎦ ⎢⎣ 2 0 4 ⎥⎦ ⎢⎣ −1⎥⎦ ⎢⎣16 ⎥⎦ 1⎤ ⎡6 5 1 0⎤ ⎡1 1 0 1 ⎤ ⎡ 1 1 0 ⎡ 1 1 0 1⎤ ⎡ 1 0 1 −5⎤ 21. ⎢ ⎥ → ⎢ 6 5 1 0 ⎥ → ⎢ 0 −1 1 −6 ⎥ → ⎢ 0 1 −1 6 ⎥ → ⎢ 0 1 −1 6 ⎥ 1 1 0 1 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ x 1 − 5 2 17 ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤ −1 ⎢ y ⎥ = A B = ⎢ −1 6 ⎥ ⎢ −3⎥ = ⎢ −20⎥ ⇒ x = 17, y = −20 ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ ⎡1 2 ⎡ 2 4 1 0 ⎤ ⎡ 1 2 12 0 ⎤ →⎢ 22. ⎢ ⎥→⎢ ⎥ ⎢0 5 ⎣ −1 3 0 1⎦ ⎢⎣ −1 3 0 1⎥⎦ ⎣ ⎡3 ⎡ x⎤ −1 ⎢ 10 = = A B ⎢ y⎥ ⎢1 ⎣ ⎦ ⎣ 10

1 2 1 2

⎡1 0 3 0⎤ 10 ⎥→⎢ ⎢0 1 1 1⎥⎦ 10 ⎣

23 ⎤ − 52 ⎤ ⎡ 5⎤ ⎡ 10 ⎥ ⎢ ⎥ = ⎢ ⎥ ⇒ x = 23 , y = 1 1 ⎥ −2 1 ⎥ 10 10 ⎣ ⎦ ⎣⎢ 10 5⎦ ⎦

1 0⎤ ⎡1 1 ⎡ 1 1 1 0⎤ ⎡ 1 1 ⎡3 1 1 0 ⎤ 3 3 3 3 3 23. ⎢ →⎢ ⎥→⎢ ⎥→⎢ ⎣3 −1 0 1⎦⎥ ⎢⎣0 1 ⎣3 −1 0 1⎦ ⎣ 0 −2 −1 1⎦ 1⎤ ⎡1 0 1 6 6⎥ →⎢ 1 1 ⎢⎣ 0 1 2 − 2 ⎥⎦

⎡1 ⎡x⎤ −1 ⎢6 = = A B ⎣⎢ y ⎦⎥ ⎢⎣ 12

− 52 ⎤ ⎥ 1⎥ 5⎦

1⎤ 6 ⎥ ⎡5 ⎤ 1 − 2 ⎥⎦ ⎣⎢7 ⎦⎥

⎡ 3 2 1 0 ⎤ ⎡ 1 23 24. ⎢ ⎥→⎢ ⎣ 4 3 0 1 ⎦ ⎢⎣ 4 3

0⎤ ⎥ − 12 ⎥⎦

⎡ 2⎤ = ⎢ ⎥ ⇒ x = 2, y = −1 ⎣ −1⎦

0⎤ ⎡ 1 3 ⎥→⎢ 0 1 ⎥⎦ ⎢⎣ 0 13 1 3

1 3 1 2

2

−

1 3 4 3

0⎤ 3 −2 ⎤ ⎡1 0 3 −2 ⎤ ⎡1 0 ⎥ →⎢ ⎥→⎢ 1 4 − 0 1 ⎥ − 0 1 4 3⎥⎦ 1 ⎢⎣ ⎥⎦ ⎣ 3 3 ⎦

⎡ x⎤ ⎡ 3 −2 ⎤ ⎡ 26 ⎤ ⎡ 4 ⎤ −1 = ⇒ x = 4, y = 7 ⎢ y ⎥ = A B = ⎢ −4 3⎥⎦ ⎢⎣37 ⎥⎦ ⎢⎣7 ⎥⎦ ⎣ ⎦ ⎣

25. The coefficient matrix is not invertible. The method of reduction yields ⎡ 2 6 2 ⎤ ⎡1 3 1 ⎤ ⎡1 3 1 ⎤ ⎢ 3 9 3 ⎥ → ⎢ 3 9 3⎥ → ⎢ 0 0 0 ⎥ . ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Thus x = –3r + 1, y = r. 26. The coefficient matrix is not invertible. The method of reduction yields ⎡ 2 6 8⎤ ⎡ 1 3 4 ⎤ ⎡ 1 3 4 ⎤ ⎢ ⎥→⎢ ⎥→⎢ ⎥. ⎣ 3 9 7 ⎦ ⎣3 9 7 ⎦ ⎣ 0 0 −5⎦ Second row indicates 0 = −5, which is never true, so there is no solution.

215



1 1 0 0⎤ ⎡1 2 1 1 0 0 ⎤ ⎡ 1 2 27. ⎢⎢3 0 1 0 1 0 ⎥⎥ → ⎢⎢ 0 −6 −2 −3 1 0 ⎥⎥ ⎢⎣1 −1 1 0 0 1 ⎥⎦ ⎢⎣ 0 −3 0 −1 0 1⎥⎦ 1 ⎡1 0 1 0 0 0⎤ ⎡1 2 1 1 3 3 ⎢ ⎢ ⎥ 1 1 1 1 1 1 ⎢ → ⎢0 1 3 2 − 6 0⎥ → 0 1 3 2 − 6 ⎢ ⎢ ⎥ ⎢0 0 1 1 − 1 0 1⎦ ⎣ 0 −3 0 −1 2 2 ⎣ ⎡1 0 0 − 1 6 ⎢ 1 → ⎢0 1 0 3 ⎢ 1 ⎢0 0 1 2 ⎣

0⎤ ⎥ 0⎥ ⎥ 1⎥⎦

− 13 ⎤ ⎥ 0 − 13 ⎥ ⎥ − 12 1⎥⎦ 1 2

1 − 1⎤ ⎡− 1 2 3 ⎥ ⎡ 4⎤ ⎡0 ⎤ ⎡ x⎤ ⎢ 6 ⎢ y ⎥ = A −1B = ⎢ 1 1 ⎥ ⎢ 2⎥ = ⎢1 ⎥ 0 − ⎢ ⎥ 3⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 3 ⎢⎣ z ⎥⎦ ⎢ ⎥ ⎢ ⎥ ⎢ 1 −1 1⎥⎦ ⎣ 1 ⎦ ⎣ 2 ⎦ 2 ⎣ 2 Thus, x = 0, y = 1, z = 2.

⎡1 1 1 1 0 0 ⎤ ⎡ 1 1 1 1 0 0⎤ 28. ⎢⎢1 −1 1 0 1 0 ⎥⎥ → ⎢⎢ 0 −2 0 −1 1 0 ⎥⎥ ⎢⎣1 −1 −1 0 0 1⎥⎦ ⎢⎣ 0 −2 −2 −1 0 1⎥⎦ 1 0 0⎤ ⎡ 1 0 ⎡1 1 1 1 ⎢ ⎢ ⎥ → ⎢0 1 0 12 − 12 0 ⎥ → ⎢ 0 1 0 ⎢ ⎥ ⎢ 0 1⎦ ⎢ 0 0 −2 ⎣ 0 −2 −2 −1 ⎣ ⎡1 0 1 ⎢ → ⎢0 1 0 ⎢ ⎢0 0 1 ⎣

1 2 1 2

0

1 2 1 −2 1 2

⎡1 0 0 0⎤ ⎥ ⎢ 0 ⎥ → ⎢0 1 0 ⎥ ⎢ ⎢0 0 1 − 12 ⎥⎦ ⎣

1 2 1 2

0⎤ ⎥ 0⎥ ⎥ −1 1⎥ ⎦

1 2 1 2

1 2 1 −2

0

1⎤ 2⎥

0

0⎥ ⎥ − 12 ⎥⎦

− 12 1 2

0

1⎤ ⎡ ⎤ ⎡1 0 2 ⎥ ⎡ 6 ⎤ ⎢ 5⎥ ⎡ x⎤ ⎢2 ⎢ y ⎥ = A −1B = ⎢ 1 − 1 0 ⎥ ⎢⎢ −1⎥⎥ = ⎢ 72 ⎥ ⎢ ⎥ 2 ⎢ ⎥ ⎢2 ⎥ ⎢⎣ z ⎥⎦ 1 − 1 ⎥ ⎢⎣ 4 ⎥⎦ ⎢ − 5 ⎥ ⎢0 2 2⎦ ⎣ ⎣ 2⎦ 7 5 Thus, x = 5, y = , z = − . 2 2

⎡1 1 1 1 0 0 ⎤ ⎡ 1 1 1 1 0 0⎤ ⎢ ⎥ 29. ⎢1 −1 1 0 1 0 ⎥ → ⎢⎢ 0 −2 0 −1 1 0 ⎥⎥ ⎢⎣1 −1 −1 0 0 1⎥⎦ ⎢⎣ 0 −2 −2 −1 0 1⎥⎦ 1 0 0⎤ ⎡ 1 0 ⎡1 1 1 1 ⎢ ⎢ ⎥ → ⎢0 1 0 12 − 12 0 ⎥ → ⎢ 0 1 0 ⎢ ⎥ ⎢ 0 1⎦ ⎢ 0 0 −2 ⎣ 0 −2 −2 −1 ⎣

1 2 1 2

0

0⎤ ⎥ 0⎥ ⎥ −1 1⎥ ⎦

1 2 − 12

216


⎡1 0 1 ⎢ → ⎢0 1 0 ⎢ ⎢0 0 1 ⎣

1 2 1 2

0

1 2 − 12 1 2

Section 6.6

⎡1 0 0 0⎤ ⎥ ⎢ 0 ⎥ → ⎢0 1 0 ⎥ ⎢ ⎢0 0 1 − 12 ⎥⎦ ⎣

1 2 1 2

1⎤ 2⎥

0 − 12

0

1 2

0⎥ ⎥ − 12 ⎥⎦

1⎤ ⎡1 ⎡ ⎤ 0 2 ⎥ ⎡ 2⎤ ⎢ 1 ⎥ ⎡ x⎤ ⎢2 ⎢ y ⎥ = A −1B = ⎢ 1 − 1 0 ⎥ ⎢⎢ 1 ⎥⎥ = ⎢ 12 ⎥ ⎢ ⎥ 2 ⎢2 ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ 1 − 1 ⎥ ⎢⎣ 0 ⎥⎦ ⎢ 1 ⎥ ⎢0 2 2⎦ ⎣ ⎣2⎦ 1 1 Thus, x = 1, y = , z = . 2 2

⎡ 2 0 8 1 0 0 ⎤ ⎡ 1 −4 0 30. ⎢⎢ −1 4 0 0 1 0 ⎥⎥ → ⎢⎢ 2 0 8 ⎢⎣ 2 1 0 0 0 1⎥⎦ ⎢⎣ 2 1 0 ⎡ 1 −4 ⎡ 1 −4 0 0 −1 0 ⎤ ⎢ ⎢ ⎥ → ⎢0 8 8 1 2 0 ⎥ → ⎢0 1 ⎢ ⎢⎣ 0 9 0 0 2 1⎥⎦ ⎣0 9

0 −1 0 ⎤ 1 0 0 ⎥⎥ 0 0 1⎥⎦ 0 −1 0 ⎤ ⎥ 1 18 14 0 ⎥ ⎥ 0 0 2 1⎦

0

1 ⎡1 0 4 0 0⎤ ⎡ 1 0 4 2 ⎢ ⎥ ⎢ 1 1 0⎥ → ⎢0 1 1 → ⎢0 1 1 8 4 ⎢ ⎥ ⎢ ⎢ 0 0 −9 − 9 − 1 1⎥ ⎢ 0 0 1 ⎢⎣ ⎥⎦ ⎣ 8 4 4⎤ ⎡0 − 1 9 9 ⎥ ⎡ 8 ⎤ ⎡0⎤ ⎡ x⎤ ⎢ ⎢ y ⎥ A −1B = ⎢ 0 2 1 ⎥ ⎢36 ⎥ = ⎢9 ⎥ ⎢ ⎥ 9 9 ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢ ⎥ ⎢ ⎥ 1 1 ⎢1 − 9 ⎥ ⎣ 9 ⎦ ⎣1 ⎦ ⎣ 8 36 ⎦ Thus, x = 0, y = 9, z = 1.

1 2 1 8 1 8

0 1 4 1 36

⎡1 0 0 0⎤ ⎥ ⎢ 0⎥ → ⎢0 1 0 ⎥ ⎢ ⎢0 0 1 − 19 ⎥ ⎦ ⎣

0 − 19 2 0 9 1 8

1 36

31. The coefficient matrix is not invertible. The method of reduction yields 7⎤ ⎡ 1 3 3 7⎤ ⎡1 3 3 7 ⎤ ⎡1 3 3 ⎡1 0 0 1 ⎤ ⎢ 2 1 1 4 ⎥ → ⎢ 0 −5 −5 −10 ⎥ → ⎢ 0 ⎥ → ⎢0 1 1 2⎥ . 1 1 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 1 1 4 ⎥⎦ ⎢⎣0 −2 −2 −3⎥⎦ ⎢⎣ 0 −2 −2 −3⎥⎦ ⎢⎣ 0 0 0 1 ⎥⎦ The third row indicates that 0 = 1, which is never true, so there is no solution. 32. The coefficient matrix is not invertible. The method of reduction yields 7⎤ ⎡ 1 3 3 7⎤ ⎡1 3 3 7 ⎤ ⎡1 3 3 ⎡1 0 0 1 ⎤ ⎢ 2 1 1 4 ⎥ → ⎢ 0 −5 −5 −10 ⎥ → ⎢ 0 ⎥ → ⎢0 1 1 2⎥ . 1 1 2 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 1 1 1 3 ⎥⎦ ⎢⎣0 −2 −2 −4 ⎥⎦ ⎢⎣ 0 −2 −2 −4 ⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦ Thus, x = 1, y = –r + 2, z = r.

217

4⎤ 9⎥ 1⎥ 9⎥ − 19 ⎥ ⎦


⎡1 0 ⎢ 1 −1 33. ⎢ ⎢2 1 ⎢ ⎢⎣ 1 2 ⎡1 ⎢0 →⎢ ⎢0 ⎢ ⎢⎣ 0


2 1 1 0 0 0⎤ ⎡1 0 2 ⎢ 0 −1 −2 0 2 0 1 0 0 ⎥⎥ →⎢ ⎢ 0 1 −4 0 1 0 0 1 0⎥ ⎥ ⎢ 1 1 0 0 0 1⎥⎦ ⎢⎣ 0 2 −1 ⎡1 0 0 2 1 1 0 0 0⎤ ⎢0 1 ⎥ 1 2 −1 1 −1 0 0 ⎥ ⎢ →⎢ 0 0 0 −6 0 −3 1 1 0 ⎥ ⎢ ⎥ 0 −5 2 −3 2 0 1⎥⎦ ⎢⎣ 0 0

⎡1 ⎢ ⎢0 →⎢ ⎢0 ⎢ ⎢0 ⎢⎣

0 0

⎡1 ⎢ ⎢0 →⎢ ⎢0 ⎢ ⎢0 ⎢⎣

0 0 0

1

0

1 0 −1

0

0

1

0

0 0

2

1 0 0 0

1 0

0 0

1

1 2 1 −2 1 4 1 −4 1 2 − 14

⎡ 1 ⎡ w⎤ ⎢ 4 ⎢x⎥ ⎢− 1 ⎢ ⎥ = A −1B = ⎢ 4 ⎢ y⎥ ⎢ 1 ⎢ ⎥ ⎢ 2 ⎢⎣ z ⎥⎦ ⎢− 1 ⎢⎣ 4

1 1 0 0 0⎤ 1 −1 1 0 0 ⎥⎥ −1 −2 0 1 0 ⎥ ⎥ 0 −1 0 0 1⎥⎦ 2 1 1 0

0 0⎤ 2 −1 1 −1 0 0 ⎥⎥ 1 0 12 − 16 − 16 0 ⎥ ⎥ −5 2 −3 2 0 1⎥⎦

1 3 − 23 − 16 7 6

1 3 1 3 1 −6 − 56

⎡1 0 0⎤ ⎥ ⎢ ⎢0 1 0⎥ ⎥ →⎢ ⎢0 0 0⎥ ⎥ ⎢ ⎢0 0 1⎥⎥ ⎣⎢ ⎦

− 14

3 4 1 − 12 − 16 5 − 12

− 12 ⎤ ⎥ 1⎥ 2⎥ 0⎥ ⎥ 1⎥ 2 ⎥⎦

1 − 12

− 16 7 12

− 14 1 − 12

− 16 7 12

3 4 1 − 12 − 16 5 − 12

0

1

0

0 −1

0

1

0

0

1

− 12 ⎤ ⎥ ⎡ 4 ⎤ ⎡ 1⎤ 1⎥⎢ ⎥ ⎢ ⎥ 2 ⎥ ⎢12 ⎥ ⎢ 3⎥ = 0 ⎥ ⎢12 ⎥ ⎢ −2 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 1 ⎥ ⎢⎣12 ⎥⎦ ⎢⎣ 7 ⎥⎦ 2 ⎥⎦

Thus, w = 1, x = 3, y = –2, z = 7.

218

1 2 1 −4

1 3 − 23 − 16 7 12

1 3 1 3 1 −6 5 − 12

0⎤ ⎥ 0⎥ ⎥ 0⎥ ⎥ 1⎥ 2 ⎦⎥


⎡ 1 1 −1 ⎢ 0 1 1 34. ⎢ ⎢ −1 1 1 ⎢ ⎢⎣ 1 −1 −1 ⎡1 ⎢0 ⎢ →⎢ 0 ⎢ ⎢0 ⎢⎣ ⎡1 ⎢0 ⎢ →⎢ 0 ⎢ ⎢0 ⎢⎣ ⎡1 ⎢ ⎢0 →⎢ ⎢0 ⎢ ⎢0 ⎢⎣

1 0 0 0⎤ ⎡ 1 1 −1 ⎥ ⎢0 1 0 1 0 0⎥ 1 1 →⎢ ⎢0 2 0 0 0 0 1 0⎥ ⎥ ⎢ 2 0 0 0 1⎥⎦ ⎢⎣ 0 −2 0 0

1 −1 1 1

1

0

1

0

1 0

0 −1 −1

1 2 1 −2

0

0

1

2

−1

1 0

1

1 1

0

1

0

1 1 − 12

1

0

0

1

1 0 0 1 0 0 1 0

0 0

1

0 0 1 2 1 2 1 −2

0 1 2

1 0

1 −1 0

0

Section 6.6

1 0 1

0 0

0⎤ ⎡ 1 0 ⎥⎥ ⎢⎢ 0 → 0⎥ ⎢0 ⎥ ⎢ 1⎥ ⎢0 ⎢⎣ 2 ⎥⎦

0⎤ ⎡ 1 ⎢ 0 0 ⎥⎥ ⎢ 0 →⎢ − 12 0 ⎥ ⎢ 0 ⎥ ⎢ 1 1⎥ ⎢⎣ 0 2 2 ⎥⎦ ⎡1 −1 − 12 ⎤ ⎢ ⎥ 1 ⎢0 ⎥ 0 2 ⎥→⎢ ⎢0 −1 − 12 ⎥ ⎢ ⎥ 1 1⎥ ⎢0 ⎢⎣ 2 2 ⎥⎦ 0

⎡ 1 1 −1 1 0 0 0⎤ ⎢0 1 1 ⎥ 1 0 1 0 0⎥ ⎢ →⎢ 0 1 0 0 1 0 1 0⎥ ⎢ ⎥ ⎢0 −1 0 2 −1 0 0 1⎥⎦ ⎣⎢ 0

1 −1 0 1 0 0 1 1 0 0 0 1 0 0

0⎤ 1 1 0 1 0 0 ⎥⎥ 1 1 − 12 1 − 12 0 ⎥ ⎥ 1 2 − 12 1 0 12 ⎥⎥⎦ −1 0 1 0 0 0⎤ ⎥ 1 1 1 0 0 1 − 2 − 2⎥ ⎥ 1 0 − 12 1 −1 − 12 ⎥ ⎥ 1 1 0 1 0 0 2 2 ⎥⎦ 0 0 0 1 − 32 − 12 ⎤ ⎥ 1 0 1 ⎥ 0 0 0 2 2 ⎥ 1 0 − 12 1 −1 − 12 ⎥ ⎥ 1 1⎥ 0 1 0 0 2 2 ⎥⎦ 1 0

0

⎡ 0 1 − 3 − 1⎤ 2 2 ⎥ ⎡1 ⎤ ⎡ −2 ⎤ ⎡ w⎤ ⎢ 1 ⎢x⎥ ⎢ 1 0 0 ⎥ ⎢ 0 ⎥ ⎢ 1⎥ 2 ⎢ ⎥ = A −1B = ⎢ 2 ⎥⎢ ⎥ = ⎢ ⎥ ⎢ y⎥ ⎢ − 1 1 −1 − 1 ⎥ ⎢1 ⎥ ⎢ −2 ⎥ 2⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 2 ⎢⎣ z ⎥⎦ 1 1 ⎥ ⎢⎣1 ⎥⎦ ⎢⎣ 1⎥⎦ ⎢ 0 0 ⎢⎣ 2 2 ⎥⎦ Thus, w = −2, x = 1, y = −2, z = 1. ⎡1 0 ⎤ ⎡5 −2 ⎤ ⎡ −4 2⎤ 35. I − A = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣0 1 ⎦ ⎣ 1 2 ⎦ ⎣ −1 −1⎦ ⎡ −4 2 1 0 ⎤ ⎡ −1 −1 0 1⎤ ⎡ 1 1 0 −1⎤ ⎡ 1 1 0 −1⎤ ⎢⎣ −1 −1 0 1⎥⎦ → ⎢⎣ −4 2 1 0 ⎥⎦ → ⎢⎣ −4 2 1 0 ⎥⎦ → ⎢⎣ 0 6 1 −4⎥⎦ ⎡ 1 1 0 −1⎤ ⎡ 1 0 − 16 − 13 ⎤ ⎥ →⎢ 1 2⎥ → ⎢ 1 −2 ⎥ ⎣ 0 1 6 − 3 ⎦ ⎢⎣ 0 1 6 3⎦ ⎡− 1 Thus, (I − A)−1 = ⎢ 16 ⎢⎣ 6

− 13 ⎤ ⎥. − 23 ⎥⎦

219

0

1 0

1

0

0

1 2 1 −2

1

0

1 0 0

1 2

0

0

0⎤ 0 ⎥⎥ 0⎥ ⎥ 1⎥ 2 ⎦⎥



⎡1 0 ⎤ ⎡ −3 2 ⎤ ⎡ 4 −2 ⎤ 36. I − A = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣ 0 1 ⎦ ⎣ 4 3⎦ ⎣ −4 −2 ⎦ ⎡ 4 −2 1 0 ⎤ ⎡ 1 − 12 14 0 ⎤ ⎥ ⎢ −4 −2 0 1⎥ → ⎢ ⎣ ⎦ ⎢⎣ −4 −2 0 1⎥⎦ ⎡1 − 1 2 →⎢ ⎢⎣ 0 −4

1 ⎡1 − 1 0⎤ 2 4 ⎢ ⎥→ 1 ⎢0 − 1 1 1⎥⎦ 4 ⎣

1 4

⎡ 1 8 Thus (I − A) −1 = ⎢ ⎢− 1 ⎣ 4

1 ⎡1 0 0⎤ 8 ⎥ →⎢ 1 1 ⎢0 1 − − 4 ⎦⎥ 4 ⎣

− 18 ⎤ ⎥ − 14 ⎥⎦

− 18 ⎤ ⎥ − 14 ⎥⎦

37. Let x = number of model A and y = number of model B. a.

The system is ⎧⎪ x + y = 100 (painting) ⎨1 ⎪⎩ 2 x + y = 80 (polishing) ⎡ 1 1⎤ Let A = ⎢ 1 ⎥ . ⎢⎣ 2 1⎥⎦ 1 0⎤ ⎡ 1 1 1 0⎤ ⎡1 1 ⎡ 1 1 1 0⎤ ⎡ 1 0 2 −2 ⎤ →⎢ ⎢1 ⎥→⎢ ⎥ →⎢ ⎥ ⎥ 1 1 ⎣ 0 1 −1 2 ⎦ ⎣ 0 1 −1 2 ⎦ ⎣⎢ 2 1 0 1 ⎦⎥ ⎣⎢0 2 − 2 1⎥⎦ ⎡ x⎤ −1 ⎡100 ⎤ ⎡ 2 −2 ⎤ ⎡100 ⎤ ⎡ 40 ⎤ ⎢ y ⎥ = A ⎢ 80 ⎥ = ⎢ −1 2 ⎥ ⎢ 80 ⎥ = ⎢ 60 ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ Thus 40 of model A and 60 of model B can be produced.

b. The system is (widgets) ⎧10 x + 7 y = 800 ⎨ x y (shims) 14 + 10 = 1130 ⎩ ⎡10 7 ⎤ Let A = ⎢ ⎥. ⎣14 10 ⎦ 7 ⎡10 7 1 0 ⎤ ⎡ 1 10 ⎢14 10 0 1⎥ → ⎢ ⎣ ⎦ ⎢⎣14 10

⎡1 →⎢ ⎢0 ⎣

7 10 1 5

1 10 − 75

0⎤ ⎡1 ⎥→⎢ ⎢⎣ 0 1⎥ ⎦

0⎤ ⎥ 0 1⎥⎦

1 10

⎡1 0 0⎤ 5 − 72 ⎤ ⎥ →⎢ ⎥ ⎢⎣ 0 1 −7 5⎥⎦ 1 −7 5⎥⎦

7 10

1 10

⎡ 5 − 7 ⎤ ⎡ 800 ⎤ ⎡ 45⎤ ⎡ x⎤ −1 ⎡ 800 ⎤ 2⎥ A = = ⎢ ⎥ ⎢ ⎥ ⎢ ⎢ ⎥=⎢ ⎥ 5⎦⎥ ⎣1130 ⎦ ⎣50 ⎦ ⎣ y⎦ ⎣1130 ⎦ ⎣⎢ −7 Thus 45 of model A and 50 of model B can be produced. ⎡1 ⎢a 38. ⎢ 0 ⎢ ⎢0 ⎣

0 1 b

0

0⎤ a 0 0 ⎤ ⎡1 0 0 ⎤ ⎥⎡ ⎢ ⎥ 0 ⎢ 0 b 0 ⎥⎥ = ⎢⎢ 0 1 0 ⎥⎥ = I ⎥ 1 ⎥ ⎢⎣ 0 0 c ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ c⎦ 220


39. a.

(B

−1 −1

A

) (AB) = B ( A A ) B = B −1

−1

Section 6.6

−1

IB = B −1B = I

Since an invertible matrix has exactly one inverse, B −1A −1 is the inverse of AB. b. From Part (a), ⎡1 1 ⎤ ⎡1 2 ⎤ ⎡ 4 6 ⎤ ( AB)−1 = B −1A −1 = ⎢ ⎥⎢ ⎥=⎢ ⎥ ⎣1 2 ⎦ ⎣3 4 ⎦ ⎣ 7 10 ⎦ ⎡1 − 1 ⎤ ⎡1 1 ⎤ 2⎥ T −1 ⎢ . 40. Left side: A T = ⎢ A We find that ( ) . = ⎥ ⎢0 1 ⎥ ⎣0 2 ⎦ 2 ⎦ ⎣ 1 ⎡ ⎤ 1 −2 ⎡ 1 0⎤ ⎥. Right side: A −1 = ⎢ 1 1 ⎥ , so ( A −1 )T = ⎢ 1⎥ ⎢0 ⎣⎢ − 2 2 ⎦⎥ 2⎦ ⎣

Thus ( A T )−1 = ( A −1 )T . ⎡ 3 5 41. P T P = ⎢ ⎢− 4 ⎣ 5

42. a.

A

−1

4⎤⎡3 5⎥⎢5 3⎥ ⎢4 5⎦ ⎣5

− 54 ⎤ ⎡1 0 ⎤ ⎥= = I, so P T = P −1 . Yes, P is orthogonal. 3 ⎥ ⎢0 1 ⎥ ⎣ ⎦ 5⎦

⎡ 14 −2 9 ⎤ = ⎢⎢ −6 1 −4 ⎥⎥ ⎢⎣ 1 0 1⎥⎦

⎡ 14 −2 9 ⎤ R1A = [33 87 70] ⎢⎢ −6 1 −4 ⎥⎥ = [10 21 19] ⎢⎣ 1 0 1⎥⎦ ⎡ 14 −2 9 ⎤ R 2 A −1 = [57 133 20] ⎢⎢ −6 1 −4 ⎥⎥ = [ 20 19 1] ⎢⎣ 1 0 1⎥⎦ ⎡ 14 −2 9 ⎤ −1 R 3 A = [38 90 33] ⎢⎢ −6 1 −4 ⎥⎥ = [ 25 14 15] ⎢⎣ 1 0 1⎥⎦ −1

b. Just say no. 43. Let x be the number of shares of D, y be the number of shares of E, and z be the number of shares of F. We get the following equations. 60x + 80y + 30z = 500,000 0.16(60x) + 0.12(80y) + 0.09(30z) = 0.1368(60x + 80y + 30z) z = 4y Simplify the first equation. 6x + 8y + 3z = 50,000 Simplify the second equation. 9.6x + 9.6y + 2.7z = 8.208x + 10.944y + 4.104z 1.392x – 1.344y – 1.404z = 0 1392x – 1344y – 1404z = 0 116x – 112y – 117z = 0 221



Simplify the third equation. 4y – z = 0 Thus we solve the following system of equations. 6x + 8y + 3z = 50,000 116x – 112y – 117z = 0 4y – z = 0 8 3⎤ ⎡ 6 ⎢ The coefficient matrix is A = ⎢116 −112 −117 ⎥⎥ . ⎢⎣ 0 4 −1⎥⎦ ⎡ 6 8 3 1 0 0⎤ ⎢ ⎥ ⎡⎣ A I ⎤⎦ = ⎢116 −112 −117 0 1 0 ⎥ ⎢ 0 −1 0 0 1 ⎥⎦ 4 ⎣ 4 1 1 0 0⎤ ⎡ 1 3 2 6 ⎢ ⎥ > ⎢116 −112 −117 0 1 0 ⎥ ⎢ ⎥ −1 0 0 1⎥ 4 ⎢⎣ 0 ⎦

1R 6 1

4 1 1 0 0⎤ ⎡1 3 2 6 ⎢ ⎥ −116R1 + R 2 ⎢ 800 58 > 0 − 3 −175 − 3 1 0 ⎥ ⎢ ⎥ ⎢0 4 −1 0 0 1⎥ ⎣ ⎦

⎡1 3 R ⎢ − 800 2 > ⎢0 ⎢ − 14 R3 ⎢0 ⎣ ⎡1 ⎢ R 2 + R3 ⎢ > 0 ⎢ ⎢0 ⎢⎣ ⎡1 4 3 32 R ⎢ 29 3 ⎢ > 0 1 ⎢ ⎢0 0 ⎢⎣

0⎤ ⎥ 3 1 − 800 0⎥ ⎥ −1 0 0 − 14 ⎥ ⎦ 4 1 1 ⎤ 0 0 3 2 6 ⎥ 29 3 21 ⎥ 1 32 − 0 400 800 ⎥ 29 29 3 1 0 32 400 − 800 − 4 ⎥⎥ ⎦ 1 1 0 0⎤ 2 6 ⎥ 29 3 21 ⎥ − 0 32 400 800 ⎥ 3 8 ⎥ 2 − 725 − 29 1 25 ⎦⎥

− 12 R 3 + R1 > 21 R + R − 32 3 2

⎡1 4 0 3 ⎢ ⎢0 1 0 ⎢ ⎢0 0 1 ⎢⎣

4 3

1 2 21 32 1 4

⎡1 0 0 − 43 R 2 + R1 ⎢ > ⎢0 1 0 ⎢ ⎢0 0 1 ⎢⎣

1 6 29 400

19 150 1 50 2 25 1 10 1 50 2 25

0

3 1450 3 − 2900 3 − 725 1 290 3 − 2900 3 − 725

3 ⎤ 1 ⎡1 − 29 290 ⎡ x ⎤ ⎢ 10 ⎥ ⎡50, 000 ⎤ ⎡ 5000 ⎤ ⎢ y⎥ = ⎢ 1 − 3 ⎥ = ⎢1000 ⎥ 21 ⎥ ⎢ 0 ⎢ ⎥ ⎢ 50 ⎥ ⎢ ⎥ 2900 116 ⎥ ⎢ ⎢⎣ z ⎥⎦ ⎢ 2 ⎢⎣ 0 ⎥⎦ ⎢⎣ 4000 ⎥⎦ 3 8 ⎥ − 725 − 29 ⎥ ⎣⎢ 25 ⎦ They should buy 5000 shares of company D, 1000 shares of company E, and 4000 shares of company F.

44. Let x be the number of shares of D, y be the number of shares of E, and z be the number of shares of F. We get the following conditions. 60x + 80y + 30z = 500,000 0.16(60x) + 0.12(80y) + 0.09(30z) = 0.1452(60x + 80y + 30z) z = 2y Simplify the first equation. 6x + 8x + 3z = 50,000 Simplify the second equation. 9.6x + 9.6y + 2.7z = 8.712x + 11.616y + 4.356z 0.888x – 2.016y – 1.656z = 0 888x – 2016y – 1656z = 0 111x – 252y – 207z = 0 Simplify the third equation. 2y – z = 0 Thus we solve the following system of equations. 6x + 8y + 3z = 50,000 111x – 252y – 207z = 0 2y – z = 0 8 3⎤ ⎡ 6 ⎢ The coefficient matrix is A = ⎢111 −252 −207 ⎥⎥ . ⎢⎣ 0 2 −1⎥⎦ 8 3 1 0 0⎤ ⎡ 6 ⎡⎣ A I ⎤⎦ = ⎢⎢111 −252 −207 0 1 0 ⎥⎥ ⎢⎣ 0 2 −1 0 0 1⎥⎦ 4 1 1 0 0⎤ ⎡ 1 3 2 6 ⎢ ⎥ > ⎢111 −252 −207 0 1 0 ⎥ ⎢ ⎥ 2 −1 0 0 1⎥ ⎢⎣ 0 ⎦

1R 6 1

4 ⎤ 29 ⎥ 21 ⎥ 116 ⎥ 8 ⎥ − 29 ⎦⎥

4 1 1 0 0⎤ ⎡1 3 2 6 ⎢ ⎥ −111R1 + R 2 ⎢ 525 37 > 0 −400 − 2 − 2 1 0 ⎥ ⎢ ⎥ ⎢0 2 0 0 1⎥ −1 ⎣ ⎦

3 ⎤ − 29 ⎥ 21 ⎥ 116 ⎥ 8 ⎥ − 29 ⎥⎦

⎡1 4 3 1 R ⎢ − 400 2 > ⎢0 1 ⎢ − 12 R3 ⎢0 −1 ⎣

222

1 2 21 32 1 2

0⎤ ⎥ 1 0⎥ − 400 ⎥ 0 0 − 12 ⎥ ⎦

1 6 37 800

0


Section 6.7 −1

1 1 ⎡1 4 0 0⎤ 3 2 6 ⎢ ⎥ R 2 + R3 ⎢ 37 21 1 ⎥ > 0 1 32 − 0 800 400 ⎢ ⎥ ⎢ 0 0 37 37 − 1 − 1 ⎥ ⎢⎣ 32 800 400 2 ⎦⎥ 1 1 ⎡1 4 0 0⎤ 3 2 6 32 R ⎢ ⎥ 37 3 37 21 1 ⎥ > ⎢ 0 1 32 − 0 800 400 ⎢ ⎥ 16 ⎥ 1 2 ⎢0 0 − − 1 ⎢⎣ 25 925 37 ⎥⎦ 8 ⎤ 1 ⎡ 1 4 0 11 3 75 925 37 ⎥ 1 ⎢ − 2 R 3 + R1 1 1 21 ⎥ > ⎢0 1 0 50 − 925 74 ⎥ 21 R + R ⎢ − 32 3 2 ⎢0 0 1 1 − 2 − 16 ⎥ 25 925 37 ⎦⎥ ⎣⎢

3 −4.7 ⎤ ⎡ 13 ⎤ ⎡ 4.78 ⎤ ⎡ x ⎤ ⎡ 0.9 ⎢ ⎥ ⎢ 48. ⎢ y ⎥ = ⎢ 2 −0.4 2 ⎥⎥ ⎢⎢ 4.7 ⎥⎥ = ⎢⎢ −1.33 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 1 −0.8 −0.5 ⎥⎦ ⎢⎣ 7.2 ⎥⎦ ⎢⎣ −2.70 ⎥⎦ x = 4.78, y = –1.33, z = –2.70 −1

1 − 3⎤ ⎡2 14 4 2 7 ⎥ ⎡ 13 ⎤ ⎡ 14.44 ⎤ ⎡ w⎤ ⎢ 5 ⎢ ⎥ ⎢ x ⎥ ⎢ 5 − 2 −4 −1⎥ ⎢ 7 ⎥ ⎢ 0.03⎥ 9 3 ⎥ ⎥ ⎢8⎥=⎢ 49. ⎢ ⎥ = ⎢ 5 ⎥ ⎢ 9 ⎥ ⎢ −0.80 ⎥ ⎢ y⎥ ⎢0 1 − 94 6⎥ ⎢ ⎥ ⎢ ⎢ 4 ⎥ ⎢⎢ 10.33⎥⎥ ⎢⎣ z ⎥⎦ ⎢ 1 ⎦ ⎥ 1 0 4 − 3 ⎥ ⎢⎣ 7 ⎥⎦ ⎣ ⎣⎢ 2 ⎦ w = 14.44, x = 0.03, y = –0.80, z = 10.33

Problems 6.7

7 6 ⎤ ⎡1 0 0 3 − 37 25 2775 4 ⎢ ⎥ − 3 R 2 + R1 1 1 21 ⎥ > ⎢0 1 0 50 − 925 74 ⎥ ⎢ ⎢0 0 1 1 − 2 − 16 ⎥ ⎢⎣ 25 925 37 ⎦⎥ 3 7 6 ⎡ − 37 ⎤ 50, 000 ⎡ x ⎤ ⎢ 25 2775 ⎤ ⎡ 6000 ⎤ ⎥⎡ ⎢ y⎥ = ⎢ 1 − 1 ⎥ = ⎢1000 ⎥ 21 ⎥ ⎢ 0 ⎢ ⎥ ⎢ 50 ⎥ ⎢ ⎥ 925 74 ⎥ ⎢ ⎢⎣ z ⎥⎦ ⎢ 1 ⎢⎣ 0 ⎥⎦ ⎢⎣ 2000 ⎥⎦ 16 2 ⎥ − 925 − 37 ⎥ ⎣⎢ 25 ⎦ They should buy 6000 shares of company D, 1000 shares of company E, and 2000 shares of company F.

45. a.

b.

46. a.

b.

⎡ 200 1200 1. A = ⎢ ⎢ 400 ⎣ 1200 ⎡ 600 ⎤ D=⎢ ⎥ ⎣805 ⎦

⎡1290 ⎤ X = (I − A)−1 D = ⎢ ⎥ ⎣1425⎦ The total value of other production costs is 600 800 PA + PB = (1290) + (1425) = 1405 1200 1500

⎡ 2.05 1.28⎤ ⎢ 0.73 1.71⎥ ⎣ ⎦ ⎡ 84 ⎢ 41 ⎢ 30 ⎣ 41

⎡ 40 200 2. A = ⎢ ⎢ 120 ⎣ 200

105 ⎤ 82 ⎥ 70 ⎥ 41 ⎦

a.

18 323 11 646 23 323

120 ⎤ 300 ⎥ 90 ⎥ 300 ⎦

⎡ 200 ⎤ D=⎢ ⎥ ⎣ 300 ⎦ ⎡812.5⎤ X = (I − A)−1 D = ⎢ ⎥ ⎣ 1125 ⎦

⎡ −0.03 0.06 −0.12 ⎤ ⎢ 0.13 0.02 0.05⎥ ⎢ ⎥ ⎢⎣ −0.10 0.07 0.01⎥⎦ ⎡ − 11 ⎢ 323 ⎢ 83 ⎢ 646 ⎢ − 32 ⎢⎣ 323

500 ⎤ 1500 ⎥ 200 ⎥ 1500 ⎦

b.

39 ⎤ − 323 ⎥ 15 ⎥ 323 ⎥ 4 ⎥ 323 ⎥⎦

⎡ 64 ⎤ D=⎢ ⎥ ⎣ 64 ⎦ ⎡ 220 ⎤ X = (I − A) −1 D = ⎢ ⎥ ⎣ 280 ⎦

⎡ 2.75 −1.59 −1.11 ⎤ 47. ⎢ −0.48 1.43 0.00 ⎥ ⎢ ⎥ ⎣ −1.22 0.32 2.22 ⎦

223


⎡ 15 ⎢ 100 25 3. A = ⎢ 100 ⎢ ⎢ 50 ⎣⎢ 100

a.


45 ⎤ 180 ⎥ 60 ⎥ 180 ⎥ 60 ⎥ 180 ⎦⎥

30 120 30 120 40 120

⎡ 400 ⎢ 1000 200 6. A = ⎢ 1000 ⎢ ⎢ 200 ⎣⎢ 1000

⎡1073⎤ X = (I − A)−1 D = ⎢⎢1016 ⎥⎥ ⎢⎣ 952 ⎥⎦ ⎡ 400 200 ⎢ 1000 1000 200 400 7. A = ⎢ 1000 1000 ⎢ ⎢ 200 100 ⎢⎣ 1000 1000 ⎡ 300 ⎤ D = ⎢⎢ 400 ⎥⎥ ⎢⎣ 500 ⎥⎦

⎡10 ⎤ D = ⎢⎢10 ⎥⎥ ⎢⎣10 ⎥⎦ ⎡ 68.59 ⎤ X = (I − A) D = ⎢⎢ 84.50 ⎥⎥ ⎢⎣108.69 ⎥⎦ −1

⎡ 100 ⎢ 1000 100 4. A = ⎢ 1000 ⎢ ⎢ 300 ⎣⎢ 1000

400 800 80 800 160 800

−1

3⎤ ⎡1 4⎥ 8. A = ⎢ 31 ⎢⎣ 4 0 ⎥⎦ ⎡300 ⎤ D=⎢ ⎣500 ⎥⎦ (I − A)X = D ⎡ 2 − 3 300 ⎤ 4 ⎥ with a calculator Reducing ⎢ 13 1 500 ⎥⎦ ⎢⎣ − 4 ⎡ 1 0 1408.70 ⎤ . results in ⎢ ⎣ 0 1 852.17 ⎥⎦ Thus 1408.70 units of agriculture and 852.17 units of milling need to be produced.

⎡1559.81 ⎤ X = (I − A) D = ⎢⎢1112.44 ⎥⎥ ⎢⎣1738.04 ⎥⎦ −1

200 1000 400 1000 100 1000

200 ⎤ 1000 ⎥ 100 ⎥ 1000 ⎥ 300 ⎥ 1000 ⎥⎦

⎡1382 ⎤ X = (I − A) D = ⎢⎢1344 ⎥⎥ ⎢⎣1301⎥⎦

240 ⎤ 1200 ⎥ 480 ⎥ 1200 ⎥ 240 ⎥ 1200 ⎦⎥

⎡500 ⎤ D = ⎢⎢150 ⎥⎥ ⎢⎣ 700 ⎥⎦

⎡ 400 ⎢ 1000 200 5. A = ⎢ 1000 ⎢ ⎢ 200 ⎣⎢ 1000

200 ⎤ 1000 ⎥ 100 ⎥ 1000 ⎥ 300 ⎥ 1000 ⎦⎥

⎡ 250 ⎤ D = ⎢⎢ 300 ⎥⎥ ⎢⎣ 350 ⎥⎦

⎡15 ⎤ D = ⎢⎢10 ⎥⎥ ⎢⎣35⎥⎦ ⎡134.29 ⎤ X = (I − A) −1 D = ⎢⎢162.25 ⎥⎥ ⎢⎣ 234.35⎥⎦

b.

200 1000 400 1000 100 1000

200 ⎤ 1000 ⎥ 100 ⎥ 1000 ⎥ 300 ⎥ 1000 ⎦⎥

⎡ 300 ⎤ D = ⎢⎢ 350 ⎥⎥ ⎢⎣ 450 ⎥⎦

1 1⎤ ⎡1 ⎢ 10 3 4 ⎥ 1 1 1 9. A = ⎢ 10 ⎥ 10 3 ⎢1 1 1⎥ ⎢⎣ 10 10 10 ⎥⎦ ⎡ 300 ⎤ D = ⎢ 200 ⎥ ⎢ ⎥ ⎣ 500 ⎦

⎡1301⎤ X = (I − A)−1 D = ⎢⎢1215⎥⎥ ⎢⎣1188⎥⎦

(I − A)X = D 224


Chapter 6 Review

⎡ 9 − 13 − 14 300 ⎤ ⎡ 1 0 0 736.39 ⎤ ⎢ 10 ⎥ 9 1 1 − − 200 Reducing ⎢ 10 ⎥ with a calculator results in ⎢ 0 1 0 563.29 ⎥ . 10 3 ⎢ ⎥ ⎢ 1 ⎥ 9 1 ⎣ 0 0 1 699.96 ⎦ ⎢⎣ − 10 − 10 10 500 ⎥⎦ Thus 736.39 units of coal, 563.29 units of steel, and 699.96 units of railroad services need to be produced.

Chapter 6 Review Problems 8⎤ ⎡ 3 4⎤ ⎡ 1 0 ⎤ ⎡ 6 8⎤ ⎡ 3 0 ⎤ ⎡ 3 − 3⎢ =⎢ −⎢ =⎢ 1. 2 ⎢ ⎥ ⎥ ⎥ ⎥ ⎥ ⎣ −5 1⎦ ⎣ 2 4 ⎦ ⎣ −10 2 ⎦ ⎣ 6 12⎦ ⎣ −16 −10 ⎦ ⎡1 2⎤ ⎡1 0 ⎤ ⎡ 8 16 ⎤ ⎡ 2 0 ⎤ ⎡ 6 16⎤ −2⎢ 2. 8 ⎢ ⎥ ⎥=⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣7 0 ⎦ ⎣0 1 ⎦ ⎣56 0 ⎦ ⎣ 0 2 ⎦ ⎣56 −2⎦ 5⎤ ⎡ 1 7⎤ ⎡ 1 + 0 0 + 42 −2 + 7 ⎤ ⎡ 1 42 ⎡ 1 0 −2 ⎤ ⎢ ⎥ = ⎢ 2 −18 −7 ⎥ = 2 + 0 0 − 18 − 4 − 3 3. ⎢⎢ 2 −3⎥⎥ ⎢ ⎥ ⎢ ⎥ 0 6 1⎥⎦ ⎢ ⎢⎣ 1 0 ⎥⎦ ⎣ ⎢⎣ 1 + 0 0 + 0 −2 + 0 ⎥⎦ ⎢⎣ 1 0 −2 ⎥⎦ ⎡ 2 3⎤ 4. [2 3 7] ⎢⎢ 0 −1⎥⎥ = [2(2) + 3(0) + 7(5) ⎢⎣ 5 2 ⎥⎦

2(3) + 3(−1) + 7(2)] = [39 17]

⎡ 2 3⎤ ⎛ ⎡ 2 3⎤ ⎡ 1 8⎤ ⎞ ⎡ 2 3⎤ ⎡ 1 −5⎤ ⎡11 −4⎤ 5. ⎢ − = ⎜ ⎟= ⎣ −1 3⎥⎦ ⎝ ⎢⎣ 7 6 ⎥⎦ ⎢⎣ 4 4 ⎥⎦ ⎠ ⎢⎣ −1 3⎥⎦ ⎢⎣3 2 ⎥⎦ ⎢⎣ 8 11⎥⎦ ⎡ 0 −5⎤ ⎪⎫ ⎪⎧ ⎡ 2 0 ⎤ ⎪⎧ ⎡ 2 0 ⎤ ⎡ 0 −10 ⎤ ⎪⎫ 6. − ⎨ ⎢ ⎥ + 2 ⎢ 6 −4 ⎥ ⎬ = − ⎨ ⎢7 8 ⎥ + ⎢12 −8⎥ ⎬ 7 8 ⎦ ⎣ ⎦ ⎭⎪ ⎦ ⎣ ⎦ ⎭⎪ ⎩⎪ ⎣ ⎩⎪ ⎣ ⎡ 2 −10 ⎤ ⎡ −2 10 ⎤ = −⎢ = 0 ⎥⎦ ⎢⎣ −19 0 ⎥⎦ ⎣19 2

⎡ 1 −2 ⎤ ⎡ −5 −4 ⎤ ⎡ 1⎤ ⎡ 3⎤ ⎡ 6 ⎤ T 1 −2] = 2 ⎢ 7. 2 ⎢ = 2⎢ ⎥ = ⎢ ⎥ [ ⎥ ⎥ ⎢ ⎥ 1⎦ ⎣3 ⎣ 6 −5⎦ ⎣ −2 ⎦ ⎣16 ⎦ ⎣32 ⎦ 2

T 2 ⎡1 0 ⎤ ⎡1 1⎤ ⎡1 0 ⎤ ⎡1 4 ⎤ ⎡1 4 ⎤ 1 ⎡3 0 ⎤ ⎧⎪ ⎡1 0 ⎤ ⎫⎪ = 8. ⎨ ⎬ ⎢3 6 ⎥ ⎢1 3⎥ ⎢1 2 ⎥ ⎢ 0 3⎥ = ⎢1 2 ⎥ ⎢ 0 9 ⎥ = ⎢1 22⎥ 3⎣ ⎦ ⎪⎩ ⎣ ⎦ ⎪⎭ ⎣ ⎦⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

⎡1 −1⎤ ⎡ 3 0 ⎤ 9. (2 A)T − 3I 2 = 2A T − 3I = 2 ⎢ ⎥−⎢ ⎥ ⎣1 2 ⎦ ⎣ 0 3⎦ ⎡ 2 −2 ⎤ ⎡ 3 0 ⎤ ⎡ −1 −2 ⎤ =⎢ ⎥−⎢ ⎥=⎢ 1⎥⎦ ⎣ 2 4⎦ ⎣0 3⎦ ⎣ 2 ⎡ 2 2⎤ 10. A(2I ) − AOT = 2( AI ) − AO = 2 A − O = 2 A = ⎢ ⎥ ⎣ −2 4 ⎦

225



3

5

⎡1 0 ⎤ ⎡1 0 ⎤ ⎡1 0 ⎤ ⎡1 0⎤ ⎡ 2 0 ⎤ 11. B3 + I 5 = ⎢ ⎥ + ⎢0 1 ⎥ = ⎢0 8 ⎥ + ⎢0 1⎥ = ⎢ 0 9⎥ 0 2 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡0 0⎤ 12. ( ABA)T − A T BT A T = A T B T A T − A T B T A T = O = ⎢ ⎥ ⎣0 0⎦ ⎡ 5 x ⎤ ⎡15⎤ 13. ⎢ ⎥ = ⎢ ⎥ ⎣7 x ⎦ ⎣ y ⎦ 5x = 15, or x = 3 7x = y, 7 · 3 = y, or y = 21 ⎡ 2 + x 2 1 + 3 x ⎤ ⎡3 4 ⎤ 14. ⎢ ⎥=⎢ ⎥ ⎣⎢ 4 + xy 2 + 3 y ⎦⎥ ⎣3 y ⎦ 2 + 3y = y, 2y = –2, or y = –1 1 + 3x = 4, 3x = 3, or x = 1

For these values of x and y, 2 + x 2 = 3 is true, and 4 + xy = 3 is true. Thus x = 1, y = –1. 4 ⎤ ⎡1 4⎤ ⎡1 4 ⎤ ⎡1 ⎡1 0 ⎤ →⎢ →⎢ →⎢ 15. ⎢ ⎥ ⎥ ⎥ ⎥ ⎣5 8 ⎦ ⎣0 −12 ⎦ ⎣ 0 1 ⎦ ⎣0 1 ⎦ ⎡0 0 7 ⎤ ⎡0 5 9 ⎤ ⎡0 1 →⎢ 16. ⎢ ⎥ ⎥→⎢ ⎣0 5 9⎦ ⎣0 0 7 ⎦ ⎢⎣ 0 0 ⎡2 4 7⎤ ⎡1 2 17. ⎢ 1 2 4 ⎥ → ⎢ 2 4 ⎢ ⎥ ⎢ ⎣ 5 8 2⎦ ⎣5 8 ⎡ 1 2 4⎤ ⎡1 → ⎢0 1 9 ⎥ → ⎢0 ⎢ ⎥ ⎢ ⎣ 0 0 −1⎦ ⎣0

9⎤ 5⎥

⎡0 1 0⎤ →⎢ ⎥ 1⎥⎦ ⎣ 0 0 1⎦

4⎤ 4⎤ ⎡1 2 ⎡1 2 7 ⎥ → ⎢0 0 −1⎥ → ⎢0 −2 ⎥ ⎢ ⎥ ⎢ 2⎦ ⎣0 −2 −18⎦ ⎣0 0 0 −14 ⎤ ⎡ 1 0 −14 ⎤ ⎡ 1 1 9⎥ → ⎢0 1 9⎥ → ⎢0 ⎥ ⎢ ⎥ ⎢ −1⎦ ⎣ 0 0 0 1⎦ ⎣ 0

4⎤ −18⎥ ⎥ −1⎦ 0 0⎤ 1 0⎥ ⎥ 0 1⎦

⎡0 0 0 1 ⎤ ⎡1 0 0 0 ⎤ 18. ⎢⎢ 0 0 0 0 ⎥⎥ → ⎢⎢ 0 0 0 1 ⎥⎥ ⎢⎣1 0 0 0 ⎥⎦ ⎢⎣ 0 0 0 0 ⎥⎦ ⎡ 2 −5 0 ⎤ ⎡ 2 −5 0 ⎤ ⎡ 1 − 52 0 ⎤ ⎡1 0 0 ⎤ →⎢ 19. ⎢ ⎥→⎢ ⎥ ⎥→⎢ ⎥ 1 0 ⎥⎦ ⎣ 0 1 0 ⎦ ⎣ 4 3 0⎦ ⎣ 0 13 0 ⎦ ⎢⎣ 0 Thus x = 0, y = 0. 3 2 3⎤ ⎡ 1 0 ⎡ 1 −1 3⎤ ⎡ 1 −1 2 3⎤ ⎡ 1 −1 2 4 ⎢ 20. ⎢ ⎥ → ⎢ 0 4 −5 −4 ⎥ → ⎢ 0 1 − 5 −1⎥ → ⎢ 5 3 1 1 5 ⎢⎣ ⎥⎦ ⎣ 0 1 − 4 ⎣ ⎦ ⎣ ⎦ 4

3 5 Thus x = − r + 2 , y = r − 1 , z = r. 4 4

226

2⎤ ⎥ −1⎥⎦


Chapter 6 Review

1⎤ 2 1⎤ ⎡ 1 1 2 1⎤ ⎡1 1 2 ⎡1 1 ⎡ 1 0 0 −1⎤ 21. ⎢⎢ 3 −2 −4 −7 ⎥⎥ → ⎢⎢0 −5 −10 −10 ⎥⎥ → ⎢⎢ 0 1 2 2 ⎥⎥ → ⎢⎢ 0 1 2 2 ⎥⎥ ⎢⎣ 2 −1 −2 2 ⎥⎦ ⎢⎣0 −3 −6 ⎢⎣ 0 0 0 6 ⎥⎦ 0 ⎥⎦ ⎢⎣ 0 −3 −6 0 ⎥⎦ Row three indicates that 0 = 6, which is never true, so there is no solution. ⎡1 0 ⎡ 1 −1 −1 1⎤ ⎡ 1 −1 −1 1⎤ ⎡ 1 −1 −1 1⎤ ⎢ ⎢ ⎥ 22. ⎢ 1 1 2 −3⎥ → ⎢0 2 3 −4 ⎥ → 0 1 32 −2 → ⎢0 1 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢0 0 ⎢⎣0 2 4 −9 ⎥⎦ ⎣ 2 0 2 −7 ⎦ ⎣0 2 4 −9 ⎦ ⎢⎣

−1⎤ ⎡ 1 0 0 32 ⎤ ⎥ ⎢ ⎥ −2 ⎥ → ⎢ 0 1 0 11 ⎥ 2 1 −5⎥⎥ ⎢⎢ 0 0 1 −5⎥⎥ ⎦ ⎣ ⎦

1 2 3 2

3 11 Thus x = , y = , z = −5. 2 2 ⎡1 5 1 0⎤ ⎡1 5 1 0 ⎤ ⎡1 5 →⎢ →⎢ 23. ⎢ ⎥ ⎥ ⎢⎣ 0 1 ⎣3 9 0 1 ⎦ ⎣ 0 −6 −3 1⎦ 5⎤ 5⎤ ⎡1 0 − 3 ⎡− 3 2 6⎥ 2 6⎥ →⎢ ⇒ A −1 = ⎢ 1 − 1⎥ ⎢0 1 ⎢ 1 − 1⎥ 2 6⎦ 6⎦ ⎣ ⎣ 2

1 1 2

0⎤

⎥ − 16 ⎥ ⎦

⎡ 0 1 1 0 ⎤ ⎡1 0 0 1 ⎤ −1 ⎡ 0 1 ⎤ 24. ⎢ ⎥ → ⎢ 0 1 1 0 ⎥ ⇒ A = ⎢1 0 ⎥ 1 0 0 1 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 −2 1 0 0⎤ ⎡ 1 3 −2 1 0 0 ⎤ ⎡ 1 ⎢ ⎥ ⎢ 25. ⎢ 4 1 0 0 1 0 ⎥ → ⎢ 0 −11 8 −4 1 0 ⎥⎥ ⎢⎣ 3 −2 2 0 0 1⎥⎦ ⎢⎣ 0 −11 8 −3 0 1⎥⎦ 0 0⎤ 3 −2 1 0 0 ⎤ ⎡ 1 3 −2 1 ⎡1 ⎢ 0 −11 8 −4 1 0 ⎥ → ⎢ 0 1 − 8 4 − 1 0 ⎥ ⎥ ⎢ ⎥ ⎢ 11 11 11 ⎥ ⎢⎣ 0 0 0 1 −1 1⎥⎦ ⎢ 0 0 − 0 1 1 1 ⎣ ⎦ 3 2 1 ⎡1 0 − 11 0⎤ 11 11 ⎢ ⎥ 8 4 − 1 0 ⎥ ⇒ no inverse exists → ⎢ 0 1 − 11 11 11 ⎢ ⎥ ⎢0 0 0 1 −1 1⎥ ⎣ ⎦ ⎡ 1 0 0 1 0 0⎤ 5 ⎡ 5 0 0 1 0 0⎤ ⎡5 0 0 1 0 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 1 ⎢ 26. ⎢ −5 2 1 0 1 0 ⎥ → ⎢0 2 1 1 1 0 ⎥ → 0 1 2 2 12 0 ⎥ ⎢ ⎥ ⎢⎣ −5 1 3 0 0 1⎥⎦ ⎢⎣0 1 3 1 0 1⎥⎦ ⎢ 0 1 3 1 0 1⎥ ⎣ ⎦ ⎡1 0 0 1 ⎤ ⎡1 ⎡1 0 0 1 ⎤ 0 0 0 0⎤ 0 0 5 5 ⎢ ⎥ ⎢5 ⎥ ⎢ ⎥ 3 − 1 ⎥ ⇒ A −1 = ⎢ 2 3 − 1 ⎥. 1 0 ⎥ → ⎢0 1 0 2 → ⎢ 0 1 12 12 2 5 5 5⎥ 5 5⎥ ⎢ ⎢5 ⎢ ⎥ 2⎥ 2⎥ ⎢0 0 1 1 − 1 ⎢1 − 1 ⎢ 0 0 5 1 − 1 1⎥ 2 2 2 5 5 5⎦ 5 5⎦ ⎣ ⎦ ⎣ ⎣5

227



⎡ 3 1 4 1 0 0 ⎤ ⎡1 0 27. ⎢⎢1 0 1 0 1 0 ⎥⎥ → ⎢⎢ 3 1 ⎢⎣ 0 2 1 0 0 1 ⎥⎦ ⎢⎣ 0 2 1 0⎤ ⎡1 0 1 0 ⎡1 ⎢ ⎥ → ⎢ 0 1 1 1 −3 0 ⎥ → ⎢⎢ 0 ⎢⎣ 0 2 1 0 0 1⎥⎦ ⎢⎣ 0

1 0 1 0⎤ 4 1 0 0 ⎥⎥ 1 0 0 1 ⎥⎦ 0 1 0 1 0⎤ 1 1 1 −3 0 ⎥⎥ 0 −1 −2 6 1⎥⎦

1 0⎤ ⎡1 0 1 0 ⎡ 1 0 0 −2 7 1⎤ ⎢ ⎥ → ⎢ 0 1 1 1 −3 0 ⎥ → ⎢⎢0 1 0 −1 3 1⎥⎥ ⎢⎣ 0 0 1 2 −6 −1⎥⎦ ⎢⎣0 0 1 2 −6 −1⎥⎦ ⎡ x⎤ ⎡ −2 7 1⎤ ⎡ 1 ⎤ ⎡ 0 ⎤ ⎢ y ⎥ = A −1B = ⎢ −1 3 1⎥ ⎢ 0 ⎥ = ⎢1 ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 2 −6 −1⎥⎦ ⎢⎣ 2 ⎥⎦ ⎢⎣ 0 ⎥⎦ Thus x = 0, y = 1, z = 0.

28. We found A −1 in Exercise 26, so ⎡3⎤ ⎡1 0 0⎤ 3 ⎡ x⎤ ⎢5 ⎥⎡ ⎤ ⎢5⎥ ⎢ y ⎥ = A −1B = ⎢ 2 3 − 1 ⎥ ⎢0⎥ = ⎢ 4 ⎥ ⎢ ⎥ 5 5⎥ ⎢ ⎥ ⎢5⎥ ⎢5 ⎢⎣ z ⎥⎦ 2 ⎥ ⎢⎣ 2 ⎥⎦ ⎢ 7 ⎥ ⎢1 − 1 ⎢⎣ 5 ⎥⎦ 5 5⎦ ⎣5 ⎡ 0 1 1⎤ ⎡ 0 29. A = AA = ⎢ 0 0 1⎥ ⎢ 0 ⎢ ⎥⎢ ⎣0 0 0⎦ ⎣0 ⎡0 0 1⎤ ⎡0 A3 = A 2 A = ⎢0 0 0 ⎥ ⎢0 ⎢ ⎥⎢ ⎣0 0 0 ⎦ ⎣0 2

1 1⎤ ⎡ 0 0 1⎥ = ⎢ 0 ⎥ ⎢ 0 0⎦ ⎣0 1 1⎤ ⎡ 0 0 1⎥ = ⎢ 0 ⎥ ⎢ 0 0⎦ ⎣0

0 1⎤ 0 0⎥ ⎥ 0 0⎦ 0 0⎤ 0 0⎥ = O ⎥ 0 0⎦

Since A3 = O, every higher power of A is also O, so A1000 = O. ⎡0 1 1 1 0 0⎤ Looking at ⎢ 0 0 1 0 1 0 ⎥ , it is clear that there is no way of transforming the left side into I3 , since there ⎢ ⎥ ⎣ 0 0 0 0 0 1⎦ is no way to get a nonzero entry in the first column. Thus A does not have an inverse. ⎡2 0⎤ 30. A T = ⎢ ⎥ ⎣0 4⎦ 1 −1 ⎡ 2 0 ⎤ T ⎢ ⎥ = A ⎢0 1⎥ 4⎦ ⎣ ⎡ 1 0⎤ ⎥ A −1 = ⎢ 2 ⎢0 1⎥ 4⎦ ⎣ 1 0⎤ ⎡ T ⎥ A −1 = ⎢ 2 ⎢0 1⎥ 4⎦ ⎣

( )

( )

Thus ( A T )−1 = ( A −1 )T . 228


Chapter 6 Review

Let x, y, and z represent the weekly doses of capsules of brand I, II, and III, respectively. Then (vitamin A) ⎧ x + y + 4 z = 13 ⎪ (vitamin B) + 2 + 7 = 22 x y z ⎨ ⎪ x + 3 y + 10 z = 31 (vitamin C) ⎩ ⎡1 1 4 13⎤ ⎢1 2 7 22 ⎥ − R1 + R 2> ⎢ ⎥ −R + R 1 3 ⎢⎣1 3 10 31⎥⎦ ⎡ 1 0 1 4⎤ −R 2 + R1 > ⎢ 0 1 3 9 ⎥⎥ −2R 2 + R 3 ⎢ ⎢⎣ 0 0 0 0 ⎥⎦

⎡ 1 1 4 13⎤ ⎢0 1 3 9⎥ ⎢ ⎥ ⎢⎣ 0 2 6 18⎥⎦

Thus x = 4 – r, y = 9 – 3r, and z = r, where r = 0, 1, 2, 3. The four possible combinations are Combination

x

y

z

1

4

9

0

2

3

6

1

3

2

3

2

4

1

0

3

b. Computing the cost of each combination, we find that they are 83, 77, 71, and 65 cents, respectively. Thus combination 4, namely x = 1, y = 0, z = 3, minimizes weekly cost. 32. a.

(A ) −1

= A −1

( ) (A A) A = (A ) ( A A ) A = A IA = A A = I

3

A 3 = A −1 −1

2

−1

−1

2

−1

2

( )

IA 2 = A −1

2

A2

−1

Thus A3 is invertible.

(

) (

)

b. AB = AC. Thus A −1 ( AB) = A −1 ( AC) , A −1A B = A −1A C , IB = IC, B = C. c.

AA = A ⇒ A −1AA = A −1A , IA = I, A = I. Thus A = I n .

⎡ 215 87 ⎤ 33. ⎢ ⎥ ⎣ 89 141⎦ −1

⎡ x ⎤ ⎡ 7.9 −4.3 2.7 ⎤ ⎡11.1⎤ ⎡ 1.57 ⎤ 34. ⎢⎢ y ⎥⎥ = ⎢⎢ 3.4 5.8 −7.6 ⎥⎥ ⎢⎢10.8⎥⎥ = ⎢⎢ −0.30 ⎥⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 4.5 −6.2 −7.4 ⎥⎦ ⎢⎣15.9 ⎥⎦ ⎢⎣ −0.95⎥⎦ Thus x = 1.57, y = −0.30, z = −0.95. ⎡ 10 34 35. A = ⎢ ⎢ 15 ⎣ 34

20 ⎤ 39 ⎥ ; 14 ⎥ 39 ⎦

⎡10 ⎤ ⎡39.7 ⎤ D = ⎢ ⎥ ; X = (I − A) −1 D = ⎢ ⎥ ⎣5⎦ ⎣ 35.1⎦

229



Mathematical Snapshot Chapter 6 ⎡ 20 40 30 10 ⎤ 1. A = ⎢⎢ 30 0 10 10 ⎥⎥ ⎢⎣ 10 0 30 50 ⎥⎦ ⎡ 7⎤ ⎢10 ⎥ T=⎢ ⎥ ⎢ 7⎥ ⎢ ⎥ ⎢⎣ 5⎥⎦ ⎡ 9⎤ C = ⎢⎢ 8⎥⎥ ⎢⎣10 ⎥⎦ ⎧ ⎡ 7⎤ ⎫ ⎡800 ⎤ ⎪ ⎡ 20 40 30 10 ⎤ ⎢ ⎥ ⎪ 10 ⎥ ⎪ ⎥ ⎥ T T ⎪⎢ T⎢ ⎢ C ( AT) = C ⎨ ⎢ 30 0 10 10 ⎥ ⎬ = C ⎢330 ⎥ ⎢ ⎥ ⎪ ⎢ 10 0 30 50 ⎥ ⎢ 7 ⎥ ⎪ ⎢⎣530 ⎥⎦ ⎦ 5 ⎪ ⎪⎣ ⎣⎢ ⎦⎥ ⎭ ⎩ ⎡800 ⎤ = [9 8 10] ⎢⎢330 ⎥⎥ = [15,140] ⎢⎣530 ⎥⎦

The cost is $151.40. 2. To the linear system, add x1 + x2 + x3 + x4 = 52. ⎡ 20 40 30 10 ⎤ ⎢ 30 0 10 10 ⎥ ⎥ A=⎢ ⎢ 10 0 30 50 ⎥ ⎢ ⎥ ⎢⎣ 1 1 1 1⎥⎦ ⎡1180 ⎤ ⎢ 580 ⎥ ⎥ B=⎢ ⎢1500 ⎥ ⎢ ⎥ ⎣⎢ 52 ⎦⎥ ⎡ 8⎤ ⎢ 10 ⎥ T = A −1B = ⎢ ⎥ ⎢ 14 ⎥ ⎢ ⎥ ⎣⎢ 20 ⎦⎥ Guest 1: 8 days; guest 2: 10 days; guest 3: 14 days; guest 4: 20 days

3. It is not possible. Different combinations of lengths of stays can cost the same. For example, guest 1 staying for 20 days and guest 3 staying for 17 days costs the same as guest 1 staying for 15 days and guest 3 staying for 21 days (each costs $214.50).

230


2.

1. Let x = the number of type A magnets and y = the number of type B magnets. The cost for producing x type A magnets and y type B magnets is 50 + 0.90x + 0.70y. The revenue for selling x type A magnets and y type B magnets is 2.00x + 1.50y. Revenue is greater than cost when 2x + 1.5y > 50 + 0.9x + 0.7y. 0.8y > –1.1x + 50 y > –1.375x + 62.5 Sketch the dashed line y = –1.375x + 62.5 and shade the half plane above the line. In order to make a profit, the number of magnets of types A and B must correspond to an ordered pair in the shaded region. Also, to take reality into account, both x and y must be positive (negative numbers of magnets are not feasible).

10

y

x 10

3.

10

y

x 10

4.

2. Since negative numbers of cameras cannot be sold, x ≥ 0 and y ≥ 0. Selling at least 50 cameras per week corresponds to x + y ≥ 50. Selling twice as many of type I as of type II corresponds to x ≥ 2y. The system of inequalities is ⎧ x + y ≥ 50, ⎪⎪ x ≥ 2 y, ⎨ x ≥ 0, ⎪ y ≥ 0. ⎪⎩

y

10

x 10

5.

5

y

The region consists of points on or above the x-axis and on or to the right of the y-axis. In addition, the points must be on or above the line x + y = 50 and on or below the line x = 2y.

x 4

Problems 7.1 1.

5

y

6.

10

y

x 5

x 10

231

Chapter 7: Linear Programming 7.

5


y

y

5

x

x

5

8.

5

5

y

y

13.

6

x 5

x 5

9.

5

y

14.

5

y

x x

5

10.

10

5

y

15.

5

y

x 10

x 5

11.

5

y

16.

5

y

x

x

5

5

232


5

Section 7.1

y

22.

10

y

x

x

5

10

y 5

18.

23.

10

y

x

x

10

5

19.

5

24.

y

20

y

x 20

x 5

20.

25. 6x + 4y ≤ 20

y 8

7

y

x

x

5

5

21.

10

y

26. 7x + 3y ≤ 25

y

x 10

2

233

x 10

Chapter 7: Linear Programming


27. Let x be the amount purchased from supplier A, and y the amount purchased from B. The system of inequalities is x + y ≤ 100, x ≥ 0, y ≥ 0. 100

25

y P = 112

1 2

⎛ 47 41 ⎞ ⎜ , ⎟ ⎝ 3 9 ⎠ x – 3y = 2

y

x 25

(2, 0) x

2. The feasible region appears below. The corner 1⎞ ⎛ 1 ⎞ ⎛ points are (0, 0), ⎜ 0, 83 ⎟ , ⎜ 62 , 0 ⎟ . 3⎠ ⎝ 2 ⎠ ⎝ Evaluating P at each corner point, we find that P 2 has a maximum value of 416 when x = 0 and 3 1 y = 83 . 3

100

28. Since negative numbers of computers cannot be produced, x ≥ 0 and y ≥ 0. Producing at most 650 computers per week corresponds to x + y ≤ 650. The system of inequalities is ⎧ x + y ≤ 650, ⎪ x ≥ 0, ⎨ ⎪⎩ y ≥ 0.

y x + 2y = 225 100

x + y = 90

1⎞ ⎛ ⎜ 0, 83 ⎟ 3⎠ ⎝

29. Since negative numbers of chairs cannot be produced, x ≥ 0 and y ≥ 0. The inequality for assembly time is 3x + 2y ≤ 240. The inequality 1 for painting time is x + y ≤ 80 . The system of 2 inequalities is ⎧3x + 2 y ≤ 240, ⎪1 ⎪ x + y ≤ 80, ⎨2 ⎪ x ≥ 0, ⎪ y ≥ 0. ⎩

4x + 3y = 250 P = 416

2 3

x ⎛ 1 ⎞ 100 ⎜ 62 , 0 ⎟ ⎝ 2 ⎠

3. The feasible region appears below. The corner ⎛ 10 ⎞ points are (2, 3), (0, 5), (0, 7) and ⎜ , 7 ⎟ . ⎝ 3 ⎠ Evaluating Z at each point, we find that Z has a maximum value of –10 when x = 2 and y = 3.

The region consists of points on or above the x-axis and on or to the right of the y-axis. In addition, the points must be on or below the line 3x + 2y = 240 and on or below the line 1 x + y = 80 (or, equivalently x + 2y = 160). 2

10

y y=7

3x − y = 3 Z = −10

x+y=5 (2, 3)

Problems 7.2

x 10

1. The feasible region appears below. The corner ⎛ 47 41 ⎞ ⎛ 45 ⎞ points are (2, 0), ⎜ , ⎟ , ⎜ , 0 ⎟ . ⎝ 3 9 ⎠ ⎝ 2 ⎠ Evaluating P at each corner point, we find that P 1 45 has a maximum value of 112 when x = 2 2 and y = 0.

4. The feasible region appears below. The corner ⎛ 12 12 ⎞ ⎛ 99 99 ⎞ points are (8, 0), (3, 0), ⎜ , ⎟ , ⎜ , ⎟ ⎝ 7 7 ⎠ ⎝ 20 20 ⎠ ⎛ 27 ⎞ and ⎜ 8, ⎟ . Evaluating Z at each point, we find ⎝ 11 ⎠ that Z has a minimum value of 3 when x = 3 and y = 0. 234


10

Section 7.2

y

9x + 11y = 99

x−y=0 Z=3

Z = 0.8

x=8

4x + 3y = 12

x

x

(3, 0)

9. The feasible region is unbounded with 3 corner points. The member (see dashed line) of the family of lines C = 3x + 2y which gives a minimum value of C, subject to the constraints, intersects the feasible region at corner point 23 ⎛7 1⎞ Thus C has a minimum ⎜ , ⎟ where C = 3 ⎝ 3 3⎠ 23 7 1 value of when x = and y = . [Note: 3 3 3 Here we chose the member of the family 1 y = (−3x + C ) whose y-intercept was closest 2 to the origin and which had at least one point in common with the feasible region.]

y 2x − y = 2

x − 4y = 4 x 10

6. The feasible region is empty, so there is no optimum solution. 10

y 2x + y = 10

5

y (0, 5) 2x + y = 5 3x + y = 4 23 C= 3 ⎛7 1⎞ , ⎜ ⎟ ⎝ 3 3⎠

8x + 7y = 56 3x + 4y = 24 x 10

x + 2y = 3

7. The feasible region is a line segment. The corner points are (0, 1) and (4, 5). Z has a minimum value of 3 when x = 0 and y = 1. 5

x

(3, 0)

5

10. The feasible region is unbounded with 4 corner points. The member (see dashed line) of the C which gives a family of lines y = − x + 2 minimum value of C, subject to the constraints, intersects the feasible region at corner point (40, 20) where C = 120. Thus C has a minimum value of 120 when x = 40 and y = 20.

y (4, 5)

(0, 1) Z=3

10

(2, 0)

10

5. The feasible region is empty, so there is no optimum solution. 10

y

10

100

x

y 5x + 2y = 200

5

8. The feasible region is a line segment. The corner ⎛ 27 21 ⎞ points are (2, 0) and ⎜ , ⎟. ⎝ 17 17 ⎠ Z has a maximum value of 0.8 for x = 2 and y = 0.

3x + 2y = 160 x + 2y = 80 C = 120

x 100

235



11. The feasible region is unbounded with 2 corner points. The family of lines given by Z = 10x + 2y has members (see dashed lines for two sample members) that have arbitrarily large values of Z and that also intersect the feasible region. Thus no optimum solution exists. 10

50

y

3x + y = 50 Maximum Profit Line

(10, 20)

y 5x + y = 70

Z = 70 x − 2y = 0

14. Let x and y be the numbers of Vista and Xtreme models made each day. Then we are to maximize P = 50x + 80y, where x≥0 ⎧ y≥0 ⎪⎪ ⎨ x + 3 y ≤ 24 (for machine A) ⎪ ⎪⎩2 x + 2 y ≤ 24 (for machine B) The feasible region is bounded. The corner points are (0, 0), (0, 8), (6, 6), and (12, 0). Evaluating P at each corner point, we find that P is maximized at corner point (6, 6) where its value is 780. Thus 6 of each model should be made each day in order to give a maximum profit of $780.

x 10

x + 2y = 4

12. The feasible region is unbounded with 3 corner points. The family of lines given by Z = y – x has members (see dashed lines for sample members) that have arbitrarily small values for Z and also intersect the feasible region. Thus no optimum solution exists. 10

y Z = −1 x − 3y = −6

x=3

25

y

Z = −7 x

x + 3y = 6

x 50

Z = 50

2x – 2y = 24

10

(6, 6)

13. Let x and y be the number of trucks and spinning tops made per week, respectively. Then we are to maximize P = 7x + 2y where x≥0 ⎧ ⎪ y≥0 ⎪ ⎨2 x + y ≤ 80 (for machine A) ⎪ 3 x + y ≤ 50 (for machine B) ⎪ 5 x + y ≤ 70 (for finishing) ⎩ The feasible region is bounded. The corner points are (0, 50), (14, 0) and (10, 20). Evaluating P at each corner point, we find that P is maximized at corner point (10, 20), where its value is 110. Thus10 trucks and 20 spinning tops should be made each week to give a maximum profit of $110.

x + 3y = 24

x

25

15. Let x and y be the numbers of units of Food A and Food B, respectively, that are purchased. Then we are to minimize C = 1.20x + 0.80y, where x ≥ 0, ⎧ ⎪⎪ y ≥ 0, ⎨2 x + 2 y ≥ 16 (for carbohydrates), ⎪ ⎪⎩ 4 x + y ≥ 20 (for protein). The feasible region is unbounded. The corner points are (8, 0), (4, 4) and (0, 20). C is minimized at corner point (4, 4) where C = 8 (see the minimum cost line). Thus 4 units of Food A and 4 units of Food B gives a minimum cost of $8.

236


30

Section 7.2

y

50

y

200x + 50y = 2500 4x + y = 20 Minimum Cost Line

2x + 2y = 16

100x + 200y = 3000 x

x 30

Minimum Cost Line

16. Let x and y be the numbers of units of Blend I and Blend II, respectively, that are bought each week. Then we are to minimize C = 8x + 10y where x ≥ 0, ⎧ ⎪ y ≥ 0, ⎪ ⎨ 2 x + 2 y ≥ 80 (for Nutrient A), ⎪ 6 x + 2 y ≥ 120 (for Nutrient B), ⎪4 x + 12 y ≥ 240 (for Nutrient C). ⎩ The feasible region is unbounded with 4 corner points. C is minimized at the corner point (30, 10) where C = 340 (see the minimum cost line). thus each week the grower should buy 30 bags of Blend I and 10 bags of Blend II. 100

50

18. Let x and y be the number of days Refinery I and Refinery II are operated, respectively. Then we are to minimize C = 25,000x + 20,000y where x ≥ 0, ⎧ ⎪ y ≥ 0, ⎪ ⎨ 2000 x + 1000 y ≥ 8000 (for low grade), ⎪3000 x + 2000 y ≥ 14, 000 (for medium grade), ⎪ 1000 x + 1000 y ≥ 5000 (for high grade). ⎩ The feasible region is unbounded with 4 corner points. Evaluating C at each corner point, we find that C is minimized at corner point (4, 1) where C = 120,000. Thus, operate Refinery I for 4 days and Refinery II for 1 day for a minimum cost of $120,000.

y

10

y

6x + 2y = 120 2000x + 1000y = 8000 2x + 2y = 80 Minimum Cost Line

3000x + 2000y = 14,000 1000x + 1000y = 5000 x

4x + 12y = 240 x 100

10

17. Let x and y be the numbers of tons of ores I and II, respectively, that are processed. Then we are to minimize C = 50x + 60y, where x ≥ 0, ⎧ y ≥ 0, ⎪⎪ ⎨100 x + 200 y ≥ 3000 (for mineral A), ⎪ ⎪⎩ 200 x + 50 y ≥ 2500 (for mineral B). The feasible region is unbounded with 3 corner points. C is minimized at the corner point (10, 10) where C = 1100 (see the minimum cost line). Thus 10 tons of ore I and 10 tons of ore II give a minimum cost of $1100.

19. Let x and y be the number of chambers of type A and B, respectively. Then we are to minimize C = 600,000x + 300,000y, where x ≥ 4, ⎧ ⎪⎪ y ≥ 4, ⎨ 10 x + 4 y ≥ 100 (for polymer P ), 1 ⎪ ⎪⎩20 x + 30 y ≥ 420 (for polymer P2 ). The feasible region is unbounded with 3 corner points. Evaluating C at each corner point, we find C is minimized at corner point (6, 10) where C = 6,600,000. Thus the solution is 6 chambers of type A and 10 chambers of type B.

237


20

ISM: Introductory Mathematical Analysis 2(300 – x) + 8(200 – y ) ≥ 300, 2200 − 2 x − 8 y ≥ 300, −2 x − 8 y ≥ –1900, 2 x + 8 y ≤ 1900. The fifth constraint reflects the fact that company A will not build more than 300 km of highway, since 300 km is the total being built; the sixth constraint is the corresponding constraint for the amount of expressway.

y x=4 10x + 4y = 100 20x + 30y = 420 y=4

x

20

20. Let x and y be the number of liters produced by the old and new processes, respectively. We want to maximize P = 0.4x + 0.15y, where x≥0 ⎧ ⎪⎪ y≥0 ⎨ 25 x + 15 y ≤ 12,525 (for carbon dioxide) ⎪ ⎩⎪50 x + 40 y ≤ 20, 000 (for particulate matter)

c.

The feasible region is bounded with three corner points. Evaluating P at each corner point, we find that P is maximized at the corner point (400, 0), where P = 160. Thus daily production of 400 liters by only the old process maximizes daily profit at $160. 1000

The feasible region (see below) is bounded. The corner points are (0, 200), (150, 200), ⎛ 650 550 ⎞ , ⎜ ⎟ , (300, 100), (300, 0), and 3 ⎠ ⎝ 3 (200, 0). Evaluating D at each corner point, we find that D is maximized at point (0, 200), where D = 2100. That is, D is maximized when x = 0, y = 200. 500

y

y

y = 200 2x + 8y = 1900 x + y = 400 x = 300

25x + 15y = 12,525

x + y = 200 50x + 40y = 20,000 x

22. Z = 2.71 when x = 1.14, y = 1.43

1000

21. a.

x

500

23. Z = 15.54 when x = 2.56, y = 6.74

A builds x km of highway and y km of expressway, so B builds (300 – x) km of highway and (200 – y) km of expressway. Thus D = 2x + 6y + 3(300 – x) + 5(200 – y) = 1900 – x + y.

24. The feasible region is empty, so there is no optimum solution. 25. Z = –75.98 when x = 9.48, y = 16.67 Principles in Practice 7.3

b. The first constraint is company A’s construction limit. The second constraint is company B’s construction limit, which arises as follows: (300 – x) + (200 – y ) ≤ 300, 500 – x – y ≤ 300, – x – y ≤ –200, x + y ≥ 200. The third constraint is the minimum contract for A. The fourth constraint is the minimum contract for B, which arises as follows:

1. Using the hint, the cost of shipping the TV sets is Z = 18x + 24(25 – x) + 9y + 15(30 – y) = 1050 – 6x – 6y. Since negative numbers of TV sets cannot be shipped, x ≥ 0, y ≥ 0, 25 – x ≥ 0, and 30 – y ≥ 0. Since warehouse C has only 45 TV sets, x + y ≤ 45. Similarly, since warehouse D has only 40 TV sets, 25 – x + 30 – y ≤ 45 or x + y ≥ 10. We need to minimize Z = 1050 – 6x – 6y subject

238


Section 7.3

to the constraints x + y ≤ 45, x + y ≥ 10, x ≤ 25, y ≤ 30, x ≥ 0, y ≥ 0.

2. The feasible region is a line segment. The corner ⎛ 2 16 ⎞ ⎛ 16 2 ⎞ points are ⎜ , ⎟ and ⎜ , ⎟ . At each of 3 3 ⎝ ⎠ ⎝ 3 3⎠ these points Z = 12. Thus Z is maximized at both corner points, as well as at all points on the line segment. Thus the solution is Z = 12 when 2 14 ⎛ 2 ⎞ 16 x = (1 − t ) ⎜ ⎟ + t = + t , 3 3 3 3 ⎝ ⎠

y

50

16 14 ⎛ 16 ⎞ 2 − t , and 0 ≤ t ≤ 1. x = (1 − t ) ⎜ ⎟ + t = 3 3 3 3 ⎝ ⎠

y = 30 C

B

x + y = 45 D x = 25

A

x

x + y = 10 F

E

50

The feasible region shown has corners A = (0, 10), B = (0, 30), C = (15, 30), D = (25, 20), E = (25, 0), and F = (10, 0). Evaluating the cost function at the corners gives Z(A) = 1050 – 6(0) – 6(10) = 990 Z(B) = 1050 – 6(0) – 6(30) = 870 Z(C) = 1050 – 6(15) – 6(30) = 780 Z(D) = 1050 – 6(25) – 6(20) = 780 Z(E) = 1050 – 6(25) – 6(0) = 900 Z(F) = 1050 – 6(10) – 6(0) = 990 The minimum value of Z is 780 which occurs at all points on the line segment joining C and D. This is x = (1 – t)(15) + t(25) = 15 + 10t and y = (1 – t)(30) + t(20) = 30 – 10t for 0 ≤ t ≤ 1. Thus, ship 10t + 15 TV sets from C to A, – 10t + 30 TV sets from C to B, 25 – (10t + 15) = –10t + 10 TV sets from D to A, and 30 – (–10t + 30) = 10t TV sets from D to B, for 0 ≤ t ≤ 1. The minimum cost is $780.

( 163 , 23 ) 5 x

3. The feasible region appears below. The corner ⎛ 8 ⎞ ⎛ 36 4 ⎞ points are (0, 0), ⎜ 0, ⎟ , ⎜ , ⎟ and (6, 0). Z ⎝ 5⎠ ⎝ 7 7⎠ ⎛ 36 4 ⎞ is maximized at ⎜ , ⎟ and (6, 0), where its ⎝ 7 7⎠ value is 84. Thus Z is also maximized at all ⎛ 36 4 ⎞ points on the line segment joining ⎜ , ⎟ and ⎝ 7 7⎠ (6, 0). The solution is Z = 84 when 6 36 ⎛ 36 ⎞ x = (1 − t ) ⎜ ⎟ + 6t = t + , 7 7 ⎝ 7 ⎠ 4 4 ⎛4⎞ y = (1 − t ) ⎜ ⎟ + 0t = − t and 0 ≤ t ≤ 1. 7 7 7 ⎝ ⎠

Problems 7.3 1. The feasible region is unbounded. Z is minimized at corner points (2, 3) and (5, 2), where its value is 33. Z is also minimized at all points on the line segment joining (2, 3) and (5, 2), so the solution is Z = 33 when x = (1 – t)(2) + 5t = 2 + 3t y = (1 – t)(3) + 2t = 3 – t and 0 ≤ t ≤ 1. 10

( 23 , 163 )

y

5

10

y

2x + 3y = 12 ⎛ 36 4 ⎞ ⎜ , ⎟ ⎝ 7 7⎠

y

y = – 3x + 6

x + 5y = 8

2

y=x–3 (2, 3)

4. Using the hint, the cost of delivering the cars is Z = 60x + 45y + 50(7 – x) + 35(4 – y) = 490 + 10x + 10y. Since negative numbers of cars is not possible, x ≥ 0, y ≥ 0, 7 – x ≥ 0, and 4 – y ≥ 0. Since

(5, 2) x

y = – 1 x + 11 3

3

x 10

10

239



the warehouse in Concord has only 6 cars, x + y ≤ 6. Similarly, since the Dublin warehouse has only 8 cars, 7 – x + 4 – y ≤ 8 or 3 ≤ x + y. We need to minimize Z = 490 + 10x + 10y subject to the constraints x + y ≤ 6, x + y ≥ 3, x ≤ 7, y ≤ 4, x ≥ 0, y ≥ 0. 10

dividing by the greatest common factor of the numbers involved. Thus, we will use 3x1 + 3 x2 + 4 x3 ≤ 300 3x1 + 3 x2 + 2 x3 ≤ 240 2 x1 + 2 x2 + 3x3 ≤ 180 x1 , x2 , x3 ≥ 0 x1 x2 x3 s1 s2 s3 P b 3 3 4 1 0 0 0 300 ⎤ s1 ⎡ ⎢ 3 3 2 0 1 0 0 240 ⎥⎥ s2 ⎢ 2 2 3 0 0 1 0 180 ⎥ s3 ⎢ ⎢ ⎥ 0⎥ P ⎣⎢ –150 –250 –200 0 0 0 1 ⎦ x1 x2 x3 s1 s2 s3 P b s1 ⎡ 0 0 2 1 –1 0 0 60 ⎤ ⎢ ⎥ 2 1 x2 ⎢ 1 1 0 0 0 80 ⎥ 3 3 ⎢ ⎥ 5 s3 ⎢ 0 0 0 – 32 1 0 20 ⎥ 3 ⎢ ⎥ P ⎢100 0 – 100 0 250 0 1 20, 000 ⎥ 3 3 ⎣⎢ ⎦⎥ x1 x2 x3 s1 s2 s3 P b

y

x=7 B A

y=4

C

x+y=6

x+y=3

x E

D

10

The feasible region shown has corners A = (0, 3), B = (0, 4), C = (2, 4), D = (6, 0), and E = (3, 0). Evaluating the cost function at the corners gives Z(A) = 490 + 10(0) + 10(3) = 520 Z(B) = 490 + 10(0) + 10(4) = 530 Z(C) = 490 + 10(2) + 10(4) = 550 Z(D) = 490 + 10(6) + 10(0) = 550 Z(E) = 490 + 10(3) + 10(0) = 520 The minimum value of Z is 520 which occurs at all points on the line segment joining A and E. This is x = (1 – t)(0) + t(3) = 3t and y = (1 – t)(3) + t(0) = –3t + 3 for 0 ≤ t ≤ 1. Thus have 3t cars delivered from Concord to Atherton, –3t + 3 delivered from Concord to Berkeley, 7 – 3t delivered from Dublin to Atherton, and 4 – (–3t + 3) = 3t + 1 delivered from Dublin to Berkeley, for 0 ≤ t ≤ 1. The minimum cost is $520.

36 ⎤ s1 ⎡ 0 0 0 1 – 15 – 65 0 ⎢ ⎥ 3 –2 0 x2 ⎢ 1 1 0 0 ⎥ 72 5 5 ⎢ ⎥ 3 0 2 ⎥ 12 x3 ⎢⎢ 0 0 1 0 – 5 5 ⎥ P ⎢100 0 0 0 70 20 1 20, 400 ⎥ ⎣ ⎦ The maximum value of P is 20,400 when x1 = 0, x2 = 72, and x3 = 12 . The maximum profit is $20,400 when 72 Type 2 players and 12 Type 3 players are produced and sold.

Problems 7.4 In these problems, the pivot entry is underlined.


1.

In these problems, the pivot entry is underlined. 1. Let x1 , x2 , and x3 be the numbers of Type 1, Type 2, and Type 3 players, respectively, that the company produces. The situation is to maximize the profit P = 150 x1 + 250 x2 + 200 x3 , subject to the constraints 300 x1 + 300 x2 + 400 x3 ≤ 30, 000 15 x1 + 15 x2 + 10 x3 ≤ 1200 2 x1 + 2 x2 + 3x3 ≤ 180 x1 , x2 , x3 ≥ 0 The constraint inequalities can be simplified by

x1 ⎡ s1 2 ⎢ s2 ⎢ 2 Z ⎢ −1 ⎣ x1 ⎡4 s1 ⎢ 3 x2 ⎢⎢ 23 ⎢1 Z ⎢⎣ 3

x2 s1 s2 Z

8 ⎤8 ⎥ 3 0 1 0 12 ⎥ 4 −2 0 0 1 0 ⎥ ⎦ x2 s1 s2 Z 0 1 – 13 0 4 ⎤ ⎥ 1 0 4⎥ 1 0 ⎥ 3 2 1 8⎥ 0 0 ⎥⎦ 3 1

1 0 0

The solution is Z = 8 when x1 = 0, x2 = 4 .

240


2.

3.

Section 7.4

x1

x1 x2 s1 s2 Z ⎡ s1 –1 1 1 0 0 4 ⎤ ⎢ ⎥ 1 0 1 0 6⎥ 6 s2 ⎢ 1 Z ⎢ –2 –1 0 0 1 0 ⎥ ⎣ ⎦ x1 x2 s1 s2 Z ⎡ s1 0 2 1 1 0 10 ⎤ ⎢ ⎥ x1 ⎢ 1 1 0 1 0 6 ⎥ Z ⎢ 0 1 0 2 1 12 ⎥ ⎣ ⎦ The solution is Z = 12 when x1 = 6, x2 = 0 . x1

x2 s1 s2 Z

s1 ⎡ 3 2 ⎢ s2 ⎢ –1 3 Z ⎢ 1 –2 ⎣ x1 x2

5 1 0 0 5⎤ 2 ⎥ 0 1 0 3⎥ 1 0 0 1 0⎥ ⎦ s1 s2 Z

s1 ⎡ 11 0 1 – 2 0 3⎤ 3 ⎢ 3 ⎥ ⎢ 1 1 0 1⎥⎥ x2 ⎢ – 3 1 0 3 ⎢ 2 1 2⎥ Z ⎢ 13 0 0 ⎥⎦ 3 ⎣ The solution is Z = 2 when x1 = 0, x2 = 1.

4.

x1 x2 s1 ⎡ 3 1 s1 2 ⎢ s2 ⎢ 1 5 0 Z ⎢ −4 –7 0 ⎣ x1 x2 s1 ⎡ 7 0 1 s1 ⎢ 5 x2 ⎢⎢ 15 1 0 ⎢ 0 0 Z ⎢ – 13 ⎣ 5 x1 x2 s1 ⎡ 5 x1 1 0 7 ⎢ x2 ⎢⎢ 0 1 – 17 ⎢ Z ⎢ 0 0 13 7 ⎣

x2 s1 s2 s3 Z

s1 ⎡ 1 –1 1 0 0 0 1 ⎤ 1 ⎢ ⎥ 5. s2 ⎢ 1 2 0 1 0 0 8 ⎥ 8 s3 ⎢ 1 1 0 0 1 0 5 ⎥ 5 ⎢ ⎥ Z ⎢⎣ –8 –2 0 0 0 1 0 ⎥⎦ x1 x2 s1 s2 s3 Z x1 ⎡1 –1 1 0 0 0 1 ⎤ ⎢ ⎥ s2 ⎢ 0 3 –1 1 0 0 7 ⎥ 73 s3 ⎢ 0 2 –1 0 1 0 4 ⎥ 2 ⎢ ⎥ Z ⎢⎣ 0 –10 8 0 0 1 8 ⎦⎥ x1 x2 s1 s2 s3 Z 1 0 1 0 ⎡ 3⎤ x1 1 0 2 2 ⎢ ⎥ s2 ⎢ 0 0 1 1 –3 0 ⎥ 1 2 2 ⎢ ⎥ 1 0 1 0 ⎢ ⎥ 0 1 – 2 x2 ⎢ 2 2 ⎥ Z ⎢0 0 3 0 5 1 28⎥ ⎣ ⎦ The solution is Z = 28 when x1 = 3, x2 = 2 .

6.

s2 Z

9⎤ 3 ⎥ 1 0 10 ⎥ 2 0 1 0⎥ ⎦ s2 Z – 53 0 3⎤ 15 ⎥ 7 1 0 2 ⎥⎥ 10 5 7 1 14 ⎥ ⎥⎦ 5 s2 Z ⎤ – 73 0 15 7⎥ 2 0 11 ⎥ 7 7⎥ 2 1 137 ⎥ 7 7 ⎥⎦ 137 15 11 The solution is Z = when x1 = , x2 = . 7 7 7 0 0

x1 x2 s1 ⎡ 1 –1 ⎢ s2 ⎢ –1 1 s3 ⎢ 1 1 ⎢ Z ⎣⎢ –2 6

s1 1 0 0 0

s2 0 1 0 0

s3 0 0 1 0

Z 0 0 0 1

x1 x2 s1 s2 s3 Z ⎡ x1 1 –1 1 0 0 0 ⎢ s2 ⎢ 0 0 1 1 0 0 s3 ⎢ 0 2 –1 0 1 0 ⎢ Z ⎢⎣ 0 4 2 0 0 1 The solution is Z = 8 when

241

4⎤ 4 4 ⎥⎥ 6⎥ 6 ⎥ 0⎥ ⎦ 4⎤ 8⎥⎥ 2⎥ ⎥ 8⎥ ⎦ x1 = 4, x2 = 0


x1

7.

x2

ISM: Introductory Mathematical Analysis Thus the maximum value of Z is 4, when x1 = 2, x2 = 0, x3 = 0 .

x3 s1 s2 Z

⎡ ⎤ s1 ⎢ 1 2 0 1 0 0 10 ⎥ 5 s2 ⎢ 2 2 1 0 1 0 10 ⎥ 5 ⎢ ⎥ Z ⎢ –3 –4 – 3 0 0 1 0 ⎥ 2 ⎣ ⎦ choosing s2 as departing variable

9. To obtain a standard linear programming problem, we write the second constraint as – x1 + 2 x2 + x3 ≤ 2 . x1

s1 ⎡ 1 1 0 1 0 0 1 ⎤ 1 ⎢ ⎥ s2 ⎢ –1 2 1 0 1 0 2 ⎥ Z ⎢ –2 –1 1 0 0 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z x1 ⎡ 1 1 0 1 0 0 1⎤ ⎢ ⎥ s2 ⎢ 0 3 1 1 1 0 3⎥ Z ⎢0 1 1 2 0 1 2⎥ ⎣ ⎦ The solution is Z = 2 when x1 = 1, x2 = 0, x3 = 0 .

x1 x2 x3 s1 s2 Z ⎡ ⎤ s1 ⎢ –1 0 –1 1 −1 0 0 ⎥ x2 ⎢⎢ 1 1 12 0 12 0 5⎥⎥ Z ⎢ 1 0 1 0 2 1 20 ⎥ ⎢⎣ ⎥⎦ 2 The solution is Z = 20 when x1 = 0, x2 = 5, x3 = 0

8. If s1 is the departing variable, then x1 x2 x3 s1 s2 Z

10. To obtain a standard linear programming problem, we write the third constraint as − x1 + x2 ≤ 3.

s1 ⎡ 2 1 –1 1 0 0 4 ⎤ 2 ⎢ ⎥ s2 ⎢ 1 1 1 0 1 0 2 ⎥ 2 Z ⎢ –2 1 –1 0 0 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z 1 1 1 ⎡ 0 0 2⎤ x1 ⎢1 2 – 2 2 ⎥ 3 1 1 0 0⎥ 0 s2 ⎢ 0 12 – 2 2 ⎢ ⎥ Z ⎢ 0 2 –2 1 0 1 4 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z ⎡ 1 1 0 2⎤ x1 1 2 0 3 3 3 ⎢ ⎥ x3 ⎢ 1 1 2 0 0⎥ 0 1 – ⎢ ⎥ 3 3 3 ⎢ ⎥ 8 1 4 1 4⎥ Z ⎢0 3 0 3 3 ⎣ ⎦ The solution is Z = 4 when x1 = 2, x2 = 0, x3 = 0 . Choosing s2 as the departing variable x1 ⎡ s1 2 ⎢ s2 ⎢ 1 Z ⎢ –2 ⎣ x1 s1 ⎡ 0 ⎢ x1 ⎢ 1 Z ⎢0 ⎣

x1 x2 s1 s2 s3 s4 s1 ⎡ 1 1 1 0 0 0 s2 ⎢ 1 −1 0 1 0 0 ⎢ s3 ⎢ −1 1 0 0 1 0 ⎢ s4 ⎢ 1 0 0 0 0 1 Z ⎢⎢ 2 −3 0 0 0 0 ⎣ x1 x2 s1 s2 s3 s4 x2 ⎡ 1 1 1 0 0 0 ⎢ s2 ⎢ 2 0 1 1 0 0 s3 ⎢ −2 0 −1 0 1 0 ⎢ s4 ⎢ 1 0 0 0 0 1 Z ⎢⎢⎣ 5 0 3 0 0 0 The solution is Z = 3 when

x2 x3 s1 s2 Z 1 –1 1 1 1 –1 x2 x3 –1 –3 1 1 3 1

x2 x3 s1 s2 Z

11.

1 0 0 4⎤ 2 ⎥ 0 1 0 2⎥ 2 0 0 1 0⎥ ⎦ s1 s2 Z 1 –2 0 0 ⎤ ⎥ 0 1 0 2⎥ 0 2 1 4⎥ ⎦

242

Z 0 0 0 0 1

1⎤ 1 2 ⎥⎥ 3⎥ 3 ⎥ 5⎥ 0 ⎥⎥ ⎦

Z 0 0 0 0 1

1⎤ 3⎥⎥ 2⎥ ⎥ 5⎥ 3⎥⎥ ⎦ x1 = 0, x2 = 1 .

x1

x2 s1 s2 s3 s4 Z

s1 ⎡ 2 ⎢ s2 ⎢ –1 s3 ⎢ 5 ⎢ s4 ⎢ 2 Z ⎢⎢⎣ –1

–1 1 0 0 0 0 2

0 1 0 0 0

3 1

0 0 1 0 0 0 0 0 1 0

–1 0 0 0 0 1

4 ⎤2 ⎥ 6⎥ 20 ⎥ 4 ⎥ 10 ⎥ 5 0 ⎥⎥ ⎦


Section 7.4

x1 x2 s1

choosing x1 as entering variable

s1 ⎡ 3 0 ⎢ 2 x2 ⎢ 1 − 1 ⎢ 2 s3 ⎢ 13 0 ⎢ 2 ⎢ s4 ⎢ 52 0 Z ⎢ −3 0 ⎢⎣ 2 x1 x2 s1 s1 ⎡ 0 0 1 ⎢ x2 ⎢ 0 1 0 ⎢ x1 ⎢ 1 0 0 ⎢ s4 ⎢ 0 0 0 ⎢ Z ⎢0 0 0 ⎢⎣

x1 x2 s1 s2 s3 s4 Z 1 1 x1 ⎡⎢1 – 2 2 0 0 0 0 2 ⎤⎥ 16 1 s2 ⎢ 0 3 1 0 0 0 8⎥ 3 2 2 ⎢ ⎥ 20 s3 ⎢ 0 11 − 5 0 1 0 0 10 ⎥ 11 2 2 ⎢ ⎥ s4 ⎢ 0 2 −1 0 0 1 0 6 ⎥ 3 ⎢ ⎥ Z ⎢0 – 3 1 0 0 0 1 2 ⎥ 2 2 ⎣⎢ ⎦⎥ x1 x2 s1 s2 s3 s4 Z 3 0 1 0 0 32 ⎤ 32 x1 ⎡ 1 0 11 11 11 ⎥ 3 ⎢ 13 1 − 3 0 0 58 ⎥ 58 s2 ⎢ 0 0 11 11 11 ⎥ 13 ⎢ x2 ⎢ 0 1 − 5 0 2 0 0 20 ⎥ ⎢ 11 11 11 ⎥ s4 ⎢ 0 0 − 1 0 − 4 1 0 26 ⎥ ⎢ 11 11 11 ⎥ 3 0 1 52 ⎥ Z ⎢0 0 − 2 0 ⎢⎣ 11 11 11 ⎥⎦ x1 x2 s1 s2 s3 s4 Z 2 22 ⎤ x1 ⎡1 0 0 − 3 0 0 13 13 13 ⎢ ⎥ s1 ⎢ 0 0 1 11 − 3 0 0 58 ⎥ 13 13 13 ⎥ ⎢ 5 50 ⎥ 1 x2 ⎢ 0 1 0 0 0 13 13 13 ⎢ ⎥ 5 1 0 36 ⎥ 1 s4 ⎢ 0 0 0 − 13 13 13 ⎥ ⎢ 3 72 ⎥ 2 Z ⎢0 0 0 0 1 13 13 13 ⎦⎥ ⎣⎢ 72 Thus the maximum value of Z is , when 13 22 50 , x2 = . If we choose x2 as the x1 = 13 13 entering variable, then we have: x1

x2 s1 s2 s3 s4 Z

s1 ⎡ 2 ⎢ s2 ⎢ –1 s3 ⎢ 5 ⎢ s4 ⎢ 2 Z ⎢⎣⎢ –1

–1 1 0 0 0 0 2

0 1 0 0 0

3 1

0 0 1 0 0 0 0 0 1 0

–1 0 0 0 0 1

s2

s3 s4 Z

1 2 1 2 3 2

0 0 0

1 0 0 –

0 – 12 0 s2 11 13 5 13 3 – 13 1 13 2 13

The solution is Z = x2 =

1 2

0 0 0 1 0 0 0

1 0

0 0

1

s3

s4 Z

3 − 13 1 13 2 13 5 − 13 3 13

0 0 0 0 0 0 1 0 0

1

7⎤ 3 ⎥ 3⎥ ⎥ 22 11⎥ 13 ⎥ ⎥ 7 ⎥ 14 5 ⎥ 3⎥ ⎦ 14

58 ⎤ 13 ⎥ 50 ⎥ 13 ⎥ 22 ⎥ 13 ⎥ 36 ⎥ 13 ⎥ 72 ⎥ 13 ⎦⎥

72 22 when x1 = , 13 13

50 . 13

12. To obtain a standard linear programming problem, we write the first constraint as 2 x1 – x2 – x3 ≤ 2 . x1 x2 x3 s1 s2 s3 W s1 ⎡ 2 –1 –1 1 0 0 0 2 ⎤ 1 ⎢ ⎥ s2 ⎢ 1 –1 1 0 1 0 0 4 ⎥ 4 s3 ⎢ 1 1 2 0 0 1 0 6 ⎥ 6 ⎢ ⎥ W ⎢⎣ –2 –1 2 0 0 0 1 0 ⎥⎦ x1 x2 x3 s1 s2 s3 W x1 ⎡1 – 12 – 12 12 0 0 0 1 ⎤ ⎢ ⎥ 3 s2 ⎢ 0 – 1 1 1 0 0 3⎥ – 2 2 2 ⎢ ⎥ 5 1 ⎢0 3 ⎥ 10 – 0 1 0 5 s3 ⎢ 2 2 2 ⎥ 3 W ⎢⎢ 0 –2 1 1 0 0 1 2 ⎥⎥ ⎣ ⎦ x1 x2 x3 s1 s2 s3 W 8⎤ 1 1 0 1 0 x1 ⎡ 1 0 3 3 3 3⎥ ⎢ 7 2 1 14 s2 ⎢ 0 0 –3 1 3 0 3⎥ 3 ⎢ ⎥ ⎢ 0 1 5 – 1 0 2 0 10 ⎥ 3 3 3 3⎥ x2 ⎢ ⎢ 13 26 1 4 0 3 1 3⎥ W ⎢0 0 3 3 ⎣ ⎦⎥

4⎤ 6 ⎥⎥ 3 20 ⎥ 20 ⎥ 3 10 ⎥ 10 0 ⎥⎥ ⎦

243


The solution is W =


26 8 10 when x1 = , x2 = , x3 = 0 . 3 3 3

13. To obtain a standard linear programming problem, we write the second constraint as − x1 – x2 + x3 ≤ 2 and the third constraint as x1 – x2 – x3 ≤ 1 . x1 x2 x3 s1 s2 s3 W s1 ⎡ 4 3 –1 1 0 0 0 1 ⎤ ⎢ ⎥ s2 ⎢ –1 –1 1 0 1 0 0 2 ⎥ 2 s3 ⎢ 1 –1 –1 0 0 1 0 1 ⎥ ⎢ ⎥ W ⎣⎢ –1 12 –4 0 0 0 1 0 ⎦⎥ x1 x2 x3 s1 3 2 0 1 ⎡ s1 ⎢ –1 –1 1 0 x3 ⎢ s3 ⎢ 0 –2 0 0 ⎢ W ⎢⎣ –5 8 0 0

s2 s3 W 1 0 0 3⎤ 1 1 0 0 2 ⎥⎥ 1 1 0 3⎥ ⎥ 4 0 1 8⎥ ⎦

x1 x2 2 x1 ⎡ 1 3 ⎢ x3 ⎢0 – 1 3 s3 ⎢ ⎢0 –2 ⎢ 34 W ⎢⎣0 3

s2 s3 W

x3 s1

1⎤ ⎥ 1 0 0 3⎥ ⎥ 0 0 1 1 0 3⎥ ⎥ 0 53 17 0 1 13⎥ 3 ⎦ The solution is W = 13 when x1 = 1, x2 = 0, x3 = 3 . x1 14. s1 ⎡ 1 ⎢ s2 ⎢ 1 s3 ⎢ 1 ⎢ W ⎣⎢ –4 x1 ⎡ s1 0 ⎢ s2 ⎢ 0 x1 ⎢1 ⎢ W ⎢⎣ 0 x1 ⎡ x2 ⎢ 0 s2 ⎢ 0 ⎢ x1 ⎢ 1 W ⎢0 ⎣⎢

1 3 1 3

0

x2 x3 1 1 –1 1 –1 –1 0 1

1 3 4 3

s1 1 0 0 0

x2 x3 s1 2 2 1 0 2 0 –1 –1 0 –4 –3 0 x2 x3 s1 s2 1 1

1 2

0 2

0 1

0 0

1 2

0

0

0

1 2 0

0 0

s2 0 1 0 0

s3 0 0 1 0

s2 0 1 0 0

s3 –1 –1 1 4

W 0 6⎤ 6 0 10 ⎥⎥ 10 0 4⎥ 4 ⎥ 1 0⎥ ⎦

W 0 2 ⎤1 0 6 ⎥⎥ 0 4⎥ ⎥ 1 16 ⎥ ⎦ s3 W – 12 0 1⎤ ⎥ –1 0 6 ⎥ ⎥ 1 0 5⎥ 2 ⎥ 2 1 20 ⎥ ⎦

The solution is W = 20 when x1 = 5, x2 = 1, x3 = 0 .

244


x1 x2 x3 x4 –2 0 0 s1 ⎡ 1 ⎢ 1 1 0 0 s2 ⎢ 15. ⎢ 0 1 1 s3 0 ⎢ 0 0 1 –2 s4 ⎢ Z ⎢⎣⎢ –60 0 –90 0 x1 x2 –2 ⎡ s1 1 ⎢ 1 1 s2 ⎢ 0 x3 ⎢ 0 ⎢ 0 s4 ⎢ 0 ⎢ –60 0 Z ⎢⎣

x3 x4 0 0 0 0 1 1 0 –3 0 90

s1 1 0 0 0 0

s1 1 0 0 0 0

s2 0 1 0 0 0

s3 0 0 1 0 0

s2 s3 0 0 1 0 0 1 0 –1 0 90

s4 0 0 0 1 0

Section 7.4

Z 0 0 0 0 1

2⎤ 5 ⎥⎥ 4⎥ 4 ⎥ 7⎥ 7 0 ⎥⎥ ⎦

s4 0 0 0 1 0

Z 0 2 ⎤2 0 5 ⎥⎥ 5 0 4 ⎥ ⎥ 0 3 ⎥ 1 360 ⎥⎥ ⎦ x1 x2 x3 x4 s1 s2 s3 s4 Z x1 ⎡1 –2 0 0 1 0 0 0 0 2 ⎤ ⎢ 3 0 0 –1 1 0 0 0 3 ⎥⎥ 1 s2 ⎢0 0 1 1 0 0 1 0 0 4 ⎥ x3 ⎢0 ⎢ ⎥ 0 0 –3 0 0 –1 1 0 3 ⎥ s4 ⎢0 Z ⎢⎢⎣0 –120 0 90 60 0 90 0 1 480 ⎥⎥⎦ x1 x2 x3 x4 s1 s2 s3 s4 Z 1 2 x1 ⎡ 1 0 0 0 0 0 0 4⎤ 3 3 ⎢ ⎥ 1 ⎥ x2 ⎢ 0 1 0 0 – 1 0 0 0 1 3 3 ⎢ ⎥ x3 ⎢ 0 0 1 1 0 0 1 0 0 4⎥ ⎥ s4 ⎢⎢ 0 0 0 –3 0 0 –1 1 0 3⎥ Z ⎢⎢ 0 0 0 90 20 40 90 0 1 600 ⎥⎥ ⎣ ⎦ The solution is Z = 600 for x1 = 4, x2 = 1, x3 = 4, x4 = 0 . x1 x2 x3 x4 s1 s2 s3 16. s ⎡ 1 0 1 −1 1 0 0 1 ⎢ 1 −1 0 1 0 1 0 s2 ⎢ s3 ⎢ 1 1 −1 1 0 0 1 ⎢ Z ⎢⎣ −3 −2 2 1 0 0 0

Z 0 3⎤ 3 0 6 ⎥⎥ 6 0 5⎥ 5 ⎥ 1 0⎥ ⎦ x1 x2 x3 x4 s1 s2 s3 Z 1 −1 1 0 0 0 3⎤ x1 ⎡ 1 0 ⎢ 0 −1 −1 2 −1 1 0 0 3⎥ s2 ⎢ ⎥ s3 ⎢ 0 1 −2 2 −1 0 1 0 2 ⎥ 2 ⎢ ⎥ Z ⎢⎣ 0 −2 5 −2 3 0 0 1 9 ⎥⎦ choosing x2 as the entering variable.

245


x1 x1 ⎡ 1 ⎢ s2 ⎢ 0 x2 ⎢ 0 ⎢ Z ⎣⎢ 0

x2 x3 x4 0 1 −1 0 −3 1 −2 0

1


x1 x2 s1 s2 R ⎡ ⎤ x2 2 1 1 0 0 2400 ⎥ ⎢ s2 ⎢ −7 0 −5 1 0 24,800 ⎥ ⎢ ⎥ 1200 ⎥ R ⎢ 14 0 12 0 1 ⎣ ⎦ Thus 0 boxes from A and 2400 from B give a maximum revenue of $1200.

s1 s2 s3 Z 1 0 0 0

3⎤ 4 −2 1 1 0 5⎥⎥ 2 −1 0 1 0 2 ⎥ ⎥ 2 1 0 2 1 13⎥ ⎦

Choosing x4 as the entering variable in the second table, we have: x1 x2 x3 x4 s1 s2 s3 Z 1 −1 1 0 0 0 3⎤ x1 ⎡ 1 0 3 ⎢ 0 −1 −1 2 −1 1 0 0 3⎥⎥ 2 s2 ⎢ 1 −2 2 −1 0 1 0 2 ⎥ 1 s3 ⎢ 0 ⎢ ⎥ 5 −2 3 0 0 1 9 ⎥ Z ⎢⎣ 0 −2 ⎦ x1 x2 x3 x4 s1 s2 s3 Z 1 1 0 1 0 x1 ⎡ 1 0 0 4⎤ 8 2 2 2 ⎢ ⎥ s2 ⎢ 0 −2 1 0 0 1 −1 0 1⎥ ⎢ ⎥ x4 ⎢ 0 12 −1 1 − 12 0 12 0 1⎥ 2 ⎢ ⎥ 2 0 1 1 11⎥ Z ⎢⎣ 0 −1 3 0 ⎦ x1 x2 x3 x4 s1 s2 s3 Z 1 −1 1 0 0 0 3⎤ x1 ⎡ 1 0 ⎢ 0 0 −3 4 −2 1 1 0 5⎥ s2 ⎢ ⎥ x2 ⎢ 0 1 −2 2 −1 0 1 0 2 ⎥ ⎢ ⎥ 1 2 1 0 2 1 13⎥ Z ⎣⎢ 0 0 ⎦ The solution is Z = 13 when x1 = 3, x2 = 2, x3 = 0, x4 = 0.

18. Let x, y, and z denote the numbers of units of X, Y, and Z produced, respectively. We want to maximize P = 6x + 8y + 12z subject to x + 2 y + 3z ≤ 900, 4 x + 4 y + 8 z ≤ 5000, x, y, z ≥ 0. x y z s1 s2 P s1 ⎡ 1 2 3 1 0 0 900 ⎤ 300 ⎢ ⎥ s2 ⎢ 4 4 8 0 1 0 5000 ⎥ 625 P ⎢ –6 –8 –12 0 0 1 0 ⎥ ⎣ ⎦ x y z s1 s2 P 2 z ⎡1 1 13 0 0 300 ⎤ 900 3 ⎢3 ⎥ ⎢ 8 4 4 s2 ⎢ 3 – 3 0 – 3 1 0 2600 ⎥⎥ 1950 P ⎢ –2 0 0 4 0 1 3600 ⎥ ⎢⎣ ⎥⎦ x y z s1 s2 P x ⎡1 2 3 1 0 0 900 ⎤ ⎢ ⎥ s2 ⎢ 0 –4 –4 –4 1 0 1400 ⎥ P ⎢ 0 4 6 6 0 1 5400 ⎥ ⎣ ⎦ P is maximum when x = 900, y = 0, z = 0. This maximum profit is $5400.

17. Let x1 and x2 denote the numbers of boxes transported from A and B, respectively. The revenue received is R = 0.75 x1 + 0.50 x2 . We want to maximize R subject to 2 x1 + x2 ≤ 2400 (volume), 3x1 + 5 x2 ≤ 36,800 (weight), x1 , x2 ≥ 0. x1 x2 s1 s2 R ⎡ ⎤ s1 ⎢ 2 1 1 0 0 2400 ⎥ 1200 s2 ⎢ 3 5 0 1 0 36,800 ⎥ 12, 266 23 ⎢ ⎥ R ⎢– 3 – 1 0 0 1 ⎥ 0 2 ⎣ 4 ⎦ x1 x2 s1 s2 R 1 1 0 0 1200 ⎤ 2400 x1 ⎡1 2 2 ⎢ ⎥ ⎢ 0 7 – 3 1 0 33, 200 ⎥ 9485 5 s2 ⎢ ⎥ 2 2 7 ⎢ ⎥ 900 ⎥ R ⎢ 0 – 18 83 0 1 ⎣ ⎦

19. Let x1 , x2 , and x3 denote the numbers of chairs, rockers, and chaise lounges produced, respectively. We want to maximize R = 21x1 + 24 x2 + 36 x3 subject to x1 + x2 + x3 ≤ 400, x1 + x2 + 2 x3 ≤ 500, 2 x1 + 3x2 + 5 x3 ≤ 1450, x1 , x2 , x3 ≥ 0. x1 x2 x3 s1 s2 s3 R 1 1 1 0 0 0 400 ⎤ 400 s1 ⎡ 1 ⎢ 1 1 2 0 1 0 0 500 ⎥⎥ 250 s2 ⎢ 3 5 0 0 1 0 1450 ⎥ 290 s3 ⎢ 2 ⎢ ⎥ 0 ⎥ R ⎢⎣ –21 –24 –36 0 0 0 1 ⎦

246


Section 7.5 so we check for multiple solutions. Treating x2 as an entering variable, the following table is obtained: x1 x2 x3 s1 s2 s3 P b 1 0 0.39 0.73 0 0.67 0 28.18 − ⎡ ⎤ x1 ⎢ 0 0 1.00 1.00 1 −2.00 0 40.00 ⎥ s2 ⎢ ⎥ 6.36 ⎥ x2 ⎢ 0 1 0.79 −0.55 0 0.67 0 ⎢ ⎥ 0 1 1727.27 ⎥ P ⎢⎣ 0 0 9.09 9.09 0 ⎦ Another optimum solution is x1 = 28, x2 = 6, x3 = 0, and P = 1727. Thus, the optimum solution is for the company to produce (1 – t)35 + 28t = 35 – 7t of device 1, (1 – t)0 + 6t = 6t of device 2, and none of device 3, for 0 ≤ t ≤ 1.

x1 x2 x3 s1 s2 s3 R 1 ⎡ 1 0 1 – 12 0 0 150 ⎤ 300 s1 ⎢ 2 2 ⎥ ⎢ 1 1 1 1 0 2 0 0 250 ⎥ 500 x3 ⎢ 2 2 ⎥ ⎢ ⎥ 5 1 1 s3 – ⎢ 2 2 0 0 – 2 1 0 200 ⎥ 400 R ⎢⎢ –3 –6 0 0 18 0 1 9000 ⎥⎥ ⎣ ⎦ x1 x2 x3 s1 s2 s3 R 300 ⎤ ⎡ x2 1 1 0 2 –1 0 0 ⎢ 0 0 1 –1 1 0 0 100 ⎥⎥ x3 ⎢ 50 ⎥ s3 ⎢ –1 0 0 –1 –2 1 0 ⎢ ⎥ R ⎢⎣ 3 0 0 12 12 0 1 10,800 ⎥⎦ The production of 0 chairs, 300 rockers, and 100 chaise lounges gives the maximum revenue of $10,800.

Problems 7.5


1. Yes; for the table, x2 is the entering variable

1. Let x1 , x2 , x3 be the numbers of device 1, device 2, and device 3, respectively, that the company produces. The situation is to maximize the profit P = 50 x1 + 50 x2 + 50 x3 subject to the constraints 5.5 x1 + 5.5 x2 + 6.5 x3 ≤ 190, 3.5 x1 + 6.5 x2 + 7.5 x3 ≤ 180, 4.5 x1 + 6.0 x2 + 6.5 x3 ≤ 165,

and the quotients smallest.

6 3 and tie for being the 2 1

2. Yes; the B.F.S. corresponding to the given table has the basic variable x2 equal to 0. x1 x2 s1 3. s ⎡ 4 –3 1 1 ⎢ s2 ⎢ 3 –1 0 s3 ⎢ 5 0 0 ⎢ Z ⎢⎣ –2 –7 0

s2 s3 Z 0 0 0 4⎤ 1 0 0 6 ⎥⎥ 0 1 0 8⎥ ⎥ 0 0 1 0⎥ ⎦ The entering variable is x2 . Since no quotients exist, the problem has an unbounded solution. Thus, no optimum solution (unbounded).

and x1 , x2 , x3 ≥ 0 . The matrices are shown rounded to 2 decimal places, although the exact values are used in the row operations. Since the indicators are equal, we choose the first column as the pivot column. x1 x2 x3 s1 s2 s3 P b s1 ⎡ 5.5 5.5 6.5 1 0 0 0 190 ⎤ ⎢ ⎥ s2 ⎢ 3.5 6.5 7.5 0 1 0 0 180 ⎥ s3 ⎢ 4.5 6.0 6.5 0 0 1 0 165⎥ ⎢ ⎥ 0⎥ P ⎢⎣ –50 –50 –50 0 0 0 1 ⎦ x1 x2 x2 s1 s2 s3 P b 1 1 1.18 0.18 0 0 0 34.55 ⎡ ⎤ x1 ⎢0 3 3.36 –0.64 1 0 0 59.09 ⎥⎥ s2 ⎢ 9.55⎥ s3 ⎢ 0 1.50 1.18 –0.82 0 1 0 ⎢ ⎥ 0 9.09 9.09 0 0 1 1727.27 ⎥ P ⎢⎣ 0 ⎦ An optimum solution is x1 = 35, x2 = 0, x3 = 0, and P = 1727. However, x2 is a nonbasic variable and its indicator is 0,

4.

247

x1 x2 s1 s2 s3 s4 ⎡ s1 1 −1 1 0 0 0 ⎢ s2 ⎢ −1 1 0 1 0 0 s3 ⎢ 8 5 0 0 1 0 ⎢ 1 0 0 0 1 s4 ⎢ 2 Z ⎢⎢⎣ −2 −1 0 0 0 0

Z 0 0 0 0 1

7⎤ 7 5⎥⎥ 40 ⎥ 5 ⎥ 6⎥ 3 0 ⎥⎥ ⎦


ISM: Introductory Mathematical Analysis Since x1 is nonbasic for the last table and its indicator is 0, there may be multiple optimum solutions. Treating x1 as an entering variable, we have x1 x2 s1 s2 s3 Z 8 − 2 0 48 ⎤ s1 ⎡ 0 0 1 7 7 7 ⎥ ⎢ 3 1 0 18 ⎥ x2 ⎢ 0 1 0 7 7 7⎥ ⎢ 8⎥ 2 0 x1 ⎢ 1 0 0 − 1 7 7 7⎥ ⎢ Z ⎢0 0 0 4 0 1 16 ⎥ ⎣ ⎦ 8 18 Here Z = 16 when x1 = , x2 = . Thus 7 7 multiple optimum solutions exist. Hence Z is 8 8 maximum when x1 = (1 − t )(0) + t = t , 7 7 18 4 x2 = (1 − t )(2) + t = 2 + t , and 0 ≤ t ≤ 1. For 7 7 the last table s3 is nonbasic and its indicator is 0. If we continue the process for determining other optimum solutions, we return to the second table.

x1 x2 s1 s2 s3 s4 Z s1 ⎡ 0 − 3 1 0 0 − 1 0 4 ⎤ 2 2 ⎢ ⎥ 3 0 1 0 1 0 ⎥ 16 s2 ⎢ 0 8 2 2 ⎢ ⎥ 3 s3 ⎢ 0 1 0 0 1 −4 0 16 ⎥ 16 ⎢ ⎥ 1 0 0 0 1 0 ⎥6 x1 ⎢ 1 3 2 2 ⎢ ⎥ Z ⎢0 0 0 0 0 1 1 6⎥ ⎣ ⎦ The maximum value of Z is 6 when x1 = 3 and x2 = 0. Since x2 is nonbasic for the last table and its indicator is 0, there may be multiple optimum solutions. Treating x2 as an entering variable and continuing, we have x1 x2 s1 s2 s3 s4 Z s1 ⎡ 0 0 1 1 0 0 0 12 ⎤ ⎢ 2 0 1 0 16 ⎥ x2 ⎢ 0 1 0 3 3 3⎥ ⎢ ⎥ 13 32 2 s3 ⎢ 0 0 0 − 1 − 0 3 3 3 ⎥ ⎢ ⎥ 1 1 x1 ⎢ 1 0 0 − 0 0 13 ⎥ 3 3 ⎢ ⎥ Z ⎢0 0 0 0 0 1 1 6⎥ ⎣ ⎦ 1 16 Here Z = 6 when x1 = and x2 = . Thus 3 3 multiple optimum solutions exist. Hence Z is a 1 8 maximum when x1 = (1 − t )(3) + t = 3 − t , 3 3 16 16 x2 = (1 − t )(0) + t = t , and 0 ≤ t ≤ 1. For the 3 3 last table, s2 is nonbasic and its indicator is 0. If we continue the process for determining other optimum solutions, we return to the second table.

6. To obtain a standard linear programming problem, we write the second constraint as – x1 + x2 + x3 ≤ 4. x1 x2 x3 s1 s2 s3 Z s1 ⎡ 1 –1 4 1 0 0 0 6 ⎤ 6 ⎢ ⎥ s2 ⎢ –1 1 1 0 1 0 0 4 ⎥ s3 ⎢ 1 –6 1 0 0 1 0 8 ⎥ 8 ⎢ ⎥ Z ⎢⎣ –8 –2 –4 0 0 0 1 0 ⎥⎦ x1 x2 x3 s1 s2 s3 Z x1 ⎡ 1 –1 4 1 0 0 0 6 ⎤ ⎢ 0 5 1 1 0 0 10 ⎥⎥ s2 ⎢ 0 s3 ⎢ 0 –5 –3 –1 0 1 0 2 ⎥ ⎢ ⎥ Z ⎢⎣ 0 –10 28 8 0 0 1 48⎥⎦ For the last table, x2 is the entering variable. Since no quotients exist, the problem has an unbounded solution. Thus, no optimum solution (unbounded).

x1 x2 s1 s2 s3 Z 5. s1 ⎡ 2 –2 1 0 0 0 4 ⎤ s2 ⎢ –1 2 0 1 0 0 4 ⎥ 2 ⎢ ⎥ 1 0 0 1 0 6⎥ 6 s3 ⎢ 3 ⎢ ⎥ Z ⎢⎣ 4 −8 0 0 0 1 0 ⎥⎦ x1 x2 s1 s2 s3 Z s1 ⎡ 1 0 1 1 0 0 8 ⎤ 8 ⎢ ⎥ x2 ⎢ − 12 1 0 12 0 0 2 ⎥ ⎢ ⎥ s3 ⎢ 72 0 0 − 12 1 0 4 ⎥ 87 ⎢ ⎥ Z ⎢ 0 0 0 4 0 1 16 ⎥ ⎣ ⎦ Z has a maximum of 16 when x1 = 0, x2 = 2. 248


x1 7. s1 ⎡ 9 ⎢ s2 ⎢ 4 s3 ⎢ 1 ⎢ Z ⎢⎣ –5 x1 ⎡3 s1 ⎢ x2 ⎢ 2 ⎢ s3 ⎢ 9 Z ⎢7 ⎣⎢

x2 x3 s1 s2 s3 3 –2 1 0 0 2 –1 0 1 0 –4 1 0 0 1 –6 –1 0 0 0 x2 x3 s1

0 0

x1 ⎡ x2 2 ⎢ s2 ⎢ 3 ⎢ ⎢4 s3 ⎢ Z ⎢⎢ 0 ⎣

5⎤ 5 3 2 ⎥⎥ 1 3⎥ ⎥ 0⎥ ⎦

s2 s3 Z 0 0 2⎤ ⎥ 1 0 0 1⎥ 2 ⎥ 2 1 0 7⎥ ⎥ 3 0 1 6⎥ ⎦

1 – 32

0 – 12 1 –

Z 0 0 0 1

Section 7.5

1 2

0

–1 0 –4 0

x2 x3 3 –3 –1 1 –1 2 –1 4

s1 1 0 0 0

s2 0 1 0 0

s3 0 0 1 0

x2 x3 s1 s2 s3 9 –9 1 –6 0 –1 1 0 1 0 1 0 0 –2 1 –3 6 0 2 0 x2 x3 s1 1 –1

s3 Z

– 23 1 3

0 0

0

0

0

1

– 19

– 43

1 0

0

3

1 3

0

0 1

Z has a maximum value of

0 0

0

0

1

0

3

0 0 0 1 0 0 0

1 0

0 0

1

10 ⎤ 3⎥ 13 ⎥ 3⎥ 46 ⎥ 3 ⎥ 10 ⎥ 3 ⎦⎥

4 26 ⎛ 4 ⎞ ⎛ 10 ⎞ x2 = (1 – t ) ⎜ ⎟ + ⎜ ⎟ t = + t , 9 9 ⎝9⎠ ⎝ 3 ⎠ x3 = (1 – t )(0) + 0t = 0, and 0 ≤ t ≤ 1. For the last table, x1 is nonbasic and its indicator is 0. if we continue the process for determining other optimum solutions, we return to the third table.

9. To obtain a standard linear programming problem, we write the second constraint as 4 x1 + x 2 ≤ 6 .

Z 0 4⎤ 4 9 0 1 ⎥⎥ 0 10 ⎥ 10 ⎥ 1 2⎥ ⎦

s2

0

Z

10 10 when x1 = 0, x2 = , x3 = 0. 3 3 Thus multiple optimum solutions exist. Hence Z is maximum when 13 13 ⎛ 13 ⎞ – t, x1 = (1 – t ) ⎜ ⎟ + 0t = 9 9 9 ⎝ ⎠

Z 0 10 ⎤ 53 0 1 ⎥⎥ 1 0 12 ⎥ 6 ⎥ 1 0⎥ ⎦

1 9 1 9

1 –1

1 3 1 3 1 3 1 3

Here Z =

For the last table, x3 is the entering variable. Since no quotients exist, the problem has an unbounded solution. Thus, no optimum solution (unbounded). x1 8. s ⎡ 6 1 ⎢ s2 ⎢ 1 s3 ⎢ 2 ⎢ Z ⎢⎣ –2 x1 s1 ⎡ 0 ⎢ x1 ⎢1 s3 ⎢ 0 ⎢ Z ⎢⎣ 0 x1 ⎡ x2 ⎢ 0 x1 ⎢1 ⎢ ⎢ s3 ⎢ 0 ⎢ Z ⎢0 ⎣

x2 x3 s1 s2 s3

x1

x2

x3 s1 s2 Z

s1 ⎡ 2 1 1 1 0 0 7 ⎤ 72 ⎢ ⎥ s2 ⎢ 4 1 0 0 1 0 6 ⎥ 3 2 Z ⎢ –6 –2 –1 0 0 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z s1 ⎡ 0 1 1 1 – 12 0 4 ⎤ 4 2 ⎢ ⎥ ⎢ 3 1 1 x1 1 0 0 4 0 2 ⎥⎥ 4 ⎢ ⎢ ⎥ 3 1 Z ⎢ 0 – 2 –1 0 2 1 9 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z x3 ⎡ 0 12 1 1 – 12 0 4 ⎤ 8 ⎢ ⎥ x1 ⎢1 14 0 0 14 0 32 ⎥ 6 ⎢ ⎥ Z ⎢0 0 0 1 1 1 13⎥ ⎣ ⎦ Z has a maximum value of 13 when 3 x1 = , x2 = 0, x3 = 4. Since x2 is nonbasic for 2 the last table and its indicator is 0, there may be multiple optimum solutions. Treating x2 as an entering variable, we have

4 ⎤ 9 ⎥ 13 ⎥ 9 ⎥ 13 3 86 ⎥ 9 ⎥ 10 ⎥ 3 ⎥⎦

10 when 3

13 4 , x2 = , x3 = 0. Since s2 is nonbasic 9 9 for the last table and its indicator is 0, there may be multiple optimum solutions. Treating s2 as an entering variable, we have x1 =

249


ISM: Introductory Mathematical Analysis choosing s3 as departing variable

x1 x2 x3 s1 s2 Z ⎡ x3 –2 0 1 1 –1 0 1⎤ ⎢ ⎥ x2 ⎢ 4 1 0 0 1 0 6 ⎥ Z ⎢ 0 0 0 1 1 1 13⎥ ⎣ ⎦ Here Z = 13 when x1 = 0, x2 = 6, x3 = 1. Thus multiple optimum solutions exist. Hence Z is maximum when 3 3 ⎛3⎞ x1 = (1 – t ) ⎜ ⎟ + 0t = – t , 2 2 ⎝2⎠

x1

x2

x3 s1 s2

s3 R

⎡ 0 1 0 – 15 0 100 ⎤ 500 s1 ⎢ ⎥ 3 2 0 ⎢ ⎥0 0 0 1 – 0 s2 5 ⎢ ⎥ ⎢ 2 ⎥ 3 1 1 0 0 5 0 300 ⎥ x3 ⎢ 5 5 750 ⎢ 24 ⎥ 16 48 R ⎢– – 5 0 0 0 5 1 14, 400 ⎥ ⎣ 5 ⎦ x1 x2 x3 s1 s2 s3 R 100 ⎤ 100 s1 ⎡0 1 0 1 –3 1 0 ⎢1 –1 0 0 5 –2 0 0 ⎥⎥ 0 x1 ⎢ 300 ⎥ 300 x3 ⎢0 1 1 0 –2 1 0 ⎢ ⎥ R ⎢⎣0 –8 0 0 24 0 1 14, 400 ⎥⎦ x1 x2 x3 s1 s2 s3 R 100 ⎤ x2 ⎡ 0 1 0 1 –3 1 0 ⎢1 0 0 1 2 –1 0 100 ⎥⎥ 50 x1 ⎢ 200 ⎥ 200 x3 ⎢ 0 0 1 –1 1 0 0 ⎢ ⎥ R ⎢⎣ 0 0 0 8 0 8 1 15, 200 ⎦⎥ The maximum value of R is 15,200 when x1 = 100, x2 = 100, x3 = 200. Since s2 is nonbasic for the last table and its indicator is 0, there may be multiple optimum solutions. Treating s2 as an entering variable, we have 3 5 1 5

x2 = (1 – t )(0) + 6t = 6t , x3 = (1 – t )(4) + (1)t = 4 – 3t , and 0 ≤ t ≤ 1. For the last table, x1 is nonbasic and its indicator is 0. If we continue the process for determining other optimum solutions, we return to the third table. x1 x2 x3 x4 s1 s2 s3 P 10. s ⎡ 1 −1 0 0 1 0 0 0 2 ⎤ 1 ⎢ ⎥ s2 ⎢ 0 1 −1 0 0 1 0 0 3⎥ 3 1 −3 1 0 0 1 0 4⎥ 4 s3 ⎢ 0 ⎢ ⎥ P ⎢⎣ −1 −2 −1 −2 0 0 0 1 0 ⎥⎦ x1 x2 x3 x4 s1 s2 s3 P s1 ⎡ 1 0 −1 0 1 1 0 0 5⎤ ⎢ ⎥ x2 ⎢ 0 1 −1 0 0 1 0 0 3⎥ 1 0 −1 1 0 1⎥ s3 ⎢ 0 0 −2 ⎢ ⎥ P ⎢⎣ −1 0 −3 −2 0 2 0 1 6 ⎦⎥ Now x3 is the entering variable but no quotients exist. Thus, the feasible region is unbounded and, hence, there is no optimum solution.

2 5 – 15

x1 x2 x3

s1 s2

5 1 0 x2 ⎡ 2 ⎢ 1 ⎢ 0 0 s2 2 ⎢ ⎢– 0 1 – 32 x3 ⎢ 8 R ⎢⎣ 0 0 0 3 2 1 2 1 2

s3 R

250 ⎤ ⎥ 0 50 ⎥ ⎥ 1 0 ⎥ 150 2 ⎥ 8 1 15, 200 ⎥ ⎦

0 – 12 1 – 12 0 0

0

Here R = 15,200 when x1 = 0, x2 = 250, x3 = 150. Thus multiple optimum solutions exist. Hence R is maximum when x1 = (1 – t )(100) + 0t = 100 – 100t , x2 = (1 – t )(100) + 250t = 100 + 150t x3 = (1 – t )(200) + 150t = 200 – 50t , and 0 ≤ t ≤ 1. For the last table, x1 is nonbasic and its indicator is 0. If we continue the process for determining other optimum solutions, we return to the fourth table. If we were to initially choose s2 as the departing variable, then

11. Let x1 , x2 , and x3 denote the numbers of chairs, rockers, and chaise lounges produced, respectively. We want to maximize R = 24 x1 + 32 x2 + 48 x3 subject to x1 + x2 + x3 ≤ 400, x1 + x2 + 2 x3 ≤ 600, 2 x1 + 3x2 + 5 x3 ≤ 1500, x1 , x2 , x3 ≥ 0. x1 x2 x3 s1 s2 s3 R 1 1 1 0 0 0 400 ⎤ 400 s1 ⎡ 1 ⎢ 1 1 2 0 1 0 0 600 ⎥⎥ 300 s2 ⎢ 3 5 0 0 1 0 1500 ⎥ 300 s3 ⎢ 2 ⎢ ⎥ 0 ⎥ R ⎣⎢ –24 –32 –48 0 0 0 1 ⎦

250


x1 x2 x3 1 1 s1 ⎡ 1 ⎢ 1 1 2 s2 ⎢ ⎢ 2 3 5 s3 ⎢ R ⎣⎢ –24 –32 –48

Section 7.6

s1 s2 s3 R 1 0 0 0

400 ⎤ 400 0 1 0 0 600 ⎥⎥ 300 0 0 1 0 1500 ⎥ 300 ⎥ 0 0 0 1 0 ⎥ ⎦ x1 x2 x3 s1 s2 s3 R 1 ⎡ s1 12 0 1 – 12 0 0 100 ⎤ 200 2 ⎢ ⎥ 1 1 ⎥ 600 x3 ⎢ 1 1 0 0 0 300 2 2 ⎢ 2 ⎥ s3 ⎢ – 1 1 0 0 – 5 1 0 0 ⎥ 0 2 ⎢ 2 2 ⎥ R ⎢⎢ 0 –8 0 0 24 0 1 14, 400 ⎥⎥ ⎣ ⎦ x1 x2 x3 s1 s2 s3 R 100 ⎤ 50 s1 ⎡ 1 0 0 1 2 –1 0 ⎢ 1 0 1 0 3 –1 0 300 ⎥⎥ 100 x3 ⎢ 0 ⎥ x2 ⎢ –1 1 0 0 –5 2 0 ⎢ ⎥ R ⎢⎣ –8 0 0 0 –16 16 1 14, 400 ⎥⎦ x1 x2 x3 s1 s2 s3 R 1 1 1 ⎡ 50 ⎤ 100 s2 ⎢ 2 0 0 2 1 – 2 0 ⎥ ⎢– 1 0 1 – 3 0 1 0 150 ⎥ x3 ⎢ 2 2 2 ⎥ ⎢ 3 5 1 x2 ⎢ 2 1 0 2 0 – 2 0 250 ⎥⎥ 500 3 R ⎢ 0 0 0 8 0 8 1 15, 200 ⎥ ⎥⎦ ⎣⎢ the maximum value of R is 15,200 when x1 = 0, x2 = 250, x3 = 150. For the last table, x1 is nonbasic and its

indicator is 0. Treating x1 as an entering variable, we have x1 x2 x3 s1 s2 s3 R 100 ⎤ x1 ⎡ 1 0 0 1 2 –1 0 ⎢ 0 0 1 –1 1 0 0 200 ⎥⎥ x3 ⎢ 100 ⎥ x2 ⎢ 0 1 0 1 –3 1 0 ⎢ ⎥ R ⎢⎣ 0 0 0 8 0 8 1 15, 200 ⎥⎦ Here R = 15,200 when x1 = 100, x2 = 100, x3 = 200. For the last table, s2 is nonbasic and its indicator is 0. If we continue the process of determining other optimum solutions, we return to the table corresponding to the solution x1 = 0, x2 = 250, x3 = 150. Thus, the maximum revenue is $15,200 when x1 = 100 – 100t , x2 = 100 + 150t , x3 = 200 – 50t , and 0 ≤ t ≤ 1 Principles in Practice 7.6 1. Using the hint, 1000 – x1 standard and 800 – x2 deluxe snowboards must be manufactured at plant II. The constraints for plant I are x1 + x2 ≤ 1200 and x2 – x1 ≤ 200. The constraints for plant II are (1000 – x1 ) + (800 – x2 ) ≤ 1000 or x1 + x2 ≥ 800. The quantity to be maximized is the profit P = 40 x1 + 60 x2 + 45(1000 – x1 ) + 50(800 – x2 ) = –5 x1 + 10 x2 + 85, 000 subject to the constraints 251



x1 + x2 ≤ 1200, – x1 + x2 ≤ 200, x1 + x2 ≥ 800,

and x1 , x2 ≥ 0 . Note that maximizing Z = –5 x1 + 10 x2 also maximizes the profit. The corresponding equations are: x1 + x2 + s1 = 1200, – x1 + x2 + s2 = 200, x1 + x2 – s3 + t = 800. The artificial objective equation is W = –5 x1 + 10 x2 – Mt. The augmented coefficient matrix is: x1 x2 s1 s2 s3 t W 1 1 0 0 0 0 1200 ⎤ ⎡ 1 ⎢ –1 1 0 1 0 0 0 200 ⎥⎥ ⎢ ⎢ 1 1 0 0 –1 1 0 800 ⎥ ⎢ ⎥ 0⎥ ⎢⎣ 5 –10 0 0 0 M 1 ⎦ The simplex tables follow. x1 x2 s1 s2 s3 t W 1 1 1 0 0 0 0 1200 ⎤ ⎡ s1 ⎢ –1 1 0 1 0 0 0 200 ⎥⎥ s2 ⎢ 1 0 0 −1 1 0 800 ⎥ t ⎢ 1 ⎢ ⎥ W ⎢⎣5 – M –10 – M 0 0 M 0 1 −800M ⎥⎦ x1 x2 s1 s2 s3 t W 2 0 1 0 0 0 1000 −1 ⎤ s1 ⎡ ⎢ –1 ⎥ 1 0 1 0 0 0 200 x2 ⎢ ⎥ ⎥ −1 1 0 2 0 0 –1 600 t ⎢ ⎢ ⎥ W ⎣⎢ –5 – 2M 0 0 10 + M M 0 1 2000 − 600 M ⎦⎥ x1 x2 s1 s2 s3 t W −1 0 400 ⎤ 0 1 s1 ⎡ 0 0 1 ⎢ ⎥ 1 1 1 0 x2 ⎢ 0 1 0 −2 500 ⎥ 2 2 ⎢ ⎥ 1 0 x1 ⎢ 1 0 0 – 1 − 1 300 ⎥ 2 2 2 ⎢ ⎥ – 52 52 + M 1 3500 ⎥ W ⎢ 0 0 0 15 2 ⎣ ⎦ Delete the t-column since t = 0 and return to Z. x1 x2 s1 s2 s3 Z s ⎡0 0 1 0 1 0 400 ⎤ 3 ⎢ ⎥ 1 0 0 700 ⎥ x2 ⎢ 0 1 12 2 ⎢ ⎥ 1 1 x1 ⎢ 1 0 2 − 2 0 0 500 ⎥ ⎢ ⎥ 5 15 0 0 4500 ⎥ Z ⎢0 0 2 2 ⎣ ⎦ Thus, x1 = 500, x2 = 700, and Z = 4500. Plant I should manufacture 500 standard and 700 deluxe snowboards. Plant II should manufacture 1000 – 500 = 500 standard and 800 – 700 = 100 deluxe snowboards. The maximum profit is P = –5(500) + 10(700) + 85,000 = $89,500. 252


Section 7.6

Problems 7.6 x1 x2 s1 s2 t2 W 1. ⎡ 1 1 1 0 0 0 6⎤ ⎢ ⎥ ⎢ –1 1 0 –1 1 0 4 ⎥ ⎢ –2 –1 0 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 s1 s2 t2 W 1 1 0 0 0 6 ⎤6 s1 ⎡ 1 ⎢ ⎥ 1 0 –1 1 0 4 ⎥4 t2 ⎢ –1 W ⎢ –2 + M –1 – M 0 M 0 1 –4M ⎥ ⎣ ⎦ x1 x2 s1 s2 t2 W –1 0 2 ⎤ 1 s1 ⎡ 2 0 1 1 ⎢ ⎥ 1 0 4⎥ x2 ⎢ –1 1 0 –1 W ⎢ –3 0 0 –1 M + 1 1 4 ⎥ ⎣ ⎦ x1 x2 s1 s2 Z ⎡ 1 0 1⎤ x1 1 0 1 2 2 ⎢ ⎥ x2 ⎢ 0 1 1 – 1 0 5⎥ 2 2 ⎢ ⎥ ⎢ ⎥ 3 1 1 7⎥ Z ⎢0 0 2 2 ⎣ ⎦ The maximum is Z = 7 when x1 = 1, x2 = 5 . x1 x2 s1 s2 t2 W 2. ⎡ 1 2 1 0 0 0 8⎤ ⎢ ⎥ ⎢ 1 6 0 –1 1 0 12 ⎥ ⎢ –3 –4 0 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 s1 s2 t2 W 2 1 0 0 0 8 ⎤4 s1 ⎡ 1 ⎢ ⎥ 6 0 –1 1 0 12 ⎥ 2 t2 ⎢ 1 W ⎢ –3 – M –4 – 6 M 0 M 0 1 –12 M ⎥ ⎣ ⎦ x1 x2 s1 s2 t2 W − 13 0 4⎤ 6 s1 ⎡ 23 0 1 13 ⎢ ⎥ ⎢ 1 1 0 −1 1 0 2 ⎥⎥ 12 x2 ⎢ 6 6 6 ⎢ 7 ⎥ 2 2 W ⎢− 3 0 0 − 3 3 + M 1 8⎥ ⎣ ⎦ x1 x2 s1 s2 Z 3 1 0 x1 ⎡ 1 0 6⎤ 2 2 ⎢ ⎥ ⎥ x2 ⎢ 0 1 – 1 – 1 0 1 4 4 ⎢ ⎥ ⎢ 7 1 1 22 ⎥ 0 0 Z ⎢ ⎥⎦ 2 2 ⎣ The maximum is Z = 22 when x1 = 6, x2 = 1 . 253



x1 x2 x3 s1 s2 t2 W 3. ⎡ 1 2 1 1 0 0 0 5⎤ ⎢ ⎥ ⎢ –1 1 1 0 –1 1 0 1⎥ ⎢ –2 –1 1 0 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t2 W ⎡ s1 1 2 1 1 0 0 0 5 ⎤ 52 ⎢ ⎥ t2 ⎢ –1 1 1 0 –1 1 0 1 ⎥ 1 W ⎢ –2 + M –1 – M 1 – M 0 M 0 1 – M ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t2 W s1 ⎡ 3 0 –1 1 2 –2 0 3⎤ 1 ⎢ ⎥ x2 ⎢ –1 1 1 0 −1 1 0 1⎥ W ⎢ –3 0 2 0 –1 1 + M 1 1⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z 2 0 1⎤ x1 ⎡ 1 0 – 1 1 3 3 3 ⎢ ⎥ 2 1 – 1 0 2⎥ x2 ⎢ 0 1 3 3 3 ⎢ ⎥ Z ⎢0 0 1 1 1 1 4⎥ ⎣ ⎦ The maximum is Z = 4 when x1 = 1, x2 = 2, x3 = 0 . x1 x2 x3 s1 s2 t2 W ⎡ 1 1 1 1 0 0 0 9⎤ ⎥ 4. ⎢ 1 0 –1 1 0 6 ⎥ ⎢ 1 –2 ⎢ –1 1 –4 0 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s1 ⎡ 1 1 1 1 ⎢ t2 ⎢ 1 –2 1 0 ⎢ W –1 – M 1 + 2M –4 – M 0 ⎣ x1 x2 x3 s1 s2 t2 W s1 ⎡ 0 3 0 1 1 –1 0 ⎢ x3 ⎢1 –2 1 0 –1 1 0 ⎢ W 3 –7 0 0 –4 4 + M 1 ⎣ x1 x2 x3 s1 s2 Z x2 ⎡ 0 1 0 ⎢ x3 ⎢ ⎢1 0 1 ⎢ Z ⎢3 0 0 ⎣

1 3

1 3

2 3 7 3

– 13 – 53

s2 t2 W 0 0 0

9 ⎤9 ⎥ –1 1 0 6 ⎥6 M 0 1 –6 M ⎥ ⎦ 3 ⎤1 ⎥ 6⎥ 24 ⎥ ⎦

1 ⎤3 ⎥ 0 8 ⎥⎥ ⎥ 1 31⎥ ⎦ 0

254


Section 7.6

x1 x2 x3 s1 s2 Z ⎡ s2 0 3 0 1 1 0 3⎤ ⎢ ⎥ x3 ⎢ 1 1 1 1 0 0 9 ⎥ Z ⎢ 3 5 0 4 0 1 36 ⎥ ⎣ ⎦ The maximum is Z = 36 when x1 = 0, x2 = 0, x3 = 9 . x1 x2 x3 s1 t2 W 5. ⎡ 1 1 1 1 0 0 10 ⎤ ⎢ ⎥ 1 0 6⎥ ⎢ 1 −1 −1 0 ⎢ −3 −2 −1 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 t2 W s1 ⎡ 1 1 1 1 0 0 10 ⎤ 10 ⎢ ⎥ −1 −1 0 1 0 t2 ⎢ 1 6⎥ 6 W ⎢ −3 − M −2 + M −1 + M 0 0 1 −6M ⎥ ⎣ ⎦ x1 x2 x3 s1 t2 W ⎡ s1 0 2 2 1 −1 0 4 ⎤ 2 ⎢ ⎥ x1 ⎢ 1 −1 −1 0 1 0 6⎥ W ⎢0 −5 −4 0 3 + M 1 18⎥ ⎣ ⎦ x1 x2 x3 s1 W ⎡ x2 0 1 1 1 0 2 ⎤ 2 ⎢ ⎥ x1 ⎢⎢ 1 0 0 12 0 8⎥⎥ Z ⎢⎢ 0 0 1 5 1 28⎥⎥ 2 ⎣ ⎦ The maximum is Z = 28 when x1 = 8, x2 = 2, and x3 = 0.

x1

x2 x3

s1

t1

t2 W

⎡ 0 1 –2 −1 1 0 0 5⎤ 6. ⎢ ⎥ 1 0 7⎥ ⎢ 1 1 1 0 0 ⎢ –2 –1 –3 0 M M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 t1 t2 W t1 ⎡ 0 ⎢ t2 ⎢ 1 W ⎢ –2 – M ⎣ x1 x2 ⎡ 0 ⎢ t2 ⎢ 1 W ⎢ –2 – M ⎣

1 1 –1 – 2 M x2

–2 1 –3 + M

x3

1 –2 0 3 0 –5 – 3M

s1

–1 1 0 0 5 ⎤5 ⎥ 0 0 1 0 7 ⎥7 M 0 0 1 –12 M ⎥ ⎦ t1 t2 W

–1 1 1 –1 –1 – M 1 + 2 M

0 0 5 ⎤ ⎥ 1 0 2 ⎥ 23 0 1 5 – 2M ⎥ ⎦

255


x1 x2 x3 ⎡ x2 ⎢ 23 ⎢ x3 ⎢ 13 ⎢ 1 W ⎢⎣⎢ − 3

s1

t1

1 0 – 13 0 1

1 3

1 3 1 –3

0 0

2 3

– 23 + M

x1 x2 x3

s1


t2 W 2 3 1 3 5 3

+M

0

19 ⎤ 19 3⎥ 2

⎥ 2⎥ 6 ⎥ ⎥ 1 25 3 ⎦⎥

0

Z

x2 ⎡ 0 1 −2 −1 0 5 ⎤ ⎢ ⎥ x1 ⎢1 0 3 1 0 2 ⎥ W ⎢0 0 1 1 1 9 ⎥ ⎣ ⎦ The maximum is Z = 9 when x1 = 2, x2 = 5, x3 = 0. x1 x2 s1 s2 s3 t3 W 7. ⎡ 1 –1 1 0 0 0 0 1⎤ ⎢ 1 2 0 1 0 0 0 8⎥ ⎢ ⎥ ⎢ 1 1 0 0 –1 1 0 5⎥ ⎢ ⎥ ⎢⎣ –1 10 0 0 0 M 1 0 ⎥⎦ x1 x2 s1 s2 s3 t3 W 1 –1 1 0 0 0 0 1 ⎤1 s1 ⎡ ⎢ 1 2 0 1 0 0 0 8 ⎥⎥ 8 s2 ⎢ 1 0 0 –1 1 0 5 ⎥5 t3 ⎢ 1 ⎢ ⎥ W ⎢⎣ –1 – M 10 – M 0 0 M 0 1 –5M ⎥⎦ x1 x2 s1 s2 s3 t3 W –1 1 0 0 0 0 1 ⎤ x1 ⎡1 ⎢0 ⎥7 3 –1 1 0 0 0 7 s2 ⎢ ⎥3 2 –1 0 –1 1 0 4 ⎥2 t3 ⎢ 0 ⎢ ⎥ W ⎢⎣ 0 9 – 2M 1 + M 0 M 0 1 1 – 4M ⎦⎥ x1 x2 s1 s2 s3 t3 W 1 0 –1 1 0 3⎤ x1 ⎡ 1 0 2 2 2 ⎢ ⎥ 3 3 0 1 1 s2 ⎢ 0 0 ⎥ – 1 2 2 2 ⎢ ⎥ 1 0 ⎥ x2 ⎢ 0 1 – 1 0 – 1 2 2 2 2 ⎢ ⎥ 9 – 9 + M 1 –17 ⎥ 0 W ⎢⎢ 0 0 11 2 2 2 ⎣ ⎦⎥ For the above table, t3 = 0. Thus W = Z. The maximum is Z = –17 when x1 = 3, x2 = 2 .

256


x1 x2 x3 s1 8. ⎡ 1 1 –1 –1 ⎢ 1 1 1 0 ⎢ ⎢ 1 –1 1 0 ⎢ ⎢⎣ –1 –4 1 0 x1 x2 t1 ⎡ 1 ⎢ s2 ⎢ 1 t3 ⎢ 1 ⎢ W ⎢⎣ –1 – 2M x1

s2

t1

s1

1 –1 –1 1 1 0 –1 1 0 –4 1 M x2

t3 W

0 1 0 1 0 0 0 0 1 0 M M x3

x3

Section 7.6

0 0 0 1

s2 t1 0 1 0 0 s1

1 0 0 0

5⎤ 3⎥⎥ 7⎥ ⎥ 0⎥ ⎦ t3 W 0 0 1 0

0 5 ⎤5 0 3 ⎥⎥ 3 0 7 ⎥7 ⎥ 1 –12M ⎥ ⎦ s2 t1 t3 W

0 –2 –1 –1 1 t1 ⎡0 ⎢1 1 1 0 1 0 x1 ⎢ ⎢ 0 –2 0 0 –1 0 t3 ⎢ W ⎢⎣0 –3 + 2 M 2 + 2M M 1 + 2 M 0 There is no solution (empty feasible region).

0 0 1 0

0 2⎤ 0 3⎥⎥ 0 4⎥ ⎥ 1 3 – 6M ⎥ ⎦

9. We write the third constraint as – x1 + x2 + x3 ≥ 6. x1 x2 x3 s1 s2 s3 t2 t3 W ⎡ 1 1 1 1 0 0 0 0 0 1⎤ ⎢ 1 –1 1 0 –1 0 1 0 0 2 ⎥⎥ ⎢ ⎢ –1 1 1 0 0 –1 0 1 0 6⎥ ⎢ ⎥ ⎢⎣ –3 2 –1 0 0 0 M M 1 0 ⎥⎦ x1 x2 x3 s1 s2 s3 t2 t3 W 1 1 1 1 0 0 0 0 0 1 ⎤1 s1 ⎡ ⎢ 1 –1 1 0 –1 0 1 0 0 2 ⎥⎥ 2 t2 ⎢ 1 0 0 –1 0 1 0 6 ⎥6 t3 ⎢ –1 1 ⎢ ⎥ W ⎢⎣ –3 2 –1 – 2 M 0 M M 0 0 1 –8M ⎥⎦ x1 x2 x3 s1 s2 s3 t2 t3 W 1 1 1 1 0 0 0 0 0 1⎤ x3 ⎡ ⎢ 0 –2 0 –1 –1 0 1 0 0 1⎥⎥ t2 ⎢ –2 0 0 –1 0 –1 0 1 0 5⎥ t3 ⎢ ⎢ ⎥ W ⎣⎢ –2 + 2 M 3 + 2 M 0 1 + 2 M M M 0 0 1 1 – 6 M ⎦⎥ There is no solution (empty feasible region).

257


x1 x2 10. ⎡ 1 2 ⎢ 1 6 ⎢ ⎢ 0 1 ⎢ –1 –4 ⎣⎢ x1 1 ⎡ s1 ⎢ 1 t2 ⎢ t3 ⎢ 0 ⎢ W ⎢⎣ –1 – M

s1 s2 s3 t2 t3 1 0 0 0 0 0 –1 0 1 0 0 0 –1 0 1 0 0 0 M M


W 0 0 0 1

x2 s1 s2 s3 2 1 0 0 6 0 –1 0 1 0 0 –1 –4 – 7 M 0 M M

8⎤ 12 ⎥⎥ 2⎥ ⎥ 0⎥ ⎦ t2 0 1 0 0

t3 0 0 1 0

W 0 8 ⎤4 0 12 ⎥⎥ 2 0 2 ⎥2 ⎥ 1 –14 M ⎥ ⎦ Here we choose t3 as the departing variable. x1 x2 s1 s2 s3 t2 t3 W 1 0 1 0 2 0 –2 0 4⎤ 2 ⎡ s1 ⎢ 1 0 0 –1 6 1 –6 0 0 ⎥⎥ 0 t2 ⎢ 1 0 0 –1 0 1 0 2⎥ x2 ⎢ 0 ⎢ ⎥ W ⎢⎣ –1 – M 0 0 M –4 – 6M 0 4 + 7 M 1 8 ⎥⎦ x1 x2 s1 s2 s3 t2 t3 W 2 1 1 ⎡ –3 0 0 4 ⎤ 12 s1 ⎢ 3 0 1 3 0 ⎥ ⎥ s3 ⎢ 1 0 0 – 1 1 1 –1 0 0 6 6 ⎢ 6 ⎥ ⎢ 1 ⎥ 1 0 1 1 0 – 0 0 2 ⎥ x2 ⎢ 6 6 6 ⎢ 1 ⎥ 2 2 W ⎢ – 3 0 0 – 3 0 3 + M M 1 8⎥ ⎣ ⎦ x1 x2 s1 s2 s3 Z ⎡ 2 0 3 1 0 0 12 ⎤ s2 ⎢ ⎥ 1 1 s3 ⎢ 2 0 2 0 1 0 2 ⎥ ⎢ ⎥ x2 ⎢ 1 1 1 0 0 0 4 ⎥ 2 2 Z ⎢ 1 0 2 0 0 1 16 ⎥ ⎢⎣ ⎥⎦ Thus the maximum value of Z is 16, when x1 = 0, x2 = 4. If we choose t2 as the original departing variable, then x1 x2 s1 s2 s3 t2 t3 W 1 2 1 0 0 0 0 0 8 ⎤4 ⎡ s1 ⎢ 1 6 0 –1 0 1 0 0 12 ⎥⎥ 2 t2 ⎢ 1 0 0 –1 0 1 0 2 ⎥2 t3 ⎢ 0 ⎢ ⎥ W ⎢⎣ –1 – M –4 – 7 M 0 M M 0 0 1 –14 M ⎥⎦

258


Section 7.6

x1 x2 s1 s2 s3 t2 t3 2 1 1 ⎡ 0 1 0 –3 0 s1 ⎢ 3 3 1 1 x2 ⎢ 1 0 – 16 0 0 6 6 ⎢ 1 ⎢ –1 0 0 –1 – 16 1 6 6 t3 ⎢ ⎢ W ⎢– 1 + 1 M 0 0 – 2 – 1 M M 2 + 7 M 0 3 6 3 6 ⎣ 3 6 x1 x2 s1 s2 s3 t2 t3 W –2 0 4⎤ 2 s1 ⎡ 1 0 1 0 2 0 ⎢ 0 1 0 0 –1 0 1 0 2 ⎥⎥ x2 ⎢ 6 0 0⎥ s2 ⎢ –1 0 0 1 –6 –1 ⎢ ⎥ W ⎢⎣ –1 0 0 0 –4 M 4 + M 1 8 ⎦⎥ x1 x2 s1 s2 s3 Z s3 ⎡ 1 0 1 0 1 0 2 ⎤ 2 ⎢2 ⎥ x2 ⎢ 1 1 1 0 0 0 4 ⎥ 2 ⎢2 ⎥ s2 ⎢ 2 0 3 1 0 0 12 ⎥ ⎢ ⎥ Z ⎢⎣ 1 0 2 0 0 1 16 ⎥⎦ The maximum is Z = 16 when x1 = 0, x2 = 4 .

W 0 4 ⎤ 12 ⎥ 0 2⎥ ⎥ 0 0⎥ ⎥ 0 ⎥ 1 8⎥ ⎦

x1 x2 s1 s3 t2 t3 W 11. ⎡ 1 –1 1 0 0 0 0 4 ⎤ ⎢ –1 1 0 0 1 0 0 4 ⎥⎥ ⎢ ⎢ 1 0 0 –1 0 1 0 6⎥ ⎢ ⎥ ⎢⎣ 3 –2 0 0 M M 1 0 ⎦⎥ x1 x2 s1 s3 t2 t3 W 1 –1 1 0 0 0 0 4 ⎤ ⎡ s1 ⎢ –1 1 0 0 1 0 0 4 ⎥⎥ 4 t2 ⎢ 0 0 –1 0 1 0 6 ⎥ t3 ⎢ 1 ⎢ ⎥ W ⎢⎣ 3 –2 – M 0 M 0 0 1 –10 M ⎥⎦ x1 x2 s1 s3 t2 t3 W 0 0 1 0 1 0 0 8 ⎤ s1 ⎡ ⎢ –1 1 0 0 1 0 0 4 ⎥⎥ x2 ⎢ 0 0 –1 0 1 0 6 ⎥6 t3 ⎢ 1 ⎢ ⎥ W ⎢⎣1 – M 0 0 M 2 + M 0 1 8 – 6M ⎥⎦ x1 x2 s1 s3 t2 t3 W 1 0 0 8⎤ s1 ⎡ 0 0 1 0 ⎢ 0 1 0 –1 1 1 0 10 ⎥⎥ x2 ⎢ 0 1 0 6⎥ x1 ⎢ 1 0 0 –1 ⎢ ⎥ W ⎣⎢ 0 0 0 1 2 + M –1 + M 1 2 ⎦⎥

For the above table, t2 = t3 = 0. Thus W = Z. The maximum is Z = 2 when x1 = 6, x2 = 10 . 259



12. We write the first constraint as − x1 + 2 x2 ≤ 12. x1 ⎡ −1 ⎢ −1 ⎢ ⎢ 1 ⎢ ⎢⎣ −2

x2 s1 2 1 1 8

s2

s3

t2

t3 W

1 0 0 0 0 0 −1 0 1 0 0 0 −1 0 1 0 0 0 M M

x1

x2 s1

2 s1 ⎡ −1 ⎢ −1 1 t2 ⎢ ⎢ 1 1 t3 ⎢ W ⎢⎣ −2 8 − 2M x1 x2 1 s1 ⎡ ⎢ 1 − x2 ⎢ 2 t3 ⎢ ⎢ W ⎣⎢ 6 − 2M x1 x2 s1

0 1 0 0

s2 s3

0 12 ⎤ 0 2 ⎥⎥ 0 10 ⎥ ⎥ 1 0⎥ ⎦ t2 t3 W

1 0 0 0 −1 0 0 0 −1 0 M M

0 1 0 0

s1

s3

s2

0 0 1 0

0 12 ⎤ 6 0 2 ⎥⎥ 2 0 10 ⎥ 10 ⎥ 1 −12 M ⎥ ⎦ t2 t3 W

1 2 0 −2 0 1 −1 0 0 1 −1 −1 0 8 − M M −8 + 2M s2

s3

t2

0 0 1 0

0 8⎤ 8 0 2 ⎥⎥ 0 8⎥ 4 ⎥ 1 −16 − 8M ⎥ ⎦

t3 W

s1 ⎡ 0 0 1 − 12 0 4⎤ ⎢ ⎥ 1 0 ⎥ x2 ⎢ 0 1 0 6 2 ⎢ ⎥ 1 0 ⎥ x1 ⎢ 1 0 0 4 2 ⎢ ⎥ W ⎢0 0 0 5 3 −5 + M −3 + M 1 −40 ⎥ ⎣ ⎦ For the above table, t2 = t3 = 0. Thus W = Z. The maximum is Z = −40 when x1 = 4 and x2 = 6. 3 2 1 −2 1 2

1 2 1 −2 − 12

− 32 1 2 1 −2

13. Let x1 and x2 denote the numbers of Standard and Executive bookcases produced, respectively, each week. We want to maximize the profit function P = 35 x1 + 40 x2 subject to 2 x1 + 3x2 ≤ 400, 3x1 + 4 x2 ≤ 500, 3x1 + 4 x2 ≥ 250, x1 , x2 ≥ 0. The artificial objective function is W = P – Mt3 . x1 x2 s1 s2 s3 t3 W 3 1 0 0 0 0 400 ⎤ ⎡ 2 ⎢ 3 4 0 1 0 0 0 500 ⎥⎥ ⎢ ⎢ 3 4 0 0 –1 1 0 250 ⎥ ⎢ ⎥ –35 –40 0 0 0 M 1 0⎥ ⎣⎢ ⎦ x1 2 s1 ⎡ ⎢ 3 s2 ⎢ ⎢ 3 t3 ⎢ W ⎣⎢ –35 − 3M

x2

s1 s2 s3

3

1 0

4 4 –40 – 4 M

t3 W

400 ⎤ 400 3 0 1 0 0 0 500 ⎥⎥ 125 0 0 –1 1 0 250 ⎥ 125 ⎥ 2 0 0 M 0 1 –250M ⎥ ⎦ 0

0 0

260


x1 x2 s1 s1 ⎡ – 1 0 1 ⎢ 4 s2 ⎢ 0 0 0 ⎢ x2 ⎢ 3 1 0 4 ⎢ W ⎢ –5 0 0 ⎣ x1 x2 s1

s2 s3

t3

0

3 4

–

3 4

1

1

–1

0

–

1 4

1 4

0 –10 10 + M

Section 7.6

W ⎤ 850 ⎥ 3 0 250 ⎥ 250 ⎥ 0 125 2 ⎥ ⎥ 1 2500 ⎥ ⎦ 0

425 2

s2 s3 P

s1 ⎡ – 1 0 1 – 3 0 0 25 ⎤ 4 ⎢ 4 ⎥ s3 ⎢ 0 0 0 1 1 0 250 ⎥ ⎢ ⎥ x2 ⎢ 34 1 0 14 0 0 125 ⎥ 500 3 ⎢ ⎥ P ⎢ –5 0 0 10 0 1 5000 ⎥ ⎣ ⎦ x1 x2 s1 s2 s3 P 200 ⎤ s1 ⎡0 1 1 – 2 0 0 3 3 3 ⎥ ⎢ s3 ⎢0 0 0 1 1 0 250 ⎥ ⎢ ⎥ 500 1 0 0 x1 ⎢ 1 4 0 3 3 3 ⎥ ⎢ ⎥ ⎥ P ⎢0 20 0 35 0 1 17,500 3 3 3 ⎥⎦ ⎢⎣

500 2 = 166 Standard and 0 Executive 3 3 bookcases. Since an integer answer is preferable, note that x1 = 167, x2 = 0 does not satisfy the constraint 3x1 + 4 x2 ≤ 500, while x1 = 166, x2 = 0 satisfies all of the constraints. Thus the company should produce 166 Standard and 0 Executive bookcases each week.

This table indicates that, to maximize profit, the company should produce

14. Let x, y and z denote the numbers of units of products X, Y, and Z produced each week, respectively. We want to maximize the profit function P = 50x + 60y + 75z subject to x + 2 y + 2 z ≤ 40, x + y + 2 z ≤ 30, z ≥ 5, x, y, z ≥ 0. The artificial objective function is W = P – Mt3 . x y z s1 s2 s3 t3 W 2 2 1 0 0 0 0 40 ⎤ ⎡ 1 ⎢ 1 1 2 0 1 0 0 0 30 ⎥⎥ ⎢ ⎢ 0 0 1 0 0 –1 1 0 5⎥ ⎢ ⎥ –50 –60 –75 0 0 0 M 1 0 ⎥ ⎣⎢ ⎦ x y z s1 2 2 1 s1 ⎡ 1 ⎢ 1 1 2 0 s2 ⎢ ⎢ 0 0 1 0 t3 ⎢ W ⎣⎢ –50 –60 –75 – M 0

s2 s3 t3 W 0 0 0 0

40 ⎤ 20 1 0 0 0 30 ⎥⎥ 15 0 –1 1 0 5 ⎥ 5 ⎥ 0 M 0 1 –5M ⎥ ⎦

261


x y 2 s1 ⎡ 1 ⎢ 1 1 s2 ⎢ ⎢ 0 0 z ⎢ W ⎣⎢ –50 –60

z s1 s2 0 1 0


s3 2

t3 –2

0 0 1

2

–2

1 0 0

–1

1

0 0 0 –75 75 + M

W 0

30 ⎤ 15 0 20 ⎥⎥ 10 0 5 ⎥ ⎥ 1 375⎥ ⎦

x y z s1 s2 s3 P 1 0 1 –1 0 0 10 ⎤ 10 ⎡ 0 s1 ⎢ ⎥ 1 1 0 0 12 1 0 10 ⎥ 20 2 s3 ⎢ 2 ⎢ 1 ⎥ 1 1 0 12 0 0 15 ⎥ 30 z⎢ 2 2 ⎢ ⎥ P ⎢ – 25 – 45 0 0 75 0 1 1125⎥ 2 2 2 ⎣ ⎦ x y z s1 s2 s3 P 0 1 0 1 –1 0 0 10 ⎤ ⎡ y ⎢ 1 ⎥ 0 0 – 12 1 1 0 5 ⎥ 10 s3 ⎢ 2 ⎢ ⎥ 1 z⎢ 1 ⎥ 20 0 1 – 1 0 0 10 2 ⎢ 2 ⎥ 25 45 P ⎢⎢ – 2 0 0 2 15 0 1 1350 ⎥⎥ ⎣ ⎦ x y z s1 s2 s3 P 0 1 0 1 –1 0 0 10 ⎤ ⎡ y ⎢ 1 0 0 –1 2 2 0 10 ⎥⎥ x⎢ 5⎥ z ⎢ 0 0 1 0 0 –1 0 ⎢ ⎥ P ⎢⎣ 0 0 0 10 40 25 1 1475⎥⎦ The production order should be 10 units of X, 10 units of Y, and 5 units of Z for a maximum profit of $1475

15. Suppose I is the total investment. Let x1 , x2 , and x3 be the proportions invested in A, AA, and AAA bonds, respectively. If Z is the total annual yield expressed as a proportion of I, then ZI = 0.08 x1I + 0.07 x2 I + 0.06 x3 I , or equivalently, Z = 0.08 x1 + 0.07 x2 + 0.06 x3 . We want to maximize Z subject to x1 + x2 + x3 = 1, x2 + x3 ≥ 0.50, x1 + x2 ≤ 0.30, x1 , x2 , x3 ≥ 0. The artificial objective function is W = Z – Mt1 – Mt2 . x1 x2 x3 s2 s3 t1 t2 W 1 1 0 0 1 0 0 1 ⎤ ⎡ 1 ⎢ 0 1 1 –1 0 0 1 0 0.5⎥⎥ ⎢ ⎢ 1 1 0 0 1 0 0 0 0.3⎥ ⎢ ⎥ ⎢⎣ –0.08 –0.07 –0.06 0 0 M M 1 0 ⎥⎦ x1 x2 x3 s2 s3 t1 t2 W 1 1 1 0 0 1 0 0 1 ⎤ 1 ⎡ t1 ⎢ 0 1 1 –1 0 0 1 0 0.5 ⎥⎥ 0.5 t2 ⎢ 1 1 0 0 1 0 0 0 0.3 ⎥ 0.3 s3 ⎢ ⎢ ⎥ W ⎢⎣ –0.08 – M –0.07 – 2M –0.06 – 2 M M 0 0 0 1 –1.5M ⎥⎦ 262


Section 7.7

x1 x2 x3 s2 s3 t1 0 0 1 0 –1 1 t1 ⎡ ⎢ –1 0 1 –1 –1 0 t2 ⎢ ⎢ 1 1 0 0 1 0 x2 ⎢ W ⎣⎢ –0.01 + M 0 –0.06 – 2 M M 0.07 + 2 M 0 x1 1 ⎡ t1 ⎢ –1 x3 ⎢ ⎢ 1 x2 ⎢ –0.07 –M W ⎢⎣

x2 x3 0 0 0 1 1 0 0 0

x1 x2 x3 0 –1 0 t1 ⎡ ⎢ 1 1 x3 ⎢0 1 0 x1 ⎢1 ⎢ W ⎢⎣0 0.07 + M 0 x1 x2 x3 s2

s2 1

s3 t1 0 1

t2 W 0 0

0.7

W 0

0.5

⎤ 0.7 ⎥ 1 0 0.2 ⎥ 0.2 ⎥ 0 0 0.3 ⎥ 0 1 0.021 – 0.9 M ⎥ ⎦

t2 –1

⎤ 0.5 ⎥ –1 –1 0 1 0 0.2 ⎥ ⎥ 0.3 0 1 0 0 0 0.3 ⎥ –0.06 – M 0.01 0 0.06 + 2 M 1 0.033 – 0.5M ⎥ ⎦ s2 s3 t1 t2 W 1 –1 1 –1 0 0.2 ⎤ 0.2 ⎥ –1 0 0 1 0 0.5 ⎥ ⎥ 0 1 0 0 0 0.3 ⎥ –0.06 – M 0.08 + M 0 0.06 + 2M 1 0.054 – 0.2 M ⎥ ⎦ s3 t1 t2 W

–1 0 1 –1 1 –1 0 0.2 ⎤ s2 ⎡0 ⎢0 0 1 0 –1 1 0 0 0.7 ⎥⎥ x3 ⎢ 1 0 0 1 0 0 0 0.3⎥ x1 ⎢ 1 ⎢ ⎥ W ⎢⎣0 0.01 0 0 0.02 0.06 + M M 1 0.066 ⎥⎦ For the above table, t1 = t2 = 0. Thus W = Z. The fund should put 30% in A bonds, 0% in AA, and 70% in AAA for a yield of 6.6%.

Problems 7.7 x1 x2 s1 s2 t1 t2 W 1. ⎡ 1 −1 –1 0 1 0 0 ⎢ 1 0 ⎢ 2 1 0 –1 0 ⎢2 5 0 0 M M 1 ⎣ x1 x2 s1 s2 t1 ⎡ t1 1 −1 –1 0 1 ⎢ t2 ⎢ 2 1 0 –1 0 ⎢ W ⎢⎣ 2 − 3M 5 M M 0 x1 x2 s1 s2 3 1 t1 ⎡0 −2 –1 2 ⎢ ⎢ 1 x1 ⎢1 0 – 12 2 W ⎢⎢0 4 + 3 M M 1 − 1 M 2 2 ⎣

7⎤ ⎥ 9⎥ 0⎥ ⎦ t2 W

⎤7 ⎥ 1 0 9 ⎥ 92 ⎥ 0 1 –16M ⎥ ⎦ t1 t2 W 0 0

7

1

− 12

0

1 2 −1 + 32

0

⎤5 ⎥ ⎥ 9 0 2 ⎥ ⎥ 1 –9 − 52 M ⎥ ⎦ 0

M

5 2

263



x1 x2 s1 s2 t1 ⎡ s2 0 −3 –2 1 2 ⎢ x1 ⎢ 1 −1 −1 0 1 ⎢ W 0 7 2 0 –2 + M ⎣ The minimum is Z = 14 when

t2 W

5⎤ ⎥ 0 0 7⎥ M 1 –14 ⎥ ⎦ x1 = 7, x2 = 0. −1 0

x1 x2 s1 s2 t1 t2 W 2. ⎡ 2 2 –1 0 1 0 0 1⎤ ⎢ ⎥ 1 0 2⎥ ⎢ 1 3 0 –1 0 ⎢ 8 12 0 0 M M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s2 s3 t1 t2

W 1

t1 ⎡ 2 2 –1 0 1 0 0 1 ⎤2 ⎢ ⎥ t2 ⎢ 1 3 0 –1 0 1 0 2 ⎥ 23 W ⎢8 – 3M 12 – 5M M M 0 0 1 –3M ⎥ ⎣ ⎦ x1 x2 s1 s2 t1 t2 W ⎡ 1 x2 ⎢ ⎢ t2 ⎢ –2 W⎢ ⎢ –4 + 2M ⎣⎢

x1 x2 s1

⎡ 1 0 x2 ⎢ s1 ⎢ – 0 1 ⎢ W⎢ 4 0 0 ⎣ 1 3 4 3

1

– 12

0

0

3 2

–1

0 6 – 32 M

s2

t1

– 13 – 23

0 –1

M

1 2 – 23

–6 + 52 M

t2 W 1 3 2 3

0 0

⎤ ⎥ ⎥1 1 0 ⎥3 ⎥ 0 1 –6 – 12 M ⎥ ⎦⎥ 1 2 1 2

0 0

2⎤ 3⎥ 1⎥ 3⎥

1 –8⎥ ⎦ 2 The minimum is Z = 8 when x1 = 0, x2 = . 3

3.

4 M

–4 + M

x1 x2 x3 s t W 1 –1 –1 –1 1 0 18⎤ ⎡ ⎢12 6 3 0 M 1 0 ⎥ ⎢⎣ ⎦⎥ x1 x2 x3 s t W 1 –1 –1 –1 1 0 18 ⎤ 18 ⎡ t ⎢12 – M 6 + M 3 + M M 0 1 –18M ⎥ W ⎣⎢ ⎦⎥ x1 x2 x3 s t W 1 0 18⎤ x1 ⎡ 1 –1 –1 –1 ⎢0 18 15 12 –12 + M 1 –216 ⎥ W ⎣⎢ ⎦⎥ The minimum is Z = 216 when x1 = 18, x2 = 0, x3 = 0 .

264


Section 7.7

x1 x2 x3 s t W 4. ⎡1 2 –1 –1 1 0 4 ⎤ ⎢1 1 2 0 M 1 0 ⎥ ⎢⎣ ⎦⎥ x1 x2 x3 s t W 2 –1 –1 1 0 4 ⎤2 t ⎡ 1 ⎢1 – M 1 – 2M 2 + M M 0 1 –4 M ⎥ W ⎢⎣ ⎥⎦ 1 0 x2 ⎡ 1 1 – 1 – 1 2⎤ 2 2 2 ⎢2 ⎥ 5 1 – 1 + M 1 –2 ⎥ ⎢1 0 2 2 2 W ⎣⎢ 2 ⎦⎥ The minimum is Z = 2 when x1 = 0, x2 = 2, x3 = 0 .

5. We write the second constraint as – x1 + x3 ≥ 4. x1 x2 x3 s1 s2 s3 t2 W ⎡ 1 1 1 1 0 0 0 0 6⎤ ⎢ –1 0 1 0 –1 0 1 0 4 ⎥⎥ ⎢ ⎢ 0 1 1 0 0 1 0 0 5⎥ ⎢ ⎥ ⎢⎣ 2 3 1 0 0 0 M 1 0 ⎥⎦ x1 x2 x3 s1 s2 s3 t2 W 1 1 1 1 0 0 0 0 6 ⎤6 s1 ⎡ ⎢ –1 0 1 0 –1 0 1 0 4 ⎥⎥ 4 t2 ⎢ 1 1 0 0 1 0 0 5 ⎥5 s3 ⎢ 0 ⎢ ⎥ W ⎢⎣ 2 + M 3 1 – M 0 M 0 0 1 –4 M ⎥⎦ x1 x2 x3 s1 s2 s3 t2 W –1 0 2 ⎤ s1 ⎡ 2 1 0 1 1 0 ⎢ –1 0 1 0 –1 0 1 0 4 ⎥⎥ x3 ⎢ –1 0 1⎥ s3 ⎢ 1 1 0 0 1 1 ⎢ ⎥ W ⎣⎢ 3 3 0 0 1 0 –1 + M 1 –4 ⎦⎥ The minimum is Z = 4 when x1 = 0, x2 = 0, x3 = 4 . 6.

x1 ⎡3 ⎢0 ⎢ ⎢1 ⎢ 5 ⎣⎢

x2 x3 s1 s2 1 −1 1 0 2 1 1

t3 W 0 0 4⎤ 2 0 1 0 0 0 5⎥⎥ 1 0 0 −1 1 0 2 ⎥ ⎥ 3 0 0 0 M 1 0⎥ ⎦

x1 3

x2 1

s3 0

x3 s1 s2 s3 t3 W −1 1 0 0 0 0

4⎤ 4 s1 ⎡ ⎢ 0 2 2 0 1 0 0 0 5⎥⎥ 52 s2 ⎢ 1 1 1 0 0 −1 1 0 2⎥ 2 t3 ⎢ ⎢ ⎥ W ⎣⎢5 − M 1 − M 3 − M 0 0 M 0 1 −2M ⎦⎥

265


x1 s1 ⎡ 2 ⎢ s2 ⎢ −2 x2 ⎢ 1 ⎢ W ⎣⎢ 4

x2 x3 s1 s2 0 −2 1 0


s3 1

t3 W −1 0

2⎤ −2 0 0 0 1 2 1⎥⎥ 1 0 0 −1 1 0 2⎥ ⎥ 2 0 0 1 −1 + M 1 −2 ⎥ ⎦

0 1 0

The minimum is Z = 2 when x1 = 0, x2 = 2, and x3 = 0. 7.

x1 x2 x3 1 ⎡1 2 ⎢0 1 1 ⎢ ⎢1 1 0 ⎢ 1 –1 –3 ⎣⎢

s3 0

x1 1 ⎡ t1 ⎢ 0 t2 ⎢ s3 ⎢ 1 ⎢ W ⎢⎣1 – M x1 t1 ⎡ 1 ⎢ x2 ⎢ 0 s3 ⎢ 1 ⎢ W ⎢⎣1 – M x1 x2 x1 ⎡1 0 ⎢ x2 ⎢ 0 1 s3 ⎢ 0 0 ⎢ W ⎣⎢ 0 0

x2 2

x1 x1 ⎡ 1 ⎢ x3 ⎢ 0 s3 ⎢ 0 ⎢ – Z ⎣⎢ 0

t1 t2 W 1 0 0 4⎤ 0 0 1 0 1⎥⎥ 1 0 0 0 6⎥ ⎥ 0 M M 1 0⎥ ⎦ x3 1

s3 t1 t2 W 0 1 0 0

4 ⎤2 1 1 0 0 1 0 1 ⎥⎥ 1 1 0 1 0 0 0 6 ⎥6 ⎥ –1 – 3M –3 – 2M 0 0 0 1 –5M ⎥ ⎦ x2 x3 s3 t1 t2 W 0 –1 0 1 –2 0 2 ⎤2 1 1 0 0 1 0 1 ⎥⎥ 0 –1 1 0 –1 0 5 ⎥5 ⎥ 0 –2 + M 0 0 1 + 3M 1 1 – 2 M ⎥ ⎦ x3 s3 t1 t2 W –1 0 1 –2 0 2⎤ 1 0 0 1 0 1 ⎥⎥ 1 0 1 –1 1 0 3⎥ ⎥ –1 0 –1 + M 3 + M 1 –1⎥ ⎦

x2 x3 s3 – Z 1 0 0 0 3⎤ 1 1 0 0 1⎥⎥ 0 0 1 0 3⎥ ⎥ 1 0 0 1 0⎥ ⎦

The minimum is Z = 0 when x1 = 3, x2 = 0, x3 = 1 .

266


8.

x1

x2

s1

t1

t2

Section 7.7

W

⎡ −1 1 −1 1 0 0 ⎢ ⎢1 1 0 0 1 0 ⎢ 1 −1 0 M M 1 ⎣ x1 x2 s1 t1

4⎤ ⎥ 1⎥ 0⎥ ⎦ t2 W

−1 1 0 0 t1 ⎡ −1 1 4 ⎤4 ⎢ ⎥ t2 ⎢ 1 1 0 0 1 0 1 ⎥1 ⎢ ⎥ W ⎢ 1 −1 − 2 M M 0 0 1 −5M ⎥ ⎣ ⎦ x1 x2 s1 t1 t2 W t1 ⎡ −2 ⎢ x2 ⎢ 1 ⎢ W ⎣2 + 2M

⎤ ⎥ 1 0 0 1 0 1 ⎥ 0 M 0 1 + 2M 1 1 − 3M ⎥ ⎦ Since all of the indicators in the last table are positive, but the artificial variable t1 is 3, the feasible region is empty. (This can also be seen graphically.)

9.

0 −1 1

x1 x2 x3 s1 s2 t1 ⎡ 1 1 1 –1 0 1 ⎢ ⎢ –1 2 1 0 –1 0 ⎢ 1 8 5 0 0 M ⎣ x1 x2 x3 1 1 t1 ⎡ 1 ⎢ t2 ⎢ –1 2 1 ⎢ W 1 8 – 3M 5 – 2 M ⎣ x1 x2 x3

−1

0

3

t2 W 0 0 8⎤ ⎥ 1 0 2⎥ M 1 0⎥ ⎦ s1 s2 t1 t2 W –1 0 1 0 0 0 M s1

⎤8 ⎥ –1 0 1 0 2 ⎥1 M 0 0 1 –10M ⎥ ⎦ s2 t1 t2 W 8

1 1 ⎤ 14 t1 ⎡ 3 0 –1 1 – 12 0 7 2 2 ⎢ 2 ⎥ 3 ⎢ ⎥ 1 1 x2 ⎢ – 12 1 0 – 12 0 0 1 2 2 ⎥ W ⎢5 – 3 M 0 1 – 1 M M 4 – 1 M 0 –4 + 3 M 1 –8 – 7 M ⎥ ⎢⎣ ⎥⎦ 2 2 2 2 x1 x2 x3 s1 s2 t1 t2 W ⎤ x1 ⎡ 1 2 1 2 – 13 0 14 ⎢1 0 3 – 3 3 3 3 ⎥ 14 ⎢ ⎥ 1 1 x2 ⎢ 0 1 23 – 13 – 13 0 10 5 3 3 3 ⎥ ⎢ ⎥ 10 7 2 ⎥ – 10 + M – 73 + M 1 – 94 W ⎢⎢⎣ 0 0 – 3 3 3 3 3 ⎥⎦ x1 x2 x3 s1 s2 t1 t2 W x1 ⎡ 1 1 1 – 12 0 – 12 – 12 0 3⎤ 2 2 ⎢ ⎥ 3 1 –1 –1 1 1 0 ⎥ x3 ⎢0 5 2 2 2 2 2 ⎢ ⎥ W ⎢0 1 0 3 2 –3 + M –2 + M 1 –28⎥ ⎣ ⎦ The minimum is Z = 28 when x1 = 3, x2 = 0, x3 = 5 .

267



x1 x2 x3 s1 s2 t2 W ⎡ 1 –1 –1 1 0 0 0 3⎤ ⎥ 10. ⎢ ⎢ 1 –1 1 0 –1 1 0 3⎥ ⎢4 4 6 0 0 M 1 0⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t2 W ⎡ s1 1 –1 –1 1 0 0 0 3 ⎤3 ⎢ ⎥ –1 1 0 –1 1 0 3 ⎥3 t2 ⎢ 1 W ⎢ 4 – M 4 + M 6 – M 0 M 0 1 –3M ⎥ ⎣ ⎦ Here we choose t2 as the departing variable. x1 x2 x3 s1 s2 t2 W ⎡ s1 0 0 –2 1 1 –1 0 0⎤ ⎢ ⎥ x1 ⎢ 1 –1 1 0 –1 1 0 3⎥ W ⎢0 8 2 0 4 –4 + M 1 –12 ⎥ ⎣ ⎦ Thus Z has a minimum value of 12 when x1 = 3, x2 = 0, x3 = 0.

If we choose s1 as the departing variable, then x1 ⎡ s1 1 ⎢ t2 ⎢ 1 W ⎢4 – M ⎣ x1 x2 ⎡ x1 1 –1 ⎢ t2 ⎢ 0 0 W ⎢0 8 ⎣ x1 x2 x1 ⎡ 1 –1 ⎢ x3 ⎢0 0 ⎢ W ⎢⎣0 8

x2

x3

s1 s2

t2 W

3 ⎤3 ⎥ –1 1 0 –1 1 0 3 ⎥3 4 + M 6 – M 0 M 0 1 –3M ⎥ ⎦ x3 s1 s2 t2 W –1 1 0 0 0 3 ⎤ ⎥ 2 –1 –1 1 0 0 ⎥ 0 10 – 2 M –4 + M M 0 1 –12 ⎥ ⎦ x3 s1 s2 t2 W 1 –1 1 0 0 3⎤ 2 2 2 ⎥ 1 0 ⎥ 1 – 12 – 12 0 2 ⎥ 0 1 5 –5 + M 1 –12 ⎥ ⎦ The minimum is Z = 12 when x1 = 3, x2 = 0, x3 = 0 . –1

–1

1

0

0 0

268


Section 7.7

11. Let x1 , x2 , and x3 denote the annual numbers of barrels of cement produced in kilns that use device A, device B, and no device, respectively. We want to minimize the annual emission control cost C (C in dollars) where 1 2 C = x1 + x2 + 0 x3 subject to 4 5 x1 + x2 + x3 = 3,300, 000, 1 1 x1 + x2 + 2 x3 ≤ 1, 000, 000, 2 4 x1 , x2 , x3 ≥ 0. x1 x2 x3 s2 ⎡ ⎢1 ⎢1 ⎢2 ⎢1 ⎣⎢ 4

t1 W

⎤ 1 0 3,300, 000 ⎥ 2 1 0 0 1, 000, 000 ⎥⎥ ⎥ 0 0 M 1 0⎥ ⎦ x2 x3 s2 t1 W

1 1 0 1 4 2 5

x1 ⎡ ⎤ 1 1 0 1 0 3,300, 000 ⎥ 3,300, 000 t1 ⎢ 1 1 s2 ⎢⎢ 12 2 1 0 0 1, 000, 000 ⎥⎥ 500, 000 4 W ⎢ 1 – M 2 – M – M 0 0 1 –3,300, 000M ⎥ ⎢⎣ 4 ⎥⎦ 5 x1 x2 x3 s2 t1 W 7 ⎡ 3 0 – 12 1 0 2,800, 000 ⎤ 3, 200, 000 8 t1 ⎢ 4 ⎥ ⎥ 1 1 x3 ⎢⎢ 14 1 0 0 500, 000 ⎥ 4, 000, 000 8 2 W ⎢ 1 – 3 M 2 – 7 M 0 1 M 0 1 –2,800, 000 M ⎥ ⎢⎣ 4 4 ⎥⎦ 5 8 2 x1 x2 x3 s2 t1 W ⎡ 6 ⎤ 8 1 0 – 74 0 3, 200, 000 ⎥ 11,200,000 x2 ⎢ 7 7 3 ⎢ ⎥ x3 ⎢ 17 0 1 74 – 17 0 100, 000 ⎥ 700, 000 ⎥ W ⎢ 13 8 16 + M 1 –1, 280, 000 ⎥ ⎢– 0 0 – 35 35 ⎢⎣ 140 ⎥⎦ x1 x2 x3 s2 – C ⎡ ⎤ x2 ⎢ 0 1 –6 –4 0 2, 600, 000 ⎥ 700, 000 ⎥ x1 ⎢ 1 0 7 4 0 ⎢ ⎥ 3 1 –1, 215, 000 ⎥ – C ⎢ 0 0 13 20 5 ⎣⎢ ⎦⎥ Thus the minimum value of C is 1,215,000 when x1 = 700, 000, x2 = 2, 600, 000, x3 = 0. The plant should install device A on kilns producing 700,000 barrels annually, and device B on kilns producing 2,600,000 barrels annually.

269



12. Let x1 = number of type A trucks rented, x2 = number of type B trucks rented. We want to minimize C = 0.40 x1 + 0.60 x2 subject to 2 x1 + 2 x2 ≥ 12, x1 + 3 x2 ≥ 12, x1 , x2 ≥ 0. x1

x2

s1

s2

t1

t2 W

⎡ 2 2 –1 0 1 0 0 12 ⎤ ⎢ ⎥ 3 0 –1 0 1 0 12 ⎥ ⎢ 1 ⎢ 0.40 0.60 0 0 M M 1 0 ⎥ ⎣ ⎦ x1 x2 s1 s2 t1 t2 W ⎡ ⎤ t1 ⎢ 2 2 –1 0 1 0 0 12 ⎥ 6 t2 ⎢ 1 3 0 –1 0 1 0 12 ⎥ 4 ⎢ ⎥ W ⎢ 2 – 3M 3 – 5M M M 0 0 1 –24 M ⎥ 5 ⎣⎢ 5 ⎦⎥ x1 x2 s1 s2 t1 t2 W 2 2 ⎡ 4 ⎤3 0 −1 1 0 4 −3 3 t1 ⎢ 3 ⎥ ⎢ ⎥ 1 1 1 1 0 0 0 4 x2 ⎢ 3 −3 ⎥ 12 3 W ⎢ 1 − 4 M 0 M 1 − 2 M 0 − 1 + 5 M 1 − 12 − 4 M ⎥ ⎢⎣ 5 3 ⎥⎦ 5 3 5 3 5 x1 x2 s1 s2 t1 t2 W ⎡ 3 3 1 – 12 0 3⎤ x1 ⎢ 1 0 – 4 2 4 ⎥ ⎢ 1 1 1 1 –2 –4 0 3⎥⎥ x2 ⎢ 0 1 4 2 W ⎢0 0 3 3 + M – 1 + M 1 –3⎥ 1 – 20 ⎢⎣ ⎥⎦ 20 10 10 The minimum value of C is 3 when x1 = 3 and x2 = 3. They should rent 3 of type A and 3 of type B. The cost per mile is $3.00.

13. Let x1 = number of DVD players shipped from Akron to Columbus, x2 = number of DVD players shipped from Springfield to Columbus, x3 = number of DVD players shipped from Akron to Dayton, x4 = number of DVD players shipped from Springfield to Dayton. We want to minimize C = 5 x1 + 3x2 + 7 x3 + 2 x4 subject to x1 + x2 x3 + x4 x1 + x3 x2 + x4 x1 , x2 , x3 , x4

= 150, = 150, ≤ 200, ≤ 150, ≥ 0.

270


x1 ⎡1 ⎢0 ⎢ ⎢1 ⎢ ⎢0 ⎢5 ⎢⎣

x2 x3 x4 s3 s4 1 0 0 0 0 0

1 1 0 0

0

1 0

1 0

t1 1

t2 W 0 0 150 ⎤ 0 1 0 150 ⎥⎥ 0 0 0 200 ⎥ ⎥ 0 0 0 150 ⎥ M M 1 0 ⎥⎥ ⎦ x3 x4 s3 s4 t1 t2 W 0 0 0 0 1 0 0

1 0

1 0

1

3 7 2 0 0

x1 x2 t1 ⎡ 1 1 t2 ⎢ 0 0 1 1 ⎢ s3 ⎢ 1 0 1 0 ⎢ 0 1 0 1 s4 ⎢ ⎢ W ⎣⎢5 − M 3 − M 7 − M 2 − M x1 x2 t1 ⎡ 1 1 ⎢ 0 0 x4 ⎢ ⎢ 1 0 s3 ⎢ 0 1 s4 ⎢ W ⎢⎢⎣5 − M 3 − M x1 t1 ⎡ 1 ⎢ 0 x4 ⎢ ⎢ s3 1 ⎢ 0 x2 ⎢ W ⎢⎢⎣5 − M

Section 7.7

0 0 0

1 0

1 0 0 0 0 0

1 0 0 0

0 0 0 0

1

x3 x4 s3 s4 t1 0 0 0 0 1

t2 W 0 0

1 1 0 0 0

1 0

1 0

1 0 0

−1 0 0

0 0 −1 0

1 0

5 0 0 0 0 −2 + M

1

x2 0

x3 x4 s3 1 0 0

s4 t1 −1 1

t2 W 1 0

0 0

1 1 0 1 0 1

0 0 0 0

1 0 0 0

1

−1 0 0

0 8− M

1 0 −1 0

0 0 −3 + M

0

1 1

150 ⎤ 150 ⎥⎥ 150 200 ⎥ ⎥ 150 ⎥ 150 −300M ⎥⎥ ⎦

150 ⎤ 150 150 ⎥⎥ 200 ⎥ ⎥ 0⎥ 0 −300 − 150 M ⎥⎥ ⎦ 150 ⎤ 150 ⎥ 150 ⎥ 200 ⎥ 200 ⎥ 0⎥ −300 − 150 M ⎥⎥ ⎦

x1 x2 x3 x4 s3 s4 t1 t2 W x1 ⎡ 1 0 1 0 0 −1 1 1 0 150 ⎤ x4 ⎢⎢ 0 0 1 1 0 0 0 1 0 150 ⎥⎥ s3 ⎢ 0 0 0 0 1 1 −1 −1 0 50 ⎥ ⎢ ⎥ x2 ⎢ 0 1 −1 0 0 1 0 −1 0 0⎥ W ⎢⎢⎣ 0 0 3 0 0 2 −5 + M −4 + M 1 −1050 ⎥⎥⎦ The retailer should ship as follows: to Columbus, 150 from Akron and 0 from Springfield; to Dayton, 0 from Akron and 150 from Springfield. The transportation cost is $1050. If s4 is chosen as the departing variable in the second table, the result is the same, although the final table is different: x1 x2 x3 x4 s3 s4 t1 t2 W x1 ⎡ 1 1 0 0 0 0 1 0 0 150 ⎤ ⎢ x3 ⎢ 0 −1 1 0 0 −1 0 1 0 0 ⎥⎥ s3 ⎢ 0 0 0 0 1 1 −1 −1 0 50 ⎥ ⎢ ⎥ x4 ⎢ 0 1 0 1 0 1 0 0 0 150 ⎥ W ⎢⎢ 0 3 0 0 0 5 −5 + M −7 + M 1 −1050 ⎥⎥ ⎣ ⎦

271

Chapter 7: Linear Programming 14. Let x A xB yA yB


= number of alternators from supplier X to plant A = number of alternators from supplier X to plant B = number of alternators from supplier Y to plant A = number of alternators from supplier Y to plant B

We want to minimize C = 300 x A + 320 xB + 340 y A + 280 yB x A + y A = 7000 xB + y B = 5000 x A + xB ≥ 3000 x A + xB ≤ 5000 y A + yB ≥ 7000 x A , xB , y A , y B ≥ 0 x A xB y A y B s3 s4 s5 t1 t2 t3 t5 W 0 1 0 0 0 0 1 0 0 0 0 ⎡ 1 ⎢ 0 1 0 1 0 0 0 0 1 0 0 0 ⎢ − 1 1 0 0 1 0 0 0 0 1 0 0 ⎢ 1 0 0 0 1 0 0 0 0 0 0 ⎢ 1 ⎢ 0 0 1 1 0 0 −1 0 0 0 1 0 ⎢300 320 340 280 0 0 0 M M M M 1 ⎢⎣ xA xB yA yB s3 1 0 1 0 0 t1 ⎡ 0 1 0 1 0 t2 ⎢ ⎢ 1 1 0 0 −1 t3 ⎢ 1 1 0 0 0 s4 ⎢ 0 0 1 1 0 t5 ⎢ ⎢ W ⎣⎢300 − 2M 320 − 2 M 340 − 2 M 280 − 2M M xA 1 t1 ⎡ 0 yB ⎢ ⎢ 1 t3 ⎢ 1 s4 ⎢ ⎢ 0 t5 ⎢ W ⎣⎢300 − 2 M

xB yA 0 1 1 0 1 0 1 0 −1 1 40 340 − 2 M

yB s3 0 0 1 1 0 −1 0 0 0 0 0 M

s4 s5 0 0 0 0 0 0 1 0 0 −1 0 M

xA xB y A yB s3 s4 −1 1 0 1 0 t1 ⎡ 0 1 0 1 0 0 yB ⎢0 ⎢ −1 0 1 0 0 xA ⎢ 1 0 0 0 1 1 s4 ⎢ 0 −1 1 0 0 0 t5 ⎢ 0 ⎢ W ⎢⎣ 0 −260 + 2M 340 − 2 M 0 300 − M 0 x A xB t1 ⎡ 0 0 yB ⎢0 1 ⎢ xA ⎢ 1 1 s4 ⎢ 0 0 y A ⎢ 0 −1 ⎢ W ⎢⎣ 0 80

yA 0 0 0 0 1 0

yB s3 0 1 1 0 0 −1 0 1 0 0 0 300 − M

s4 s5 0 1 0 0 0 0 1 0 −1 0 0 340 − M

s5 0 0 0 0 −1 M

subject to

7000 ⎤ 5000 ⎥ ⎥ 3000 ⎥ 5000 ⎥ 7000 ⎥ 0 ⎥⎥⎦

s4 s5 0 0 0 0 0 0 1 0 0 −1 0 M

t1 t2 1 0 0 1 0 0 0 0 −1 0 0 −280 + 2 M

t1 1 0 0 0 0 0 t3 0 0 1 0 0 0

t2 0 1 0 0 0 0 t5 0 0 0 0 1 0

t3 0 0 1 0 0 0

t5 0 0 0 0 1 0

W 0 7000 ⎤ 0 5000 ⎥ 5000 ⎥ 0 3000 ⎥ 0 5000 ⎥ 0 7000 ⎥ 7000 1 −22, 000 M ⎥⎦⎥

W 0 7000 ⎤ 7000 0 5000 ⎥ ⎥ 0 3000 ⎥ 3000 0 5000 ⎥ 5000 0 2000 ⎥ 1 −1, 400, 000 − 12, 000M ⎦⎥⎥

t1 t2 t3 t5 W −1 0 0 1 0 4000 ⎤ 4000 0 1 0 0 0 5000 ⎥ ⎥ 0 0 1 0 0 3000 ⎥ −1 0 0 0 0 2000 ⎥ −1 0 0 1 0 2000 ⎥ 2000 0 −280 + 2M −300 + 2 M 0 1 −2,300, 000 − 6000M ⎥⎦⎥

t1 t2 t3 t5 W 1 1 2000 ⎤ 2000 −1 −1 0 0 1 0 0 0 5000 ⎥ ⎥ 0 0 1 0 0 3000 ⎥ 0 0 0 0 2000 ⎥ 2000 −1 0 −1 0 1 0 2000 ⎥ 0 60 −300 + 2 M −340 + 2M 1 2,980, 000 − 2000 M ⎥⎥⎦

We choose t1 as the departing variable. 272


Section 7.7

x A xB y A yB s3 s4 s5 t1 t2 t3 t5 W 1 1 1 −1 2000 ⎤ −1 0 s3 ⎡ 0 0 0 0 1 0 0 1 0 0 0 5000 ⎥ yB ⎢0 1 0 1 0 0 0 ⎢ ⎥ 1 1 1 0 5000 ⎥ −1 0 xA ⎢ 1 1 0 0 0 0 1 0 0⎥ −1 −1 0 s4 ⎢ 0 0 0 0 0 1 −1 −1 0 0 1 0 2000 ⎥ y A ⎢ 0 −1 1 0 0 0 −1 ⎢ ⎥ W ⎢⎣ 0 80 0 0 0 0 40 −300 + M −240 + M M −40 + M 1 −3,580, 000 ⎦⎥ The manufacturer should order 5000 alternators from X to be shipped to A, 2000 from Y to A, and 5000 from Y to B. The minimum cost is $3,580,000. (Note that the same result is obtained if s4 is chosen as the departing variable in the fifth table.)

15. a.

Roll width

⎧15" 3 2 1 0 ⎨10" 0 1 3 4 ⎩ Trim loss 3 8 3 8

b. We want to minimize L = 3x1 + 8 x2 + 3x3 + 8 x4 subject to 3x1 + 2 x2 + x3 ≥ 50, x2 + 3x3 + 4 x4 ≥ 60, x1 , x2 , x3 , x4 ≥ 0. x1 ⎡3 ⎢ ⎢0 ⎢3 ⎣

x2 x3 x4 s1

s2

t1

t2 W

0 0 50 ⎤ ⎥ 1 3 4 0 –1 0 1 0 60 ⎥ 8 3 8 0 0 M M 1 0⎥ ⎦ x1 x2 x3 x4 s1

2 1 0 –1

0

1

t1 ⎡ 3 2 1 0 ⎢ t2 ⎢ 0 1 3 4 ⎢ W 3 – 3M 8 – 3M 3 – 4 M 8 – 4 M ⎣ x1 x2 x3 x4 s1 ⎡ 3 5 0 – 43 –1 t1 ⎢ 3 1 4 1 0 x3 ⎢⎢ 0 3 3 W ⎢3 – 3M 7 – 5 M 0 4 + 4 M M ⎢⎣ 3 3 x1 x2 x3 x4 s1 s2 t1 ⎡1 5 0 – 4 – 1 1 1 x1 ⎢ 9 9 3 9 3 ⎢ 1 4 1 0 –3 0 x3 ⎢0 3 1 3 W ⎢0 16 0 16 2 –1 + M 1 ⎢⎣ 3 3 3 x1 = 10, x2 = 0, x3 = 20, x4 = 0.

c.

s2

t1 t2 W

⎤ 50 ⎥ 0 –1 0 1 0 60 ⎥ 20 M M 0 0 1 –110 M ⎥ ⎦ s2 t1 t2 W –1

0

1 0 0

1 3 – 13 1 – 13 M

1

– 13

0

1 3 –1 + 43

0

t2 W –

1 9 1 3

– 23 + M

90 in.

273

10 ⎤ ⎥ 0 20 ⎥⎥ ⎥ 1 –90 ⎥ ⎦ 0

50

⎤ ⎥ 10 ⎥ 0 20 ⎥ ⎥ 1 –60 – 30M ⎥ ⎦ 0

M

30



Principles in Practice 7.8 1. Let x1 , x2 , and x3 be the numbers respectively, of Type 1, Type 2, and Type 3 gadgets produced. The original problem is to maximize P = 300 x1 + 200 x2 + 200 x3 , subject to 300 x1 + 220 x2 + 180 x3 ≤ 60, 000, 20 x1 + 40 x2 + 20 x3 ≤ 2000, 3 x1 + x2 + 2 x3 ≤ 120, and x1 , x2 , x3 ≥ 0. The dual problem is to minimize W = 60, 000 y1 + 2000 y2 + 120 y3 , subject to 300 y1 + 20 y2 + 3 y3 ≥ 300, 220 y1 + 40 y2 + y3 ≥ 200, 180 y1 + 20 y2 + 2 y3 ≥ 200, and y1 , y2 , y3 ≥ 0. 2. Let x1 and x2 be the amounts, respectively of supplement 1 and supplement 2. The original problem is to minimize C = 6 x1 + 2 x2 , subject to 20 x1 + 6 x2 ≥ 98, 8 x1 + 16 x2 ≥ 80, and x1 , x2 ≥ 0. The dual problem is to maximize W = 98 y1 + 80 y2 , subject to 20 y1 + 8 y2 ≤ 6, 6 y1 + 16 y2 ≤ 2, and y1 , y2 ≥ 0.

3. Let x1 , x2 , and x3 be the numbers, respectively, of devices 1, 2, and 3 produced. The original problem is to maximize P = 30 x1 + 20 x2 + 20 x3 , subject to 30 x1 + 15 x2 + 10 x3 ≤ 300, 20 x1 + 30 x2 + 20 x3 ≤ 400, 40 x1 + 30 x2 + 25 x3 ≤ 600, and x1 , x2 , x3 ≥ 0. The dual problem is to minimize W = 300 y1 + 400 y2 + 600 y3 , subject to 30 y1 + 20 y2 + 40 y3 ≥ 30, 15 y1 + 30 y2 + 30 y3 ≥ 20, 10 y1 + 20 y2 + 25 y3 ≥ 20, and y1 , y2 , y3 ≥ 0. The tablex to maximize Z = −W = −300 y1 − 400 y2 − 600 y3 follow. y1 y2 y3 s1 s2 s3 t1 t2 t3 Z 1 0 0 0 30 ⎤ ⎡ 30 20 40 −1 0 0 ⎢ 15 30 30 0 −1 0 0 1 0 0 20 ⎥⎥ ⎢ ⎢ 10 20 25 0 0 −1 0 0 1 0 20 ⎥ ⎢ ⎥ 300 400 600 0 0 0 M M M 1 0 ⎥ ⎣⎢ ⎦

274


y1 30 t1 ⎡ ⎢ 15 t2 ⎢ ⎢ 10 t3 ⎢ Z ⎣⎢300 – 55M

Section 7.8

y2 20

y3 40

s1 s2 s3 t1 t2 t3 Z –1 0 0 1 0 0 0

30

30

0

20

25

0

400 – 70 M

600 – 95M

M

30 ⎤ −1 0 0 1 0 0 20 ⎥⎥ 0 −1 0 0 1 0 20 ⎥ ⎥ M M 0 0 0 1 –70M ⎥ ⎦

y1 y2 y3 s1 s2 s3 t1 t2 t3 Z 10 4 4 ⎡ 10 ⎤ −20 −3 0 −1 0 1 0 0 3 3 ⎥ t1 ⎢ 1 1 2 ⎢ 1 ⎥ − 1 1 0 0 0 0 0 y3 ⎢ 2 30 30 3 ⎥ 5 5 10 ⎥ t3 ⎢ − 5 − − − 5 0 0 1 0 1 0 2 6 6 3 ⎢ ⎥ Z ⎢ 15 13 19 20 − M –200 + 25M 0 M 20 − 6 M M 0 −20 + 6 M 0 1 −400 − 3 M ⎥ ⎢⎣ 2 ⎥⎦ y1 y2 y3 s1 s2 s3 t1 t2 t3 Z 1 2 1 2 1 y1 ⎡1 ⎤ −2 − 15 0 − 10 0 0 0 15 10 3 ⎢ ⎥ ⎢0 ⎥ 1 1 1 1 1 − 10 2 1 20 0 − 20 0 0 y3 ⎢ 10 2 ⎥ ⎥ 7 7 25 1 1 t3 ⎢ 0 − − − − 10 0 1 1 0 ⎢ ⎥ 4 6 4 6 6 ⎢ 7M M 3 M −20 + 13 M 0 1 −400 − 25 M ⎥ 1 − + M M − 0 200 10 0 20 Z ⎢ ⎥⎦ 4 6 4 6 6 ⎣ y1 y2 y3 s1 s2 s3 t1 t2 t3 Z 15 3 3 5 s2 ⎡ ⎤ −15 −4 −1 0 0 0 1 0 2 4 2 ⎢ ⎥ 3 3 1 1 1 ⎢ ⎥ − 1 0 0 0 0 0 4 2 40 40 4 y3 ⎢ ⎥ 15 5 5 5 ⎥ − 35 − − 0 0 1 0 1 0 t3 ⎢ 4 2 8 8 4 ⎢ ⎥ ⎢ ⎥ 35 15 5 13 5 Z ⎢ −150 + 4 M 100 − 2 M 0 15 − 8 M 0 M −15 + 8 M M 0 1 −450 − 4 M ⎥ ⎣ ⎦ y1 y2 y3 s1 s2 s3 t1 t2 t3 Z 1 1 s2 ⎡ −10 0 0 −2 −1 1 −2 2 0 5 ⎤ 2 ⎢ ⎥ 1 0 1 1 1 2 ⎥ ⎢ 4 − − 0 1 0 0 15 15 15 15 3 ⎥ y3 ⎢ 3 ⎥ 1 2 1 2 1 y2 ⎢ − 7 1 0 − 12 0 − 15 0 0 ⎢ 6 12 15 6 ⎥ ⎢ ⎥ − 20 + M M − 40 + M 1 − 1400 0 0 20 0 40 Z ⎢ − 100 ⎥⎦ 3 3 3 3 3 3 ⎣ The t1 , t2 , and t3 columns are no longer needed. y1 y2 y3 s1 s2 s3 Z 15 0 1 − 32 0 10 ⎤ s2 ⎡ 0 0 2 ⎢ ⎥ 1 1⎥ y1 ⎢ 1 0 3 − 1 0 0 4 20 20 2⎥ ⎢ 7 3 3 1 ⎢ ⎥ y2 0 1 0 − 40 0 8 40 4⎥ ⎢ Z ⎢ 0 0 25 5 0 15 1 −450 ⎥ ⎣ ⎦ From this table, the maximum profit of $450 corresponds to x1 = 5, x2 = 0, and x3 = 15. The company should produce 5 of device 1 and 15 of device 3.

275



1. Minimize W = 5 y1 + 3 y2 subject to y1 − y2 ≥ 1 y1 + y2 ≥ 2 y1 , y2 ≥ 0

8. The second constraint can be written as −8 x1 + 10 x2 ≥ −80. Maximize W = −10 y1 − 80 y2 subject to −4 y1 − 8 y2 ≤ 5 3 y1 + 10 y2 ≤ 4 y1 , y2 ≥ 0

2. Minimize W = 3 y1 + 5 y2 subject to

9. The dual is: Maximize W = 2 y1 + 3 y2 subject to

Problems 7.8

2 y1 – y2 2 y1 + 4 y2 2 y2 y1 , y2

≥ 2, ≥ 1, ≥ –1, ≥ 0.

y1 – y2 ≤ 2, – y1 + 2 y2 ≤ 2, 2 y1 + y2 ≤ 5, y1 , y2 ≥ 0. y1 y2 s1 s1 ⎡ 1 –1 1 ⎢ s2 ⎢ –1 2 0 s3 ⎢ 2 1 0 ⎢ W ⎢⎣ –2 –3 0 y1 y2 s1 s1 ⎡ 1 0 1 ⎢ 2 y2 ⎢ – 1 1 0 ⎢ 2 s3 ⎢ 52 0 0 ⎢ ⎢ W ⎢ – 72 0 0 ⎣ y1 y2 s1

3. Maximize W = 8 y1 + 2 y2 subject to y1 – y2 y1 + 2 y2 y1 + y2 y1 , y2

≤ 1, ≤ 8, ≤ 5, ≥ 0.

4. Maximize W = y1 + 2 y2 subject to 2 y1 + y2 ≤ 8, 2 y1 + 3 y2 ≤ 12, y1 , y2 ≥ 0.

5. The second and third constraints can be written as x1 − x2 ≤ −3 and − x1 − x2 ≤ −11. Minimize W = 13 y1 – 3 y2 –11 y3 subject to

s1 ⎡ 0 ⎢ y2 ⎢ 0 ⎢ y1 ⎢ 1 ⎢ W ⎢⎢ 0 ⎣

– y1 + y2 – y3 ≥ 1, 2 y1 – y2 – y3 ≥ –1, y1 , y2 , y3 ≥ 0.

6. The second constraint can be written as − x1 + 2 x2 − x3 ≤ −6. Minimize W = 9 y1 – 6 y2 subject to y1 – y2 ≥ 1, y1 + 2 y2 ≥ –1, y1 – y2 ≥ 4, y1 , y2 ≥ 0.

0 1 1 0 0 0 0 0

s2 0 1 0 0

s3 0 0 1 0

W 0 0 0 1

2⎤ 2 ⎥⎥ 1 5⎥ 5 ⎥ 0⎥ ⎦

s2 s3 W 0 0 3⎤ 6 ⎥ 0 0 1⎥ ⎥ 1 0 4 ⎥ 85 ⎥ ⎥ 0 1 3⎥ ⎦ s3 W

1 2 1 2 – 12 3 2

s2

3 5 2 5 – 15 4 5

− 15

0

1 5 2 5 7 5

0

The minimum is Z =

0 1

11 ⎤ 5⎥ 9⎥ 5⎥ 8⎥ 5⎥ 43 ⎥ 5 ⎥⎦

43 4 when x1 = 0, x2 = , 5 5

7 x3 = . 5

10. The dual is: Maximize W = 28 y1 + 2 y2 + 16 y3 subject to y1 + 2 y2 – 3 y3 ≤ 2, 4 y1 – y2 + 8 y3 ≤ 2, y1 , y2 , y3 ≥ 0.

7. The first constraint can be written as − x1 + x2 + x3 ≥ −3. Maximize W = –3 y1 + 3 y2 subject to – y1 + y2 ≤ 4, y1 – y2 ≤ 4, y1 + y2 ≤ 6, y1 , y2 ≥ 0.

y1 y2 y3 ⎡ 2 −3 s1 1 ⎢ –1 8 s2 ⎢ 4 W ⎢ –28 –2 –16 ⎣ 276

s1 s2 W 1 0 0 2⎤ 2 ⎥ 0 1 0 2 ⎥ 12 0 0 1 0⎥ ⎦


Section 7.8

y1 y2 y3 s1 s2 W s1 ⎡ 0 94 –5 1 – 14 0 23 ⎤ 2 ⎢ ⎥3 ⎢ 1 1 1 y1 1 – 4 2 0 4 0 2 ⎥ ⎢ ⎥ W ⎢ 0 –9 40 0 7 1 14 ⎥ ⎢⎣ ⎥⎦ y1 y2 y3 s1 s2 W y2 ⎡ 20 0 1 − 9 94 – 91 0 32 ⎤ ⎢ ⎥ 13 1 2 0 2⎥ y1 ⎢ 1 0 9 9 9 3⎥ ⎢ W ⎢0 0 20 4 6 1 20 ⎥ ⎣⎢ ⎦⎥ The minimum is Z = 20 when x1 = 4, x2 = 6 .

11. The dual is: Minimize W = 8 y1 + 12 y2 subject to y1 + y2 ≥ 3, 2 y1 + 6 y2 ≥ 8, y1 , y2 ≥ 0. y1 y2 s1 s2 t1 t2 U ⎡ 1 1 –1 0 1 0 0 3⎤ ⎢ ⎥ 1 0 8⎥ ⎢ 2 6 0 –1 0 ⎢ 8 12 0 0 M M 1 0 ⎥ ⎣ ⎦ y1 y2 s1 s2 t1 t2 U 1 –1 0 1 0 0 3 ⎤3 t1 ⎡ 1 ⎢ ⎥ 6 0 –1 0 1 0 8 ⎥ 43 t2 ⎢ 2 U ⎢8 – 3M 12 – 7 M M M 0 0 1 –11M ⎥ ⎣ ⎦ y1 y2 s1 s2 t1 t2 U 5 1 ⎡ 2 ⎤5 0 –1 1 – 16 0 6 3 t1 ⎢ 3 ⎥2 ⎥4 1 4 y2 ⎢⎢ 13 1 0 – 16 0 0 ⎥ 6 3 U ⎢ 4 – 2 M 0 M 2 – 1 M 0 –2 + 7 M 1 –16 – 5 M ⎥ ⎢⎣ ⎥⎦ 3 6 6 3 y1 y2 s1 s2 t1 t2 U y1 ⎡ 3 5⎤ 1 1 0 – 32 – 14 0 4 2 2⎥ ⎢ 1 –1 1 1 0 1⎥ – y2 ⎢ 0 1 2 4 2 4 2⎥ ⎢ U ⎢0 0 6 1 –6 + M –1 + M 1 –26 ⎥ ⎣ ⎦ The maximum is Z = 26 when x1 = 6, x2 = 1 .

277



12. The dual is: Minimize W = 12 y1 + 8 y2 subject to 3 y1 + y2 ≥ 2, y1 + y2 ≥ 6, y1 , y2 ≥ 0. y1 ⎡ 3 ⎢ ⎢ 1 ⎢12 ⎣

y2 s1

s2

t1

t2 U

1 –1 0 1 0 0 2⎤ ⎥ 1 0 –1 0 1 0 6⎥ 8 0 0 M M 1 0⎥ ⎦ y1 y2 s1 s2 t1 t2 U

2 t1 ⎡ 3 1 –1 0 1 0 0 2 ⎤3 ⎢ ⎥ t2 ⎢ 1 1 0 –1 0 1 0 6 ⎥6 U ⎢12 – 4M 8 – 2M M M 0 0 1 –8M ⎥ ⎣ ⎦ y1 y2 s1 s2 t1 t2 U 1 1 2 y1 ⎡1 ⎤2 – 13 0 0 0 3 3 3 ⎢ ⎥ ⎢ ⎥ 16 2 1 1 t2 ⎢ 0 –1 –3 1 0 ⎥8 3 3 3 U ⎢ 0 4 – 2 M 4 – 1 M M –4 + 4 M 0 1 –8 – 16 M ⎥ ⎢⎣ ⎥⎦ 3 3 3 3 y1 y2 s1 s2 t1 t2 U ⎡ ⎤ 3 1 –1 0 1 0 0 2 y2 ⎢ ⎥ t2 ⎢ –2 0 1 –1 –1 1 0 4 ⎥4 U ⎢ –12 + 2 M 0 8 – M M –8 + 2M 0 1 –16 – 4M ⎥ ⎣ ⎦ y1 y2 s1 s2 t1 t2 U 1 0 6⎤ y2 ⎡ 1 1 0 –1 0 ⎢ ⎥ s1 ⎢ –2 0 1 –1 –1 1 0 4⎥ U ⎢ 4 0 0 8 M –8 + M 1 –48⎥ ⎣ ⎦ The maximum is Z = 48 when x1 = 0, x2 = 8 .

13. The first constraint can be written as x1 − x2 ≥ −1. The dual is: Maximize W = – y1 + 3 y2 subject to y1 + y2 ≤ 6, – y1 + y2 ≤ 4, y1 , y2 ≥ 0. y1 ⎡ s1 1 ⎢ s2 ⎢ –1 W⎢1 ⎣ y1 ⎡ s1 2 ⎢ y2 ⎢ –1 W ⎢ –2 ⎣

y2 s1 s2 W

1 1 0 0 6⎤ 6 ⎥ 1 0 1 0 4⎥ 4 –3 0 0 1 0 ⎥ ⎦ y2 s1 s2 W 0 1 –1 0 2 ⎤ 1 ⎥ 1 0 1 0 4⎥ 0 0 3 1 12 ⎥ ⎦

278


y1 y2 s1

Section 7.8

s2 W

y1 y2 s1 s2 W 1 ⎤ 1 ⎡ 80 1 – 54 0 s1 ⎢ 0 5 ⎥ 400 1 1 1 ⎥ 1 y1 ⎢1 0 0 2 50 50 ⎥ 25 ⎢ W ⎢ 0 –20, 000 0 1600 1 1600 ⎥ ⎣ ⎦ y1 y2 s1 s2 W 1 1 1 ⎤ ⎡ – 100 0 y2 0 1 80 400 ⎥ ⎢ 3 ⎥ 1 1 y1 ⎢ 1 0 – 160 0 40 160 ⎥ ⎢ W ⎢ 0 0 250 1400 1 1650 ⎥ ⎣ ⎦ The firm should spend $250 on newspaper advertising and $1400 on radio advertising for a cost of $1650.

y1 ⎡ 1 0 12 – 12 0 1⎤ ⎢ ⎥ 1 0 ⎥ y2 ⎢ 0 1 12 5 2 ⎢ ⎥ W ⎢0 0 1 2 1 14 ⎥ ⎣ ⎦ The minimum is Z = 14 when x1 = 1, x2 = 2 .

14. The first constraint can be written as −2 x1 + x2 + x3 ≥ −2. The dual is: Maximize W = −2 y1 + 4 y2 subject to −2 y1 − y2 ≤ 2 y1 − y2 ≤ 1 y1 + 2 y2 ≤ 1 y1 , y2 ≥ 0 y1 y2 s1 s2 s3 W s1 ⎡ −2 −1 1 0 0 0 2 ⎤ s2 ⎢ 1 −1 0 1 0 0 1⎥ ⎢ ⎥ s3 ⎢ 1 2 0 0 1 0 1⎥ 12 W ⎢⎣ 2 −4 0 0 0 1 0 ⎥⎦

16. Let x1 = number of type A trucks rented, x2 = number of type B trucks rented. We want to minimize C = 0.40 x1 + 0.60 x2 subject to 2 x1 + 2 x2 ≥ 12, x1 + 3 x2 ≥ 12, x1 , x2 ≥ 0.

y1 y2 s1 s2 s3 W s1 ⎡ − 3 ⎢ 2 s2 ⎢ 3 2 ⎢ y2 ⎢ 1 2 W ⎢⎣ 4

0 1 0 0 0 1 1 0 0 0 0 0

1 2 1 2 1 2

0 0

5⎤ 2⎥ 3 ⎥ 2 1⎥ 2⎥

The dual is: Maximize W = 12 y1 + 12 y2 subject to 2 y1 + y2 ≤ 0.40, 2 y1 + 3 y2 ≤ 0.60, y1 , y2 ≥ 0.

0 2 1 2 ⎥⎦

The minimum is Z = 2 when x1 = 0, x2 = 0, x3 = 2.

y1 y2 s1 s2 W s1 ⎡ 2 1 1 0 0 52 ⎤ 15 ⎢ ⎥ 3 s2 ⎢ 2 3 0 1 0 53 ⎥ 10 ⎢ ⎥ W ⎢ –12 –12 0 0 1 0 ⎥ ⎣ ⎦ Here we choose y1 as the entering variable.

15. Let x1 = amount spent on newspaper advertising, x2 = amount spent on radio advertising. We want to minimize C = x1 + x2 subject to 40 x1 + 50 x2 ≥ 80, 000, 100 x1 + 25 x2 ≥ 60, 000, x1 , x2 ≥ 0. The dual is: Maximize W = 80, 000 y1 + 60, 000 y2 subject to

y1 y2 s1 s2 W ⎡1 1 1 0 0 15 ⎤ 52 y1 ⎢ 2 2 ⎥ ⎢ 1 1 s2 ⎢ 0 2 –1 1 0 5 ⎥⎥ 10 W ⎢ 0 –6 6 0 1 12 ⎥ ⎢⎣ 5 ⎥⎦ y1 y2 s1 s2 W y1 ⎡ 3 –1 0 3 ⎤ 1 0 4 4 20 ⎥ ⎢ 1 0 1⎥ y2 ⎢ 0 1 – 12 2 10 ⎥ ⎢ W ⎢0 0 3 3 1 3⎥ ⎣ ⎦ Thus the maximum value of W, and hence the minimum value of C, is 3.

40 y1 + 100 y2 ≤ 1, 50 y1 + 25 y2 ≤ 1, y1 , y2 ≥ 0. y1

y2

s1 s2 W

1

100 1 0 0 1 ⎤ 40 s1 ⎡ 40 ⎢ ⎥ 1 25 0 1 0 1 ⎥ 50 s2 ⎢ 50 W ⎢ –80, 000 –60, 000 0 0 1 0 ⎥ ⎣ ⎦ 279



If we choose y2 as the entering variable, then y1 y2 s1 s2 W ⎡ 1 1 0 0 52 ⎤ 52 s1 ⎢ 2 ⎥ 3 0 1 0 53 ⎥ 15 s2 ⎢ 2 ⎢ ⎥ W ⎢ –12 –12 0 0 1 0 ⎥ ⎣ ⎦ y1 y2 s1 s2 W ⎡ 4 0 1 –1 0 1 ⎤ 3 3 5 ⎥ 20 s1 ⎢ 3 ⎢ 3 2 1 1 y2 ⎢ 3 1 0 3 0 5 ⎥⎥ 10 W ⎢ –4 0 0 4 1 12 ⎥ ⎢⎣ 5 ⎥⎦ y1 y2 s1 s2 W 3 –1 0 3 ⎤ ⎡ y1 ⎢ 1 0 4 4 20 ⎥ 1 0 1⎥ y2 ⎢ 0 1 – 12 2 10 ⎥ ⎢ W ⎢0 0 3 3 1 3⎥ ⎣ ⎦ The minimum total cost per mile is $3. 17. Let y1 = number of shipping clerk apprentices, y2 = number of shipping clerks, y3 = number of semiskilled workers, y4 = number of skilled workers. We want to minimize W = 6 y1 + 9 y2 + 8 y3 + 14 y4 subject to y1 + y2 ≥ 60, –2 y1 + y2 ≥ 0, y3 + y4 ≥ 90, y3 – 2 y4 ≥ 0, y1 , y2 , y3 , y4 ≥ 0. The dual is: Maximize Z = 60 x1 + 0 x2 + 90 x3 + 0 x4 subject to x1 – 2 x2 ≤ 6, x1 + x2 ≤ 9, x3 + x4 ≤ 8, x3 – 2 x4 ≤ 14, x1 , x2 , x3 , x4 ≥ 0. x1 x2 x3 x4 0 s1 ⎡ 1 –2 0 ⎢ 1 1 0 0 s2 ⎢ ⎢ 0 0 1 1 s3 ⎢ 0 1 –2 s4 ⎢ 0 ⎢ Z ⎢⎣ –60 0 –90 0

s1 1 0 0 0 0

s2 0 1 0 0 0

s3 0 0 1 0 0

s4 0 0 0 1 0

Z 0 6⎤ 0 9 ⎥⎥ 0 8⎥ 8 ⎥ 0 14 ⎥ 14 1 0 ⎥⎥ ⎦

280


x1 x2 x3 x4 s1 –2 0 0 1 s1 ⎡ 1 ⎢ 1 0 0 0 s2 ⎢ 1 ⎢ 0 1 1 0 x3 0 ⎢ 0 0 0 –3 0 s4 ⎢ ⎢ Z ⎣⎢ –60 0 0 90 0

s2 s3 0 0 1 0 0 1 0 –1 0 90

Chapter 7 Review

s4 0 0 0 1 0

Z 0 6 ⎤6 0 9 ⎥⎥ 9 0 8 ⎥ ⎥ 0 6 ⎥ 1 720 ⎥⎥ ⎦ x1 x2 x3 x4 s1 s2 s3 s4 Z 1 –2 0 0 1 0 0 0 0 6 ⎤ ⎡ x1 ⎢0 3 0 0 –1 1 0 0 0 3 ⎥⎥ 1 s2 ⎢ 0 1 1 0 0 1 0 0 8 ⎥ x3 ⎢0 ⎢ ⎥ 0 0 –3 0 0 –1 1 0 6 ⎥ s4 ⎢0 Z ⎢⎢⎣0 –120 0 90 60 0 90 0 1 1080 ⎥⎥⎦ x1 x2 x3 x4 s1 s2 s3 s4 Z 1 2 ⎡ 0 0 0 8⎤ x1 ⎢ 1 0 0 0 3 3 ⎥ 1 ⎥ x2 ⎢ 0 1 0 0 – 1 0 0 0 1 3 3 ⎢ ⎥ x3 ⎢ 0 0 1 1 0 0 1 0 0 8⎥ ⎥ s4 ⎢ 0 0 0 –3 0 0 –1 1 0 6⎥ ⎢ Z ⎢ 0 0 0 90 20 40 90 0 1 1200 ⎥ ⎥⎦ ⎣⎢ The company should employ 20 shipping clerk apprentices, 40 shipping clerks, 90 semiskilled workers, and 0 skilled workers for a total hourly wage of $1200.

Chapter 7 Review Problems 1.

5

y

x 5

2.

10

y

x 10

281

Chapter 7: Linear Programming 3.

5


y

8.

–5 3

5

y

x

x

5

4.

5

5

y

9.

10

y

x

x

10

5

5.

10

y

10.

10

y

x 10

6.

5

x 10

y

11. Feasible region follows. Corner points are (0, 0), (0, 2), (1, 3), (3, 1), (3, 0). Z is maximized at (3, 0) where its value is 3. Thus Z = 3 when x = 3 and y = 0.

x

5

5

y y−x=2 x+y=4

7.

5

y

x=3 (3, 0)

x 5

282

x 5


Chapter 7 Review

12. Feasible region follows. Corner points are (0, 1), (0, 5), (4, 3), and (4, 1). Z is maximized at (4, 3) where its value is 22. Thus Z = 22 when x = 4 and y = 3.

10

y

x+y=5

y

10

⎛ 20 10 ⎞ ⎜ , ⎟ ⎝ 9 9⎠

x + 2y = 10

2x + 5y = 10 x

(4, 3)

(4, 0) 5x + 8y = 20

x=4 x

(5, 0)

16. Feasible region follows. Corner points are (0, 4), (0, 6), (6, 8), (6, 0), and (4, 0). Z is minimized at (0, 4) and (4, 0) where its value is 8. Thus Z is minimized at all points on the line segment joining (0, 4) and (4, 0). The solution is Z = 8 when x = (1 – t)(0) + 4t = 4t, y = (1 – t)(4) + 0t = 4 – 4t, and 0 ≤ t ≤ 1.

10 y=1

13. Feasible region is unbounded. Z is minimized at the corner point (0, 2) where its value is –2. Thus Z = –2 when x = 0 and y = 2. 5

10

y

y

10

x − y = −2

−x + 3y = 18

(6, 8)

(0, 6) (0, 2)

(0, 4) x+y=1

x − 2y = 2 x

17. Feasible region follows. Corner points are (0, 0), (0, 4), (2, 3), and (4, 0). Z is maximized at (2, 3) and (4, 0) where its value is 36. Thus Z is maximized at all points on the line segment joining (2, 3) and (4, 0). The solution is Z = 36 when x = (1 – t)(2) + 4t = 2 + 2t, y = (1 – t)(3) + 0t = 3 – 3t, and 0 ≤ t ≤ 1.

y

5

4x + 3y = 15

⎛ 10 5 ⎞ ⎜ , ⎟ ⎝ 3 9⎠ (0, 0)

x – 6y = 0

x 10

(4, 0)

14. Feasible region follows. Corner points are (0, 0), ⎛ 10 5 ⎞ ⎜ , ⎟ , and (0, 5). Z is minimized at (0, 0) ⎝ 3 9⎠ where its value is 0. Thus Z = 0 when x = 0 and y = 0.

(0, 5)

(6, 0)

x+y=4

5

10

x=6

(0, 4)

x 10 (0, 0)

y x + 2y = 8 (2, 3)

3x + 2y = 12 x (4, 0) 5

15. Feasible region follows. Corner points are ⎛ 20 10 ⎞ ⎜ , ⎟ , (5, 0), and (4, 0). Z is minimized at ⎝ 9 9⎠ 70 70 ⎛ 20 10 ⎞ . Thus Z = ⎜ , ⎟ where its value is 9 9 ⎝ 9 9⎠ 20 10 when x = and y = . 9 9

18. Feasible region is unbounded. The family of lines given by Z = 4x + y has members having arbitrarily large values of Z and that also intersect the feasible region. Thus no optimum solution exists.

283


20

y


x1 x2 3 s1 ⎡ 2 ⎢ 4 3 20. s2 ⎢ ⎢ 0 1 s3 ⎢ Z ⎣⎢ –18 –20

s1 s2 s3 Z 1 0 0 0 18 ⎤ 6 0 1 0 0 24 ⎥⎥ 8 0 0 1 0 5 ⎥5 ⎥ 0 0 0 1 0⎥ ⎦ x1 x2 s1 s2 s3 Z 3 s1 ⎡ 2 0 1 0 –3 0 3 ⎤ 2 ⎢ ⎥ s2 ⎢ 4 0 0 1 –3 0 9 ⎥ 9 4 x2 ⎢ 0 1 0 0 1 0 5 ⎥ ⎢ ⎥ Z ⎣⎢ –18 0 0 0 20 1 100 ⎦⎥

Z = 40

(0, 12) (4, 6)

3x + 2y = 24

x + 2y = 16

Z = 16

(16, 0) x 20

x1 ⎡ s1 1 ⎢ 19. s2 ⎢ 1 Z ⎢ –4 ⎣ x1

x2 s1

1 0 0 12 ⎤ 2 ⎥ 2 0 1 0 8 ⎥4 –5 0 0 1 0 ⎥ ⎦ x2 s1 s2 Z 6

x2 ⎡ 1 ⎢ 6 ⎢ s2 ⎢ 23 ⎢ 19 Z ⎢⎣⎢ – 6 x1 x2 ⎡ 0 ⎢ x1 ⎢⎢1 Z ⎢0 ⎣⎢

s2 Z

0

0 0

5 6

0

x2 s1

s2

Z

1 4

1 4

0

1

0 – 12 0 – 34

–

3 2 19 4

x1 x2 s1 s2 ⎡ s1 0 4 1 –1 ⎢ x1 ⎢ 1 2 0 1 Z ⎢0 3 0 4 ⎣ Thus Z = 32 when

s3

Z

0 – 0 32 ⎤ x1 ⎡1 0 ⎢ ⎥ s2 ⎢ 0 0 –2 1 3 0 3 ⎥ 1 x2 ⎢⎢ 0 1 0 0 1 0 5 ⎥⎥ 5 Z ⎢ 0 0 9 0 –7 1 127 ⎥ ⎣ ⎦ x1 x2 s1 s2 s3 Z 1 1 0 0 3⎤ x1 ⎡⎢ 1 0 – 2 2 ⎥ 1 1 0 ⎥ s3 ⎢ 0 0 – 2 1 3 3 ⎢ ⎥ 2 –1 0 0 ⎢0 1 ⎥ 4 x2 ⎢ 3 3 ⎥ 7 0 1 134 ⎥ Z ⎢ 0 0 13 3 3 ⎣⎢ ⎦⎥ Thus Z = 134 when x1 = 3 and x2 = 4 . 1 2

⎤ 2 ⎥ 12 ⎥ 1 0 4⎥ 6 ⎥ 0 1 10 ⎥ ⎦⎥

1 6 – 13

1

x1 x2 s1 s2

1 ⎤4 ⎥ 0 6 ⎥⎥ ⎥ 1 29 ⎥ ⎦

Z 4⎤ ⎥ 0 8⎥ 1 32 ⎥ ⎦ x1 = 8 and x2 = 0.

0

284

3 2


x1 x2 x3 s1 t1 W 21. ⎡ 1 2 3 –1 1 0 5⎤ ⎢3 2 1 0 M 1 0 ⎥ ⎢⎣ ⎦⎥ x1 x2 x3 2 3 t1 ⎡ 1 ⎢3 − M 2 − 2M 1 − 3M W ⎢⎣ x1 x2 x3 s1 t1 1 2 1 1 ⎡ x3 ⎢ 3 3 1 − 3 3 8 4 1 1 ⎢ W 0 −3+M 3 ⎢⎣ 3 3 5 Thus Z = when x1 = 0, x2 3 x1 ⎡3 22. ⎢ ⎢1 ⎢1 ⎣

x2

s1

s2 t1

t1 ⎡ 3 ⎢ t2 ⎢ 1 W ⎢1 – 4M ⎣ x1

5 0 2 – 5M

Chapter 7 Review

s1 t1 W 5 −1 1 0 5 ⎤3 M 0 1 −5M ⎥⎥ ⎦ W 5⎤ 0 3⎥ 5 1 − 3⎥ ⎥⎦ 5 = 0, and x3 = . 3

t2 W

5 –1 0 1 0 0 20 ⎤ ⎥ 0 0 –1 0 1 0 5⎥ 2 0 0 M M 1 0⎥ ⎦ x1 x2 s1 s2 t1 t2 W

–1 0 1 0 0 20 ⎤ 4 ⎥ 0 –1 0 1 0 5 ⎥ M M 0 0 1 –25M ⎥ ⎦ s1 s2 t1 t2 W

x2

x2 ⎡ 3 1 – 15 ⎢ 5 t2 ⎢ 1 0 0 ⎢ 1 W ⎢ − 5 − M 0 52 ⎣ x1 x2 s1 s2 x2 ⎡ 0 ⎢ x1 ⎢1 ⎢ W ⎢0 ⎢⎣ x1 ⎡ s2 0 ⎢ x1 ⎢⎢1 ⎢ Z ⎢0 ⎣

⎤ 20 ⎥ 3 –1 0 1 0 5 ⎥ 5 ⎥ M − 52 + M 0 1 –8 − 5M ⎥ ⎦ t1 t2 W 1 5

0

0 0

1 – 15

3 5

1 5

− 53

0

0

–1

0

1

0

2 5

− 15

− 52 + M

x2

s1 s2 Z

5 3 5 3 1 3

Thus Z =

− 13 1 0 − 13

0 0

1 3

0 1

1 5

+M

4

1 ⎤5 ⎥3 0 5⎥ ⎥ 1 –7 ⎥ ⎥⎦ 0

5 ⎤ 3 ⎥ 20 ⎥ 3 ⎥ ⎥ − 20 3 ⎥⎦

20 20 , x2 = 0. when x1 = 3 3

285


x1 x2 s1 s2 s3 t2 ⎡ 1 1 1 0 0 0 ⎢ 1 23. ⎢ 1 1 0 –1 0 ⎢ 1 0 0 0 1 0 ⎢ –1 –2 0 0 0 M ⎣⎢ x1 1 ⎡ s1 ⎢ 1 t2 ⎢ s3 ⎢ 1 ⎢ W ⎢⎣ –1 – M


W 0 12 ⎤ 0 5⎥⎥ 0 10 ⎥ ⎥ 1 0⎥ ⎦

x2 s1 s2 1 1 0 1 0 –1 0 0 0 –2 – M 0 M

s3 0 0 1 0

t2 0 1 0 0

W 0 12 ⎤ 12 0 5 ⎥⎥ 5 0 10 ⎥ ⎥ 1 –5M ⎥ ⎦

x1 x2 s1 s2 s3 t2 W 0 0 1 1 0 –1 0 7 ⎤7 s1 ⎡ ⎢1 1 0 –1 0 1 0 5 ⎥⎥ x2 ⎢ 0 0 10 ⎥ s3 ⎢1 0 0 0 1 ⎢ ⎥ W ⎢⎣1 0 0 –2 0 2 + M 1 10 ⎥⎦ x1 x2 s1 s2 s3 Z s2 ⎡ 0 0 1 1 0 0 7 ⎤ ⎢ ⎥ x2 ⎢ 1 1 1 0 0 0 12 ⎥ s3 ⎢ 1 0 0 0 1 0 10 ⎥ ⎢ ⎥ Z ⎢⎣ 1 0 2 0 0 1 24 ⎦⎥ Thus Z = 24 when x1 = 0 and x2 = 12. x1 ⎡1 24. ⎢ ⎢1 ⎢2 ⎣

x2 s1

s2

t2 W

2 1 0 0 0 6⎤ ⎥ 1 0 –1 1 0 1⎥ 1 0 0 M 1 0⎥ ⎦ x1 x2 s1 s2 t2 W

s1 ⎡ 1 2 1 0 0 0 6 ⎤3 ⎢ ⎥ t2 ⎢ 1 1 0 –1 1 0 1 ⎥ 1 W ⎢2 – M 1 – M 0 M 0 1 – M ⎥ ⎣ ⎦ s1 ⎡ –1 0 1 2 –2 0 4 ⎤ ⎢ ⎥ x2 ⎢ 1 1 0 –1 1 0 1⎥ W ⎢ 1 0 0 1 –1 + M 1 –1⎥ ⎣ ⎦ Thus Z = 1 when x1 = 0 and x2 = 1.

25. We write the first constraint as – x1 + x2 + x3 ≥ 1. x1

x2 x3

s1

t1

t2 W

⎡ –1 1 1 –1 1 0 0 1⎤ ⎢ ⎥ 1 0 12 ⎥ ⎢ 6 3 2 0 0 ⎢ 1 2 1 0 M M 1 0⎥ ⎣ ⎦ 286


Chapter 7 Review

x1 x2 x3 s1 t1 t2 W ⎡ t1 –1 1 1 –1 1 0 0 1 ⎤ ⎢ ⎥ t2 ⎢ 6 3 2 0 0 1 0 12 ⎥ 2 W ⎢1 – 5M 2 – 4 M 1 – 3M M 0 0 1 –13M ⎥ ⎣ ⎦ x1 x2 x3 s1 t1 t2 W 3 4 1 t1 ⎡0 –1 1 0 3 ⎤2 2 3 6 ⎢ ⎥ ⎥4 1 1 1 x1 ⎢⎢1 0 0 0 2 ⎥ 2 3 6 W ⎢0 3 – 3 M 2 – 4 M M 0 – 1 + 5 M 1 –2 – 3M ⎥ ⎢⎣ ⎥⎦ 2 2 3 3 6 6 x1 x2 x3 s1 t1 t2 W 2 1 x2 ⎡ 0 1 8 – 2 0 2 ⎤ 94 9 3 3 9 ⎢ ⎥ ⎥ 1 1 1 x1 ⎢⎢1 0 – 19 – 0 1 ⎥ 3 3 9 W ⎢ 0 0 – 2 1 –1 + M – 1 + M 1 –5⎥ ⎢⎣ ⎥⎦ 3 3 x1 x2 x3 s1 – Z x3 ⎡ 9⎤ 0 98 1 – 34 0 4⎥ ⎢ ⎢ ⎥ 5 1 1 x1 ⎢ 1 8 0 0 4 4⎥ – Z ⎢0 3 0 1 1 – 7⎥ ⎢⎣ 4 2 2 ⎥⎦ 7 5 9 when x1 = , x2 = 0, and x3 = . Thus Z = 2 4 4 x1

x2 x3

s1

s2 t1 W

⎡ 1 1 3 –1 0 1 0 5⎤ ⎥ 26. ⎢ 1 4 0 1 0 0 5⎥ ⎢ 2 ⎢ –2 –3 –5 0 0 M 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1

s2 t1 W

t1 ⎡ 1 1 3 −1 0 1 ⎢ s2 ⎢ 2 1 4 0 1 0 ⎢ W −2 − M −3 − M −5 − 3M M 0 0 ⎣ x1 x2 x3 s1 s2 1 1 ⎡ –2 0 –1 − 34 4 t1 ⎢ 1 1 1 1 x3 ⎢⎢ 0 2 4 4 ⎢1 1 7−1M 0 M 5+3M M – + 4 4 4 4 W ⎢⎣ 2 2 We choose t1 as the departing variable.

5

0 5⎤ 3 ⎥ 0 5⎥ 54 1 −5 M ⎥ ⎦ t1 W 5⎤5 4⎥

1 0

5⎥ 4⎥5

0 0 0 1

25 4

287

⎥ – 54 M ⎥ ⎦


ISM: Introductory Mathematical Analysis Since s2 is nonbasic for the last table and its indicator is 0, there may be multiple optimum solutions. Treating s2 as an entering variable, deleting the t2 -column, changing W to –Z, and continuing, we have x1 x2 x3 s1 s2 – Z s2 ⎡ 1 1 0 12 1 0 1⎤ ⎢2 2 ⎥ x3 ⎢ 12 12 1 12 0 0 2 ⎥ ⎢ ⎥ – Z ⎢ 1 1 0 0 0 1 0⎥ ⎣ ⎦ Here Z = 0 when x1 = 0, x2 = 0, and x3 = 2. Thus multiple optimum solutions exist. Hence Z is minimum when x1 = (1 – t )(0) + 0t = 0, x2 = (1 – t )(0) + 0t = 0, x3 = (1 – t )(1) + 2t = 1 + t ,

x1 x2 x3 s1 s2 t1 W ⎡ x2 −2 1 0 −4 −3 4 0 5⎤ ⎢ ⎥ x3 ⎢ 1 0 1 1 −1 0 0 ⎥ 0 1 W ⎢ −3 0 0 −7 −4 7 + M 1 15⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z ⎡ x2 2 1 4 0 1 0 5⎤ ⎢ ⎥ s1 ⎢ 1 0 1 1 1 0 0 ⎥ Z ⎢ 4 0 7 0 3 1 15⎥ ⎣ ⎦ Thus Z = 15 when x1 = 0, x2 = 5, and x3 = 0. Note that choosing x3 as the departing variable results in the same solution. x1 x2 x3 s1 s2 Z ⎡ s1 4 –1 0 1 0 0 2 ⎤ ⎢ ⎥ 27. s2 ⎢ –8 2 5 0 1 0 2⎥ 1 ⎢ ⎥ Z ⎢ –1 –4 –2 0 0 1 0 ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 Z s1 ⎡ 0 0 5 1 1 0 3⎤ 2 2 ⎢ ⎥ 5 1 ⎢ x2 –4 1 2 0 2 0 1⎥ ⎢ ⎥ Z ⎢ –17 0 8 0 2 1 4 ⎥ ⎣ ⎦ For the last table, x1 is the entering variable. Since no quotients exist, the problem has an unbounded solution. That is, no optimum solution (unbounded). x1 ⎡1 28. ⎢ ⎢0 ⎢1 ⎣

x2 x3 s1

s2

and 0 ≤ t ≤ 1. For the last table, s1 is nonbasic and its indicator is 0. If we continue the process for determining other optimum solutions, we return to a table corresponding to the third table. 29. The dual is: Maximize W = 35 y1 + 25 y2 subject to y1 + y2 ≤ 2, 2 y1 + y2 ≤ 7, 3 y1 + y2 ≤ 8, y1 , y2 ≥ 0. y1

y2

1 s1 ⎡ 1 ⎢ 2 1 s2 ⎢ 1 s3 ⎢ 3 ⎢ W ⎢⎣ –35 –25

t2 W

1 2 1 0 0 0 4⎤ ⎥ 0 1 0 –1 1 0 1⎥ 1 0 0 0 M 1 0⎥ ⎦ x1 x2 x3 s1 s2 t2 W s1 ⎡1 1 2 1 0 0 0 4 ⎤ 2 ⎢ ⎥ t2 ⎢0 0 1 0 –1 1 0 1 ⎥ 1 W ⎢1 1 – M 0 M 0 1 – M ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t2 W ⎡ s1 1 1 0 1 2 –2 0 2 ⎤ 1 ⎢ ⎥ x3 ⎢0 0 1 0 –1 1 0 1 ⎥ W ⎢1 1 0 0 0 M 1 0 ⎥ ⎣ ⎦ The minimum value of Z is 0 for x1 = 0, x2 = 0, and x3 = 1.

s1 s2 s3 W 1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

2⎤ 2 7 ⎥⎥ 72 8⎥ 8 ⎥3 0⎥ ⎦

y1 y2 s1 s2 s3 W ⎡ y1 1 1 1 0 0 0 2 ⎤ ⎢ ⎥ s2 ⎢ 0 –1 –2 1 0 0 3⎥ s3 ⎢ 0 –2 –3 0 1 0 2 ⎥ ⎢ ⎥ W ⎢⎣ 0 10 35 0 0 1 70 ⎥⎦ Thus Z = 70 when x1 = 35, x2 = 0, and x3 = 0.

288


Chapter 7 Review

30. We write the third constraint as −4 x1 − x2 ≤ −2. The dual is: Minimize W = 3 y1 + 4 y2 – 2 y3 subject to y1 + y2 – 4 y3 ≥ 1, – y1 + 2 y2 – y3 ≥ –2, y1 , y2 , y3 ≥ 0.

We write the second constraint of the dual as y1 – 2 y2 + y3 ≤ 2. y1 y2 y3 s1 s2 t1 U ⎡ 1 1 –4 –1 0 1 0 1⎤ ⎢ ⎥ 1 0 1 0 0 2⎥ ⎢ 1 –2 ⎢3 4 –2 0 0 M 1 0 ⎥ ⎣ ⎦ y1 y2 y3 s1 s2 t1 ⎡ t1 1 1 –4 –1 0 1 ⎢ s2 ⎢ 1 –2 1 0 1 0 ⎢ U 3 – M 4 – M –2 + 4M M 0 0 ⎣ y1 y2 y3 s1 s2 t1 U ⎡ y1 1 1 –4 –1 0 1 0 1⎤ ⎢ ⎥ s2 ⎢ 0 –3 5 1 1 –1 0 1⎥ U ⎢0 1 10 3 0 –3 + M 1 –3⎥ ⎣ ⎦ Thus Z = 3 when x1 = 3 and x2 = 0.

U 0 1 ⎤1 ⎥ 0 2 ⎥2 1 –M ⎥ ⎦

31. Let x, y, and z denote the numbers of units of X, Y, and Z produced weekly, respectively. If P is the total profit obtained, we want to maximize P = 10x + 15y + 22z subject to x + 2 y + 2 z ≤ 40, x + y + 2 z ≤ 34, x, y, z ≥ 0. x y z s1 s2 P ⎡ s1 1 2 2 1 0 0 40 ⎤ 20 ⎢ ⎥ s2 ⎢ 1 1 2 0 1 0 34 ⎥ 17 P ⎢ –10 –15 –22 0 0 1 0 ⎥ ⎣ ⎦ x y z s1 s2 P ⎡ s1 0 1 0 1 –1 0 6 ⎤ 6 ⎢ ⎥ z ⎢ 12 12 1 0 12 0 17 ⎥ 34 ⎢ ⎥ P ⎢ 1 –4 0 0 11 1 374 ⎥ ⎣ ⎦ 1 –1 0 6⎤ y ⎡0 1 0 ⎢ ⎥ z ⎢ 12 0 1 – 12 1 0 14 ⎥ ⎢ ⎥ P⎢ 1 0 0 4 7 1 398⎥ ⎣ ⎦ Thus 0 units of X, 6 units of Y, and 14 units of Z give a maximum profit of $398.

289



32. We want to maximize P = 10x + 15y + 22z subject to x + 2 y + 2 z ≤ 40, x + y + 2 z ≤ 34, x + y + z ≥ 24, x, y, z ≥ 0. x y z s1 s2 s3 t3 W 2 2 1 0 0 0 0 40 ⎤ ⎡ 1 ⎢ 1 1 2 0 1 0 0 0 34 ⎥⎥ ⎢ ⎢ 1 1 1 0 0 –1 1 0 24 ⎥ ⎢ ⎥ –10 –15 –22 0 0 0 M 1 0 ⎥ ⎣⎢ ⎦ x y z s1 s2 s3 t3 W 1 2 2 1 0 0 0 0 40 ⎤ 20 s1 ⎡ ⎢ 1 1 2 0 1 0 0 0 34 ⎥⎥ 17 s2 ⎢ 1 1 1 0 0 –1 1 0 24 ⎥ 24 t3 ⎢ ⎢ ⎥ W ⎢⎣ –10 – M –15 – M –22 – M 0 0 M 0 1 −24M ⎦⎥ x y z s1 s2 s3 t3 W 0 1 0 1 –1 0 0 0 6 ⎡ ⎤ 6 s1 ⎢ 1 ⎥ 1 1 z⎢ 1 0 0 0 0 17 ⎥ 34 2 2 2 ⎢ 1 ⎥ 14 1 1 0 0 –2 –1 1 0 7 ⎢ 2 ⎥ 2 t3 ⎢ ⎥ M M M W ⎢⎣1 – 2 –4 – 2 0 0 11 + 2 M 0 1 374 – 7 M ⎥⎦ x y z s1 s2 s3 t3 W 1 0 1 –1 0 0 0 6 ⎤ y⎡ 0 ⎢ 1 ⎥ 1 0 1 –2 1 0 0 0 14 z⎢ 2 ⎥ 28 ⎢ 1 ⎥ 0 0 – 12 0 –1 1 0 4 ⎢ 2 ⎥8 t3 ⎢ ⎥ M M W ⎢⎢1 – 2 0 0 4 + 2 7 M 0 1 398 – 4M ⎥⎥ ⎣ ⎦ x y z s1 s2 s3 t3 W 0 0 6⎤ y ⎡0 1 0 1 –1 0 ⎢0 0 1 0 1 1 –1 0 10 ⎥⎥ z⎢ 2 0 8⎥ x ⎢ 1 0 0 –1 0 –2 ⎢ ⎥ W ⎢⎣0 0 0 5 7 2 –2 + M 1 390 ⎥⎦ The company should produce 8 units of X, 6 units of Y, 10 units of Z, for a profit of $390.

33. Let x AC , x AD , xBC , and xBD denote the amounts (in hundreds of thousands of gallons) transported from A to C, A to D, B to C, and B to D, respectively. If C is the total transportation cost in thousands of dollars, we want to minimize C = x AC + 2 x AD + 2 xBC + 4 xBD subject to x AC + x AD xBC + xBD x AC + xBC x AD + xBD x AC , x AD , xBC , xBD

≤ 6, ≤ 6, = 5, = 5, ≥ 0.

290


x AC ⎡1 ⎢0 ⎢ ⎢1 ⎢ ⎢0 ⎢1 ⎢⎣

x AD 1 0 0 1 2

x AC 1 ⎡ s1 ⎢ 0 s2 ⎢ t3 ⎢ 1 ⎢ t4 ⎢ 0 W ⎢⎢⎣1 – M

xBC 0 1 1 0 2

xBD 0 1 0 1 4

x AD 1 0 0 1 2–M

s1 s2 1 0 0 1 0 0 0 0 0 0 xBC 0 1 1 0 2–M

t3 0 0 1 0 M

Chapter 7 Review

t4 0 0 0 1 M

xBD s1 0 1 1 0 0 0 1 0 4–M 0

W 0 0 0 0 1 s2 0 1 0 0 0

t3 0 0 1 0 0

6⎤ 6 ⎥⎥ 5⎥ ⎥ 5⎥ 0 ⎥⎥ ⎦ t4 0 0 0 1 0

W 0 6 ⎤6 0 6 ⎥⎥ 0 5 ⎥5 ⎥ 0 5 ⎥ 1 –10M ⎥⎥ ⎦ W 0 1⎤ 1 0 6 ⎥⎥ 0 5⎥ ⎥ 0 5⎥ 5 1 –5 – 5M ⎥⎥ ⎦ W 0 1 ⎤ ⎥ 0 6 ⎥6 ⎥5 0 5 ⎥ 0 4 ⎥4 1 –7 – 4M ⎥⎥ ⎦

x AC x AD xBC xBD s1 s2 t3 t4 1 –1 0 1 0 –1 0 s1 ⎡ 0 ⎢0 0 1 1 0 1 0 0 s2 ⎢ 0 1 0 0 0 1 0 x AC ⎢ 1 ⎢ 1 0 1 0 0 0 1 t4 ⎢ 0 ⎢ 1 4 – M 0 0 –1 + M 0 W ⎢⎣ 0 2 – M x AC x AD xBC xBD s1 s2 t3 t4 –1 0 1 0 –1 0 x AD ⎡ 0 1 ⎢0 0 1 1 0 1 0 0 s2 ⎢ ⎢ 1 0 1 0 0 0 1 0 x AC ⎢ 1 1 –1 0 1 1 t4 ⎢ 0 0 ⎢ W ⎢⎣ 0 0 3 – M 4 – M –2 + M 0 1 0 x AC x AD xBC xBD s1 s2 t3 t4 W 0 1 0 5⎤ x AD ⎡ 0 1 0 1 0 0 ⎢0 0 0 0 1 1 –1 –1 0 2 ⎥⎥ s2 ⎢ 0 –1 0 1⎥ x AC ⎢ 1 0 0 –1 1 0 ⎢ ⎥ 1 1 0 4⎥ xBC ⎢ 0 0 1 1 –1 0 W ⎢⎢⎣ 0 0 0 1 1 0 –2 + M –3 + M 1 –19 ⎥⎦⎥ The minimum value of C is 19, when x AC = 1, x AD = 5, xBC = 4, and xBD = 0. Thus 100,000 gal from A to C, 500,000 gal from A to D, and 400,000 gal from B to C give a minimum cost of $19,000.

34. Let x and y be the weekly sales of Space Traders and Green Dwarf, respectively. We want to maximize P = 5x + 9y subject to the constraints 30 x + 10 y ≤ 300 20 x + 10 y ≤ 200 10 x + 50 y ≤ 500 x, y ≥ 0 The constraints can be written as

291



3 x + y ≤ 30 2 x + y ≤ 20 x + 5 y ≤ 50 x, y ≥ 0 970 ⎛ 50 80 ⎞ The feasible region has corner points (0, 0), (0, 10), ⎜ , at ⎟ , and (10, 0). P has a maximum of 9 ⎝ 9 9 ⎠ ⎛ 50 80 ⎞ ⎛ 5 8 ⎞ ⎜ , ⎟ = ⎜ 5 , 8 ⎟ . The possible integer values are (5, 8), (5, 9), (6, 8), and (6, 9). However, the point (6, 9) ⎝ 9 9 ⎠ ⎝ 9 9⎠ does not satisfy the second or third constraints. Evaluating P at the other three points gives that Jason should sell 5 copies of Space Trader and 9 copies of Green Dwarf, for a weekly profit of $106. 50

y

3x + y = 30 (0, 10) 2x + y = 20

⎛ 50 80 ⎞ ⎜ , ⎟ ⎝ 9 9 ⎠ x + 5y = 50

x

50

(10, 0)

35. Let x and y represent daily consumption of foods A and B in 100-gram units. We want to minimize C = 8x + 22y subject to the constraints 8 x + 4 y ≥ 176, 16 x + 32 y ≥ 1024, 2 x + 5 y ≥ 200, x ≥ 0, y ≥ 0. ⎛5 ⎞ The feasible region is unbounded with corner points (100, 0), ⎜ , 39⎟ and (0, 44). C has a minimum value at ⎝2 ⎠ (100, 0). Thus the animals should be fed 100 × 100 = 10, 000 grams = 10 kilograms of food A each day. 100

y

8x + 4y = 176 2x + 5y = 200 x 100 20

36.

0 0

50

Z = 0.89 when x = 4.78, y = 9.14

292



10

37.

0 −2

10

Z = 129.83 when x = 9.38, y = 1.63 Mathematical Snapshot Chapter 7 CURATIVE UNITS

TOXIC UNITS

RELATIVE DISCOMFORT

Drug (per ounce)

500

400

1

Radiation (per min)

1000

600

1

≥ 2000

≤1400

1.

Requirement

Let x1 = number of ounces of drug and let x2 = number of minutes of radiation. We want to minimize the discomfort D, where D = x1 + x2 , subject to 500 x1 + 1000 x2 ≥ 2000, 400 x1 + 600 x2 ≤ 1400, where x1 , x2 ≥ 0. 5

x2

400x1 + 600x2 = 1400 (2, 1)

500x1 + 1000x2 = 2000 x1 5

⎛ 7⎞ The corner points are (0, 2), ⎜ 0, ⎟ , and (2, 1). ⎝ 3⎠ At (0, 2), D = 0 + 2 = 2; 7 7 ⎛ 7⎞ at ⎜ 0, ⎟ , D = 0 + = ; 3 3 ⎝ 3⎠ at (2, 1), D = 2 + 1 = 3. Thus D is minimum at (0, 2). The patient should get 0 ounces of drug and 2 minutes of radiation.

293



CURATIVE UNITS

TOXIC UNITS

RELATIVE DISCOMFORT

Drug A (per ounce)

600

500

1

Drug B (per ounce)

500

100

2

Radiation (per min)

1000

1000

1

≥ 3000

≤ 2000

2.

Requirement

Let x1 = number of ounces of drug A, x2 = number of ounces of drug B, and x3 = number of minutes of radiation. We want to minimize the discomfort D, where D = x1 + 2 x2 + x3 , subject to 600 x1 + 500 x2 + 1000 x3 ≥ 3000, 500 x1 + 100 x2 + 1000 x3 ≤ 2000, x1 , x2 , x3 ≥ 0 To minimize D, we maximize –D by considering the artificial objective function W = –D – Mt1 . x1 x2 x3 s1 s2 t1 W ⎡ 600 500 1000 –1 0 1 0 3000 ⎤ ⎢ ⎥ ⎢ 500 100 1000 0 1 0 0 2000 ⎥ ⎢ 1 2 1 0 0 M 1 0⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t1 W t1 ⎡ 600 500 1000 –1 0 1 0 3000 ⎤ 3 ⎢ ⎥ s2 ⎢ 500 100 1000 0 1 0 0 2000 ⎥ 2 W ⎢1 – 600M 2 – 500M 1 –1000 M M 0 0 1 –3000M ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t1 W ⎤ 2.5 t1 ⎡ 100 400 0 −1 1 1 0 1000 ⎢ ⎥ x3 ⎢ 0.5 0.1 1 0 0.001 0 0 2 ⎥ 20 ⎢ W 0.5 –100M 1.9 – 400 M 0 M –0.001 + M 0 1 −2 − 1000 M ⎥ ⎣ ⎦ x1 x2 x3 s1 s2 t1 W x2 ⎡ 0.25 1 0 –0.0025 –0.0025 0.0025 0 2.5⎤ ⎢ ⎥ x3 ⎢ 0.475 0 1 0.00025 0.00125 –0.00025 0 1.75⎥ W ⎢ 0.025 0 0 0.00475 0.00375 –0.00475 + M 1 –6.75⎥ ⎣ ⎦ The minimum value of D is 6.75 when x1 = 0, x2 = 2.5, and x3 = 1.75. The patient should get 0 ounces of drug A, 2.5 ounces of drug B, and 1.75 minutes of radiation.

3. Answers may vary.

294


3. Assembly Line

1.

A Start

Finishing Line D

Production Route AD

E

AE

D

BD

E

BE

D

CD

E

CE

B C

Red Die 1

2

6 possible production routes

2.

BTU's 6000 Start

8000 10,000

Fan Model Speeds Type 1 6000 – 1

Green Die 1

Result 1, 1

2

1, 2

3

1, 3

4

1, 4

5

1, 5

6

1, 6

1

2, 1

2

2, 2

3

2, 3

4

2, 4

5

2, 5

6

2, 6

2

6000 – 2

1

8000 – 1

1

3, 1

2

8000 – 2

2

3, 2

1

10,000 – 1

3

3, 3

2

10,000 – 2

4

3, 4

5

3, 5

6

3, 6

3

6 model types Start

4

5

6

1

4, 1

2

4, 2

3

4, 3

4

4, 4

5

4, 5

6

4, 6

1

5, 1

2

5, 2

3

5, 3

4

5, 4

5

5, 5

6

5, 6

1

6, 1

2

6, 2

3

6, 3

4

6, 4

5

6, 5

6

6, 6

36 possible results

295

Chapter 8: Introduction to Probability and Statistics


4. Toss 1

Toss 2

Toss 3 H

H T H H T T Start H H T T H T T

Toss 4 H

Result H, H, H, H

T

H, H, H, T

H

H, H, T, H

T

H, H, T, T

H

H, T, H, H

T

H, T, H, T

H

H, T, T, H

T

H, T, T, T

H

T, H, H, H

T

T, H, H, T

H

T, H, T, H

T

T, H, T, T

H

T, T, H, H

T

T, T, H, T

H

T, T, T, H

T

T, T, T, T

16 possible results

95! 95! = = 95 (95 − 1)! 94!

13.

6 P6

=

6! 6! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = = = 720 (6 – 6)! 0! 1

14.

9 P4

=

9! 9! = = 9 ⋅ 8 ⋅ 7 ⋅ 6 = 3024 (9 − 4)! 5!

15.

4 P2 ⋅ 5 P3

There are 5 roads from A to B, and 5 roads from B to A. By the basic counting principle, the number of possible routes for a round trip is 5 · 5 = 25. 18.

99 P4

=

99 ⋅ 98 ⋅ 97 ⋅ 96 ⋅ 95 = 95 99 ⋅ 98 ⋅ 97 ⋅ 96

1000! 1000 ⋅ 999! = = 1000 999! 999! For most calculators, attempting to evaluate 1000! results in an error message (because of 999! the magnitude of the numbers involved). n Pr

n!

=

n! ( n – r )!

n!

=

1 (n – r )!

20. The number of ways to arrange 6 teams in an order is 6 P6 = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 720 . 21. The number of ways of selecting 3 of 8 contestants in an order is 8 P3 = 8 ⋅ 7 ⋅ 6 = 336 .

8. For each of the 6 questions, there are 4 choices. By the basic counting principle, the number of ways to answer the questions is

22. Six out of eight items in column 2 must be selected in an order. Thus the number of ways the matching can be done is 8 P6 = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 20,160 .

4 · 4 · 4 · 4 · 4 · 4 = 46 = 4096 . 9. For each of the 10 questions, there are 2 choices. By the basic counting principle, the number of ways to answer the examination is 2 · 2 · ... · 2 = 210 = 1024 .

23. On each roll of a die, there are 6 possible outcomes. By the basic counting principle, on 4 rolls the number of possible results is 6 ⋅ 6 ⋅ 6 ⋅ 6 = 64 = 1296.

10. Since there are 26 letters, there are 26 choices for the first, third and fifth symbols. There are 10 possible digits (0 through 9) for the second, fourth, and sixth symbols. By the basic counting principle, the number of codes is 26 · 10 · 26 · 10 · 26 · 10 = 17,576,000. =

99 P5

= (4 ⋅ 3)(5 ⋅ 4 ⋅ 3) = (12)(60) = 720

19. A name for the firm is an ordered arrangement of the three last names. Thus the number of possible firm names is 3 P3 = 3! = 3 ⋅ 2 ⋅1 = 6 .

7. There are 2 appetizers, 4 entrees, 4 desserts, and 3 beverages. By the basic counting principle, the number of possible complete dinners is 2 · 4 · 4 · 3 = 96.

6 P3

=

17.

b. There are 5 possible roads from A to B. Since a different road is to be used for the return trip, there are only 4 possible roads from B to A. By the basic counting principle, the number of possible round-trip routes is 5 · 4 = 20.

11.

95 P1

16.

5. There are 5 science courses and 4 humanities courses. By the basic counting principle, the number of selections is 5 · 4 = 20. 6. a.

12.

24. On each toss there are 2 possible outcomes. By the basic counting principle, the number of possible results on 8 tosses is 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 28 = 256 .

25. The number of ways of selecting 3 of the 12 students in an order is 12 P3 = 12 ⋅11 ⋅10 = 1320 .

6! 6! = = 6 ⋅ 5 ⋅ 4 = 120 (6 − 3)! 3! 296


Section 8.1

26. Three of the 26 letters must be selected (without repetition) in an order. Thus the number of possible lock combinations is 26 P3 = 15, 600 .

35. The number of ways the waitress can place five of the five different sandwiches (and order is important) is 5 P5 = 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 120.

27. The number of ways a student can choose 4 of the 6 items in an order is 6 P4 = 6 ⋅ 5 ⋅ 4 ⋅ 3 = 360.

36. Because order is important, the number of ways that the 5 people can line up is 5 P5 = 5! = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 120 . If a woman is to be at each end, then the number of ways to place one of the two women on the left side is 2 P1 . Once a woman is chosen for the left side, the other woman must be on the right side. The number of ways to line the three men in the middle is 3 P3 . By the basic counting principle, the number of line ups is 2 P1 ⋅ 3 P3 = (2)(3 ⋅ 2 ⋅1) = 12 .

28. On the second roll, there are 2 possible outcomes (a 1 or a 2). For each of the other two rolls, there are 6 possible outcomes. By the basic counting principle, the number of possible results for the three rolls is 6 · 2 · 6 = 72. 29. The number of ways to select six of the six different letters in the word MEADOW in an order is 6 P6 = 6! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 720 .

37. a.

30. The number of ways to select four of the four different letters in the word DISC in an order is 4 P4 = 4! = 4 ⋅ 3 ⋅ 2 ⋅1 = 24 . 31. For an arrangement of books, order is important. The number of ways to arrange 5 of 7 books is 7 P5 = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 2520 . All 7 books can be arranged in 7 P7 = 7! = 5040 ways. 32. a.

To fill the four offices by different people, 4 of 12 members must be selected, and order is important. This can be done in 12 P4 = 12 ⋅11 ⋅10 ⋅ 9 = 11,880 ways.

b. If the president and vice president must be different members, then there are 12 choices for president, 11 for vice president, 12 for secretary, and 12 for treasurer. By the basic counting principle, the offices can be filled in 12 · 11 · 12 · 12 = 19,008 ways.

A student can enter by any of 5 doors. After a door is chosen, the student can exit by any of the 4 remaining doors. By the basic counting principle, the number of ways to enter by one door and exit by a different door is 5 · 4 = 20.

38. a.

There are 24 possibilities for each of the three letters in a name. By the basic counting principle, the number of names is 24 · 24 · 24 = 243 = 13,824 .

b. There are 5 doors by which to enter and 5 doors by which to exit. By the basic counting principle, the total number of ways to enter and exit is 5 · 5 = 25.

b. Since the order of letters is important and no letter is used more than one time, the number of names is 24 P3 = 24 ⋅ 23 ⋅ 22 = 12,144 .

33. After a “four of a kind” hand is dealt, the cards can be arranged so that the first four have the same face value, and order is not important, There are 13 possibilities for the first four cards (all 2’s, all 3’s, ..., all aces). The fifth card can be any one of the 48 cards that remain. By the basic counting principle, the number of “four of a kind” hands is 13 · 48 = 624.

39. There are 2 choices for the center position. After that choice is made, to fill the remaining four positions (and order is important), there are 4 P4 ways. By the basic counting principle, to assign positions to the five-member team there are 2 ⋅ 4 P4 = 2(4!) = 2(24) = 48 ways. 40. For the first letter there are two possibilities. For the second and third letters there are 26 possibilities, and for the last letter there are 25 possibilities. By the basic counting principle, the number of possible identifications is 2 · 26 · 26 · 25 = 33,800.

34. Five colors are available, and two are selected so that order is important. Thus the number of ways of placing an order is 5 P2 = 5 ⋅ 4 = 20.

297



41. There are 3 P3 ways to select the first three batters (order is important) and there are 6 P6 ways to select the remaining batters. By the basic counting principle, the number of possible batting orders is 3 P3 ⋅ 6 P6 = 3! ⋅ 6! = 6 ⋅ 720 = 4320. 42. a.

6.

= (4 ⋅ 3)

7.

Four of four flags can be arranged (order is important) in 4 P4 = 4! = 24 ways. Thus 24 different signals are possible.

=

6! 6! 6 ⋅ 5 ⋅ 4! 6 ⋅ 5 = = = = 15 4!(6 – 4)! 4! ⋅ 2! 4!(2 ⋅1) 2 ⋅1

2.

6 C2

=

6! 6! 6 ⋅ 5 ⋅ 4! 6 ⋅ 5 = = = = 15 2!(6 – 2)! 2! ⋅ 4! (2 ⋅1)4! 2 ⋅1

3.

100 C100

4.

1001 C1

5.

=

8.

5! 3!(5 – 3)!

5 ⋅ 4 ⋅ 3! = (12) ⋅10 = 120 3!⋅ 2!

n! r !(n – r )! =

n! n! = ⋅ (n – r )![n – (n – r )]! (n – r )!r !

n Cn

=

n! 1 1 = = =1. n !(n – n)! 0! 1

9. The number of ways of selecting 4 of 17 people so that order is not important is 17! 17! = 17 C4 = 4!(17 – 4)! 4! ⋅13! =

17 ⋅16 ⋅15 ⋅14 ⋅13! = 2380 4 ⋅ 3 ⋅ 2 ⋅1(13!)

10. If horses A, B, and C finish in the money, then it does not matter if A finishes in first, second, or third place. Similarly for B and C. Thus order is not important. The number of ways in which 3 of 8 horses finish in the money is the number of ways of selecting 3 of the 8 without regard to order, namely 8 ⋅ 7 ⋅ 6 ⋅ 5! 8! 8! = = 56 . = 8 C3 = 3!(8 – 3)! 3! ⋅ 5! 3 ⋅ 2 ⋅1 ⋅ 5!

100! 1 1 = = =1 100!(100 – 100)! 0! 1

11. The number of ways of selecting 9 out of 13 questions (without regard to order) is 13! 13! 13 ⋅12 ⋅11 ⋅10 ⋅ 9! = = 13 C9 = 9!(13 − 9)! 9! ⋅ 4! 9!⋅ 4 ⋅ 3 ⋅ 2 ⋅1 = 715.

1001! 1001! = 1!(1001 − 1)! 1!⋅1000! 1001 ⋅1000! = = 1001 1000! =

5 P3 ⋅ 4 C2

=

= (4 ⋅ 3)

Thus n Cr = n Cn – r .

Problems 8.2 6 C4

n Cr

n Cn – r

b. If only one of the four flags is used, there are 4 P1 possible signals. If exactly two flags are used, there are 4 P2 possible signals. Similarly, for exactly three and exactly four flags, there are 3 P4 and 4 P4 possible signals, respectively. Thus if at least one flag is used, the number of possible signals is 4 P1 + 4 P2 + 4 P3 + 4 P4 =4+4·3+4·3·2+4·3·2·1 = 4 + 12 + 24 + 24 = 64.

1.

4 P2 ⋅ 5C3

12. In a deck of 52 cards, 26 of the cards are red. In a four-card hand, the order is not important. Thus, the number of four-card hands from the 26 red cards is 26! 26 C4 = 4!(26 − 4)! 26 ⋅ 25 ⋅ 24 ⋅ 23 ⋅ 22! = 4!22! = 14,950

4! 2!(4 − 2)! 4 ⋅ 3 ⋅ 2! = 5⋅ 4⋅3 2!2! = 60 ⋅ 6 = 360

= 5⋅ 4⋅3

298


Section 8.2

13. The order of selecting 10 of the 74 dresses is of no concern. Thus the number of possible 74! 74! . = samples 74 C10 = 10! ⋅ (74 – 10)! 10! ⋅ 64!

18. The word STREETSBORO has 11 letters with repetition: two S’s, two T’s, two R’s, two E’s, one B, and two O’s. Thus the number of distinguishable arrangements is 11! 11! = = 1, 247, 400. 2! ⋅ 2! ⋅ 2! ⋅ 2! ⋅1! ⋅ 2! 32

14. This situation can be considered as a two-stage process. In the first stage, one of the 3 boxes is selected. In the second stage, 4 of the 7 types of jelly are selected (and order is not important), which can be done in 7 C4 ways. By the basic counting principle, the number of different gift boxes that are possible is 7! 7! 3 ⋅ 7 C4 = 3 ⋅ = 3⋅ 4!(7 – 4)! 4! ⋅ 3! = 3⋅

19. The number of ways 4 heads and 3 tails can occur in 7 tosses of a coin is the same as the number of distinguishable permutations in the “word” HHHHTTT, which is 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! = = 35 . 4! ⋅ 3! 4!(6) 20. The number of ways for the given outcome to occur is the number of distinguishable permutations of six numbers such that two are 2’s, three are 3’s, and one is 4, which is 6! 6 ⋅ 5 ⋅ 4 ⋅ 3! = = 60 . 2! ⋅ 3! ⋅1! (2)3!

7 ⋅ 6 ⋅ 5 ⋅ 4! = 3 ⋅ 35 = 105 . 4!(3 ⋅ 2 ⋅1)

15. To score 80, 90, or 100, exactly 8, 9, or 10 questions must be correct, respectively. The number of ways in which 8 of 10 questions can be correct is 10! 10! 10 ⋅ 9 ⋅ 8! = = = 45 . 10 C8 = 8!(10 – 8)! 8! ⋅ 2! 8! ⋅ 2 ⋅1 For 9 of 10 questions, the number of ways is 10! 10! 10 ⋅ 9! = = = 10 , 10 C9 = 9!(10 – 9)! 9! ⋅1! 9! ⋅1 and for 10 of 10 questions, it is 10! 10! = =1. 10 C10 = 10!(10 – 10)! 10! ⋅ 0! Thus the number of ways to score 80 or better is 45 + 10 + 1 = 56.

21. Since the order in which the calls are made is important, the number of possible schedules for the 6 calls is 6 P6 = 6! = 720 . 22. The number of ways to place the 12 members in three specific cars (cells), with 4 members in 12! = 34, 650 . each car, is 4! ⋅ 4! ⋅ 4! 23. The number of ways to assign 9 scientists so 3 work on project A, 3 work on B, and 3 work on 9! = 1680 . C is 3!3!3!

16. Each of the 11 games can be assigned to one of three cells: a win cell, a loss cell, or a tie cell. The number of ways to have 4 wins, 5 losses, and 3 ties is 11! 11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 6930. 4! ⋅ 5! ⋅ 2! 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 5!⋅ 2 ⋅1

24. There are 9 holly bushes, 5 of which are female, and 4 of which are male. Then the number of possible distinguishable arrangements is 9! 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 126. 5! ⋅ 4! 5! ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

17. The word MISSISSAUGA has 11 letters with repetition: one M, two I’s, four S’s, two A’s, one U, and one G. Thus the number of distinguishable arrangements is 11! 11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! = 1! ⋅ 2! ⋅ 4! ⋅ 2! ⋅1! ⋅1! (2)4!(2) = 415,800.

25. A response to the true-false questions can be considered an ordered arrangement of 10 letters, 5 of which are T’s and 5 of which are F’s. The number of different responses is 10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 252 . 5!⋅ 5! 5!(5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1)

299



26. The order in which the 7 food items are placed is important. However, there are 3 hamburgers (type 1), 2 cheeseburgers (type 2), and 2 steak sandwiches (type 3). Then the number of possible distinguishable ways of placing the 7! = 210 . items is 3! ⋅ 2! ⋅ 2!

12 C8 ⋅ 7 C4

12! 7! ⋅ 8!(12 – 8)! 4!(7 – 4)!

12! 7! 12 ⋅11 ⋅10 ⋅ 9 ⋅ 8! 7 ⋅ 6 ⋅ 5 ⋅ 4! ⋅ = ⋅ 8! ⋅ 4! 4! ⋅ 3! 8! ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 4! ⋅ 3 ⋅ 2 ⋅1 = 495 ⋅ 35 = 17,325 .

=

32. Suppose the possible games are numbered 1, 2, 3, ..., 7. The order in which four games are won is not important. The number of ways that 4 of the possible 7 games can be won is 7! 7! = = 35 . 7 C4 = 4!(7 – 4)! 4! ⋅ 3!

27. The number of ways to assign 15 clients to 3 caseworkers (cells) with 5 clients to each 15! = 756, 756. caseworker is 5! ⋅ 5! ⋅ 5! 28. The number of ways of selecting 5 of the 10 remaining members so that order is important is 10! 10! 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = = 10 P5 = (10 − 5)! 5! 5! = 30, 240. 29. a.

=

33. a.

Selecting 3 of the 3 males can be done in only 1 way.

b. Selecting 4 of the 4 females can be done in only 1 way.

Seven flags must be arranged: two are red (type 1), three are green (type 2), and two are yellow (type 3). Thus the number of distinguishable arrangements (messages) is 7! = 210. 2! ⋅ 3! ⋅ 2!

c.

b. If exactly two yellow flags are used, then seven flags are involved and the number of 7! = 210. If all different messages is 2! ⋅ 3! ⋅ 2! three yellow flags are used, then eight flags are involved and the number of different 8! = 560. Thus if at least messages is 2! ⋅ 3! ⋅ 3! two yellow flags are used, the number of different messages is 210 + 560 = 770.

Selecting 2 males and 2 females can be considered as a two-stage process. In the first stage, 2 of the 3 males are selected (and order is not important), which can be done in 3 C2 ways. In the second stage, 2 of the 4 females are selected, which can be done in 4 C2 ways. By the basic counting principle, the ways of selecting the subcommittee is 3! 4! ⋅ 3 C2 ⋅ 4 C2 = 2!(3 – 2)! 2!(4 – 2)! =

3! 4! ⋅ = 3 ⋅ 6 = 18 2! ⋅1! 2! ⋅ 2!

34. Exactly 2, 3, or 4 females can serve on the subcommittee. Following the procedure in Problem 33(c), the number of ways exactly 2 females can serve is 4 C2 ⋅ 4C2 . The number of ways exactly 3 females can serve is 4 C3 ⋅ 4 C1 . The number of ways exactly four females can serve is 1. Thus the number of ways that at least 2 females can serve on the subcommittee is 4 C2 ⋅ 4 C2 + 4 C3 ⋅ 4 C1 + 1

30. Of the 10 applicants, 4 will be hired for the assembly department (cell 1), 2 for the shipping department (cell 2), and 4 will not be hired (cell 3). Thus the number of ways to fill the 10! = 3150 . positions is 4! ⋅ 2! ⋅ 4!

4! 4! 4! 4! ⋅ + ⋅ +1 2! ⋅ 2! 2! ⋅ 2! 3! ⋅1! 1! ⋅ 3! = 6 · 6 + 4 · 4 + 1 = 36 + 16 + 1 = 53. =

31. The order in which the securities go into the portfolio is not important. The number of ways to select 8 of 12 stocks is 12 C8 . The number of ways to select 4 of 7 bonds is 7 C4 . By the basic counting principle, the number of ways to create the portfolio is

35. There are 4 cards of a given denomination and the number of ways of selecting 3 cards of that denomination is 4 C3 . Since there are 13 denominations, the number of ways of selecting 3 cards of one denomination is 300


Section 8.3

13 ⋅ 4C3 . After that selection is made, the 2 other cards must be of the same denomination (of which 12 denominations remain). Thus for the remaining 2 cards there are 12 ⋅ 4 C2 selections. By the basic counting principle, the number of possible full-house hands is 4! 4! 13 ⋅ 4C3 ⋅12 ⋅ 4C2 = 13 ⋅ ⋅12 ⋅ 3! ⋅1! 2! ⋅ 2! = 13 · 4 · 12 · 6 = 3744.

Principles in Practice 8.3 1. This is a combination problem because the order in which the videos are selected is not important. The number of possible choices is the number of ways 3 videos can be selected from 400 without regard to order. 400! 400! = 400 C3 = 3!(400 – 3)! 3!397! 400 ⋅ 399 ⋅ 398 ⋅ 397! 3!397! 400 ⋅ 399 ⋅ 398 = 3⋅ 2 = 10,586,800 =

36. There are 13 denominations and four cards of each denomination. The number of ways to get a pair of 8’s is 4 C2 . For the other pair, there are 12 denominations left to choose from, so 12 C1 possibilities, with 4 C2 ways to get such a pair. For the last card there are 11 denominations left, with 4 cards in each denomination. By the basic counting principle the number of two-pair hands where one pair is 8’s is 4 C2 ⋅ 12 C1 ⋅ 4 C2 ⋅11 ⋅ 4 4! 12! 4! = ⋅ ⋅ ⋅11 ⋅ 4 2! ⋅ 2! 1! ⋅11! 2! ⋅ 2! = 19, 008.

Problems 8.3 1. {9D, 9H, 9C, 9S] 2. {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT} 3. {1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T}

37. This situation can be considered as placing 18 tourists into 3 cells: 7 tourist go to the 7-passenger tram, 8 go to the 8-passenger tram, and 3 tourists remain at the bottom of the mountain. This can be done in 18! = 5,250,960 ways. 7! ⋅ 8! ⋅ 3!

4. {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

38. a.

7. a.

5. [64, 69, 60, 61, 46, 49, 40, 41, 96, 94, 90, 91, 06, 04, 09, 01, 16, 14, 19, 10] 6. {BBBB, BBBG, BBGB, BBGG, BGBB, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG}

The 10 students are to be placed in 3 groups, with 4 in group A, 3 in group B, and 3 in group C. This can be done in 10! = 4200 ways. 4! ⋅ 3! ⋅ 3!

{RR, RW, RB, WR, WW, WB, BR, BW, BB};

b. {RW, RB, WR, WB, BR, BW} 8. {ADF, ADG, AEF, AEG, BDF, BDG, BEF, BEG, CDF, CDG, CEF, CEG}

b. For a given assignment of students to the three groups, the number of ways of selecting a group leader and a secretary for group A (order is important) is 4 P2 ; for group B, it is 3 P2 ; and for group C it is 3 P2 . Thus the number of ways that the instructor can split the class into 3 groups and designate a group leader and secretary in each group is 4200 ⋅ 4 P2 ⋅ 3 P2 ⋅ 3 P2 = 4200(4 ⋅ 3)(3 ⋅ 2)(3 ⋅ 2) = 1,814,400.

9. Sample space consists of ordered sets of six elements and each element is H or T. Since there are two possibilities for each toss (H or T), and there are six tosses, by he basic counting principle, the number of sample points is 2 · 2 · 2 · 2 · 2 · 2 = 26 = 64 . 10. Sample space consists of ordered sets of five elements where each element is an integer between 1 and 6 inclusive. Since there are six possibilities for each die, and there are 5 dice, by

301


ISM: Introductory Mathematical Analysis 22. ( E ∪ G ) ∩ F ′ = {1, 2,3, 4,5, 6,8} ∩ {1, 2, 4, 6,8,10} = {1, 2, 4, 6, 8}

the basic counting principle, the number of sample points is 6 · 6 · 6 · 6 · 6 = 6 = 7776 . 5

11. Sample space consists of ordered pairs where the first element indicates the card drawn (52 possibilities) and the second element indicates the number on the die (6 possibilities). By the basic counting principle, the number of sample points is 52 · 6 = 312.

23. E1 ∩ E2 ≠ ∅ ; E1 ∩ E3 ≠ ∅ ; E1 ∩ E4 = ∅ ; E2 ∩ E3 = ∅ ; E2 ∩ E4 ≠ ∅ ; E3 ∩ E4 = ∅ . Thus E1 and E4 , E2 and E3 , and E3 and E4 are mutually exclusive.

12. Sample space consists of ordered sets of four elements where the elements and their position indicate the rabbit selected on the respective draw. Since the rabbits are not replaced, for the first draw there are 9 possibilities, for the second draw there are 8 possibilities, and for the third and fourth there are 7 and 6 possibilities, respectively. By the basic counting principle, the number of sample points is 9 · 8 · 7 · 6 = 3024.

24. If both cards are jacks, then both cards can neither be clubs nor 3’s. Thus E J ∩ EC = ∅ and E J ∩ E3 = ∅. If both cards are clubs, then both cards cannot be 3’s. Thus EC ∩ E3 = ∅. E J and EC , E J and E3 , EC and E3 are mutually exclusive. 25. E ∩ F ≠ ∅ , E ∩ G = ∅ , E ∩ H ≠ ∅, E ∩ I ≠ ∅ , F ∩G ≠ ∅ , F ∩ H ≠ ∅ F ∩ I = ∅, G ∩ H = ∅ , G ∩ I = ∅, H ∩ I ≠ ∅. Thus E and G, F and I, G and H, and G and I are mutually exclusive.

13. Sample space consists of combinations of 52 cards taken 10 at a time. Thus the number of sample points is 52 C10 . 14. Sample space consists of all four letter “words.” For each of the four letters there are 26 possibilities. By the basic counting principle, the number of sample points is

26. E ∩ F = ∅ , E ∩ G = ∅ , E ∩ H ≠ ∅ , E ∩ I ≠ ∅ , F ∩G ≠ ∅ , F ∩ H ≠ ∅ , F ∩I =∅, G∩H =∅, G∩I =∅, H ∩I =∅. Thus E and F, E and G, F and I, G and H, G and I, H and I are mutually exclusive.

26 · 26 · 26 · 26 = 264 = 456,976 . 15. The sample points that are either in E, or in F, or in both E and F are 1, 3, 5, 7, and 9. Thus E ∪ F = {1, 3, 5, 7, 9}.

27. a.

16. The sample points in S that are not in G are 1, 3, 5, 7, 9, and 10. Thus G ′= {1, 3, 5, 7, 9, 10}.

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

b.

E1 = {HHH, HHT, HTH, HTT, THH, THT, TTH}

c.

E2 = {HHT, HTH, HTT, THH, THT, TTH, TTT}

d.

18. F ′ = {1, 2, 4, 6, 8, 10} and G ′ = {1, 3, 5, 7, 9, 10}, so F ′ ∩ G ′ = {1, 10}.

E1 ∪ E2 = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} = S

e.

E1 ∩ E2 = {HHT, HTH, HTT, THH, THT, TTH}

19. The sample points in S that are not in F are 1, 2, 4, 6, 8, and 10. Thus F′ = {1, 2, 4, 6, 8, 10}.

f.

( E1 ∪ E2 )′ = S ′ = ∅

20. ( E ∪ F )′ = {1,3,5, 7,9}′ = {2, 4, 6,8,10}

g.

( E1 ∩ E2 )′ = {HHT, HTH, HTT, THH,

17. The sample points in S that are not in E are 2, 4, 6, 7, 8, 9, and 10. Thus E ′ = {2, 4, 6, 7,8,9,10}. The sample points common to both E ′ and F are 7 and 9. Thus E ′ ∩ F = {7, 9}.

THT, TTH}′ = {HHH, TTT}

21. ( F ∩ G )′ = ∅′ = S

28. a. 302

{BB, BG, GB, GG}


Section 8.4

b. {BG, GB, GG} c.

Problems 8.4

{BB, BG, GB}

1. 3000P(E) = 3000(0.25) = 750

d. No; {BG, GB, GG}′ = {BB} ≠ event in (c) 29. a.

2. 3000P(E) = 3000[1 – P(E′)] = 3000(1 – 0.45) = 3000(0.55) = 1650

{ABC, ACB, BAC, BCA, CAB, CBA}

3. a.

b. {ABC, ACB} c. 30. a.

b.

{BAC, BCA, CAB, CBA} {UUV, UUW, UUX, UUZ, UVV, UVW, UVX, UVZ, UXV, UXW, UXX, UXZ, UYV, UYW, UYX, UYZ, VUV, VUW, VUX, VUZ, VVV, VVW, VVX, VVZ, VXV, VXW, VXX, VXZ, VYV, VYW, VYX, VYZ, WUV, WUW, WUX, WUZ, WVV, WVW, WVX, WVZ, WXV, WXW, WXX, WXZ, WYV, WYW, WYX, WYZ}

4. a. b.

P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ) = 0.2 + 0.3 – 0.1 = 0.4 1 3 = 4 4

P(E´) = 1 – P(E) = 1 –

P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ) =

1 1 1 5 + – = 4 2 8 8

5. If E and F are mutually exclusive, then E∩F =∅ . Thus P ( E ∩ F ) = P (∅) = 0 . Since it is given that P ( E ∩ F ) = 0.831 ≠ 0 , E and F are not mutually exclusive.

b. {VVV} c.

P(E′) = 1 – P(E) = 1 – 0.2 = 0.8

{UUV, UUW, UUX, UUZ, UVV, UVW, UVX, UVZ, UXV, UXW, UXX, UXZ, UYV, UYW, UYX, UYZ, VUV, VUW, VUX, VUZ, VVW, VVX, VVZ, VXV, VXW, VXX, VXZ, VYV, VYW, VYX, VYZ, WUV, WUW, WUX, WUZ, WVV, WVW, WVX, WVZ, WXV, WXW, WXX, WXZ, WYV, WYW, WYX, WYZ} More than one supplier is used.

6. P ( E ∪ F ) = P ( E ) + P( F ) − P( E ∩ F ) Thus P ( F ) = P ( E ∪ F ) + P( E ∩ F ) − P( E ) 13 1 1 1 = + − = . 20 10 2 4

31. Using the properties in Table 8.1, we have ( E ∩ F ) ∩ ( E ∩ F ′) = ( E ∩ F ∩ E ) ∩ F ′ [property 15] = ( E ∩ E ∩ F ) ∩ F ′ [property 11] = ( E ∩ E ) ∩ ( F ∩ F ′) [porperty 15] [property 5] = E ∩∅ [property 9] =∅ Thus ( E ∩ F ) ∩ ( E ∩ F ′) = ∅, so E ∩ F and E ∩ F ′ are mutually exclusive.

7. a.

E8 = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} P ( E8 ) =

b.

n( E8 ) 5 = n( S ) 36

E2 or 3 = {(1, 1), (1, 2), (2, 1)} P ( E2 or 3 ) =

c.

32. Using the properties in Table 8.1, we have ( E ∩ F ) ∪ ( E ∩ F ′) [property 16] = E ∩ ( F ∪ F ′) [property 4] = E∩S [property 7] =E

n( E2 or 3 ) 3 1 = = 36 12 n( S )

E3, 4, or 5 = {(1, 2), (2,1), (1,3), (2, 2), (3,1), (1, 4), (2,3), (3, 2), (4,1)}

P ( E3, 4, or 5 ) =

d.

n( E3, 4,

or 5 )

n( S )

=

9 1 = 36 4

E12 or 13 = E12 , since E13 is an impossible

event. E12 = {(6, 6)} P ( E12 or 13 ) =

303

n( E12 or 13 ) 1 = n( S ) 36

Chapter 8: Introduction to Probability and Statistics e.


E2 = {(1,1)}

e.

E4 = {(1,3), (2, 2), (3,1)} E6 = {(1,5), (2, 4), (3,3, ), (4, 2), (5,1)} E8 = {(2, 6), (3,5), (4, 4), (5,3), (6, 2)} E10 = {(4, 6), (5,5), (6, 4)} E12 = {(6, 6)}

Because a heart is not a club, Eheart ∩ Eclub = ∅ . Thus P ( Eheart or club ) = P( Eheart ∪ Eclub ) = P ( Eheart ) + P ( Eclub ) =

n( Eheart ) n( Eclub ) 13 13 + = + n( S ) n( S ) 52 52

=

26 1 = 52 2

P ( Eeven ) = P( E2 ) + P ( E4 )

+ P ( E6 ) + P ( E8 ) + P ( E10 ) + P ( E12 ) =

1 3 5 5 3 1 18 1 + + + + + = = 36 36 36 36 36 36 36 2

f.

P ( Eclub and 4 ) =

f.

1 1 P ( Eodd ) = 1 – P ( Eeven ) = 1 – = 2 2

g.

′ than 10 = E10 ∪ E11 ∪ E12 Eless = {(4, 6), (5,5), (6, 4)} ∪ {(5, 6), (6,5)} ∪ {(6, 6)} = {(4, 6), (5,5), (6, 4), (5, 6), (6,5), (6, 6)} . ′ than 10 ) P ( Eless than 10 ) = 1 – P ( Eless =1–

6 30 5 = = . 36 36 6

g.

P(club or 4) = P(club) + P(4) – P(club and 4) 13 4 1 16 4 = + – = = 52 52 52 52 13

h.

Ered and king = {KH, KD}

=

i.

b. P(diamond) =

n( S )

=

1 52

n( E jack )

=

b.

d

n( Ered ) 26 1 P(red) = = = n( S ) 52 2

n( S )

Espade and heart = ∅

EH,5 = {H5}

c.

304

n( S )

=

1 12

n( Ehead ) 6 1 = = n( S ) 12 2

n( E3 ) = 2 ⋅1 = 2

P(3) = d.

n( EH,5 )

n( Ehead ) = 1 ⋅ 6 = 6 .

P(head) =

4 1 = 52 13

P(jack) =

2 1 = 52 26

P(head and 5) =

n( E diamond ) 13 1 = = n( S ) 52 4

c.

n( S )

10. n(S) = 2 · 6 = 12 a.

n( Eking of hearts )

n(Ered and king )

Thus P(spade and heart) = 0

9. n(S) = 52. P(king of hearts) =

n( Eclub and 4 ) 1 = n( S ) 52

P (red and king)=

8. E2 or 3 shows = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 2), (4, 2), (5, 2), (6, 2), (1, 3), (4, 3), (5, 3), (6, 3)} n( E2 or 3 shows ) 20 5 = = P ( E2 or 3 shows ) = n( S ) 36 9

a.

Eclub and 4 = {4C}

n( E3 ) 2 1 = =− n( S ) 12 6

n( Ehead and even ) = 1 ⋅ 3 = 3 P(head and even) n( Ehead and even ) 3 1 = = = n( S ) 12 4


Section 8.4

11. n(S) = 2 · 6 · 52 = 624 a.

14. n(S) = 52 · 52 = 2704

P(tail, 3, queen of hearts) n( ET,3,QH ) 1 ⋅1 ⋅1 1 = = = n( S ) 624 624

a.

=

b. P(tail, 3, queen) n( ET,3,Q ) 1 ⋅1 ⋅ 4 1 = = = n( S ) 624 156 c.

P(head, 2 or 3, queen) n( EH,2 or 3,Q ) 1 ⋅ 2 ⋅ 4 1 = = = n( S ) 624 78

b.

a.

E3 heads = {HHH} n( E3 heads ) 1 = n(S ) 8

d.

E1 tail = {HHT,HTH,THH} .

b.

P(no more than 2 heads) = 1 – P(3 heads) 1 7 = 1– = 8 8

c.

b.

P (all hearts) =

n( E3 girls ) n( S )

=

1 8

E1 boy = {BGG,GBG,GGB} n( E1 boy )

=

3 8

n( Eno girl )

=

n( S )

Eno girl = {BBB}

P(no girl) =

E no more than 1 tail = E0 tails ∪ E1 tail = {HHH} ∪ {HHT,HTH,THH} = {HHH, HHT, HTH, THH}. P(no more than 1 tail) n( Eno more than 1 tail ) 4 1 = = = 8 2 n( S )

P (all kings) =

E3 girls = {GGG}

P(1 boy) =

n( S )

1 8

d. P(at least 1 girl) = 1 – P(no girl) 1 7 =1– = 8 8 16. The sample space consists of 18 jelly beans. Thus n(S) = 18.

13. n(S) = 52 · 51 · 50 = 132,600 a.

1 169

P(3 girls) =

n( E1 tail ) 3 P(1 tail) = = n( S ) 8

c.

4⋅4 2704

15. n(S) = 2 · 2 · 2 = 8

12. n(S) = 8

P(3 heads) =

n( S )

=

b. Number of ways both cards are king of hearts: 1. Number of ways either first card is king of hearts and second card is a different heart, or vice versa: 2(1 · 12) = 24. Number of ways either first card is king of diamonds, clubs, or spades, and second card is a heart, or vice versa: 2(3 · 13) = 78. Thus, number ways one card is a king and the other is a heart is 1 + 24 + 78 = 103, so probability of 103 . given event is 2704

d. P(head, even, diamond) n( EH,E,D ) 1 ⋅ 3 ⋅13 1 = = = n( S ) 624 16

a.

n( Eboth kings )

P(both kings) =

a.

4⋅3⋅ 2 1 = 132, 600 5525

P(blue) =

n( Eblue ) 8 4 = = n( S ) 18 9

b. P(not red) = 1 – P(red) n( Ered ) 7 11 =1– =1– = n( S ) 18 18

13 ⋅12 ⋅11 11 = 132, 600 850

305

Chapter 8: Introduction to Probability and Statistics c.


The events of drawing a red jelly bean and drawing a white jelly bean are mutually exclusive. Thus P(red or white) = P(red) + P(white) 7 3 10 5 + = = = 18 18 18 9

d. P(neither red nor blue) = P(white) = e.

Eyellow = ∅ . Thus P(yellow) = 0

f.

Ered ∩ Eyellow = ∅

d. P(no F) = 1 – P(F) = 1 –

3 1 = 18 6

=1–

e.

Thus P(red or yellow) = P(red) + P(yellow) 7 7 . +0 = = 18 18 17. The sample space consists of 60 stocks. Thus n(S) = 60. a.

P(6% or more) = =

n( E6% or more ) n( S )

48 4 = 60 5

b. P(less than 6%) = 1 – P(6% or more) 4 1 =1– = 5 5 18. Let N = number of ties. Then the number of pure silk ties is 0.4N. a. b.

P (100% pure silk) =

0.4 N = 0.4 N

a.

P (not 100% silk) = 1 − P(100% pure silk) = 1 − 0.4 = 0.6

P(A) =

=

2 38 = = 0.95 40 40

Let N = number of students. Then n(S) = N. Of the N students, 0.10N received an A, 0.25N a B, 0.35N a C, 0.25N a D, 0.05N an F. 0.10 N = 0.1 P(A) = N 0.10 N + 0.25 N P(A or B) = N 0.35 N = = 0.35 N P(neither D nor F) = P(A, B, or C) 0.10 N + 0.25 N + 0.35 N = N 0.70 N = = 0.7 N P(no F) = 1 – P(F) 0.05 N = 1 – 0.05 = 0.95 =1– N

P(both red) =

n( ER,R ) n( S )

=

3⋅ 4 4 = 45 15

b. P(one red and other green) n( ER,G ) + n( EG , R ) 3 ⋅ 5 + 2 ⋅ 4 = = n( S ) 45 =

n( EA ) 4 1 = = = 0.1 n( S ) 40 10

b. P(A or B) =

n ( EF ) n( S )

20. Bag 1 contains 5 jelly beans, and Bag 2 contains 9. n(S) = 5 · 9 = 45.

19. n(S) = 40 Of the 40 students, 4 received an A, 10 a B, 14 a C, 10 a D, and 2 an F. a.

P(neither D nor F) = P(A, B, or C) n( EA,B, or C ) 4 + 10 + 14 28 = = = 0.7 = n( S ) 40 40

n( EA or B ) 4 + 10 = n( S ) 40

14 = 0.35 40

306

15 + 8 23 = 45 45


Section 8.4

21. The sample space consists of combinations of 2 people selected from 5. Thus 5! 5⋅4 = = 10 . Because there n(S) = 5 C2 = 2! ⋅ 3! 2 are only 2 women in the group, the number of possible 2-woman committees is 1. Thus n( E2 women ) 1 = . P(2 women) = n( S ) 10

4

22. Because there are 3 men and 2 women, the number of possible committees consisting of a man and a woman is 3 · 2 = 6. Thus n( Eman and woman ) . P(man and woman) = n( S ) =

34 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4! 34 2 ⋅ 7 ⋅ 5 ⋅ ⋅ = 1 48 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 4! 48 2835 . = 32, 768

=

6 3 = . 10 5

25. A poker hand is a 5-card deal from 52 cards. Thus n( S ) = 52 C5 . In 52 cards, there are 4 cards of a particular denomination. Thus, for a four of a kind, the number of ways of selecting 4 of 4 cards of a particular denomination is 4 C4 . Since there are 13 denominations, 4 cards of the same denomination can be dealt in 13 ⋅ 4C4 ways. For the remaining card, there are 12 denominations that are possible, and for each denomination there are 4 C1 ways of dealing a card. Thus

23. Number of ways to answer exam is 210 = 1024 = n( S ) .

a.

There is only one way to achieve 100 points, namely to answer each question correctly. Thus n( E100 points ) 1 = . P(100 points) = n( S ) 1024

n( Efour of a kind ) n( S ) 13 ⋅ 4C4 ⋅12 ⋅ 4C1 = 52 C5 13 ⋅12 ⋅ 4 = 52 C5

P(four of a kind) =

b. Number of ways to score 90 points = number of ways that exactly one question is answered incorrectly = 10. Thus P(90 or more points) = P(90 points) + P(100 points) 10 1 11 + = . = 1024 1024 1024

26. a.

24. Number of ways to answer exam is 48 = 65,536 = n( S ) .

a.

4

34 ⎛1⎞ ⎛3⎞ . Since there incorrectly is ⎜ ⎟ ⎜ ⎟ = ⎝4⎠ ⎝4⎠ 48 are 8 C4 distinguishable orders in which one can arrange 4 correct and 4 incorrect answers, and since each arrangement has the same overall probability of occurring, the probability of 4 correct and 4 incorrect 34 34 8! ⋅ 8 C4 = ⋅ answers is 48 48 4!4!

b.

n( Eall correct ) 1 = P(all correct) = n( S ) 65,536

b. The probability of answering one question correctly when answering in a random 1 and the probability of fashion is 4 3 answering incorrectly is . Thus, the 4 probability of answering the first four questions correctly and the last four

P( E ∪ F ) = P( E ) + P( F ) − P( E ∩ F ) Thus P ( F ) = P ( E ∪ F ) + P ( E ∩ F ) − P ( E ) 5 1 1 1 = + − = . 14 7 4 4

P ( E ′ ∪ F ) = P ( E ′) + P( F ) – P( E ′ ∩ F ) ⎛ 1⎞ 1 = ⎜1 − ⎟ + − P( E ′ ∩ F ) ⎝ 4⎠ 4 = 1 − P( E ′ ∩ F ) Since F = ( E ∩ F ) ∪ ( E ′ ∩ F ) and E ∩ F and E ′ ∩ F are mutually exclusive P ( F ) = P( E ∩ F ) + P( E ′ ∩ F ) , 1 1 = + P( E ′ ∩ F ) 4 7

307


ISM: Introductory Mathematical Analysis 30. Here Shiloh needs to win 5 more rounds to win the game and Caitlin needs to win 8 more rounds. Shiloh’s probability of winning is 7 C 3302 1651 ∑ 12212k = 4096 = 2048 . Thus Shiloh’s share k =0

1 1 3 − = . Hence, 4 7 28 3 25 P( E ′ ∪ F ) = 1 − = . 28 28

Thus P ( E ′ ∩ F ) =

27. n(S) =

a.

100 C3

=

100! = 161, 700 3! ⋅ 97!

of the pot is

31. Let p = P(1) = P(3) = P(5). Then 2p = P(2) = P(4) = P(6). Since P(S) = 1, then 1 3(p) + 3(2p) = 1, 9p = 1, p = p (1) = . 9

35! = 6545 3!⋅ 32! n( E3 females ) P ( E3 females ) = n( S )

n( E3 females ) = 35C3 =

=

32. Let p1 = P (a) = P(b) = P(c) = P(d ) = P(e), and p2 = P ( f ) = P( g ). Then

6545 ≈ 0.040 161, 700

1 5 − p1. 2 2 Since p1 is not known, it is not possible to determine P ( f ) = p2 . If it is also known that

P ( S ) = 5( p1 ) + 2( p2 ) = 1, p2 =

b. The number of ways of selecting one professor is 15; the number of ways of selecting two associate professors is 24 C2 . Thus n( E1 professor & 2 associate professors )

1 P ({a, f }) = , then we have 3

24! = 15 ⋅ 276 = 4140 . 2! ⋅ 22! Therefore, P ( E1 professor & 2 associate professers ) = 15 ⋅

=

1 P ({a, f }) = P(a ) + P( f ) = p1 + p2 = . 3 1 1 5⎛1 ⎞ Thus p1 = − p2 and p2 = − ⎜ − p2 ⎟ . 3 2 2⎝3 ⎠

4140 ≈ 0.026 . 161, 700

−

28. P(even number) = P(2) + P(4) + P(6) 2 1 1 4 2 = + + = = 10 10 10 10 5

33. a.

29. Shiloh needs to win 3 more rounds to win the game and Caitlin needs to win 5 more rounds. Shiloh’s probability of winning is 4

∑

7 Ck 7

3 1 2 2 p2 = − or p2 = and so P ( f ) = . 2 3 9 9

Of the 100 voters, 51 favor the tax increase. 51 Thus P(favors tax increase) = = 0.51 . 100

b. Of the 100 voters, 44 oppose the tax increase. Thus 44 P(opposes tax increase) = = 0.44 . 100

1 4

∑ 7 Ck 27 k = 0 1 = (7 C0 + 7 C1 + 7 C2 + 7 C3 + 7 C4 ) 27 1 = (1 + 7 + 21 + 35 + 35) 27 99 = 128 Shiloh’s share of the pot is then 99 ($25) ≈ $19.34. 128 k =0

=

1651 ($50) ≈ $40.31. 2048

2

308

c.

Of the 100 voters, 3 are Republican with no opinion. Thus 3 P(is a Republican with no opinion) = 100 = 0.03 .

34. a.

For the chain, the total average number of sales is 170 units. For brand B, 65 units per month are sold. Thus 65 13 = ≈ 0.38 . P(sale is for brand B) = 170 34


Section 8.5

b. Since 95 units per month are sold at the Exton store, and 30 are of brand C, P(sale is for brand C given that it is at Exton 30 6 = ≈ 0.32 . store) = 95 19

Problems 8.5 1. a.

P(E F ) =

n( E ∩ F ) 1 = n( F ) 5

b. Using the result of part (a), 35.

P( E ) P( E ) = = P( E ′) 1 – P( E ) 1 –

4 5

( 54 )

=

4 5 1 5

=

P( E ′ | F ) = 1 – P( E | F ) = 1 –

4 1

The odds are 4:1.

c. 1

P( E ) P( E ) = = 6 = 36. P( E ′) 1 – P( E ) 1 – 1 6

( )

37.

38.

1 6 5 6

1 = 5

F ′ = {3, 7,8, 9} so P ( E | F ′) =

n( E ∩ F ′) 1 = . n( F ′) 4 n( F ∩ E ) 1 = n( E ) 2

The odds are 1:5.

d.

P( F | E ) =

P( E ) P( E ) 0.7 0.7 7 = = = = P( E ′) 1 – P( E ) 1 – 0.7 0.3 3 The odds are 7:3.

e.

F ∩ G = {5, 6} so P( E | F ∩ G ) =

P( E ) P( E ) 0.001 0.001 1 = = = = P( E ′) 1 – P( E ) 1 – 0.001 0.999 999 The odds are 1:999.

2. a.

1 4 = . 5 5

P( E ) =

n( E ∩ ( F ∩ G )) 0 = = 0. n( F ∩ G ) 2

n( E ) 2 = n( S ) 5

39. P ( E ) =

7 7 = 7 + 5 12

b.

P( E | F ) =

n( E ∩ F ) 0 = =0 n( F ) 2

40. P ( E ) =

100 100 = 100 + 1 101

c.

P( E | G ) =

n( E ∩ G ) 2 = n(G ) 3

41. P( E ) =

4 4 2 = = 4 + 10 14 7

d.

P (G | E ) =

n(G ∩ E ) 2 = =1 n( E ) 2

42. P( E ) =

a a 1 = = a + a 2a 2

e.

F ′ = {1, 2,5} P (G | F ′) =

43. Odds that it will rain tomorrow P (rain) 0.75 0.75 = = = =3. P(no rain) 1 − 0.75 0.25 The odds are 3:1.

f.

P( E ) = P ( E ′)

1 P ( E ′) P( E )

=

1 3 5

E ′ = {3, 4,5} P ( E ′ | F ′) =

44. The odds of E not occurring are the odds of P ( E ′) P( E ′) 3 = = . Then event E ′ which is P ( E ′′) P( E ) 5

n( E ′ ∩ F ′) 1 = n( F ′) 3

3. P( E | E ) =

P( E ∩ E ) P( E ) = =1 P( E ) P( E )

4. P(∅ | E ) =

P(∅ ∩ E ) P(∅) 0 = = =0 P( E ) P( E ) P( E )

5 = , so the odds that E does 3

occur are 5:3. In general, if the odds of E not occurring are a:b, then the odds that E does occur are b:a.

n(G ∩ F ′) 2 = n( F ′) 3

5. P ( E ′ | F ) = 1 – P( E | F ) = 1 – 0.57 = 0.43 309


6. P ( F | G ) =

7. a.

b.

ISM: Introductory Mathematical Analysis 10. P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ), so P ( E ∩ F ) = P( E ) + P( F ) – P( E ∪ F )

P ( F ∩ G ) P(∅) 0 = = =0 P (G ) P (G ) P (G )

3 3 7 1 + – = 5 10 10 5 P( E ∩ F ) 1/ 5 2 Then P(E|F) = = = . P( F ) 3 /10 3 =

P ( E ∩ F ) 1/ 6 1 P( E | F ) = = = P( F ) 1/ 3 2 P( F | E ) =

P ( F ∩ E ) 1/ 6 2 = = P( E ) 1/ 4 3

11. a.

P( F ) =

8. First we find P ( E ∩ F ) : P( E | F ) =

P( E ∩ F ) , P( F )

P( E ∩ F ) = P ( E F ) P( F ) =

3 1 1 ⋅ = . 4 3 4

Then P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ) 1 1 1 = + − 4 3 4 1 = . 3 9. a. b.

P( F | E ) =

P ( F ∩ E ) 1/ 6 2 = = P( E ) 1/ 4 3

c.

d.

P ( F | II) =

c.

P(O | I) =

d.

P (III) =

e.

P (III | O) =

n(III ∩ O) 10 = n(O) 47

f.

P (II | N ′) =

n(II ∩ N ′) n( N ′)

=

12. a.

From part (b) P( F ) =

b.

P ( E ) = P( E ∩ F ) + P( E ∩ F ′)

=

c.

d.

n(Public ∩ Middle) n(Middle)

55 11 = 80 16

n(High ∩ Private) n(Private)

14 2 = 49 7

n(Private ∩ High) n(High)

14 25

P (Public ∪ Low) =P (Public) + P (Low) – P (Public ∩ Low) =

310

35 + 15 50 25 = = 125 + 47 172 86

P(Private|High)= =

1/12 1/12 1 = = . 1 − 1/ 2 1/ 2 6

64 8 = 200 25

P(High|Private)= =

1 1 = + P ( E ∩ F ′) 4 6 1 1 1 so P ( E ∩ F ′) = – = . 4 6 12 P ( E ∩ F ′) Then P( E | F ′) = P( F ′)

n(O ∩ I) 22 11 = = n(I) 78 39

P (Public|Middle)= =

1 . 2 P ( E ∩ F ) 1/ 6 1 Then P( E | F ) = = = . P( F ) 1/ 2 3

n( F ∩ II) 35 = n(II) 58

b.

P ( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ) 7 1 1 = + P( F ) – 12 4 6 7 1 1 1 Thus P ( F ) = − + = .. 12 4 6 2

125 5 = 200 8

126 70 60 136 + – = 175 175 175 175


Section 8.5 18. S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. Let E = {four tails } = {TTTT}, F = {first toss is a tail} = {THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}. n( E ∩ F ) 1 = , the Since P ( E | F ) = n( F ) 8 corresponding odds are P( E | F ) 1/ 8 1 = = ; that is, 1 to 7. P( E ′ | F ) 1 – (1/ 8) 7

13. a.

P(A | B) =

P(A ∩ B) 0.20 1 = = P(B) 0.40 2

b.

P(B | A) =

P(B ∩ A) 0.20 4 = = P (A) 0.45 9

14. P (scratched screen|def. ear pieces) P(scratched screen ∩ def. ear pieces) = P(def. ear pieces) 0.13 13 = = 0.19 19

19. P (< 4 | odd) =

15. S = {BB, BG, GG, GB} Let E = {at least one girl} = {BG, GG, GB}, F = {at least one boy} = {BB, BG, GB}. n( E ∩ F ) 2 = . Thus P ( E | F ) = n( F ) 3

20. Let F denote face card. There are 3 face cards for each suit. Let R denote red card. Half the cards are red, so there are 26. n( F ∩ R ) 6 3 P ( F R) = = = . n( R ) 26 13

16. S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} Let E = {at least two girls} = {BGG, GBG, GGB, GGG}, F = {at least one boy} = {BBB, BBG, BGB, BGG, GBB, GBG, GGB}, G = {oldest is a girl} = {GBB, GBG, GGB, GGG}. a.

P( E | F ) =

n( E ∩ F ) 3 = n( F ) 7

b.

P( E | G ) =

n( E ∩ G ) 3 = n(G ) 4

21. Method 1. The usual sample space has 36 outcomes, where the event “two 1’s” is {(1, 1)}. Note that {at least one 1}′ ={no 1's} , and the event “no 1’s” occurs in 5 · 5 = 25 ways. Thus P(two 1’s | at least one 1) n(two 1's ∩ at least one 1) n({(1,1)}) 1 = = = 36 – 25 11 n(at least one 1) Method 2. From the usual sample space, we find that the reduced sample space for “at least one 1” (which has 11 outcomes) is {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (3, 1), (4, 1), (5, 1)}. 1 . Thus P(two 1’s | at least one 1) = 11

17. S = {HHH, HHT, HTH, THH, THT, TTH, TTT}. Let E = {exactly two tails} = {HTT, THT, TTH}, F = {second toss is a tail} = {HTH, HTT, TTH, TTT}, G = {second toss is a head} = {HHH, HHT, THH, THT}. a.

P( E | F ) =

n( E ∩ F ) 2 1 = = n( F ) 4 2

b.

P( E | G ) =

n( E ∩ G ) 1 = n(G ) 4

n(< 4 ∩ odd) n({1,3}) 2 = = n(odd) n({1,3,5}) 3

22. Method 1. The reduced sample space, having 6 outcomes, is {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}, where, in each pair, the outcome 5 on the red die is given first. Two pairs have a sum greater than 9, namely (5, 5) and 2 1 (5, 6). Thus P (sum>9|5 on red) = = . 6 3 Method 2. The usual sample space has 36 outcomes. Let E = {5 on red}. Then n(E) = 6. Let F = {sum > 9}. Then n( E ∩ F ) = 2 , namely (red 5, green 5) and (red 5, green 6). Thus n( E ∩ F ) 2 1 P( F | E ) = = = . n( E ) 6 3 311


ISM: Introductory Mathematical Analysis 29. Let E = {second card is not a face card} and F = {first card is a face card}. 51−11 40 n( E ∩ F ) 12 ⋅ 51 P( E | F ) = = = 12 51 n( F )

23. The usual sample space consists of ordered pairs (R, G), where R = no. on red die and G = no. on green die. Now, n(green is even) = 6 · 3 = 18, because the red die can show any of six numbers and the green any of three: 2, 4, or 6. Also, n(total of 7 ∩ green even) = n({(5, 2), (3, 4), (1, 6)}) = 3. Thus P (total of 7|green even) =

n(total of 7 ∩ green even) n(green even)

=

3 1 = . 18 6

30. a.

=

b.

P( E | F ) =

b.

P( E ∩ F ) =

12 12 3 3 9 ⋅ = ⋅ = 52 52 13 13 169

31. P ( K1 ∩ Q2 ∩ J 3 ) = P ( K1 ) P(Q2 | K1 ) P ( J 3 | ( K1 ∩ Q2 )) =

4 4 4 8 ⋅ ⋅ = 52 51 50 16,575

32. P ( AS1 ∩ AH 2 ∩ AD2 ) = P ( AS1 ) P ( AH 2 AS1 ) P ( AD2 ( AS1 ∩ AH 2 ) ) 1 1 1 1 = ⋅ ⋅ = . 52 51 50 132, 600

n( E ∩ F ) 3 1 = = n( F ) 24 8

a.

12 11 11 ⋅ = 52 51 221

P ( F1 ∩ F2 ) = P ( F1 ) P( F2 | F1 ) =

24. The usual sample space S consists of 36 ordered pairs. Let E = {sum is 6} and F = {second toss is neither 2 nor 4}. Then n(F) = 6 · 4 = 24 and n( E ∩ F ) = n{(5, 1), (3, 3), (1, 5)} = 3.

n( E ∩ F ) 3 1 = = n( S ) 36 12

33. P ( J1 ∩ J 2 ∩ J 3 ) = P ( J1 ) P ( J 2 | J1 ) P( J 3 | ( J1 ∩ J 2 ))

25. The usual sample space consists of 36 ordered pairs. Let E = {total > 7} and F = {first toss > 3}. Then n( F ) = 3 ⋅ 6 = 18 and n( E ∩ F ) = n({(4, 4), (4, 5), (4, 6), (5, 3), (5, 4), (5, 5), (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}) = 12 n( E ∩ F ) 12 2 = = . Thus P ( E F ) = n( F ) 18 3

=

4 3 2 1 ⋅ ⋅ = 52 51 50 5525

34. Using a probability tree, we find that there are two possible paths such that the second card is a heart, namely, a heart followed by a heart, or a nonheart followed by a heart. Thus P ( H 2 ) = P( H1 ∩ H 2 ) + P ( H1′ ∩ H 2 ) = P ( H1 ) P( H 2 | H1 ) + P ( H1′ ) P ( H 2 | H1′ )

26. Let the sample space consist of ordered pairs (c, d), where c is T or H, and d is the number showing on the die. Let E = {tails shows} and F = {die shows odd number). Then N(F) = 2 · 3 = 6 and n( E ∩ F ) = 1 ⋅ 3 = 3 . Thus

=

13 12 39 13 1 ⋅ + ⋅ = . 52 51 52 51 4

35. Let D = {two diamonds} and R = {first card red}. We have D ∩ R = {two diamonds} = D and

n( E ∩ F ) 3 1 P( E | F ) = = = . n( F ) 6 2

P(D) =

n( K ∩ H ) 1 27. P ( K | H ) = = n( H ) 13

28. P ( H | F ) =

P ( F1 ∩ F2 ) = P ( F1 ) P( F2 | F1 )

13 12 ⋅ . 52 51

Thus P ( D | R ) =

n( H ∩ F ) 3 1 = = n( F ) 12 4

312

P( D ∩ R) = P( R)

13 ⋅ 12 52 51 26 52

=

2 . 17


Section 8.5

36. Using a probability tree, we find that there are two possible paths such that she will be on time, namely, she gets the call and she is on time, or she doesn’t get the call and she is on time. P (T ) = P (C ∩ T ) + P(C ′ ∩ T ) = P (C ) P (T | C ) + P (C ′) P(T | C ′) = (0.9)(0.9) + (0.1)(0.4) = 0.85 37. a.

b. 38. a.

P (U ) = P( F ∩ U ) + P (O ∩ U ) + P ( N ∩ U ) = P(F)P(U|F) + P(O)P(U|O) + P(N)P(U|N) = (0.60)(0.45) + (0.30)(0.55) + (0.10)(0.35) 47 = 0.47 = 100 P( F | U ) =

P( F ∩ U ) (0.60)(0.45) 27 = = P (U ) 0.47 47

P (contact ∩ purchase) =P (contact)P (purchase|contact) = (0.02)(0.014) = 0.00028

b. 100,000(0.00028) = 28 39. a.

After the first draw, if the rabbit drawn is red, then 4 rabbits remain, 3 of which are yellow. 3 P(second is yellow | first is red) = 4

b. After red rabbit is replaced, 5 rabbits remain, 3 of which are yellow. 3 P(second is yellow | first is red) = 5 40. P (G2 ) = P (G1 ∩ G2 ) + P ( R1 ∩ G2 ) = P (G1 ) P (G2 | G1 ) + P( R1 ) P (G2 | R1 ) =

4 4 3 3 25 ⋅ + ⋅ = 7 7 7 7 49

41. P (W ) = P (Box 1 ∩ W) + P (Box 2 ∩ W) = P(Box 1)P(W | Box 1) + P(Box 2)P(W | Box 2) = 42. a.

P (W ) = P ( B1 ∩ W ) + P ( B 2 ∩ W ) + P( B3 ∩ W ) = P(B1)P(W | B1) + P(B2)P(W | B2) + P(B3)P(W | B3) 1 3 1 4 1 2 158 = ⋅ + ⋅ + ⋅ = 3 5 3 7 3 6 315

b.

P ( R ) = P ( B1 ∩ R) + P( B 2 ∩ R) + P( B3 ∩ R) = P(B1)P(R | B1) + P(B2)P(R | B2) + P(B3)P(R | B3) 1 2 1 3 1 2 122 = ⋅ + ⋅ + ⋅ = 3 5 3 7 3 6 315

c.

1 1 1 P (G ) = P ( B3 ∩ G ) = P ( B3) P(G | B3) = ⋅ = 3 3 9

313

1 2 1 2 9 ⋅ + ⋅ = 2 5 2 4 20



43. P (W2 ) = P( B1 ∩ G1 ∩ W2 ) + P ( B1 ∩ R1 ∩ W2 ) + P( B 2 ∩ W1 ∩ W2 )

= P ( B1) P ( G1 B1) P (W2 (G1 ∩ B1) ) + P( B1) P ( R1 B1) P (W2 ( R1 ∩ B1) ) + P( B 2) P (W1 B 2 ) P (W2 (W1 ∩ B 2) )

=

1 1 1 1 1 1 1 1 1 1 ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = 2 2 3 2 2 3 2 2 3 4

44. P ( D1 ∩ D2 ∩ D3 ∩ D4 ) = P ( D1 ) P ( D2 D1 ) P ( D3 ( D1 ∩ D2 ) ) P ( D4 ( D1 ∩ D2 ∩ D3 ) ) 5 4 3 2 1 = ⋅ ⋅ ⋅ = 10 9 8 7 42 45. P (Und.) = P (MS ∩ Und.) + P(DS ∩ Und.) = P (MS)P (Und.|MS) + P(DS)P(Und|DS) 20, 000 1 40, 000 3 = ⋅ + ⋅ 60, 000 100 60, 000 100 7 = 300 46. P (5000) = P ( B1 ∩ 5000) + P ( B 2 ∩ 5000) + P( B3 ∩ 5000) = P(B1)P(5000|B1) + P(B2)P(5000|B2) + P(B3)P(5000|B3) 1 1 1 2 1 1 11 = ⋅ + ⋅ + ⋅ = 3 2 3 8 3 6 36 47. P (Def) = P(A ∩ Def)+P (B ∩ Def)+P(C ∩ Def) = P(A)P(Def | A) + P(B)P(Def | B) + P(C)P(Def | C) = (0.10)(0.06) + (0.20)(0.04) + (0.70)(0.05) = 0.049 48. P (Def) = P(A ∩ Def) +P(B ∩ Def)+P(C ∩ Def)+P(D ∩ Def) = P(A)P(Def | A) + P(B)P(Def | B) + P(C)P(Def | C) + P(D)P(Def | D) = (0.30)(0.06) + (0.20)(0.03) + (0.35)(0.02) + (0.15)(0.05) = 0.0385 49. a. b.

P ( D ∩ V ) = P( D ) P(V | D) = (0.40)(0.15) = 0.06 P (V ) = P ( D ∩ V ) + P ( R ∩ V ) + P ( I ∩ V ) = P(D)P(V | D) + P(R)P(V | R) + P(I)P(V | I) = (0.40)(0.15) + (0.35)(0.20) + (0.25)(0.10) = 0.155

50. Because Richard was not hired, the number of sample points in the reduced sample space is 7 C4 = 35, of which Allison, Lesley, Tom, and Bronwyn form one sample point. Thus 1 P (Allison, Lesley, Tom, and Bronwyn were hired) = . 35

314


Section 8.6 3. P ( E ∩ F ) = P ( E ) P ( F ) ,

51. P(3 Fem|at least one Fem) P (3 Fem ∩ at least one Fem) = P (at least one Fem) 6 C3

1 2 1 7 7 = ⋅ P ( F ) so P ( F ) = ⋅ = 9 7 9 2 18 4

P (3 Fem) 4 C = = 11 3 = 33 = 2 C 5 3 1–P (no Fem) 1 – 1 – 33 31 C

4. P ( E ) = P ( E | F ) =

1 , 3

11 3

so P ( E ′) = 1 – P ( E ) = 1 –

Problems 8.6 1. a. b.

P( E ∩ F ) = P( E ) P( F ) =

3 8 2 ⋅ = = P( E ∩ F ) 4 9 3 Since P ( E ) P ( F ) = P ( E ∩ F ) , events E and F are independent.

1 3 1 ⋅ = 3 4 4

5. P ( E ) P ( F ) =

P( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F ) =

1 3 1 5 + – = 3 4 4 6

6. P(E)P(F) = (0.28)(0.15) = 0.042 ≠ P ( E ∩ F ) , so E and F are dependent events. 1 3

c.

P( E | F ) = P( E ) =

d.

P( E ′ | F ) = 1 – P( E | F ) = 1 –

e.

P ( E ∩ F ′) = P ( E ) P ( F ′) =

f.

P ( E ∪ F ′) = P ( E ) + P ( F ′) – P( E ∩ F ′)

7. Let F = {full service} and I = {increase in value}. 400 2 = P( F ) = 600 3 n( F ∩ I ) 320 2 and P ( F | I ) = = = n( I ) 480 3 Since P(F | I) = P(F), events F and I are independent.

1 2 = 3 3

1 1 1 ⋅ = 3 4 12

8. Let M = {male} and C = {cruncher}. 130 26 and P(M ) = = 175 35 n( M ∩ C ) 55 11 P(M | C ) = = = n(C ) 80 16 Since P(M | C) ≠ P(M), events M and C are dependent.

1 1 1 1 = + – = 3 4 12 2

g.

P ( E | F ′) =

P ( E ∩ F ′) 1/12 1 = = P ( F ′) 1/ 4 3

2. a.

P ( E ∩ F ) = P ( E ) P ( F ) = (0.1)(0.3) = 0.03

b.

P ( F ∩ G ) = P ( F ) P(G ) = (0.3)(0.6) = 0.18

c.

P ( E ∩ F ∩ G ) = P ( E ) P ( F ) P (G ) = (0.1)(0.3)(0.6) = 0.018

d.

e.

1 2 = . 3 3

9. Let S be the usual sample space consisting of ordered pairs of the form (R, G), where the first component of each pair represents the number showing on the red die, and the second component represents the number on the green die. Then n( S ) = 6 ⋅ 6 = 36. For E, any number of four can occur on the red die, and any number on the green die. Thus n( E ) = 4 ⋅ 6 = 24. For F we have F = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, so n(F) = 5. Also, E ∩ F = {(4, 4), (5, 3), (6, 2)}, so

P( E ∩ F ∩ G ) P( F ∩ G) 0.018 = = 0.1 0.18

P ( E (F ∩ G) ) =

P ( E ′ ∩ F ∩ G ′) = P ( E ′) P ( F ) P(G ′) = (0.9)(0.3)(0.4) = 0.108

n( E ∩ F ) = 3. Thus P ( E ) P ( F ) =

and P ( E ∩ F ) =

315

3 1 = . Since 36 12

24 5 5 ⋅ = 36 36 54



P ( E ) P ( F ) ≠ P ( E ∩ F ), events E and F are dependent.

a.

E ∩ F = {(3,3)}, so P(E ∩ F )=

1 . Since 49

7 7 1 ⋅ = = P( E ∩ F ) , 49 49 49 events E and F are independent. P( E ) P( F ) =

26 1 = 52 2 12 3 6 3 P( F ) = = , and P ( E ∩ F ) = = . 52 13 52 26 1 3 3 Because P ( E ) P ( F ) = ⋅ = = P( E ∩ F ) , 2 13 26 events E and F are independent.

10. P ( E ) =

b.

E ∩ G = {(3, 2), (3, 4), (3, 6)}, 3 . Since 49 7 24 24 P ( E ) P (G ) = ⋅ = ≠ P( E ∩ G ) , 49 49 343 events E and G are dependent. so P( E ∩ G ) =

11. S = {HH, HT, TH, TT}, E = {HT, TH, TT}, F = {HT, TH}, and E ∩ F = {HT, TH} .

c.

3 Thus P ( E ) = 4 2 1 P ( F ) = = , and 4 2 2 1 P ( E ∩ F ) = = . We have 4 2 3 1 3 P ( E ) P ( F ) = ⋅ = ≠ P ( E ∩ F ) , so events E 4 2 8 and F are dependent.

F ∩ G = {(2,3), (4,3), (6,3)} so P( F ∩ G ) =

3 . 49

Since 7 24 24 ⋅ = ≠ P( F ∩ G ) . 49 49 343 Events F and G are dependent. P ( F ) P(G ) =

d.

12. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and n(S) = 8. E = {HTT, THT, TTH, TTT} and n(E) = 4. F = {HHT, HTH, THH, HTT, THT, TTH} and n(F) = 6. E ∩ F = {HTT, THT, TTH} and n( E ∩ F ) = 3 .

14. a.

4 6 3 ⋅ = = P( E ∩ F ) , so E 8 8 8 and F are independent.

Thus P ( E ) P ( F ) =

b.

E ∩ F ∩ G = ∅, so P ( E ∩ F ∩ G ) = 0 . However, P ( E ) P ( F ) P(G ) ≠ 0 = P ( E ∩ F ∩ G ) , so events E, F and G are not independent.

E = {3} F = {5} E ∩ F = ∅ , so E and F are mutually exclusive. P( E ) = P( F ) =

1 6

P( E ∩ F ) = 0 1 1 1 ⋅ = ≠ P( E ∩ F ) 6 6 36 Thus E and F are not independent. P( E ) P( F ) =

13. Let S be the set of ordered pairs whose first (second) component represents the number on the first (second) chip. Then n(S) = 7 · 7 = 49, n(E) = 1 · 7 = 7, and n(F) = 7 · 1 = 7. For G, if the first chip is 1, 3, 5 or 7, then the second chip must be 2, 4 or 6; if the first chip is 2, 4 or 6, the second must be 1, 3, 5 or 7. Thus n(G) = 4 · 3 + 3 · 4 = 24.

15. P ( E ∩ F ) = P ( E ) P ( F | E ), thus P ( E ∩ F ) 0.3 = = 0.75 P( F | E ) 0.4 Since P(E) = 0.75 ≠ 0.5 = P(E | F), E and F are dependent.

P(E) =

316


Section 8.6

16. P ( E ∩ F ) = P( F ) P ( E | F ) , thus P ( F ) =

P( E ∩ F ) = P( E | F )

5 9 2 3

=

5 6

P ( E ∪ F ) = P ( E ) + P ( F ) – P ( E ∩ F ), so P( E ) = P( E ∪ F ) – P( F ) + P( E ∩ F ) =

Since P ( E ) =

17 5 5 2 – + = 18 6 9 3

2 = P ( E | F ) , events E and F are independent. 3

17. Let E = {red 4} and F = {green > 4}. Assume E and F are independent. 1 1 1 P( E ∩ F ) = P( E ) P( F ) = ⋅ = 6 3 18 18. Ei = {2 or 3 shows on ith roll}, where i = 1, 2, 3. Assume the Ei 's are independent. ⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ 1 P ( E1 ∩ E2 ∩ E3 ) = P ( E1 ) P ( E2 ) P ( E3 ) = ⎜ ⎟ ⎜ ⎟⎜ ⎟ = ⎝ 3 ⎠ ⎝ 3 ⎠⎝ 3 ⎠ 27

19. Let F = {first person attends regularly} and S = {second person attends regularly}. 1 1 1 Then P ( F ∩ S ) = P ( F ) P( S ) = ⋅ = . 5 5 25 6 1 = 36 6 Assume that the throws are independent. P(double on all three throws) = P(double on 1st) · P(double on 2nd) · P(double on 3rd) 1 1 1 1 . = ⋅ ⋅ = 6 6 6 216

20. P(double on any throw) =

21. Because of replacement, assume the cards selected on the draws are independent events. P(ace, then face card, then spade) = P(ace) · P(face card) · P(spade) 4 12 13 3 ⋅ ⋅ = = 52 52 52 676 22. Assume the outcomes on the rolls are independent events. a.

P (> 4, > 4, > 4, > 4, > 4, > 4, > 4) =

2 2 2 2 2 2 2 1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 6 6 6 6 6 6 6 2187

b.

P (< 4, < 4, < 4, < 4, < 4, < 4, < 4) =

3 3 3 3 3 3 3 1 ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 6 6 6 6 6 6 6 128

23. a.

P (Bill gets A ∩ Jim gets A ∩ Linda gets A) = P(Bill gets A) · P(Jim gets A) · P(Linda gets A) 3 1 4 3 . = ⋅ ⋅ = 4 2 5 10

317

Chapter 8: Introduction to Probability and Statistics b.

P (Bill no A ∩ Jim no A ∩ Linda no A) = P(Bill no A) · P(Jim no A) · P(Linda no A) 1 1 1 1 = ⋅ ⋅ = 4 2 5 40

c.

P (Bill no A ∩ Jim no A ∩ Linda gets A) = P(Bill no A) · P(Jim no A) · P(Linda gets A) 1 1 4 1 = ⋅ ⋅ = 4 2 5 10


24. Assume independence of rolls. ⎛ 5 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎞ 91 P(at least one 6) = 1 – P(no 6’s) = 1 – ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎝ 6 ⎠ ⎝ 6 ⎠ ⎝ 6 ⎠ 216

25. Let A = {A survives 15 more years}, B = {B survives 15 more years}. 2 3 2 ⋅ = 3 5 5

a.

P ( A ∩ B ) = P ( A) P ( B) =

b.

P ( A′ ∩ B) = P( A′) P ( B ) =

c.

A ∩ B ′ and A′ ∩ B are mutually exclusive.

1 3 1 ⋅ = 3 5 5

P[( A ∩ B ′) ∪ ( A′ ∩ B)] = P ( A) P ( B ′) + P ( A′) P( B ) =

2 2 1 3 7 ⋅ + ⋅ = 3 5 3 5 15

d. P(at least one survives) = P(exactly one survives) + P(both survive) =

e.

P(neither survives) = 1 – P(at least one survives) = 1 –

7 2 13 . + = 15 5 15

13 2 . = 15 15

26. Assume that drawing a particular size of paper and a particular size of envelope are independent events. P (paper A ∩ envelope A) + P (paper B ∩ envelope B) = (0.63)(0.57) + (0.37)(0.43) ≈ 0.52 27. Assume the colors selected on the draws are independent events. 7 6 7 ⋅ = 18 18 54

a.

P (W1 ∩ G2 ) = P (W1 ) P (G2 ) =

b.

P[( R1 ∩ W2 ) ∪ (W1 ∩ R2 )] = P ( R1 ) P (W2 ) + P (W1 ) P ( R2 ) =

318

5 7 7 5 35 ⋅ + ⋅ = 18 18 18 18 162


Section 8.6

28. Assume the rolls are independent. P(7 on a roll) = P{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} = P(12 on a roll) = P{(6, 6)} =

6 1 = 36 6

1 36

P(7 on one roll and 12 on the other) =

1 1 1 1 1 ⋅ + ⋅ = 6 36 36 6 108

29. Assume that the selections are independent. 3 3 7 7 9 9 139 P (both red ∪ both white ∪ both green) = ⋅ + ⋅ + ⋅ = 19 19 19 19 19 19 361 30. Assume the throws are independent. For a particular number, 3

1 1 1 ⎛1⎞ . ⋅ ⋅ = 6 6 6 ⎜⎝ 6 ⎟⎠ Since the particular number can be any of 6 numbers,

P(particular number on three throws) =

3

1 ⎛1⎞ . P(same number in 3 throws) = 6 ⎜ ⎟ = 6 36 ⎝ ⎠

31. Assume that the draws are independent. P (particular 1st ticket ∩ particular 2nd ticket) 1 1 1 ⋅ = 20 20 400 P(sum is 35) = P{(20, 15), (19, 16), (18, 17), (17, 18), (16, 19), (15, 20)} 3 ⎛ 1 ⎞ = 6⎜ = ⎟ ⎝ 400 ⎠ 200 =

32. a.

P({TT33}) = P(T on 1st coin) P(T on 2nd coin) P(3 on 1st die) P(3 on 2nd die) 1 1 1 1 1 = ⋅ ⋅ ⋅ = 2 2 6 6 144

b. P(two heads, one 4 and one 6) = P(H on 1st coin) P(H on 2nd coin) P(4 on 1st die) P(6 on 2nd die) + P(H on 1st coin)P(H on 2nd coin) P(6 on 1st die) P(4 on 2nd die) 1 1 ⎛ 1 1 1⎞ = ⎜ ⋅ ⋅ ⋅ ⎟⋅2 = 72 ⎝2 2 6 6⎠ 33. a.

1 1 1 1 ⋅ ⋅ = 12 12 12 1728

b. To get exactly one even, there are 3 C1 = 3 ways. P(one even and two odd) = 3[P(even 1st spin) · P(odd 2nd spin)·P(odd 3rd spin)] ⎛ 6 6 6⎞ 3 = 3⎜ ⋅ ⋅ ⎟ = . ⎝ 12 12 12 ⎠ 8 34. a.

4 13 2 1 ⋅ ⋅ = 52 52 52 1352 319



b.

4 4 4 1 ⋅ ⋅ = 52 52 52 2197

c.

The queen, spade, and black ace can be drawn in any order, so there are 3! = 6 orders, thus 4 13 2 3 6⋅ ⋅ ⋅ = . 52 52 52 676

d. The ace can come first, second, or third, so 3 ⋅ 35. a.

4 48 48 432 ⋅ ⋅ = . 52 52 52 2197

The number of ways of getting exactly four correct answers out of five is 5 C4 = 5 . Each of these ways has a 1 1 1 1 3 3 . Thus ⋅ ⋅ ⋅ ⋅ = 4 4 4 4 4 1024 3 15 P(exactly 4 correct) = 5 ⋅ . = 1024 1024

probability of

b. P(at least 4 correct) = P(exactly 4) + P(exactly 5) 15 1 1 1 1 1 1 = + ⋅ ⋅ ⋅ ⋅ = 1024 4 4 4 4 4 64 c.

36. a.

The number of ways of getting exactly three correct answers out of five is 1 1 1 3 3 9 , so ⋅ ⋅ ⋅ ⋅ = 5 C3 = 10 . Each of these ways has a probability of 4 4 4 4 4 1024 9 45 P(exactly 3 correct) = 10 ⋅ . Thus = 1024 512 P(3 or more correct) = P(exactly 3) + P(at least 4) 45 1 53 . = + = 512 64 512 P(none hit) = (0.5)(0.6)(0.3) = 0.09

b. P(only Linda hits) = (0.5)(0.6)(0.7) = 0.21 c.

P(exactly one hits target) = P(only Bill) + P(only Jim) + P(only Linda) = (0.5)(0.6)(0.3) + (0.5)(0.4)(0.3) + (0.5)(0.6)(0.7) = 0.36

d. P(exactly 2) = P(not Bill) + P(not Jim) + P(not Linda) = (0.5)(0.4)(0.7) + (0.5)(0.6)(0.7) + (0.5)(0.4)(0.3) = 0.41 e.

P(all hit) = (0.5)(0.4)(0.7) = 0.14

37. A wrong majority decision can occur in one of two mutually exclusive ways: exactly two wrong recommendations, or three wrong recommendations. Exactly two wrong recommendations can occur in 3 C2 = 3 mutually exclusive ways. Thus P(wrong majority decision) = [(0.04)(0.05)(0.9) + (0.04)(0.95)(0.1) + (0.96)(0.05)(0.1)] + (0.04)(0.05)(0.1) = 0.0106.

320


Section 8.7

Problems 8.7 1. P ( E | D) =

P( E ) P( D | E ) = P( E ) P( D | E ) + P( F ) P( D | F )

2⋅ 1 5 10 2⋅ 1 + 3⋅1 5 10 5 5

For the second part, P ( D′ | F ) = 1 – P ( D | F ) = 1 – P( D′ | E ) = 1 – P( D | E ) = 1 – P ( F | D ′) =

2. P ( E1 | S ) =

P ( E3 | S ′) =

1 4

1 4 = , and 5 5

1 9 . Then = 10 10

P( F ) P( D′ | F ) = P( E ) P( D′ | E ) + P( F ) P( D ′ | F )

3⋅4 5 5 2⋅ 9 + 3⋅4 5 10 5 5

=

4 . 7

P ( E1 ) P ( S | E1 ) = P ( E1 ) P ( S | E1 ) + P( E2 ) P( S | E2 ) + P( E3 ) P( S | E3 )

1⋅2 5 5 1⋅2+ 3 ⋅ 7 5 5 10 10

P ( E3 ) P ( S ′ | E3 ) = P ( E1 ) P ( S ′ | E1 ) + P ( E2 ) P ( S ′ | E2 ) + P( E3 ) P( S ′ | E3 )

3. D = {is Democrat}, R = {is Republican}, I = {is Independent}, V = {voted}. P ( D) P (V | D) P ( D) P(V | D) + P( R) P(V | R) + P( I ) P(V | I ) (0.42)(0.25) = (0.42)(0.25) + (0.33)(0.27) + (0.25)(0.15) 175 = ≈ 0.453 386

P( D | V ) =

4. D = {tire is domestic} I = {tire is imported} S = {tire is all-season} 2000 2 1000 1 = and P ( I ) = = . P( D) = 3000 3 3000 3 2 1 Note: 40% = and 10% = . 5 10 P( I ) P( S | I ) P(I|S) = P( I ) P( S | I ) + P( D) P( S | D) =

=

1⋅ 1 3 10 1⋅ 1 + 2⋅2 3 10 3 5

=

1 9

321

+

1⋅1 2 2 1⋅3+ 3 ⋅ 3 5 5 10 10

1⋅1 2 2

+

=

1⋅1 2 2

4 . 27

=

25 . 46



5. D = {has the disease} D′ = {does not have the disease} R = {positive reaction} N = {negative reaction} = R ′ a.

P( D | R) =

P( D) P( R | D) (0.03)(0.86) 258 = = ≈ 0.275 P ( D) P( R | D) + P ( D′) P ( R | D′) (0.03)(0.86) + (0.97)(0.07) 937

b.

P( D | N ) =

P( D) P( N | D) (0.03)(0.14) 14 = = ≈ 0.005 P ( D) P ( N | D) + P ( D′) P ( N | D′) (0.03)(0.14) + (0.97)(0.93) 3021

6. I = {increase in earnings} D = {declare a dividend} 3 1 and 10% = . Note: 60% = 5 10 P( I | D) =

P( I ) P( D | I ) = P ( I ) P ( D | I ) + P ( I ′) P ( D | I ′)

1⋅3 3 5 1⋅3+ 2⋅ 1 3 5 3 10

=

3 = 75% 4

7. B1 = {first bag selected} B2 = {second bag selected} R = {red jelly bean drawn} 1 P ( B1 ) = P ( B2 ) = . 2 P( B1 ) P( R | B1 ) = P ( B1 ) P ( R | B1 ) + P ( B2 ) P ( R | B2 )

P ( B1 | R ) =

1⋅4 2 6 1⋅4+1⋅2 2 6 2 5

=

5 . 8

8. B1 = {Bowl I selected} B2 = {Bowl II selected} B3 = {Bowl III selected} W = {white ball selected} P ( B1 ) = P ( B2 ) = P ( B3 ) = P ( B1 | W ) =

1 3

P ( B1 ) P (W | B1 ) = P( B1 ) P (W | B1 ) + P ( B2 ) P(W | B2 ) + P( B3 ) P(W | B3 )

9. A = {unit from line A} B = {unit from line B} D = {defective unit}. 300 3 = P ( A) = 800 8 500 5 P( B) = = 800 8 P( A | D) =

P ( A) P ( D | A) = P ( A) P ( D | A) + P ( B) P( D | B )

3⋅ 2 8 100 3⋅ 2 + 5⋅ 5 8 100 8 100

322

=

6 31

1⋅3 3 5

+

1⋅3 3 5 1⋅3 3 7

+

1⋅2 3 6

=

63 143


Section 8.7

10. A = {unit from line A} B = {unit from line B} C = {unit from line C} D = {unit from line D} F = {defective unit} a.

P( A | F ) =

P ( A) P ( F | A) P ( A) P ( F | A) + P ( B ) P ( F | B ) + P (C ) P ( F | C ) + P ( D) P( F | D)

(0.35)(0.02) 7 = (0.35)(0.02) + (0.20)(0.05) + (0.30)(0.03) + (0.15)(0.04) 32 Parts (b), (c), and (d) are similarly determined.

=

b.

10 5 = 32 16

c.

9 32

d.

6 3 = 32 16

11. C = {call made} T = {on time for meeting} P (C ) P (T | C ) P (C | T ) = P (C ) P (T | C ) + P (C ′) P (T | C ′) =

(0.95)(0.9) 114 = ≈ 0.958 (0.95)(0.9) + (0.05)(0.75) 119

12. J D = {jar with dark chocolate only selected} J M = {jar with dark and milk chocolates selected} D = {dark chocolate selected} 1 P( J D ) = P( J M ) = 2 P( J D | D) =

P( J D ) P( D | J D ) = P( J D ) P( D | J D ) + P( J M ) P( D | J M )

1 ⋅ 50 2 50 1 ⋅ 50 + 1 ⋅ 20 2 50 2 50

=

5 7

13. W = {walking reported} B = {bicycling reported} R = {running reported} C = {completed requirement} P (W | C ) =

P(W ) P (C | W ) = P(W ) P (C | W ) + P ( B ) P (C | B) + P ( R ) P (C | R)

55.1% would be expected to report walking.

323

1⋅ 9 2 10 1⋅ 9 +1⋅4 2 10 4 5

+

1⋅2 4 3

=

27 ≈ 0.551 49



14. C = {charges battery} S = {car starts} P (C ′ | S ′) =

P (C ′) P ( S ′ | C ′) = P (C ′) P( S ′ | C ′) + P(C ) P( S ′ | C )

1 ⋅4 10 5 1 ⋅4 + 9 ⋅1 10 5 10 8

=

32 ≈ 0.416 77

15. J = {had Japanese-made car} E = {had European-made car} A = {had American-made car} B = {buy same make again} P( J | B) =

P( J ) P( B | J ) = P ( J ) P ( B | J ) + P ( E ) P ( B | E ) + P( A) P( B | A)

3 ⋅ 85 5 100 3 ⋅ 85 + 1 ⋅ 50 5 100 10 100

3 ⋅ 40 + 10 100

=

3 4

16. D = {dalhousium is present} P = {positive test} N = {negative test} = P ′ a.

P( D | P) =

P( D) P( P | D) (0.005)(0.80) 400 = = ≈ 0.0261 P ( D) P( P | D) + P ( D′) P( P | D′) (0.005)(0.80) + (0.995)(0.15) 15,325

b.

P( D | N ) =

P( D) P( N | D) (0.005)(0.20) 100 = = ≈ 0.0012 P ( D) P ( N | D) + P( D′) P ( N | D′) (0.005)(0.20) + (0.995)(0.85) 84, 675

17. P = {pass the exam} A = {answer every question} P ( A) P ( P | A) (0.75)(0.8) 24 = = ≈ 0.828 P( A | P) = P ( A) P( P | A) + P( A′) P ( P | A′) (0.75)(0.8) + (0.25)(0.50) 29 18. P = {predicted smoking} S = {smoking now} P( P) P( S ′ | P) (0.75)(0.7) 7 = = = 70% P ( P | S ′) = P( P ) P ( S ′ | P ) + P ( P′) P ( S ′ | P′) (0.75)(0.7) + (0.25)(0.9) 10 19. S = {signals sent} D = {signals detected} P( S | D) =

P( S ) P( D | S ) = P ( S ) P ( D | S ) + P ( S ′) P ( D | S ′)

2⋅3 5 5 2⋅3+ 3⋅ 1 5 5 5 10

=

4 5

20. AM = {A average at midterm} A = {A for course} ′ ) P ( A′M ) P( A | AM (0.4)(0.6) 4 P ( A′M | A) = = = ≈ 0.364 P( A′M ) P ( A | A′M ) + P ( AM ) P( A | AM ) (0.4)(0.6) + (0.6)(0.7) 11 21. S = {movie is a success} U = {“Two Thumbs Up”} P(S | U ) =

P( S ) P (U | S ) = P ( S ) P (U | S ) + P ( S ′) P(U | S ′)

8 ⋅ 70 10 100 8 ⋅ 70 + 2 ⋅ 20 10 100 10 100

324

=

14 ≈ 0.933 15


Section 8.7

22. G1 = {green ball drawn from Bowl 1} R1 = {red ball drawn from Bowl 1} G2 = {green ball drawn from Bowl 2} P (G1 ) P (G2 | G1 ) = P (G1 | G2 ) = P (G1 ) P(G2 | G1 ) + P( R1 ) P (G2 | R1 )

5⋅4 9 8 5⋅4 + 4⋅3 9 8 9 8

=

5 8

23. S = {is substandard request} C = {is considered substandard request by Blackwell} a.

P (C ) = P ( S ) P(C | S ) + P( S ′) P(C | S ′) = (0.20)(0.75) + (0.8)(0.15) = 0.27 =

b.

P(S | C ) =

c.

P(Error) = P (C ′ ∩ S ) + P (C ∩ S ′) = P ( S ) P(C ′ | S ) + P( S ′) P(C | S ′)

27 100

(0.20)(0.75) 0.15 15 P ( S ) P (C | S ) = = = ≈ 0.556 0.27 0.27 27 P ( S ) P (C | S ) + P ( S ′) P (C | S ′)

= (0.20)(0.25) + (0.80)(0.15) = 0.17 =

17 100

24. I = {first chest selected} II = {second chest selected} III = {third chest selected} G = {gold coin found}. For the coin in the other drawer to be silver, we want the probability that the third chest was selected given that a gold coin was found. 1⋅1 P ( III ) P (G | III ) 1 3 2 = = P ( III | G ) = P( I ) P (G | I ) + P ( II ) P(G | II ) + P( III ) P(G | III ) 1 ⋅1 + 1 ⋅ 0 + 1 ⋅ 1 3 3

25. a.

P( L | E ) =

=

b.

(0.25)(0.49) ≈ 0.18 (0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)

P(M | E ) =

=

c.

P( L) P( E | L) P ( L) P ( E | L) + P ( M ) P ( E | M ) + P ( H ) P( E | H )

(0.25)(0.64) ≈ 0.23 (0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)

P( H | E ) =

=

P( M ) P( E | M ) P( L) P( E | L) + P( M ) P( E | M ) + P( H ) P( E | H )

P( H ) P( E | H ) P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )

(0.5)(0.81) ≈ 0.59 (0.25)(0.49) + (0.25)(0.64) + (0.5)(0.81)

d. High quality

325

3

3 2


26. a.

P( L | E ) =

(a)

=


(0.25)(0.44) ≈ 0.39 (0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18)

P(M | E ) =

(b)

=

=


(0.25)(0.32) ≈ 0.29 (0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18) P( H ) P( E | H ) P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )

P( H | E ) =

(c)


(0.5)(0.18) ≈ 0.32 . (0.25)(0.44) + (0.25)(0.32) + (0.5)(0.18)

(d) Low quality P( L | E ) =

b. (a)

=


(0.25)(0.07) ≈ 0.54 (0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01)

P(M | E ) =

(b)

=

(0.25)(0.04) ≈ 0.31 (0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01)

P( H | E ) =

(c)


P( H ) P( E | H ) P( L) P ( E | L) + P( M ) P( E | M ) + P ( H ) P ( E | H )

(0.5)(0.01) ≈ 0.15 (0.25)(0.07) + (0.25)(0.04) + (0.5)(0.01) (d) Low quality

=

27. F = {fair weather} I = {inclement weather} W = {predict fair weather}. P ( F ) P(W | F ) (0.6)(0.7) 7 = = ≈ 0.78 P( F | W ) = P( F ) P (W | F ) + P ( I ) P (W | I ) (0.6)(0.7) + (0.4)(0.3) 9 Chapter 8 Review Problems = 8 ⋅ 7 ⋅ 6 = 336

1.

8 P3

2.

20 P1

= 20

3.

9 C7

=

9! 9! 9 ⋅ 8 ⋅ 7! 9 ⋅ 8 = = = = 36 7!(9 – 7)! 7!⋅ 2! 7!⋅ 2 ⋅1 2 326


4.

12 C5

=

Chapter 8 Review of the same face value, another two with a different face value, and the last with yet another face value is 4! 4! ⋅12 ⋅ ⋅ 44 13 ⋅ 4C2 ⋅12 ⋅ 4C 2 ⋅ 44 = 13 ⋅ 2!⋅ 2! 2!2! = 13 ⋅ 6 ⋅12 ⋅ 6 ⋅ 44 = 247,104.

12! 12 ⋅11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7! = = 792 5!(12 − 5)! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 7!

5. For each of the first 3 characters there are 26 choices, while for each of the last 3 characters there are 10 choices. By the basic counting principle, the number of license plates that are possible is 26 ⋅ 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 = 17,576,000.

13. a.

6. The number of choices for appetizers is 2, for the entrée it is 4, and for the dessert it is 3. By the basic counting principle, the number of complete dinners that are possible is 2 · 4 · 3 = 24.

Three bulbs are selected from 24, and the order of selection is not important. Thus the number of possible selections is 24! 24! = 24 C3 = 3!(24 – 3)! 3!⋅ 21! =

7. Each of the five switches has 2 possible positions. By the basic counting principle, the number of different codes is

24 ⋅ 23 ⋅ 22 ⋅ 21! 24 ⋅ 23 ⋅ 22 = = 2024 . 3 ⋅ 2 ⋅1 ⋅ 21! 3 ⋅ 2 ⋅1

b. Only one bulb is defective and that bulb must be included in the selection. The other two bulbs must be selected from the 23 remaining bulbs and there are 23 C2 such selections possible. Thus the number of ways of selecting three bulbs such that one is defective is 23! 23! = 1 ⋅ 23C2 = 23C2 = 2!(23 – 2)! 2!⋅ 21!

2 · 2 · 2 · 2 · 2 = 25 = 32 . 8. A batting order consists of nine names selected from nine names such that order is important. The number of such selections is 9 P9 = 9! = 362,880 .

23 ⋅ 22 ⋅ 21! 23 ⋅ 22 = = 253 . 2 ⋅1 ⋅ 21! 2 ⋅1

9. A possibility for first, second, and third place is a selection of three of the seven teams so that order is important. Thus the number of ways the season can end is 7 P3 = 7 ⋅ 6 ⋅ 5 = 210 .

14. To score 90, exactly nine questions must be correct; to score 100, all ten questions must be correct. If exactly nine questions are answered correctly, there are three ways of answering the tenth question incorrectly. But the number of ways of selecting nine of ten items is 10 C9 . Thus the number of ways to score 90 is 3 ⋅ 10C9 . The number of ways to answer all ten questions correctly is 10 C10 , or more simply, 1. Thus the number of ways to score 90 or better is 10! +1 3 ⋅ 10C9 + 1 = 3 ⋅ 9! ⋅1! = 3 · 10 + 1 = 31.

10. Nine of the nine trophies can be arranged so that order is important. The first two can be placed on the top shelf, the next three on the middle shelf, and the last four on the bottom shelf. The number of such arrangements is 9 P9 = 9! = 362,880 . 11. The order of the group is not important. Thus the number of groups that can board is 11! 11 ⋅10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6! = = 462. 11 C6 = 6! ⋅ 5! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 ⋅ 6! 12. There are four cards with a particular face value and there are 4 C2 ways of selecting two of them. Because there are 13 different face values, the number of ways of selecting two cards with the same face value is 13 ⋅ 4C2 . There are 12 remaining face values, so there are 12 ⋅ 4 C2 ways of selecting two cards having a different face value. After making these selections, there are 44 cards available with a different face value. Thus the number of 5-card hands with two cards

15. In the word MISSISSIPPI, there are 11 letters with repetition: 1 M, 4 I’s, 4 S’s, and 2 P’s. Thus the number of distinguishable permutations is 11! = 34, 650 . 1!⋅ 4!⋅ 4!⋅ 2!

327



16. Nine flags must be arranged: two are red (type 1), three are green (type 2) and four are white (type 3). Thus the number of distinguishable 9! = 1260. permutations is 2! ⋅ 3! ⋅ 4!

c.

22. P ( E1 ∪ E2 ) = P( E1 ) + P( E2 ) – P( E1 ∩ E2 ) 0.7 = 0.6 + P ( E2 ) – 0.2 P ( E2 ) = 0.3

17. Of the nine professors, four go to Dalhousie University (Cell A), three go to St. Mary’s (Cell B), and two are not assigned (Cell C). The number of possible assignments is 9! = 1260. 4! ⋅ 3! ⋅ 2!

10! 10 ⋅ 9 ⋅ 8! 10 ⋅ 9 = = = 45 2!⋅ 8! 2 ⋅1 ⋅ 8! 2 ⋅1 Let E be the event that box is rejected. If box is rejected, the one defective chip must be in the two-chip sample and there are nine possibilities for the other chip. Thus n(E) = 9 n( E ) 9 1 = = = 0.2 . and P(E) = n( S ) 45 5

23. n(S) =

18. Two of the three vans can be selected in 3 C2 ways. After two vans are chosen, the operator must assign 14 people so that 7 go to one van (cell 1) and 7 go to the other van (cell 2). This 14! ways. By the basic can be done in 7! ⋅ 7! counting principle, the number of ways to assign the people to two vans is 14! 3! 14! = ⋅ = 10, 296 . 3 C2 ⋅ 7! ⋅ 7! 2! ⋅1! 7! ⋅ 7! 19. a.

E1 ∩ E2 = {4, 5, 6}

c.

E1′ ∪ E2 = {7,8} ∪ {4,5, 6, 7} = {4,5, 6, 7,8}

d. The intersection of any event and its complement is ∅. e.

( E1 ∩ E2′ )′ = ({1, 2,3, 4,5, 6} ∩ {1, 2,3,8})′ = {1, 2, 3}′ = {4, 5, 6, 7, 8}

f.

From (b), E1 ∩ E2 ≠ ∅ , so E1 and E2 are not mutually exclusive.

20. a.

{1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}

21. a.

b.

=

25. Number of ways to answer exam is 45 = 1024 = n( S ) . Let E = {exactly two questions are incorrect). The number of ways of selecting two of the five 5! = 10 . questions that are incorrect is 5 C2 = 2! ⋅ 3! However, there are three ways to answer a question incorrectly. Since two questions are incorrect n(E) = 10 · 3 · 3 = 90. Thus n( E ) 90 45 = = . P(E) = n( S ) 1024 512

b. {2H, 2T} c.

10 C2

24. Percentage of rats given drug D = 100 – (35 + 25 + 15) = 25%. Number of rats given C = 100(0.15) = 15. Number of rats given D = 100(0.25) = 25. If E = event that rat was injected with C or D, n( E ) 15 + 25 = = 0.40. then P(E) = n( S ) 100 If the experiment is repeated on a larger group of 300 rats but with the drugs given in the same proportion, then the number of rats given drug C is 300(0.15) = 45 and the number of rats given drug D is 300(0.25) = 75 and n( E ) 45 + 75 = = 0.40. Thus there is no P( E ) = n( S ) 300 effect on the previous probability.

E1 ∪ E2 = {1, 2, 3, 4, 5, 6, 7}

b.

{R1R 2 R 3 , G1G 2 G 3 }

{2H, 4H, 6H}

26. a.

{R1R 2 R 3 , R1R 2 G 3 , R1G 2 R 3 , R1G 2 G 3 , G1R 2 R 3 , G1R 2 G 3 , G1G 2 R 3 , G1G 2 G 3 } {R1R 2 G 3 , R1G 2 R 3 , G1R 2 R 3 } 328

Of the 200 cola drinkers, 35 like both A and B. Thus 35 7 . = P(likes both A and B) = 200 40


Chapter 8 Review

b. If a person likes A but not B, then the person likes A only, and conversely. Thus 70 7 = . P(likes A, but not B) = 200 20 27. a.

There are 10 jelly beans in the bag. n(S) = 10 · 10 = 100 n( Eboth red ) = 4 ⋅ 4 = 16 Thus P ( Eboth red ) =

n( Eboth red ) 16 4 = = . n( S ) 100 25

b. n(S) = 10 · 9 = 90 n( Eboth red ) = 4 ⋅ 3 = 12 Thus P ( Eboth red ) =

12 2 = . 90 15

28. n(S) = 6 · 6 = 36 a.

E2 or 7 = {(1, 1), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3)} P ( E2 or 7 ) =

n( E2 or 7 ) 7 = n( S ) 36

b.

Emultiple of 3 = E3, 6, 9 or 12 = {(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (3, 6), (6, 3), (4, 5), (5, 4), (6, 6)} n( Emultiple of 3 ) 12 1 = = P ( Emultiple of 3 ) = 36 3 n( S )

c.

Eno less than 7 = E7, 8, 9, 10, 11, or 12 = {(1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (2, 6), (6, 2), (3, 5), (5, 3),

(4, 4), (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)} n( E7, 8, 9, 10, 11, or 12 ) 21 7 = = P ( Eno less than 7 ) = 36 12 n( S ) 29. n(S) = 52 · 52 ⋅ 52. a.

There are 26 black cards in a deck. Thus n( Eall black ) = 26 ⋅ 26 ⋅ 26 and P ( Eall black ) =

26 ⋅ 26 ⋅ 26 1 = . 52 ⋅ 52 ⋅ 52 8

b. There are 13 diamonds in a deck, none of which are black. If E = event that two cards are black and the other is a diamond, then E occurs if the diamond is the first, second, or third card. Thus 3 ⋅13 ⋅ 26 ⋅ 26 3 n(E) = 13 ⋅ 26 ⋅ 26 + 26 ⋅ 13 ⋅ 26 + 26 ⋅ 26 ⋅ 13 = 3 ⋅ 13 ⋅ 26 ⋅ 26 and P ( E ) = = . 52 ⋅ 52 ⋅ 52 16

329


30. n(S) = a.

52 C2

=


52! = 1326 2!⋅ 50!

There are 13 hearts in a deck. Thus 13! n( Eboth hearts ) = 13C2 = = 78 2! ⋅11! and P ( Eboth hearts ) =

78 1 = . 1326 17

b. Out of 36 sample points, the event {getting a total of 7 and having a 4 show} is {(4, 3), (3, 4)}. Thus the probability of this 2 1 = . event is 36 18

b. There are four aces and two red kings, and no red king is an ace. If E = event that one card is an ace and the other is a red king, then n(E) = 4 · 2 = 8 and 8 4 = ≈ 0.006. P(E) = 1326 663 3

3 8 5 8

31.

P( E ) = 8 = P( E ′) 1 – 3 8

32.

P( E ) 0.92 0.92 92 23 = = = = or 23:2 P( E′) 1 – 0.92 0.08 8 2

()

33. P ( E ) =

6 6 = 6 +1 7

34. P ( E ) =

3 3 = 3+ 4 7

35. P ( F ′ H ) =

=

40. The reduced sample space consists of {(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}. Out of these 10 points, only one has a first toss that is less than 4. Thus the 1 . conditional probability is 10

3 or 3:5 5

P( F ′ ∩ H ) = P( H )

10 52 1 4

=

41. The second number must be a 1 or 2, so the reduced sample space has 6 · 2 = 12 sample points. Of these, the event {first number ≤ second number} consists of (1, 1), (1, 2), and (2, 2). Thus the conditional 3 1 = . probability is 12 4 42. It does not matter whether the first two cards are drawn or are left in place. Thus, imagine that they are merely lifted high enough for the third card to be drawn. The probability that this card is 1 a heart is . 4

10 13

36. The reduced sample space consists of {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6)}. In none of these 11 points, is the sum less than 7. Thus P(sum < 7 | a 6 shows) = 0. 37. P ( S ∩ M ) = P ( S ) P ( M | S ) = (0.6)(0.7) = 0.42

43. a.

b.

38. P (Q ∩ H ∩ AC ) = P (Q) P ( H ) P ( AC ) =

The reduced sample space consists of {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (1, 4), (2, 4), (3, 4), (5, 4), (6, 4)}. In two of these 11 points, the sum of the components is 7. Thus 2 . P(sum = 7 | a 4 shows) = 11

4 13 1 1 ⋅ ⋅ = 52 52 52 2704

P ( L′ | F ) =

n( L′ ∩ F ) 160 1 = = n( F ) 480 3

400 2 = and 600 3 n( L ∩ M ) 80 2 = = . P(L|M) = n( M ) 120 3 Since P(L|M) = P(L), events L and M are independent. P ( L) =

44. E = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)} F = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)} a.

330

Since E ∩ F = {(4, 4)} ≠ ∅ , E and F are not mutually exclusive.


b.

P( E ) =

6 1 P( E ∩ F ) = = and P ( E | F ) = 36 6 P( F )

Chapter 8 Review

1 36 6 36

=

1 . 6

Since P(E) = P(E | F), events E and F are independent. 45. P = {attend public college} M = {from middle-class family} 125 5 = P( P) = 175 7 n( P ∩ M ) 55 11 = = P( P | M ) = n( M ) 80 16 Since P(P | M) ≠ P(P), events P and M are dependent. 46. P ( E | F ) =

P( E ∩ F ) P( F )

so P ( E ∩ F ) = P ( E F ) P ( F ) =

1 1 1 ⋅ = , thus 6 3 18 P( E ∪ F ) = P( E ) + P( F ) – P( E ∩ F )

=

47. a.

1 1 1 19 + − = . 4 3 18 36

P(none take root) = (0.3)(0.3)(0.3)(0.3) = 0.0081

b. The probability that a particular two shrubs take root and the remaining two do not is (0.7)(0.7)(0.3)(0.3). The number of ways the two that take root can be chosen from the four shrubs is 4 C2 . Thus P(exactly two take root) = 4 C2 (0.7)2 (0.3)2 = 0.2646 . c.

For at most two shrubs to take root, either none does, exactly one does, or exactly two do. P(none) + P(exactly one) + P(exactly two) = 0.0081 + 4C1 (0.7)(0.3)3 + 0.2646 = 0.0081 + 0.0756 + 0.2646 = 0.3483

48. Being effective for at least three of the persons means that it is effective for exactly three of them or for all four of them. Thus P(exactly three) + P(all four) = 4C3 (0.75)(0.75)(0.75)(0.25) + (0.75)(0.75)(0.75)(0.75) ≈ 0.738 49. P ( RII ) = P (GI ) P ( RII | GI ) + P ( RI ) P( RII | RI ) 3 4 2 5 22 . = ⋅ + ⋅ = 5 9 5 9 45

50. a.

P (W ) = P ( BI ) P (W | BI ) + P ( BII ) P(W | BII )

=

1 2 1 3 1 3 7 ⋅ + ⋅ = + = 2 6 2 5 6 10 15

331

Chapter 8: Introduction to Probability and Statistics b.

ISM: Introductory Mathematical Analysis Mathematical Snapshot Chapter 8

P ( BII | W )

=

P ( BII ) P (W | BII ) P ( BI ) P (W | BI ) + P( BII ) P (W | BII )

=

1⋅3 2 5 7 15

51. P (G | A) =

=

1. Trial and error should yield a critical value of around 0.645.

9 14

2. Possible answers: One could use cellular automata to model disease spread. The rules would be similar to the fad model, since a person who recovers from a disease is generally immune for some time afterward. One could also use cellular automata to model the formation of political opinion blocks. Each cell could be in one of three of four states, and a cell could be influenced by its neighbors. Some cells could be highly subject to neighbor influence while others were relatively immune.

P (G ∩ A) 0.1 1 = = P ( A) 0.4 4

52. S = {live within the state} and F = {first time attending}. P( S ′) P( F ′ | S ′) P ( S ′ | F ′) = P ( S ) P ( F ′ | S ) + P ( S ′) P ( F ′ | S ′) =

53. a.

b.

98 ⋅ 27 507 100 409 ⋅ 60 + 98 ⋅ 27 507 100 507 100

=

441 ≈ 0.097 4531

F = {produced by first shift} S = {produced by second shift} D = {scratched} P(D) = P(F)P(D|F) + P(S)P(D|S) 3000 5000 = ⋅ (0.01) + ⋅ (0.02) 8000 8000 = 0.00375 + 0.0125 = 0.01625 P( F ) P( D | F ) P( F ) P( D | F ) + P( S ) P( D | S ) 0.00375 3 = = ≈ 0.23 0.01625 13

P( F | D) =

54. E = {passed the exam} S = {satisfactory performance}. P( E ) P( S | E ) P( E | S ) = P( E ) P( S | E ) + P ( E′) P ( S | E′) =

(0.35)(0.8) 0.28 56 = = ≈ 0.59 (0.35)(0.8) + (0.65)(0.3) 0.475 95

332

Chapter 9 Problems 9.1 1. µ =

∑ x f ( x) = 0(0.1) + 1(0.4) + 2(0.2) + 3(0.3) = 1.7 x

Var( X ) =

∑ x2 f ( x) − µ 2 = [02 (0.1) + 12 (0.4) + 22 (0.2) + 32 (0.3)] − (1.7)2 = 1.01 x

σ = Var( X ) = 1.01 ≈ 1.00 0.5

f(x)

0.4 0.3 0.2 0.1 x 0

2. µ =

1

2

3

∑ x f ( x) = 4(0.4) + 5(0.6) = 4.6 x

Var(X) = [42 (0.4) + 52 (0.6)] − (4.6)2 = 0.24

σ = 0.24 ≈ 0.49 1.0

f(x)

0.8 0.6 0.4 0.2 x 5

3. µ =

⎛1⎞

⎛1⎞

⎛1⎞

∑ x f ( x) = 1⎜⎝ 4 ⎟⎠ + 2 ⎜⎝ 4 ⎟⎠ + 3 ⎜⎝ 2 ⎟⎠ = 4 = 2.25 9

x

Var( X ) =

⎡ ⎛1⎞

⎛1⎞

⎛ 1 ⎞⎤ ⎛ 9 ⎞

∑ x2 f ( x) − µ 2 = ⎢⎣12 ⎜⎝ 4 ⎟⎠ + 22 ⎜⎝ 4 ⎟⎠ + 32 ⎜⎝ 2 ⎟⎠⎥⎦ − ⎜⎝ 4 ⎟⎠ x

σ=

11 11 = ≈ 0.83 16 4

333

2

=

11 = 0.6875 16

Chapter 9: Additional Topics in Probability

⎛1⎞

⎛2⎞

⎛1⎞


⎛2⎞

⎛1⎞

∑ x f ( x) = 0 ⎜⎝ 7 ⎟⎠ + 1⎜⎝ 7 ⎟⎠ + 2 ⎜⎝ 7 ⎟⎠ + 3 ⎜⎝ 7 ⎟⎠ + 4 ⎜⎝ 7 ⎟⎠ = 7

4. µ =

14

=2

x

⎡ ⎛1⎞ 12 ⎛2⎞ ⎛1⎞ ⎛2⎞ ⎛ 1 ⎞⎤ ≈ 1.71 Var( X ) = ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ + 32 ⎜ ⎟ + 42 ⎜ ⎟ ⎥ − 22 = 7 7 7 7 7 7 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎝ ⎠ 12 ≈ 1.31 7

σ=

P(X = 3) = 1 − [P(X = 5) + P(X = 6) + P(X = 7)] = 1 − [0.3 + 0.2 + 0.4]= 0.1

5. a.

µ=

b.

∑ x f ( x) = 3(0.1) + 5(0.3) + 6(0.2) + 7(0.4) = 5.8 x

σ2 =

c.

∑ x2 f ( x) − µ 2 = [32 (0.1) + 52 (0.3) + 62 (0.2) + 72 (0.4)] − (5.8)2 = 1.56 x

6a + 2a + 0.2 = 1 ⇒ a = 0.1 Thus P(X = 2) = 6(0.1) = 0.6, and P(X = 4) = 2(0.1) = 0.2.

6. a.

µ = 2(0.6) + 4(0.2) + 6(0.2) = 3.2.

b.

7. Distribution of X: 1 3 3 1 f (0) = , f (1) = , f (2) = , f (3) = 8 8 8 8 ⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 3 ⎞ ⎛ 1 ⎞ 12 3 = = 1.5 E( X ) = x f ( x ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟ + 3 ⎜ ⎟ = ⎝8⎠ ⎝8⎠ ⎝8⎠ ⎝8⎠ 8 2

∑ x

σ = Var( X ) = 2

∑ x2 f ( x) − [ E ( x)]2 x

⎡ ⎛1⎞ ⎛3⎞ ⎛3⎞ ⎛ 1 ⎞⎤ ⎛ 3 ⎞ = ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ + 32 ⎜ ⎟ ⎥ − ⎜ ⎟ ⎝8⎠ ⎝8⎠ ⎝ 8 ⎠⎦ ⎝ 2 ⎠ ⎣ ⎝8⎠ 24 9 6 3 = − = = = 0.75 8 4 8 4

σ=

2

3 3 = ≈ 0.87 4 2

8. Distribution of X: f (1) =

4 2 2 1 = , f (2) = = 6 3 6 3

⎛2⎞ ⎛1⎞ 4 E ( X ) = 1⎜ ⎟ + 2 ⎜ ⎟ = ≈ 1.33 ⎝3⎠ ⎝3⎠ 3

σ2 =

⎡ ⎛2⎞

⎛ 1 ⎞⎤ ⎛ 4 ⎞

∑ x2 f ( x) − [ E ( x)]2 = ⎢⎣12 ⎜⎝ 3 ⎟⎠ + 22 ⎜⎝ 3 ⎟⎠⎥⎦ − ⎜⎝ 3 ⎟⎠

2

= 2−

x

σ=

2 ≈ 0.47 9

334

16 2 = ≈ 0.22 9 9


Section 9.1

9. The number of outcomes in the sample space is 5 C2 = 10 . Distribution of X: C ⋅ C C 1 3 f (0) = 2 2 = , f (1) = 2 1 3 1 = , 10 10 10 5 C 3 f (2) = 3 2 = 10 10 ⎛ 1 ⎞ ⎛3⎞ ⎛ 3⎞ E( X ) = x f ( x ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟ ⎝ 10 ⎠ ⎝ 5 ⎠ ⎝ 10 ⎠

13. a.

If X is the gain (in dollars), then X = –2 or 4998. Distribution of X: 7999 1 , f (4998) = f (−2) = 8000 8000 E ( x) = ∑ xf ( x) x

∑

7999 1 + 4998 ⋅ 8000 8000 11, 000 =− ≈ −$1.38 (a loss) 8000 = −2 ⋅

x

6 = = 1.2 5

σ2 =

∑ x2 f ( x) − [ E ( x)]2

b. Here X = –4 or 4996. Distribution of X: 7998 2 f (−4) = , f (4996) = 8000 8000

x

⎡ ⎛ 1 ⎞ ⎛3⎞ ⎛ 3 ⎞⎤ ⎛ 6 ⎞ = ⎢ 02 ⎜ ⎟ + 12 ⎜ ⎟ + 22 ⎜ ⎟ ⎥ − ⎜ ⎟ 10 5 ⎝ ⎠ ⎝ 10 ⎠ ⎦ ⎝ 5 ⎠ ⎣ ⎝ ⎠ 9 36 9 = − = = 0.36 5 25 25

σ=

2

E ( X ) = ∑ xf ( x) x

7998 2 + 4996 ⋅ 8000 8000 = −$2.75 (a loss) = −4 ⋅

9 3 = = 0.6 25 5

14. If X is the gain (in dollars) per game, then X = 10 or –6. Distribution of X: 2 1 6 3 f (10) = = , f (−6) = = 8 4 8 4 1 3 E( X ) = x f ( x) = 10 ⋅ + (−6) ⋅ 4 4

10. Distribution of X: 9 12 4 f (0) = , f (1) = , f (2) = 25 25 25 ⎛ 9 ⎞ ⎛ 12 ⎞ ⎛ 4 ⎞ 20 4 E ( X ) = 0 ⎜ ⎟ + 1⎜ ⎟ + 2 ⎜ ⎟ = = 25 25 ⎝ ⎠ ⎝ ⎠ ⎝ 25 ⎠ 25 5 = 0.8 ⎡

⎛ 9 ⎞ 2 ⎛ 12 ⎞ 2 ⎛ 4 ⎞ ⎤ ⎛ 4 ⎞ ⎟ + 1 ⎜ 25 ⎟ + 2 ⎜ 25 ⎟ ⎥ − ⎜ 5 ⎟ ⎝ ⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎣ ⎝ 25 ⎠ 28 16 12 = − = = 0.48 25 25 25

σ 2 = ⎢ 02 ⎜

σ=

∑

2

x

= −$2 (a loss)

15. Let X = daily earnings (in dollars). Distribution of X: 4 3 f (200) = , f (−30) = 7 7

12 2 3 = ≈ 0.69 25 5

E( X ) =

11.

C 1 f (0) = P ( X = 0) = 2 2 = 10 5 C2

x

4 3 = 200 ⋅ + (−30) ⋅ 7 7 710 = ≈ $101.43 7

C ⋅ C 6 3 f (1) = P ( X = 1) = 3 1 2 1 = = 10 5 5 C2 C 3 f (2) = P( X = 2) = 3 2 = 10 5 C2

12. P ( X = x) =

∑ x f ( x)

16. Let X = gain (in dollars) to the chain of a restaurant in a shopping center. Distribution of X: f (75, 000) = 0.65, f (−20, 000) = 0.35 E(X) = 75,000(0.65) + (–20,000)(0.35) = $41,750.

⋅ 6C3− x 10 C3

4 Cx

335



17. The probability that a person in the group is not hospitalized is 1 – (0.001 + 0.002 + 0.003 + 0.004 + 0.008) = 0.982. Let X = gain (in dollars) to the company from a policy. Distribution of X: f (10) = 0.982, f (−90) = 0.001, f (−190) = 0.002, f (−290) = 0.003, f (−390) = 0.004, f (−490) = 0.008 E ( X ) = 10(0.982) + (−90)(0.001) + (−190)(0.002) + (−290)(0.003) + (–390)(0.004) + (–490)(0.008) = $3.00 18. E(X) = 0(0.05) + 1(0.10) + 2(0.15) + 3(0.20) + 4(0.15) + 5(0.15) + 6(0.10) + 7(0.05) + 8(0.05) = 3.70 19. Let p = the annual premium (in dollars) per policy. If X = gain (in dollars) to the company from a policy, then either X = p or X = –(180,000 – p). We set E(X) = 50: −(180, 000 − p )(0.002) + p (0.998) = 50 −360 + 0.002 p + 0.998 p = 50 −360 + p = 50 p = $410 20. Let X = player’s gain (in dollars) per play. Distribution of X: 1 36 f (35) = , f (−1) = 37 37 1 36 1 E ( X ) = 35 ⋅ + (−1) ⋅ =− ≈ −$0.03 (a loss) 37 37 37 21. Let X = gain (in dollars) on a play. 5 If 0 heads show, then X = 0 − 1.25 = − . 4 1 If exactly 1 head shows, then X = 1.00 − 1.25 = − . 4 3 If 2 heads show, then X = 2.00 − 1.25 = . 4 Distribution of X: ⎛ 5⎞ 1 ⎛ 1⎞ 1 ⎛3⎞ 1 f ⎜− ⎟ = , f ⎜− ⎟ = , f ⎜ ⎟ = ⎝ 4⎠ 4 ⎝ 4⎠ 2 ⎝4⎠ 4 1 ⎛ 5 ⎞⎛ 1 ⎞ ⎛ 1 ⎞⎛ 1 ⎞ ⎛ 3 ⎞⎛ 1 ⎞ E ( X ) = ⎜ − ⎟ ⎜ ⎟ + ⎜ − ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ = − = −0.25 4 ⎝ 4 ⎠⎝ 4 ⎠ ⎝ 4 ⎠⎝ 2 ⎠ ⎝ 4 ⎠⎝ 4 ⎠ Thus there is an expected loss of $0.25 on each play. For a fair game, let p = amount (in dollars) paid to play. Distribution of X: 1 1 1 f (− p ) = , f (1 − p ) = , f (2 − p ) = 4 2 4 We set E(X) = 0: 1 1 1 (− p ) + (1 − p ) + (2 − p ) = 0 4 2 4 p 1 p 1 p − + − + − =0 4 2 2 2 4 1− p = 0 p =1 Thus you should pay $1 for a fair game. 336


Section 9.2


0

2.

1. Here p = 0.30, q = 1 – p = 0.70, and n = 4.

1

P ( X = 0) = 4C0 (0.3)0 (0.7)4 = 0.2401 2401 10, 000 4116 10, 000 2646 10, 000

3.

756 10, 000 0

81 10, 000

3

0

2

2! 16 ⎛1⎞ ⎛4⎞ f (0) = 2 C0 ⎜ ⎟ ⎜ ⎟ = ⋅1 ⋅ 5 5 0! ⋅ 2! 25 ⎝ ⎠ ⎝ ⎠ 16 16 = 1⋅1⋅ = 25 25 1

2

1

3

0

3! 8 ⎛2⎞ ⎛1⎞ ⋅ ⋅1 f (3) = 3C3 ⎜ ⎟ ⎜ ⎟ = 3!⋅ 0! 27 ⎝ 3⎠ ⎝3⎠ 8 8 = 1 ⋅ ⋅1 = 27 27

1

2! 1 4 ⎛1⎞ ⎛4⎞ f (1) = 2C1 ⎜ ⎟ ⎜ ⎟ = ⋅ ⋅ ⎝ 5 ⎠ ⎝ 5 ⎠ 1!⋅1! 5 5 1 4 8 = 2⋅ ⋅ = 5 5 25 2

2

3! 4 1 ⎛2⎞ ⎛1⎞ ⋅ ⋅ f (2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 2!⋅1! 9 3 ⎝ 3⎠ ⎝3⎠ 4 1 4 = 3⋅ ⋅ = 9 3 9

Problems 9.2

µ = np = 3 ⋅

0

2! 1 ⎛1⎞ ⎛4⎞ f (2) = 2C2 ⎜ ⎟ ⎜ ⎟ = ⋅ ⋅1 5 5 2! ⋅ 0! 25 ⎝ ⎠ ⎝ ⎠ 1 1 . = 1 ⋅ ⋅1 = 25 25 1 2 µ = np = 2 ⋅ = 5 5 1 4 σ = npq = 2 ⋅ ⋅ 5 5

=

0

1 1 ⎛ 2⎞ ⎛1⎞ f (0) = 3C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ = 27 27 ⎝ 3⎠ ⎝ 3⎠ 3! 2 1 ⎛ 2⎞ ⎛1⎞ ⋅ ⋅ f (1) = 3C1 ⎜ ⎟ ⎜ ⎟ = 1!⋅ 2! 3 9 ⎝ 3⎠ ⎝3⎠ 2 1 2 = 3⋅ ⋅ = 3 9 9

P ( X = 4) = 4 C4 (0.3) (0.7) = 0.0081

1.

0

1

4

=

3

1 1 ⎛1⎞ ⎜ 2 ⎟ = 1 ⋅ 8 ⋅1 = 8 ⎝ ⎠ 3 2 1 1 3 σ = npq = 3 ⋅ ⋅ = 2 2 2

P ( X = 3) = 4C3 (0.3)3 (0.7)1 = 0.0756 =

1

⎛1⎞ f (3) = 3C3 ⎜ ⎟ ⎝2⎠ 1 µ = np = 3 ⋅ = 2

P ( X = 2) = 4C2 (0.3) 2 (0.7) 2 = 0.2646 =

2

1 1 3 ⎛1⎞ ⎛1⎞ f (2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ = 2 2 4 2 8 ⎝ ⎠ ⎝ ⎠

P ( X = 1) = 4C1 (0.3)1 (0.7)3 = 0.4116 =

2

1 1 3 ⎛1⎞ ⎛1⎞ f (1) = 3C1 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ = 2 4 8 ⎝2⎠ ⎝2⎠

P ( X = x) = n C x p x q n − x , x = 0, 1, 2, 3, 4

=

3

1 1 ⎛1⎞ ⎛1⎞ f (0) = 3C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ = 2 2 8 8 ⎝ ⎠ ⎝ ⎠

=

4.

2 2 1 = 2; σ = npq = 3 ⋅ ⋅ 3 3 3

2 6 = 3 3

f (0) = 4C0 (0.4)0 (0.6)4 =

4! ⋅1 ⋅ (0.6)4 0!⋅ 4!

= 1 ⋅1 ⋅ (0.6)4 = 0.1296 f (1) = 4C1 (0.4)1 (0.6)3 = = 4(0.4)(0.6)3 = 0.3456

8 2 2 = 25 5

337

4! (0.4)(0.6)3 1!⋅ 3!


f (2) = 4C2 (0.4)2 (0.6) 2 =

ISM: Introductory Mathematical Analysis 11. Let X = number of heads that occurs. 1 p = , n = 11 2

4! (0.4) 2 (0.6) 2 2!⋅ 2!

= 6(0.4) 2 (0.6) 2 = 0.3456

8

⎛1⎞ ⎛1⎞ P ( X = 8) = 11C8 ⎜ ⎟ ⎜ ⎟ ⎝2⎠ ⎝2⎠ 1 1 = 165 ⋅ ⋅ 256 8 165 = ≈ 0.081 2048

4! f (3) = 4C3 (0.4) (0.6) = (0.4)3 (0.6) 3!⋅1! 3

1

= 4(0.4)3 (0.6) = 0.1536 f (4) = 4C4 (0.4)4 (0.6)0 =

4! (0.4) 4 ⋅1 4!⋅ 0!

= 1(0.4)4 ⋅1 = 0.0256 µ = np = 4(0.4) = 1.6

12. Let X = number of correct answers. p =

σ = npq = 4(0.4)(0.6) ≈ 0.98

3

5

1

2

3

⎛1⎞ ⎛ 2⎞ 6. P ( X = 2) = 5C2 ⎜ ⎟ ⎜ ⎟ ⎝3⎠ ⎝ 3⎠ 1 8 80 = 10 ⋅ ⋅ = ≈ 0.3292 9 27 243

2

14. Let X = number of aces selected. The probability 4 1 of selecting an ace on any draw is p = = . 52 13 n=3

9. P ( X < 2) = P ( X = 0) + P ( X = 1) 1

⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ = 5C0 ⎜ ⎟ ⎜ ⎟ + 5C1 ⎜ ⎟ ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ ⎝2⎠ ⎝2⎠ 1 1 1 6 3 = 1⋅1⋅ + 5 ⋅ ⋅ = = 32 2 16 32 16

2

⎛ 7 ⎞ ⎛ 5 ⎞ P ( X = 2) = 4 C2 ⎜ ⎟ ⎜ ⎟ ⎝ 12 ⎠ ⎝ 12 ⎠ 49 25 1225 = 6⋅ ⋅ = ≈ 0.3545 144 144 3456

2

8. P ( X = 4) = 7 C4 (0.2)4 (0.8)3 = 35(0.0016)(0.512) = 0.028672

5

3

13. Let X = number of green marbles drawn. The probability of selecting a green marble on any 7 draw is , n = 4. 12

16 1 ⎛4⎞ ⎛1⎞ 7. P ( X = 2) = 4 C2 ⎜ ⎟ ⎜ ⎟ = 6 ⋅ ⋅ 5 5 25 25 ⎝ ⎠ ⎝ ⎠ 96 = = 0.1536 625

0

1 ,n =6 4

27 ⎛1⎞ ⎛3⎞ P ( X = 3) = 6C3 ⎜ ⎟ ⎜ ⎟ = 20 ⋅ ⎝4⎠ ⎝4⎠ 46 540 = ≈ 0.132 4096

5. P(X = 5) = 6 C5 (0.2) (0.8) = 6(0.00032)(0.8) = 0.001536

2

3

2

1

1 12 ⎛ 1 ⎞ ⎛ 12 ⎞ P ( X = 2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ 169 13 ⎝ 13 ⎠ ⎝ 13 ⎠ 36 = ≈ 0.016 2197

4

15. Let X = number of defective switches selected. The probability that a switch is defective is p = 0.02, n = 4.

10. P ( X ≥ 2) = 1 − [ P ( X = 0) + P ( X = 1)] 0 6 1 5 ⎡ ⎛ 2⎞ ⎛1⎞ ⎛ 2⎞ ⎛1⎞ ⎤ = 1 − ⎢ 6 C0 ⎜ ⎟ ⎜ ⎟ + 6 C1 ⎜ ⎟ ⎜ ⎟ ⎥ ⎢⎣ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎥⎦ 2 ⎤ 13 ⎡ 1 = 1 − ⎢1 ⋅ + 6⋅ ⎥ = 1 − 729 729 729 ⎣ ⎦ 716 = ≈ 0.982 729

P ( X = 2) = 4C2 (0.02) 2 (0.98) 2 = 6(0.0004)(0.9604) ≈ 0.002

16. p = 0.2, n = 3 P ( X = x) = 3C x (0.2) x (0.8)3− x

338


Section 9.2

17. Let X = number of heads that occurs. p =

a.

b.

2

1

3

0

1 ,n =3 4

9 1 3 ⎛1⎞ ⎛3⎞ P ( X = 2) = 3C2 ⎜ ⎟ ⎜ ⎟ = 3 ⋅ ⋅ = 64 4 4 16 4 ⎝ ⎠ ⎝ ⎠ 1 1 ⎛1⎞ ⎛3⎞ P ( X = 3) = 3C3 ⎜ ⎟ ⎜ ⎟ = 1 ⋅ ⋅1 = 64 4 4 64 ⎝ ⎠ ⎝ ⎠ Thus 9 1 10 5 P ( X = 2) + P ( X = 3) = + = = 64 64 64 32

18. Let X = number of hearts selected. 13 1 p= = ,n =7 52 4 4

3

a.

1 27 945 ⎛1⎞ ⎛3⎞ P ( X = 4) = 7 C4 ⎜ ⎟ ⎜ ⎟ = 35 ⋅ = ≈ 0.058 ⋅ 256 64 16,384 ⎝4⎠ ⎝4⎠

b.

P ( X ≥ 4) = P ( X = 4) + P ( X = 5) + P( X = 6) + P( X = 7) 5

2

6

1

7

945 ⎛1⎞ ⎛3⎞ ⎛1⎞ ⎛3⎞ ⎛1⎞ ⎛3⎞ + 7 C5 ⎜ ⎟ ⎜ ⎟ + 7 C6 ⎜ ⎟ ⎜ ⎟ + 7 C7 ⎜ ⎟ ⎜ ⎟ 16,384 4 4 4 4 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝4⎠ ⎝4⎠ 945 1 9 1 3 1 = + 21 ⋅ ⋅ + 7⋅ ⋅ + 1⋅ ⋅1 16,384 1024 16 4096 4 16,384 1156 289 = = ≈ 0.071 16,384 4096 =

19. Let X = number of defective in sample. 1 p= , n=6 5 P ( X ≤ 1) = P ( X = 0) + P ( X = 1) 0

6

1

⎛1⎞ ⎛4⎞ ⎛1⎞ ⎛4⎞ = 6C0 ⎜ ⎟ ⎜ ⎟ + 6 C1 ⎜ ⎟ ⎜ ⎟ 5 5 ⎝ ⎠ ⎝ ⎠ ⎝5⎠ ⎝5⎠ 4096 1 1024 = 1 ⋅1 ⋅ + 6⋅ ⋅ 15, 625 5 3125 10, 240 2048 = = ≈ 0.655 15, 625 3125

5

20. Let X = number of persons with computer. p = 0.7, n = 5 P ( X ≥ 3) = P ( X = 3) + P ( X = 4) + P ( X = 5) = 5C3 (0.7)3 (0.3) 2 + 5C4 (0.7) 4 (0.3)1 + 5C5 (0.7)5 (0.3)0 = 10(0.343)(0.09) + 5(0.2401)(0.3) + 1(0.16807)(1) = 0.3087 + 0.36015 + 0.16807 = 0.83692

339

0



21. Let X = number of hits in four at-bats. p = 0.300, n = 4 P ( X ≥ 1) = 1 − P( X = 0) = 1 − 4C0 (0.300)0 (0.700) 4 = 1 − 1 ⋅1 ⋅ (0.2401) = 0.7599

22. Let X = number of stocks that increase in value. The probability that a stock increases in value is p = 0.6. Here n = 4. We must find P(X ≥ 2) = 1 – P(X < 2) = 1 − [P(X = 0) + P(X = 1)]. P ( X = 0) = 4 C0 (0.6)0 (0.4) 4 = 1 ⋅1 ⋅ (0.0256) = 0.0256 P ( X = 1) = 4C1 (0.6)1 (0.4)3 = 4(0.6)(0.064) = 0.1536 P ( X ≥ 2) = 1 − P ( X < 2) = 1 − [0.0256 + 0.1536] = 1 − 0.1792 ≈ 0.82

23. Let X = number of girls. The probability that a child is a girl is p = P(X ≥ 2) = 1 – P(X < 2) = 1 – [P(X = 0) + P(X = 1)]. 0

1 . Here n = 5. We must find 2

5

1 1 ⎛1⎞ ⎛1⎞ P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ = 2 2 32 32 ⎝ ⎠ ⎝ ⎠ 1

4

1 1 5 ⎛1⎞ ⎛1⎞ P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ = 2 16 32 ⎝2⎠ ⎝2⎠ Thus, 5⎤ 3 13 ⎡1 = P ( X ≥ 2) = 1 − [ P( X = 0) + P ( X = 1)] = 1 − ⎢ + ⎥ = 1 − 32 32 16 16 ⎣ ⎦ 2 2 3 , n = 50, q = 1 − p = 1 − = 5 5 5 2 3 σ 2 = npq = 50 ⋅ ⋅ = 12 5 5

24. p =

25. µ = 3, σ 2 = 2 Since µ = np, then np = 3. Since σ 2 = npq, then (np )q = 2, or 3q = 2, so q = Since np = 3, then n ⋅ 2

1 = 3, or n = 9. Thus 3 7

1 128 512 ⎛1⎞ ⎛2⎞ P ( X = 2) = 9C2 ⎜ ⎟ ⎜ ⎟ = 36 ⋅ ⋅ = ≈ 0.234. 9 2187 2187 ⎝3⎠ ⎝ 3⎠

26. a.

E ( X ) = µ = np = 15(0.06) = 0.9

b.

Var( X ) = σ 2 = npq = 15(0.06)(0.94) = 0.846

c.

P ( X ≤ 1) = P( X = 0) + P ( X = 1) = 15C0 (0.06)0 (0.94)15 + 15C1 (0.06)1 (0.94)14 = 1 ⋅1 ⋅ (0.94)15 + 15(0.06)(0.94)14 ≈ 0.77

340

2 2 1 . Thus, p = 1 − q = 1 − = . 3 3 3


Section 9.3

Problems 9.3

⎡a 8. ⎢ 5 ⎣⎢ 12

⎡ 1 2⎤ 2 3⎥ 1. ⎢ ⎢− 3 1 ⎥ ⎣ 2 3⎦ No, since the entry at row 2 column 1 is negative.

b⎤ ⎥ a ⎦⎥

5 5 7 = 1, so a = 1 − = . 12 12 12 7 5 b + a = 1, so b = 1 − = . 12 12

a+

⎡ 0.1 1 ⎤ 2. ⎢ ⎥ ⎣ 0.9 0 ⎦ Yes, since all entries are nonnegative and the sum of the entries in each column is 1.

⎡ 0.4 a a ⎤ 9. ⎢⎢ a 0.1 b ⎥⎥ ⎢⎣ 0.3 b c ⎥⎦ 0.4 + a + 0.3 = 1, so a = 0.3. a + 0.1 + b = 1, 0.3 + 0.1 + b = 1, so b = 0.6. a + b + c = 1, 0.3 + 0.6 + c = 1, so c = 0.1.

⎡ 1 1 1⎤ ⎢ 2 8 3⎥ 3. ⎢ − 14 85 13 ⎥ ⎢ ⎥ ⎢ 3 1 1⎥ ⎣⎢ 4 4 3 ⎦⎥ No, since there is a negative entry.

⎡a ⎢ 10. ⎢ a ⎢ ⎣⎢ a

⎡ 0.2 0.6 0 ⎤ 4. ⎢⎢ 0.7 0.2 0 ⎥⎥ ⎢⎣ 0.1 0.2 0 ⎥⎦ No, since the sum of the entries in column 3 is not 1.

a b 1 4

a⎤ ⎥ b⎥ ⎥ c ⎦⎥

a + a + a = 1, 3a = 1, a =

1 3

1 1 1 5 = 1, + b + = 1, b = 4 3 4 12 1 5 1 a + b + c = 1, + + c = 1, c = 3 12 4

a+b+

⎡ 0.4 0 0.5 ⎤ 5. ⎢⎢ 0.2 0.1 0.3⎥⎥ ⎢⎣ 0.4 0.9 0.2 ⎥⎦ Yes, since all entries are nonnegative and the sum of the entries in each column is 1.

⎡ 0.4 ⎤ 11. ⎢ ⎥ ⎣ 0.6 ⎦ Yes, all entries are nonnegative and their sum is 1. ⎡1 ⎤ 12. ⎢ ⎥ ⎣0⎦ Yes, all entries are nonnegative and their sum is 1.

⎡ 0.5 0.1 0.3⎤ 6. ⎢⎢ 0.4 0.3 0.3⎥⎥ ⎢⎣ 0.6 0.6 0.4 ⎥⎦ No, since the sum of the entries in column 1 is not 1.

⎡ 0.2 ⎤ 13. ⎢ 0.7 ⎥ ⎢ ⎥ ⎣ 0.5 ⎦ No, the sum of the entries is not 1.

⎡ 2 b⎤ 3 ⎥ 7. ⎢ ⎢a 1 ⎥ 4⎦ ⎣ 2 1 + a = 1, so a = . 3 3 1 3 b + = 1, so b = . 4 4

⎡ 0.9 ⎤ 14. ⎢ −0.1⎥ ⎢ ⎥ ⎣ 0.2 ⎦ No, the entry in the second row is negative.

341



⎡ 2 1⎤ ⎡ 1 ⎤ ⎡ 11 ⎤ ⎥ ⎢ 4 ⎥ = ⎢ 12 ⎥ 15. X1 = TX0 = ⎢ 31 1 ⎢⎣ 3 0 ⎥⎦ ⎢⎣ 34 ⎥⎦ ⎢⎣ 12 ⎥⎦ ⎡ 2 1⎤ ⎡ 11 ⎤ ⎡ 25 ⎤ ⎥ ⎢ 12 ⎥ = ⎢ 36 ⎥ X 2 = TX1 = ⎢ 31 1 0 ⎥ ⎢⎣ 3 ⎥⎦ ⎢⎣ 12 ⎥⎦ ⎢⎣ 11 36 ⎦ ⎡ 2 1⎤ ⎡ 25 ⎤ ⎡ 83 ⎤ ⎥ ⎢ 36 ⎥ = ⎢ 108 ⎥ X3 = TX2 = ⎢ 31 ⎥ ⎢ 25 ⎥ ⎢⎣ 3 0 ⎥⎦ ⎢⎣ 11 36 ⎦ ⎣ 108 ⎦ ⎡1 16. X1 = TX0 = ⎢ 12 ⎢⎣ 2

1⎤ ⎡1⎤ 4⎥ ⎢2⎥ 3 1 ⎥⎢ ⎥ 4⎦ ⎣2⎦

⎡3⎤ = ⎢ 85 ⎥ ⎢⎣ 8 ⎥⎦

⎡1 X 2 = TX1 = ⎢ 12 ⎢⎣ 2

1 ⎤ ⎡3⎤ 4 ⎥ ⎢8 ⎥ 3 5 ⎥⎢ ⎥ 4 ⎦ ⎣8 ⎦

⎡ 11 ⎤ ⎥ = ⎢ 32 21 ⎢⎣ 32 ⎥⎦

⎡1 X3 = TX2 = ⎢ 12 ⎢⎣ 2

1 ⎤ ⎡ 11 ⎤ 4 ⎥ ⎢ 32 ⎥ 3 ⎥ ⎢ 21 ⎥ 4 ⎦ ⎣ 32 ⎦

⎡ 0.3 17. X1 = TX0 = ⎢ ⎣ 0.7 ⎡ 0.3 X 2 = TX1 = ⎢ ⎣ 0.7

⎡ 43 ⎤ ⎥ = ⎢ 128 85 ⎢⎣ 128 ⎥⎦

0.5⎤ ⎡ 0.4 ⎤ ⎡ 0.42 ⎤ = 0.5⎥⎦ ⎢⎣ 0.6 ⎥⎦ ⎢⎣ 0.58 ⎥⎦ 0.5⎤ ⎡ 0.42 ⎤ ⎡0.416 ⎤ = 0.5⎥⎦ ⎢⎣ 0.58 ⎥⎦ ⎢⎣0.584 ⎥⎦

⎡ 0.3 0.5⎤ ⎡0.416 ⎤ ⎡ 0.4168 ⎤ = X3 = TX2 = ⎢ ⎣0.7 0.5⎥⎦ ⎢⎣0.584 ⎥⎦ ⎢⎣ 0.5832 ⎥⎦

⎡ 0.1 18. X1 = TX0 = ⎢ ⎣ 0.9 ⎡ 0.1 X 2 = TX1 = ⎢ ⎣ 0.9 ⎡ 0.1 X3 = TX2 = ⎢ ⎣0.9

0.9 ⎤ ⎡ 0.2 ⎤ ⎡ 0.74 ⎤ = 0.1⎥⎦ ⎢⎣ 0.8⎥⎦ ⎢⎣ 0.26 ⎥⎦ 0.9 ⎤ ⎡ 0.74 ⎤ ⎡ 0.308⎤ = 0.1⎥⎦ ⎢⎣ 0.26 ⎥⎦ ⎢⎣0.692 ⎥⎦ 0.9 ⎤ ⎡ 0.308 ⎤ ⎡0.6536 ⎤ = 0.1⎥⎦ ⎢⎣0.692 ⎥⎦ ⎢⎣ 0.3464 ⎥⎦

⎡ 0.1 19. X1 = TX0 = ⎢ 0.2 ⎢ ⎣ 0.7 ⎡ 0.1 X 2 = TX1 = ⎢ 0.2 ⎢ ⎣ 0.7 ⎡ 0.1 X3 = TX2 = ⎢ 0.2 ⎢ ⎣ 0.7

0 0.4 0.6 0 0.4 0.6 0 0.4 0.6

0.3⎤ ⎡0.2 ⎤ ⎡ 0.26 ⎤ 0.3⎥ ⎢ 0 ⎥ = ⎢ 0.28 ⎥ ⎥⎢ ⎥ ⎢ ⎥ 0.4 ⎦ ⎣ 0.8⎦ ⎣ 0.46 ⎦ 0.3⎤ ⎡ 0.26 ⎤ ⎡ 0.164 ⎤ 0.3⎥ ⎢ 0.28⎥ = ⎢ 0.302⎥ ⎥⎢ ⎥ ⎢ ⎥ 0.4 ⎦ ⎣ 0.46⎦ ⎣ 0.534⎦ 0.3⎤ ⎡ 0.164 ⎤ ⎡ 0.1766 ⎤ 0.3⎥ ⎢ 0.302 ⎥ = ⎢ 0.3138 ⎥ ⎥ ⎢ ⎥ ⎥⎢ 0.4 ⎦ ⎣ 0.534 ⎦ ⎣ 0.5096 ⎦

342


Section 9.3

⎡ 0.4 ⎢ 0 20. X1 = TX0 = ⎢ ⎢ 0.4 ⎢⎣ 0.2

0.1 0.1 0.7 0.1

0.2 0.3 0.4 0.1

0.1⎤ ⎡ 0.1⎤ ⎡ 0.17 ⎤ 0.3⎥ ⎢ 0.3⎥ ⎢ 0.21⎥ ⎥⎢ ⎥ = ⎢ ⎥ 0.4 ⎥ ⎢0.4 ⎥ ⎢ 0.49 ⎥ 0.2 ⎥⎦ ⎢⎣0.2 ⎥⎦ ⎢⎣ 0.13⎥⎦

⎡ 0.4 ⎢ 0 X 2 = TX1 = ⎢ ⎢ 0.4 ⎢⎣ 0.2

0.1 0.1 0.7 0.1

0.2 0.3 0.4 0.1

0.1⎤ ⎡0.17 ⎤ ⎡ 0.200 ⎤ 0.3⎥ ⎢ 0.21⎥ ⎢0.207 ⎥ ⎥⎢ ⎥=⎢ ⎥ 0.4 ⎥ ⎢ 0.49 ⎥ ⎢ 0.463⎥ 0.2 ⎥⎦ ⎢⎣ 0.13⎥⎦ ⎢⎣ 0.130 ⎥⎦

⎡0.4 ⎢ 0 X3 = TX2 = ⎢ ⎢0.4 ⎣⎢0.2

0.1 0.1 0.7 0.1

0.2 0.3 0.4 0.1

0.1⎤ ⎡ 0.200 ⎤ ⎡ 0.2063⎤ 0.3⎥ ⎢ 0.207 ⎥ ⎢ 0.1986 ⎥ ⎥⎢ ⎥=⎢ ⎥ 0.4 ⎥ ⎢ 0.463 ⎥ ⎢ 0.4621⎥ 0.2 ⎦⎥ ⎣⎢ 0.130 ⎦⎥ ⎣⎢ 0.1330 ⎦⎥

21. a.

⎡1 T2 = ⎢ 4 ⎢3 ⎣4

3⎤ ⎡1 4⎥⎢4 1⎥⎢3 4⎦⎣4

⎡5 8 T3 = T2 T = ⎢ ⎢3 ⎣8

3⎤ 4⎥ 1⎥ 4⎦

⎡5 8 =⎢ ⎢3 ⎣8

3⎤ ⎡1 8⎥ ⎢4 5⎥ ⎢3 8⎦ ⎣4

3⎤ 4⎥ 1⎥ 4⎦

3⎤ 8⎥ 5⎥ 8⎦

⎡7 16 =⎢ ⎢9 ⎣ 16

9⎤ 16 ⎥ . 7⎥ 16 ⎦

3 . 8

b. Entry in row 2, column 1, of T2 is

c.

22. a.

Entry in row 1, column 2 of T3 is ⎡1 3 T2 = ⎢ ⎢2 ⎣3

1⎤⎡1 2⎥⎢3 1⎥⎢2 2⎦⎣3

⎡4 9 T3 = T2 T = ⎢ ⎢5 ⎣9

⎡4 5 ⎤ 9 12 ⎥ =⎢ ⎢5 7 ⎥ ⎣ 9 12 ⎦ 5 ⎤ ⎡ 1 1 ⎤ ⎡ 23 12 ⎥ ⎢ 3 2 ⎥ ⎢ 54 = 7 ⎥ ⎢ 2 1 ⎥ ⎢ 31 12 ⎦ ⎣ 3 2 ⎦ ⎣ 54 1⎤ 2⎥ 1⎥ 2⎦

b. Entry in row 2, column 1, of T2 is

c.

23. a.

9 . 16

Entry in row 1, column 2 of T3 is

31 ⎤ 72 ⎥ 41 ⎥ 72 ⎦

5 . 9 31 . 72

⎡ 0 0.5 0.3⎤ ⎡0 0.5 0.3⎤ ⎡ 0.50 0.23 0.27 ⎤ T = ⎢⎢1 0.4 0.3⎥⎥ ⎢⎢1 0.4 0.3⎥⎥ = ⎢⎢ 0.40 0.69 0.54 ⎥⎥ ⎢⎣ 0 0.1 0.4 ⎥⎦ ⎢⎣ 0 0.1 0.4 ⎥⎦ ⎢⎣0.10 0.08 0.19 ⎥⎦ ⎡ 0.50 0.23 0.27 ⎤ ⎡0 0.5 0.3⎤ ⎡ 0.230 0.369 0.327 ⎤ 3 2 T = T T = ⎢⎢ 0.40 0.69 0.54 ⎥⎥ ⎢⎢1 0.4 0.3⎥⎥ = ⎢⎢ 0.690 0.530 0.543 ⎥⎥ ⎢⎣ 0.10 0.08 0.19 ⎥⎦ ⎢⎣0 0.1 0.4 ⎥⎦ ⎢⎣0.080 0.101 0.130 ⎥⎦ 2

b. Entry in row 2, column 1, of T2 is 0.40. 343


c.


Entry in row 1, column 2 of T3 is 0.369. ⎡ 0.1 0.1 0.1⎤ ⎡ 0.1 0.1 0.1⎤ ⎡ 0.10 0.10 0.10 ⎤ T = ⎢⎢ 0.2 0.1 0.1⎥⎥ ⎢⎢ 0.2 0.1 0.1⎥⎥ = ⎢⎢ 0.11 0.11 0.11⎥⎥ ⎢⎣ 0.7 0.8 0.8⎥⎦ ⎢⎣ 0.7 0.8 0.8⎥⎦ ⎢⎣0.79 0.79 0.79⎥⎦ ⎡0.10 0.10 0.10 ⎤ ⎡ 0.1 0.1 0.1⎤ ⎡ 0.10 0.10 0.10 ⎤ 3 2 T = T T = ⎢⎢ 0.11 0.11 0.11⎥⎥ ⎢⎢ 0.2 0.1 0.1⎥⎥ = ⎢⎢ 0.11 0.11 0.11⎥⎥ . ⎢⎣0.79 0.79 0.79 ⎥⎦ ⎢⎣0.7 0.8 0.8⎥⎦ ⎢⎣ 0.79 0.79 0.79 ⎥⎦ 2

24. a.

b. Entry in row 2, column 1, of T2 is 0.11. c.

Entry in row 1, column 2 of T3 is 0.10.

2⎤ ⎡ 1 2 ⎤ ⎡1 0 ⎤ ⎡ − 1 2 3⎥ 2 3⎥ ⎢ ⎢ 25. T − I = − = ⎢ 1 1 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 1 − 2 ⎥ 3⎦ ⎣2 3⎦ ⎣ 2 ⎡ 1 ⎡1 0 4 ⎤ 1 1⎤ 7⎥ ⎢ ⎥ ⎢ 2 0⎥ → " → ⎢0 1 3 ⎥ ⎢− 1 3 7⎥ ⎢ 2 ⎥ ⎢ ⎢ 1 − 2 0⎥ ⎢0 0 0 ⎥ 3 ⎣ 2 ⎦ ⎣ ⎦

⎡4⎤ Q = ⎢ 73 ⎥ ⎢⎣ 7 ⎥⎦ ⎡1 2 26. T − I = ⎢ ⎢1 ⎣2 ⎡ 1 ⎢ ⎢− 1 ⎢ 2 ⎢ 1 ⎣ 2

1 1 4 − 14

1⎤ 4 ⎥ ⎡1 − 3 ⎥ ⎢0 ⎣ 4⎦

1 0⎤ ⎡ − 2 ⎢ = 1 ⎥⎦ ⎢ 1 ⎣ 2

1 ⎤ 4 ⎥ − 14 ⎥⎦

⎡1 0 1 ⎤ 1⎤ 3⎥ ⎢ ⎥ 2 0⎥ → " → ⎢0 1 3 ⎥ ⎢ ⎥ ⎥ ⎢0 0 0 ⎥ 0 ⎥⎦ ⎣ ⎦

⎡1⎤ 3 Q=⎢ ⎥ ⎢1⎥ ⎣2⎦ ⎡ 1 3 ⎤ ⎡1 0 ⎤ ⎡ − 4 3 ⎤ 5 5 ⎥ 5 5⎥ ⎢ 27. T − I = −⎢ =⎢ ⎥ 4 2 4 ⎢ ⎢ ⎥ 0 1⎦ − 53 ⎥ ⎣5 5⎦ ⎣ ⎣ 5 ⎦ ⎡ 1 ⎡1 0 3 ⎤ 1 1⎤ 7⎥ ⎢ ⎥ ⎢ ⎢ − 4 3 0⎥ → " → ⎢0 1 4 ⎥ 7⎥ ⎢ 5 5 ⎥ ⎢ ⎢ 4 − 3 0⎥ ⎢0 0 0 ⎥ ⎢⎣ 5 ⎥⎦ 5 ⎣ ⎦ ⎡3⎤ 7 Q=⎢ ⎥ ⎢4⎥ ⎣7⎦ 344


Section 9.3

⎡ 1 1 ⎤ ⎡1 0 ⎤ ⎡ − 3 1 ⎤ 4 3⎥ 4 3 ⎥ ⎢ 28. T − I = −⎢ =⎢ ⎥ 3 3 2 ⎢ ⎥ 0 1⎦ ⎢ − 13 ⎥ ⎣4 3⎦ ⎣ ⎣ 4 ⎦ ⎡ 1 ⎡1 0 4 ⎤ 1 1⎤ 13 ⎥ ⎢ ⎥ ⎢ ⎢ − 3 1 0 ⎥ → " → ⎢0 1 9 ⎥ 13 ⎥ ⎢ 4 3 ⎥ ⎢ ⎢ 3 − 1 0⎥ ⎢0 0 0 ⎥ ⎢⎣ 4 ⎥⎦ 3 ⎣ ⎦ ⎡4⎤ 13 Q=⎢ ⎥ ⎢9⎥ ⎣ 13 ⎦ ⎡ 0.4 29. T − I = ⎢⎢ 0.3 ⎢⎣ 0.3 1 ⎡ 1 ⎢ −0.6 0.6 ⎢ ⎢ 0.3 −0.7 ⎢ 0.1 ⎢⎣ 0.3

0.6 0.6 ⎤ ⎡1 0 0 ⎤ ⎡ −0.6 0.6 0.6 ⎤ 0.3 0.1⎥⎥ − ⎢⎢0 1 0 ⎥⎥ = ⎢⎢ 0.3 −0.7 0.1 ⎥⎥ 0.1 0.3⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.3 0.1 −0.7 ⎥⎦ 1 1⎤ ⎡1 0 0 0.5 ⎤ ⎢0 1 0 0.25⎥ 0.6 0 ⎥⎥ ⎥ →" → ⎢ ⎢0 0 1 0.25⎥ 0.1 0 ⎥ ⎥ ⎢ ⎥ −0.7 0 ⎥⎦ 0 ⎥⎦ ⎢⎣0 0 0

⎡ 0.5 ⎤ Q = ⎢⎢ 0.25⎥⎥ ⎢⎣ 0.25⎥⎦ ⎡ 0.1 0.4 0.3⎤ ⎡1 0 0 ⎤ ⎡ −0.9 0.4 0.3 ⎤ 30. T − I = ⎢⎢ 0.2 0.2 0.3⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ 0.2 −0.8 0.3 ⎥⎥ ⎢⎣ 0.7 0.4 0.4 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.7 0.4 −0.6 ⎥⎦ 1 1 1⎤ ⎡ 1 ⎡1 0 0 0.2707 ⎤ ⎢ −0.9 0.4 0.3 0 ⎥ ⎢ ⎥ ⎢ ⎥ → " → ⎢ 0 1 0 0.2481⎥ ⎢ 0.2 −0.8 0.3 0 ⎥ ⎢0 0 1 0.4812 ⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎥⎦ ⎢⎣ 0.7 0.4 −0.6 0 ⎥⎦ ⎢⎣ 0 0 0 ⎡ 0.2707 ⎤ Q ≈ ⎢⎢ 0.2481⎥⎥ ⎢⎣ 0.4812 ⎥⎦ Flu

31. a.

b.

No flu

Flu ⎡ 0.1 0.2 ⎤ T= No flu ⎢⎣ 0.9 0.8 ⎥⎦ ⎡ 120 ⎤ ⎡ 0.6 ⎤ 200 X0 = ⎢ ⎥ = ⎢ ⎥ . ⎢ 80 ⎥ ⎣ 0.4 ⎦ ⎣ 200 ⎦ If a period is 4 days, then 8 days corresponds to 2 periods, and 12 days corresponds to 3 periods. The state vector corresponding to 8 days from now is ⎡ 0.19 0.18⎤ ⎡ 0.6 ⎤ ⎡ 0.186 ⎤ X 2 = T2 X0 = ⎢ ⎥⎢ ⎥ = ⎢ ⎥. ⎣ 0.81 0.82 ⎦ ⎣ 0.4 ⎦ ⎣ 0.814 ⎦ 345



Thus 0.186(200) ≈ 37 students can be expected to have the flu 8 days from now. The state vector corresponding to 12 days from now is ⎡ 0.181 0.182 ⎤ ⎡ 0.6 ⎤ X3 = T3 X0 = ⎢ ⎥⎢ ⎥ ⎣0.819 0.818⎦ ⎣ 0.4 ⎦ ⎡0.1814 ⎤ =⎢ ⎥. ⎣0.8186 ⎦ Thus 0.1814(200) ≈ 36 students can be expected to have the flu 12 days from now.

b.

35% to location 1, 45.8% to location 2, 19.2% to location 3

H L H ⎡0.55 0.25⎤ 32. T = ⎢ L ⎣0.45 0.75⎥⎦ ⎡ 0.65⎤ X0 = ⎢ ⎥ ⎣ 0.35⎦

35. a.

2

c.

A B A ⎡0.7 0.4 ⎤ T= ⎢ B ⎣ 0.3 0.6 ⎥⎦

b. Wednesday corresponds to step 2. ⎡ 0.61 0.52 ⎤ T2 = ⎢ ⎥. ⎣ 0.39 0.48⎦ The probability is 0.61. 34. a.

D R O D ⎡0.8 0.1 0.3 ⎤ T= ⎢ R ⎢0.1 0.8 0.2 ⎥⎥ O ⎢⎣0.1 0.1 0.5 ⎥⎦

⎡ 0.68 0.19 0.41 ⎤ b. T = ⎢⎢ 0.18 0.67 0.29 ⎥⎥ ⎢⎣ 0.14 0.14 0.30 ⎥⎦ The probability is 0.19.

⎡ 0.415 0.325⎤ ⎡ 0.65⎤ X 2 = T X0 = ⎢ ⎥⎢ ⎥ ⎣ 0.585 0.675⎦ ⎣ 0.35⎦ ⎡ 0.3835⎤ =⎢ ⎥ ⎣ 0.6165⎦ 38.35% of the members will be performing highimpact exercising. 2

33. a.

X 2 = TX1 ⎡0.7 0.2 0.2 ⎤ ⎡ 0.30 ⎤ = ⎢⎢ 0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.48⎥⎥ ⎢⎣ 0.2 0 0.6 ⎥⎦ ⎢⎣0.22 ⎥⎦ ⎡0.350 ⎤ = ⎢⎢ 0.458 ⎥⎥ ⎢⎣0.192 ⎥⎦

X1 = TX0 ⎡0.8 0.1 0.3 ⎤ ⎡ 0.40 ⎤ = ⎢⎢0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.40 ⎥⎥ ⎢⎣0.1 0.1 0.5 ⎥⎦ ⎢⎣ 0.20 ⎥⎦ ⎡0.42 ⎤ = ⎢⎢0.40 ⎥⎥ ⎢⎣ 0.18⎥⎦

40% are expected to be Republican. U S R U ⎡0.7 0.1 0.1⎤ 36. T = ⎢ S ⎢0.1 0.8 0.1⎥⎥ R ⎢⎣0.2 0.1 0.8⎥⎦

X1 = TX0 ⎡0.7 0.2 0.2 ⎤ ⎡ 0.2 ⎤ = ⎢⎢ 0.1 0.8 0.2 ⎥⎥ ⎢⎢ 0.5⎥⎥ ⎢⎣ 0.2 0 0.6 ⎥⎦ ⎢⎣ 0.3⎥⎦ ⎡0.30 ⎤ = ⎢⎢ 0.48⎥⎥ ⎢⎣0.22 ⎥⎦

a.

30% to location 1, 48% to location 2, 22% to location 3

15 years corresponds to step 3. ⎡ 0.412 0.196 0.196 ⎤ T3 = ⎢⎢ 0.219 0.562 0.219 ⎥⎥ ⎢⎣ 0.369 0.242 0.585 ⎥⎦ The entry in row 3, column 2 of T3 is 0.242, so the probability is 0.242.

346


b.

Section 9.3

X3 = T3 X0

⎡0.412 0.196 0.196 ⎤ ⎡ 0.50 ⎤ ⎡ 0.304 ⎤ = ⎢⎢0.219 0.562 0.219 ⎥⎥ ⎢⎢ 0.25⎥⎥ = ⎢⎢0.30475⎥⎥ ⎣⎢0.369 0.242 0.585 ⎦⎥ ⎣⎢ 0.25⎦⎥ ⎣⎢0.39125⎦⎥ The population is expected to be 30.4% urban, 30.475% suburban, 39.125% rural.

37. a.

b.

c.

A Compet. A ⎡ 0.8 0.3⎤ T= Compet. ⎢⎣ 0.2 0.7 ⎥⎦ ⎡ 0.8 0.3⎤ ⎡ 0.70 ⎤ X1 = TX0 = ⎢ ⎥⎢ ⎥ ⎣ 0.2 0.7 ⎦ ⎣ 0.30 ⎦ ⎡0.65⎤ =⎢ ⎥ ⎣0.35⎦ A is expected to control 65% of the market. ⎡ 0.8 T−I = ⎢ ⎣ 0.2 1 ⎡ 1 ⎢ −0.2 0.3 ⎢ ⎢⎣ 0.2 −0.3

0.3 ⎤ ⎡1 0 ⎤ ⎡ −0.2 0.3⎤ − = 0.7 ⎥⎦ ⎢⎣ 0 1 ⎥⎦ ⎢⎣ 0.2 −0.3⎥⎦ 1⎤ ⎡ 1 0 0.6 ⎤ ⎥ 0 ⎥ → … → ⎢⎢ 0 1 0.4⎥⎥ ⎢⎣ 0 0 0 ⎥⎦ 0 ⎥⎦

⎡0.6 ⎤ Q=⎢ ⎥ ⎣0.4 ⎦ In the long run, A can expect to control 60% of the market.

38. a.

b.

Fords Non-Fords Fords ⎡ 0.75 0.35⎤ T= Non-fords ⎢⎣ 0.25 0.65⎥⎦ ⎡ 0.75 0.35⎤ ⎡1 0 ⎤ ⎡ −0.25 0.35⎤ T−I = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣ 0.25 0.65⎦ ⎣ 0 1 ⎦ ⎣ 0.25 −0.35⎦ 1 1⎤ ⎡ 1 ⎡1 0 0.5833⎤ ⎢ −0.25 0.35 0 ⎥ → … → ⎢ 0 1 0.4167 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0.25 −0.35 0 ⎥⎦ ⎢⎣ 0 0 0 ⎥⎦ ⎡ 0.5833⎤ Q≈⎢ ⎣ 0.4167 ⎥⎦ In the long run, 58.33% of car purchases in the region are expected to be Fords.

39. a.

T=

1 2 5 3⎤ ⎡ 1 7 7 ⎢ ⎥ 2⎢2 4⎥ ⎣7 7⎦

347



⎡ 31 27 ⎤ ⎡ 1 ⎤ ⎡ 29 ⎤ ⎡ 0.5918 ⎤ 49 49 ⎥ ⎢ 2 ⎥ ⎢ 49 ⎥ b. X 2 = T X0 = ⎢ = ≈ 18 22 ⎥ ⎢ 1 ⎥ ⎢ 20 ⎥ ⎢ 0.4082 ⎥⎦ ⎢ ⎣ 49 49 ⎦ ⎣ 2 ⎦ ⎣ 49 ⎦ ⎣ About 59.18% in compartment 1 and 40.82% in compartment 2. 2

c.

40. a.

b.

c.

3⎤ ⎡ 5 3 ⎤ ⎡1 0 ⎤ ⎡ − 2 7 7⎥ 7 7⎥ ⎢ ⎢ T−I = − = ⎢ 2 4 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 2 − 3 ⎥ 7⎦ ⎣7 7⎦ ⎣ 7 ⎡ 1 ⎡1 0 3⎤ 1 1⎤ 5⎥ ⎢ ⎥ ⎢ ⎢ − 2 3 0⎥ → … → ⎢0 1 2 ⎥ 5⎥ ⎢ 7 7 ⎥ ⎢ ⎢ 2 − 3 0⎥ ⎢0 0 0⎥ ⎢⎣ 7 ⎥⎦ 7 ⎣ ⎦ ⎡ 3 ⎤ ⎡ 0.6 ⎤ 5 Q=⎢ ⎥=⎢ ⎥ ⎢ 2 ⎥ ⎣ 0.4 ⎦ ⎣5⎦ In the long run, there will be 60% in compartment 1 and 40% in compartment 2.

Doesn't Works Work T= Works ⎡ 0.8 0.1⎤ ⎢ Doesn't Work ⎣ 0.2 0.9 ⎥⎦ ⎡ 0.562 0.219 ⎤ T3 = ⎢ ⎥ ⎣ 0.438 0.781 ⎦ The probability is 0.562. ⎡ 0.8 0.1⎤ ⎡1 0 ⎤ ⎡ −0.2 0.1⎤ T−I = ⎢ ⎥−⎢ ⎥=⎢ ⎥ ⎣ 0.2 0.9 ⎦ ⎣ 0 1 ⎦ ⎣ 0.2 −0.1⎦ ⎡1 0 1 ⎤ 1 1⎤ 3⎥ ⎡ 1 ⎢ ⎢ −0.2 0.1 0 ⎥ → … → ⎢0 1 2 ⎥ ⎢ ⎥ 3⎥ ⎢ ⎢⎣ 0.2 −0.1 0 ⎥⎦ ⎢0 0 0 ⎥ ⎣ ⎦ ⎡1⎤ 3 Q=⎢ ⎥ ⎢2⎥ ⎣3⎦ ⎛1⎞ In the long run, the number of machines working properly is ⎜ ⎟ (42) = 14 . ⎝3⎠

348


41. a.

⎡3 ⎢ T−I = 4 ⎢1 ⎣4

1⎤ 2 ⎥ ⎡1 − 1 ⎥ ⎢0 ⎣ 2⎦

⎡ 1 ⎢ ⎢− 1 ⎢ 4 ⎢ 1 ⎣ 4

⎡1 0 1⎤ ⎢ ⎥ 0⎥ → … → ⎢0 1 ⎢ ⎥ ⎢0 0 0 ⎥⎦ ⎣

1 1 2 − 12

0⎤ ⎡ − 4 =⎢ 1 ⎥⎦ ⎢ 1 ⎣ 4 1

Section 9.3

1⎤ 2⎥ 1 − 2 ⎥⎦ 2⎤ 3⎥ 1⎥ 3⎥

0⎥ ⎦

⎡2⎤ 3 Q=⎢ ⎥ ⎢1⎥ ⎣3⎦

b. Presently, A accounts for 50% of sales and in long run A will account for percentage increase in sales above the present level is

42. a.

b.

c.

66 23 − 50 50

⋅100% =

2 2 , or 66 % , of sales. Thus the 3 3

16 23 50

1 ⋅100% = 33 % . 3

A B C A ⎡ 0.8 0.2 0.2 ⎤ T= ⎢ B ⎢ 0.1 0.7 0.1 ⎥⎥ C ⎢⎣ 0.1 0.1 0.7 ⎥⎦ ⎡ 0.68 0.32 0.32 ⎤ T2 = ⎢⎢ 0.16 0.52 0.16 ⎥⎥ ⎢⎣ 0.16 0.16 0.52 ⎥⎦ The probability is 0.52.

Initially 500 customers are to be considered. The probability that a customer goes to branch A is ⎡ 0.4 ⎤ 200 100 to branch B, = 0.4 ; and to branch C, = 0.2 . Thus X0 = ⎢⎢ 0.4 ⎥⎥ . 500 500 ⎢⎣ 0.2 ⎥⎦ ⎡ 0.8 0.2 0.2 ⎤ ⎡0.4 ⎤ ⎡ 0.44 ⎤ X1 = TX0 = ⎢⎢ 0.1 0.7 0.1⎥⎥ ⎢⎢0.4 ⎥⎥ = ⎢⎢ 0.34 ⎥⎥ ⎢⎣ 0.1 0.1 0.7 ⎥⎦ ⎢⎣0.2 ⎥⎦ ⎢⎣ 0.22 ⎥⎦

Thus 0.44(500) = 220 customers can be expected to go to A on their next visit, 0.34(500) = 170 to B, and 0.22(500) = 110 to C.

349

200 = 0.4 ; 500


d.


⎡ 0.8 0.2 0.2 ⎤ ⎡1 0 0 ⎤ ⎡ −0.2 0.2 0.2 ⎤ T − I = ⎢⎢ 0.1 0.7 0.1 ⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥ = ⎢⎢ 0.1 −0.3 0.1⎥⎥ ⎢⎣ 0.1 0.1 0.7 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ ⎢⎣ 0.1 0.1 −0.3⎥⎦ 1 1 1⎤ ⎡ 1 ⎡1 0 0 0.50 ⎤ ⎢ −0.2 0.2 0.2 0 ⎥ ⎢ ⎥ ⎢ ⎥ → … → ⎢0 1 0 0.25⎥ ⎢ 0.1 −0.3 0.1 0 ⎥ ⎢0 0 1 0.25⎥ ⎢ ⎥ ⎢ ⎥ 0 ⎦⎥ ⎣⎢ 0.1 0.1 −0.3 0 ⎦⎥ ⎣⎢0 0 0 In the long run, 0.50(500) = 250 can be expected to go to A, 0.25(500) = 125 to B, and 0.25(500) to C.

⎡1 43. T2 = TT = ⎢ 2 ⎢1 ⎣2

1 ⎤ ⎡ 12 ⎥⎢ 0 ⎥⎦ ⎢⎣ 12

1 ⎤ ⎡ 34 ⎥=⎢ 0 ⎥⎦ ⎢⎣ 14

1⎤ 2⎥ 1⎥ 2⎦

Since all entries of T2 are positive, T is regular. ⎡0 1 ⎤ 2 n n 44. For the matrix A = ⎢ ⎥ , A = I (the 2 × 2 identity matris). Thus A = I if n is even, and A = A if n is odd. 1 0 ⎣ ⎦ In either case there are nonpositive entries, and thus A is not regular.

Chapter 9 Review Problems 1. µ =

∑ xf ( x) = 1⋅ f (1) + 2 ⋅ f (2) + 3 ⋅ f (3) = 1(0.7) + 2(0.1) + 3(0.2) = 1.5 x

Var( X ) =

∑ x2 f ( x) − µ 2 = ⎡⎣12 (0.7) + 22 (0.1) + 32 (0.2)⎤⎦ − (1.5)2 = 0.65 x

σ = Var( X ) = 0.65 ≈ 0.81 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

f(x)

x 1

2. µ =

2

3

∑ xf ( x) = 0 ⋅ 6 + 1⋅ 2 + 2 ⋅ 3 = 6 1

x

Var( X ) =

1

1

7

∑ x 2 f ( x) − µ 2 x

1 1 1⎤ ⎛ 7 ⎞ ⎡ = ⎢ 02 ⋅ + 12 ⋅ + 22 ⋅ ⎥ − ⎜ ⎟ 6 2 3⎦ ⎝ 6 ⎠ ⎣ 11 49 17 = − = 6 36 36

σ=

2

17 17 = ≈ 0.69 36 6

350


1/2

Chapter 9 Review

f(x)

192 32 , = 1326 221 6 1 . f (2) = P ( E2 aces ) = = 1326 221

f (1) = P ( E1 ace ) =

1/3 1/6

b.

E( X ) =

x 0

3. a.

1

=

n(S) = 2 · 6 = 12 E0 = {H1}, E1 = {T1, H2} , E2 = {T 2, H 3} , E3 = {T3, H4} , E4 = {T4, H5} , E5 = {T5, H6} , E6 = {T6}

f (1) = P ( E1 ) =

n ( E0 ) n( S )

n ( E1 ) n( S )

=

=

b.

E( X ) =

n ( E6 ) n( S )

2 1 = 12 6

=

1 . 6

1 12

n ( Etwo 10's ) = 4 ⋅ 4 .

∑ xf ( x)

Dist. of X: ⎛ 1 ⎞ 48 ⋅ 48 144 , f ⎜− ⎟ = = ⎝ 4 ⎠ 52 ⋅ 52 169

1 1+ 2 + 3 + 4 + 5 1 + + 6⋅ 12 6 12 15 6 36 = 0+ + = =3 6 12 12 = 0⋅

⎛ 3 ⎞ 2 ⋅ 4 ⋅ 48 24 , f ⎜ ⎟= = ⎝ 4 ⎠ 52 ⋅ 52 169 4⋅4 1 ⎛7⎞ . f ⎜ ⎟= = ⎝ 4 ⎠ 52 ⋅ 52 169 1 144 3 24 7 1 E( X ) = − ⋅ + ⋅ + ⋅ 4 169 4 169 4 169 −144 + 72 + 7 65 5 = =− =− ≈ −0.10 4 ⋅169 676 52 There is a loss of $0.10 per play.

52! 52 ⋅ 51 n( S ) = 52 C2 = = = 1326 . In a 2!⋅ 50! 2 deck there are 4 aces and 48 non-aces. Thus 48! 48 ⋅ 47 n ( E0 aces ) = 48 C2 = = 2!⋅ 46! 2 = 1128 . For E1 ace to occur, one card is an ace and the other is non-ace. Thus n ( E1 ace ) = 4 ⋅ 48 = 192 . n ( E2 aces ) = 4 C2 =

6. Let X = gain (in dollars) to company. Dist. of X: f(40,000) = 0.45, f(–10,000) = 1 – 0.45 = 0.55 E(X) = (40,000)(0.45) + (–10,000)(0.55) = 18,000 – 5500 = $12,500 per station

4! 4⋅3 = =6. 2!⋅ 2! 2

Therefore, f (0) = P ( E0 aces ) =

1

34 2 = 221 13

Eone 10 occurs if the first card is a 10 and the second is a non-10, or vice versa. Thus n ( Eone 10 ) = 4 ⋅ 48 + 48 ⋅ 4 = 2 ⋅ 4 ⋅ 48 .

x

4. a.

32

5. Let X = gain (in dollars) on a play. If no 10 1 1 appears, then X = 0 − = − ; if exactly one 10 4 4 1 3 appears, then X = 1 − = ; if two 10’s appear, 4 4 1 7 then X = 2 − = . 4 4 n(S) = 52 · 52. In a deck, there are 4 10’s and 48 non 10’s. Thus n ( Eno 10 ) = 48 ⋅ 48 . The event

1 12

Similarly, f(2), f(3), f(4), and f(5) equal f (6) = P ( E6 ) =

188

x

2

f (0) = P ( E0 ) =

∑ xf ( x) = 0 ⋅ 221 + 1⋅ 221 + 2 ⋅ 221

1128 188 , = 1326 221

351

Chapter 9: Additional Topics in Probability 7. a.


Let X = gain (in dollars) on each unit shipped. Then P(X = –100) = 0.08 and P(X = 200) = 1 – 0.08 = 0.92. E(X) = –100f(–100) + 200f(200) = –100(0.08) + 200(0.92) = $176 per unit

0

10.

1

2

5

0

⎛1⎞ f (5) = 5C5 ⎜ ⎟ ⎝ 3⎠ 1 µ = np = 5 ⋅ = 3

1 1 ⎛2⎞ ⋅1 = ⎜ ⎟ = 1⋅ 243 243 ⎝3⎠ 5 3 1 2 10 10 = = ≈ 1.05 3 3 9 3

σ = npq = 5 ⋅ ⋅

11. P(X ≤ 1) = P(X = 0) + P(X = 1) 0

5

1

4

⎛3⎞ ⎛1⎞ ⎛3⎞ ⎛1⎞ = 5C0 ⎜ ⎟ ⎜ ⎟ + 5C1 ⎜ ⎟ ⎜ ⎟ 4 4 ⎝ ⎠ ⎝ ⎠ ⎝4⎠ ⎝4⎠ 1 3 1 16 1 = 1⋅1⋅ + 5⋅ ⋅ = = 1024 4 256 1024 64

⎛ 40,999,999 ⎞ – 1.00 ⎜ ⎟ ≈ –0.63 ⎝ 41, 000, 000 ⎠ There is a loss of about $0.63 per play.

= 0.522

3

1 4 40 ⎛1⎞ ⎛ 2⎞ f (3) = 5C3 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅ = 27 9 243 ⎝3⎠ ⎝ 3⎠ 4 1 1 2 10 ⎛1⎞ ⎛ 2⎞ f (4) = 5C4 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ = 3 3 81 3 243 ⎝ ⎠ ⎝ ⎠

8. There are 41 million combinations from which to choose. Let x = gain (in dollars) per play. If the player wins, then x = 15,000,000 – 1.00 = 14,999,999 and 1 . If the player P(X = 14,999,999) = 41, 000, 000 loses, then X = –1.00 and 1 40,999,999 . P(X = –1.00) = 1 – = 41, 000, 000 41, 000, 000 E(X) = 14,999,999f (14,999,999) – 1.00f (–1.00) ⎛ 1 ⎞ = 14,999,999 ⎜ ⎟ ⎝ 41, 000, 000 ⎠

f (0) = 4C0 (0.15)0 (0.85)4 ≈

4

1 16 80 ⎛1⎞ ⎛ 2⎞ f (1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ = 3 81 243 ⎝3⎠ ⎝ 3⎠ 2 3 1 8 80 ⎛1⎞ ⎛ 2⎞ = f (2) = 5C2 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅ 9 27 243 ⎝ 3⎠ ⎝ 3⎠

b. Since the expected gain per unit is $176 and 4000 units are shipped per year, then expected annual profit is 4000(176) = $704,000.

9.

5

32 32 ⎛1⎞ ⎛2⎞ f (0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ = 3 3 243 243 ⎝ ⎠ ⎝ ⎠

0

4! ⋅1(0.522) 0!4!

6

6! ⎛ 2⎞ ⎛1⎞ ⎛ 1 ⎞ (1) ⎜ 12. P ( X = 0) = 6 C0 ⎜ ⎟ ⎜ ⎟ = ⎟ 3 3 0! ⋅ 6! ⎝ ⎠ ⎝ ⎠ ⎝ 729 ⎠ 1 ⎛ 1 ⎞ = 1(1) ⎜ ⎟= ⎝ 729 ⎠ 729

f (1) = 4 C1 (0.15)1 (0.85)3 4! ≈ ⋅ (0.15)(0.614) = 0.368 1!3!

1

5

6! ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛2⎞ ⎛1⎞ P ( X = 1) = 6C1 ⎜ ⎟ ⎜ ⎟ = 1!⋅ 5! ⎜⎝ 3 ⎟⎠ ⎜⎝ 243 ⎟⎠ ⎝ 3⎠ ⎝3⎠

f (2) = 4 C2 (0.15)2 (0.85)2 4! = ⋅ (0.0225)(0.7225) ≈ 0.098 2!2!

⎛ 2 ⎞ ⎛ 1 ⎞ 12 = 6⎜ ⎟⎜ ⎟= ⎝ 3 ⎠ ⎝ 243 ⎠ 729

f (3) = 4C3 (0.15)3 (0.85)1 4! = ⋅ (0.003375)(0.85) ≈ 0.011 3!1!

2

4

6! ⎛ 4 ⎞⎛ 1 ⎞ ⎛2⎞ ⎛1⎞ P ( X = 2) = 6C2 ⎜ ⎟ ⎜ ⎟ = ⎟ 2!⋅ 4! ⎜⎝ 9 ⎟⎜ ⎝ 3⎠ ⎝3⎠ ⎠⎝ 81 ⎠ 6 ⋅ 5 ⋅ 4! ⎛ 4 ⎞⎛ 1 ⎞ ⎛ 4 ⎞ ⎛ 1 ⎞ 60 = 15 ⎜ ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟ 2 ⋅1 ⋅ 4! ⎝ 9 ⎠⎝ 81 ⎠ ⎝ 9 ⎠ ⎝ 81 ⎠ 729 P(X > 2) = 1 – P(X ≤ 2) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)] 12 60 ⎤ 73 656 ⎡ 1 = 1− ⎢ + + = 1− = ⎥ 729 729 ⎣ 729 729 729 ⎦

f (4) = 4C4 (0.15) 4 (0.85)0 4! ≈ ⋅ (0.000506)1 = 0.0005 4!0! µ = np = 4(0.15) = 0.6

=

σ = npq = 4(0.15)(0.85) ≈ 0.71

352


Chapter 9 Review

13. The probability that a 2 or 3 results on one roll is 2 1 = . Let X = number of 2’s or 3’s that appear 6 3 1 on 4 rolls. Then X is binomial with p = and 3 n = 4. 3

2

1

1 2 8 ⎛1⎞ ⎛2⎞ P ( X = 3) = 4C3 ⎜ ⎟ ⎜ ⎟ = 4 ⋅ ⋅ = 27 3 81 ⎝3⎠ ⎝ 3⎠

=

P ( X = 0) = 4C0 (0.9)0 (0.1) 4 = 0.0001

15. Let X = number of heads that occur. Then X is binomial. 5

243 243 ⎛ 2⎞ ⎛3⎞ P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ = 3125 3125 ⎝5⎠ ⎝5⎠ 1 4 2 81 810 ⎛ 2⎞ ⎛3⎞ P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ = 5 625 3125 ⎝ 5⎠ ⎝5⎠ P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] 810 ⎤ 1053 2072 ⎡ 243 = 1− ⎢ + = 1− = ⎥ 3125 3125 ⎣ 3125 3125 ⎦

18. From column 1, a + a + 0.2 = 1, so 2a = 0.8, or a = 0.4. From column 3, b + b + a = 1, so 2b = 1 – a, or 1 − a 1 − 0.4 b= = = 0.3. 2 2 From column 2, a + b + c = 1, so c = 1 – a – b = 1 – 0.4 – 0.3 = 0.3. ⎡ 0.1 0.3 0.1⎤ ⎡ 0.5⎤ ⎡ 0.10 ⎤ 19. X1 = TX0 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0 ⎥ = ⎢ 0.15 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0.7 0.3 0.8⎦ ⎣ 0.5⎦ ⎣ 0.75⎦ ⎡ 0.1 0.3 0.1⎤ ⎡0.10 ⎤ ⎡ 0.130 ⎤ X 2 = TX1 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0.15⎥ = ⎢ 0.155 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0.7 0.3 0.8⎦ ⎣ 0.75⎦ ⎣ 0.715 ⎦

16. On any draw, the probability of selecting a red 2 1 = . Let jelly bean is 10 5 X = number of red jelly beans selected in five 1 draws. Then X is binomial with p = and 5 n = 5. 0

⎡ 0.1 0.3 0.1⎤ ⎡ 0.130 ⎤ X3 = TX2 = ⎢ 0.2 0.4 0.1⎥ ⎢ 0.155⎥ ⎢ ⎥⎢ ⎥ ⎣ 0.7 0.3 0.8⎦ ⎣ 0.715⎦ ⎡0.1310 ⎤ = ⎢ 0.1595⎥ ⎢ ⎥ ⎣ 0.7095⎦

5

1024 ⎛1⎞ ⎛4⎞ P ( X = 0) = 5C0 ⎜ ⎟ ⎜ ⎟ = 1 ⋅1 ⋅ 5 5 3125 ⎝ ⎠ ⎝ ⎠ 1024 = 3125 1

1024 1280 640 2944 + + = = 0.94208 3125 3125 3125 3125

17. From column 1, 0.1 + a + 0.6 = 1, so a = 0.3. From column 2, 2a + b + b = 1, so 2b = 1 – 2a, 1 − 2a 1 − 2(0.3) = = 0.2. or b = 2 2 From column 3, a + b + c = 1, so c = 1 – a – b, or c = 1 – 0.3 – 0.2 = 0.5.

14. Let X = number of bushes that live. Then X is binomial.

0

3

1 64 ⎛1⎞ ⎛4⎞ P ( X = 2) = 5C2 ⎜ ⎟ ⎜ ⎟ = 10 ⋅ ⋅ 5 5 25 125 ⎝ ⎠ ⎝ ⎠ 640 = 3125 P ( X ≤ 2) = P ( X = 0) + P( X = 1) + P( X = 2)

4

1 256 ⎛1⎞ ⎛4⎞ P ( X = 1) = 5C1 ⎜ ⎟ ⎜ ⎟ = 5 ⋅ ⋅ 5 625 ⎝5⎠ ⎝5⎠ 1280 = 3125

353



⎡ 0.4 0.1 0.1⎤ ⎡ 0.1⎤ ⎡ 0.13⎤ 20. X1 = TX0 = ⎢ 0.2 0.6 0.5 ⎥ ⎢ 0.3⎥ = ⎢ 0.50 ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0.4 0.3 0.4 ⎦ ⎣ 0.6 ⎦ ⎣ 0.37 ⎦ ⎡ 0.4 0.1 0.1⎤ ⎡ 0.13⎤ ⎡0.139 ⎤ X 2 = TX1 = ⎢ 0.2 0.6 0.5 ⎥ ⎢ 0.50 ⎥ = ⎢ 0.511⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎣ 0.4 0.3 0.4 ⎦ ⎣0.37 ⎦ ⎣0.350 ⎦

c.

⎡ 1 2 ⎤ ⎡1 0 ⎤ ⎡ − 2 2 ⎤ 3 3⎥ 3 3 ⎥ − =⎢ 23. T − I = ⎢ ⎢ 2 1 ⎥ ⎢⎣ 0 1 ⎥⎦ ⎢ 2 − 2 ⎥ 3⎦ ⎣3 3⎦ ⎣ 3 1 ⎡ 1 ⎤ ⎡ ⎤ 1 0 2 1 1 ⎢ ⎥ ⎢ ⎥ ⎢ − 2 2 0⎥ → … → ⎢0 1 1 ⎥ 2⎥ ⎢ 3 3 ⎥ ⎢ ⎢ 2 − 2 0⎥ ⎢0 0 0 ⎥ 3 ⎣ ⎦ ⎣ 3 ⎦ 1 ⎡ ⎤ Q = ⎢2⎥ ⎢1⎥ ⎣2⎦

⎡ 0.4 0.1 0.1⎤ ⎡ 0.139⎤ X3 = TX2 = ⎢ 0.2 0.6 0.5⎥ ⎢ 0.511⎥ ⎢ ⎥⎢ ⎥ ⎣ 0.4 0.3 0.4 ⎦ ⎣ 0.350 ⎦ ⎡0.1417 ⎤ = ⎢ 0.5094 ⎥ ⎢ ⎥ ⎣ 0.3489 ⎦

21. a.

⎡1 7 T2 = TT = ⎢ ⎢6 ⎣7 ⎡ 19 49 T =T T=⎢ ⎢ 30 ⎣ 49 ⎡ 109 117 ⎤ 343 343 ⎥ =⎢ ⎢ 234 226 ⎥ ⎣ 343 343 ⎦ 3

2

3⎤ ⎡1 7⎥ ⎢7 4⎥⎢6 7⎦ ⎣7

3⎤ 7⎥ 4⎥ 7⎦

15 ⎤ ⎡ 1 49 ⎥ ⎢ 7 34 ⎥ ⎢ 6 49 ⎦ ⎣ 7

⎡ 19 49 =⎢ ⎢ 30 ⎣ 49 3⎤ 7⎥ 4⎥ 7⎦

15 ⎤ 49 ⎥ 34 ⎥ 49 ⎦

b. From T2 , entry in row 1, column 2, is c.

22. a.

⎡ 0.4 0.4 0.3⎤ ⎡1 0 0 ⎤ 24. T − I = ⎢⎢ 0.3 0.2 0.3⎥⎥ − ⎢⎢ 0 1 0 ⎥⎥ ⎢⎣ 0.3 0.4 0.4 ⎥⎦ ⎢⎣ 0 0 1 ⎥⎦ 0.3⎤ ⎡ −0.6 0.4 ⎢ = ⎢ 0.3 −0.8 0.3⎥⎥ ⎢⎣ 0.3 0.4 −0.6 ⎥⎦

1 1 1⎤ ⎡ 1 ⎢ −0.6 0.4 0.3 0 ⎥ ⎢ ⎥ →… ⎢ 0.3 −0.8 0.3 0 ⎥ ⎢ ⎥ ⎣⎢ 0.3 0.4 −0.6 0 ⎦⎥ ⎡ 1 0 0 0.36 ⎤ ⎢ 0 1 0 0.27 ⎥ ⎥ →⎢ ⎢ 0 0 1 0.36 ⎥ ⎢ ⎥ ⎣⎢ 0 0 0 0 ⎦⎥

15 . 49

From T3 , entry in row 2, column 1, is 234 . 343

⎡ 0.36 ⎤ Q ≈ ⎢ 0.27 ⎥ ⎢ ⎥ ⎣ 0.36 ⎦

T2 = TT ⎡ 0 0.4 0.3⎤ ⎡ 0 0.4 0.3⎤ = ⎢⎢ 0 0.3 0.5⎥⎥ ⎢⎢ 0 0.3 0.5⎥⎥ ⎢⎣1 0.3 0.2 ⎥⎦ ⎢⎣1 0.3 0.2 ⎥⎦ = ⎡ 0.3 0.21 0.26 ⎤ ⎢ 0.5 0.24 0.25 ⎥ ⎢ ⎥ ⎣⎢ 0.2 0.55 0.49 ⎦⎥ T3 = T2 T ⎡0.3 = ⎢⎢0.5 ⎢⎣0.2 ⎡0.26 = ⎢⎢0.25 ⎢⎣0.49

From T3 , entry in row 2, column 1, is 0.25.

Japanese Non-Japanese Japanese ⎡ 0.8 0.6⎤ 25. T = Non-Japanese ⎢⎣ 0.2 0.4⎥⎦

a. 0.21 0.26 ⎤ ⎡ 0 0.4 0.3⎤ 0.24 0.25 ⎥⎥ ⎢⎢0 0.3 0.5 ⎥⎥ 0.55 0.49 ⎥⎦ ⎢⎣1 0.3 0.2 ⎥⎦ 0.261 0.247 ⎤ 0.347 0.32 ⎥⎥ 0.392 0.433 ⎥⎦

b. From T2 , entry in row 1, column 2, is 0.21. 354

⎡ 0.8 0.6 ⎤ ⎡ 0.8 0.6 ⎤ T2 = ⎢ ⎥⎢ ⎥ ⎣ 0.2 0.4 ⎦ ⎣ 0.2 0.4⎦ ⎡0.76 0.72 ⎤ =⎢ ⎥ ⎣0.24 0.28 ⎦ From row 1, column 1, the probability that a person who currently owns a Japanese car will buy a Japanese car two cars later is 0.76. Thus 76% of people who currently own Japanese cars will own Japanese cars two cars later.



⎡ 0.76 0.72 ⎤ ⎡ 0.6 ⎤ ⎡ 0.744 ⎤ X 2 = T2 X0 = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣ 0.24 0.28⎦ ⎣ 0.4 ⎦ ⎣ 0.256 ⎦ Two cars from now, we expect 74.4% Japanese, 25.6% non-Japanese.

b.

⎡ 0.8 0.6 ⎤ ⎡1 0 ⎤ ⎡ −0.2 0.6 ⎤ T−I = ⎢ ⎥−⎢ ⎥ =⎢ ⎥ ⎣ 0.2 0.4 ⎦ ⎣ 0 1 ⎦ ⎣ 0.2 −0.6 ⎦ 1 1⎤ ⎡ 1 ⎡1 0 0.75⎤ ⎢ −0.2 0.6 0 ⎥ → … → ⎢0 1 0.25⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣ 0.2 −0.6 0 ⎥⎦ ⎢⎣0 0 0 ⎥⎦ ⎡ 0.75⎤ Q=⎢ ⎣ 0.25⎥⎦ In the long run, 75% Japanese cars, 25% non-Japanese cars.

c.

⎡0.7 0.4 0.1⎤ ⎡ 0.5 ⎤ ⎡ 0.49 ⎤ X1 = TX0 = ⎢⎢ 0.2 0.5 0.1⎥⎥ ⎢⎢ 0.3 ⎥⎥ = ⎢⎢ 0.27 ⎥⎥ ⎢⎣ 0.1 0.1 0.8⎥⎦ ⎢⎣ 0.2 ⎥⎦ ⎢⎣ 0.24 ⎥⎦ 49% are expected to vote for party 1, 27% for party 2, 24% for party 3.

26. a.

⎡ 0.7 0.4 0.1⎤ ⎡ 1 T − I = ⎢ 0.2 0.5 0.1⎥ − ⎢ 0 ⎢ ⎥ ⎢ ⎣ 0.1 0.1 0.8⎦ ⎣ 0

b.

1 1 ⎡ 1 ⎢ −0.3 0.4 0.1 ⎢ ⎢ 0.2 −0.5 0.1 ⎢ ⎢⎣ 0.1 0.1 −0.2

0 1 0

0 ⎤ ⎡ −0.3 0.4 0.1 ⎤ 0 ⎥ = ⎢ 0.2 −0.5 0.1 ⎥ ⎥ ⎢ ⎥ 1 ⎦ ⎣ 0.1 0.1 −0.2 ⎦

⎡1 1⎤ ⎢ ⎢0 0 ⎥⎥ →… → ⎢ 0⎥ ⎢0 ⎥ ⎢ 0 ⎥⎦ ⎢⎣ 0

0 0 1 0 0 1 0 0

3⎤ 7 ⎥ 5⎥ 21 ⎥ 1 ⎥ 3⎥

0 ⎥⎦

⎡0.429 ⎤ Q ≈ ⎢0.238 ⎥ ⎢ ⎥ ⎣0.333 ⎦ In the long run, 43% will vote for party 1, 24% for party 2, and 33% for party 3.

Mathematical Snapshot Chapter 9 ⎡0 ⎤ ⎡0⎤ ⎡0⎤ ⎢1 ⎥ ⎢0⎥ ⎢ 1⎥ 1. For X0 = ⎢ ⎥ or ⎢ ⎥ , the first entry of the state vector is greater than 0.5 for n = 7 or greater. If X0 = ⎢ ⎥ , then ⎢0⎥ ⎢0 ⎥ ⎢ 1⎥ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎢⎣ 0 ⎥⎦ ⎡ 0.5217 ⎤ ⎢ 0.0000 ⎥ T X0 ≈ ⎢ ⎥. ⎢ 0.4783⎥ ⎢⎣ 0.0000 ⎥⎦ 7

355


⎡ 1 0.1 0.1 0.01 ⎤ ⎡1 ⎢ 0 0 0.9 0.09 ⎥ ⎢0 ⎥−⎢ 2. T − I = ⎢ ⎢ 0 0.9 0 0.09 ⎥ ⎢0 ⎢ ⎥ ⎢ 0 0.81 ⎥⎦ ⎢⎣0 ⎢⎣ 0 0 1 1 1⎤ ⎡1 1 ⎢ 0 0.1 0.1 0.01 0 ⎥ ⎢ ⎥ ⎢ 0 −1 0.9 0.09 0 ⎥ → … → ⎢ ⎥ ⎢ 0 0.9 −1 0.09 0 ⎥ ⎢⎣ 0 0 0 −0.19 0 ⎥⎦

0 1 0 0 ⎡1 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎣


0 ⎤ ⎡0 0.1 0.1 0.01 ⎤ 0 ⎥⎥ ⎢⎢ 0 −1 0.9 0.09 ⎥⎥ = 0 ⎥ ⎢ 0 0.9 −1 0.09 ⎥ ⎥ ⎢ ⎥ 1 ⎥⎦ ⎢⎣ 0 0 0 −0.19 ⎥⎦

0 0 1 0 0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

1⎤ 0 ⎥⎥ 0⎥ . ⎥ 0⎥ 0 ⎥⎦

3. Against Always Defect, 1 2 3 4 1 ⎡0 0 0 0 ⎤ 2 ⎢1 0.1 1 0.1⎥ . T= ⎢ ⎥ 3 ⎢0 0 0 0 ⎥ 4 ⎢⎣ 0 0.9 0 0.9 ⎥⎦ Against Always Cooperate, 1 2 3 4 1 ⎡1 0.1 1 0.1⎤ 2 ⎢ 0 0 0 0 ⎥⎥ . T= ⎢ 3 ⎢ 0 0.9 0 0.9 ⎥ ⎢ ⎥ 4 ⎣⎢ 0 0 0 0 ⎦⎥ Against regular Tit-for-tat, 1 2 3 4 1 ⎡1 0.1 0 0 ⎤ 2 ⎢ 0 0 1 0.1⎥⎥ . T= ⎢ 3 ⎢ 0 0.9 0 0 ⎥ ⎢ ⎥ 4 ⎣⎢ 0 0 0 0.9 ⎦⎥ 4. With Player 2 always defecting, after one round the game is in a stable pattern of Player 1 cooperating with ⎡ 0 ⎤ ⎢ 0.1⎥ probability 0.1 and defecting with probability 0.9. The steady state vector in this case is ⎢ ⎥ . ⎢ 0 ⎥ ⎢⎣ 0.9 ⎥⎦ With Player 2 always cooperating, after one round the game settles into steady mutual cooperation. ⎡1 ⎤ ⎢0⎥ With Player 2 playing standard Tit-for-tat, the probabilities gradually tilt toward mutual cooperation: ⎢ ⎥ is the ⎢0⎥ ⎣⎢ 0 ⎦⎥ steady state vector. In this case, it takes only one “forgiving” Tit-for-tat-er to guarantee mutual cooperation in the long run.

356



1. a.

1. The graph of the greatest integer function is shown.

b. 0

10

10

–10

lim f ( x) does not exist when a is an integer

x →a

since the limits are different depending on the side from which you approach the integer. lim f ( x) exists for all numbers which are not integers.

2. a.

2

c.

2

3. a.

1

c. 4. a.

4 4 2. lim V ( r ) = lim π r 3 = π lim r 3 3 r →1 r →1 r →1 3 4 4 = π (1)3 = π 3 3

(

3. lim R ( x) = lim 500 x − 6 x 2 x →8

x →8

c.

)

x →8

= 500 lim x − 6 lim x 2 = 500(8) − 6(8) 2 x →8

x →8

= 4000 − 384 = 3616

4. lim p = lim

) = lim ⎡⎣⎢50 (t + 4t )⎤⎦⎥ + 3t + 20 lim ( t + 3t + 20 ) (

50 t 2 + 4t

t →2 t 2

3 –1

b. does not exist

= lim 500 x − lim 6 x 2

=

1

b. does not exist

x →a

t →2

c.

b. 1

–10

x →8

1

2

t →2

2

t →2

50 ⎡ 22 + 4(2) ⎤ 600 ⎣ ⎦= = 20 2 30 2 + 3(2) + 20

5. As h → 0 , both the numerator and denominator approach 0. For h ≠ 0, 125 + 2( x + h) − (125 + 2 x) lim h h →0 125 + 2 x + 2h − 125 − 2 x 2h = lim = lim h h→0 h →0 h = lim 2 = 2 .

1

5. f(−0.9) = −3.7 f(−0.99) = −3.97 f(−0.999) = −3.997 estimate of limit: −4

f(−1.1) = −4.3 f(−1.01) = −4.03 f(−1.001) = −4.003

6. f(−3.1) = −6.1 f(−3.01) = −6.01 f(−3.001) = −6.001 estimate of limit: −6

f(−2.9) = −5.9 f(−2.99) = −5.99 f(−2.999) = −5.999

7. f(−0.1) ≈ 0.9516 f(−0.01) ≈ 0.9950 f(−0.001) ≈ 0.9995 estimate of limit: 1

f(0.1) ≈ 1.0517 f(0.01) ≈ 1.0050 f(0.001) ≈ 1.0005

8. f(−0.1) ≈ 0.5132 f(−0.01) ≈ 0.5013 f(−0.001) ≈ 0.5001 estimate of limit: 0.5

f(0.1) ≈ 0.4881 f(0.01) ≈ 0.4988 f(0.001) ≈ 0.4999

9. lim 16 = 16 x→2

10. lim 2 x = 2(3) = 6 x →3

h→0

357

Chapter 10: Limits and Continuity

11.

12. 13.

(


)

lim t 2 − 5 = (−5) 2 − 5 = 25 − 5 = 20

t →−5

21.

16 ⎛1⎞ lim (5t − 7) = 5 ⎜ ⎟ − 7 = − 3 3 ⎝ ⎠

22.

t →1/ 3

lim (3 x3 − 4 x 2 + 2 x − 3)

24. lim

(

lim x 2 + 6 lim ( x − 6)

) = (−6)

x →−6

+6 (−6) − 6 2

t2 − 4 (t + 2)(t − 2) = lim = lim(t + 2) = 4 t−2 t →2 t − 2 t →2 t →2

26. lim

0 − 7(0) + 1

=

(

h →0

)

x2 − 2 x x( x − 2) = lim = lim ( x − 2) = −2 x x x →0 x →0 x →0

lim h 2 − 7h + 1

h→0

28. lim

=0

z 2 − 5z − 4

z →0

z +1 2

29. lim = z →0 lim ( z 2 + 1)

lim

p2 + p + 5 =

= lim

2

z →0

02 − 5(0) − 4

= −4

p →4

x 2 − 9 x + 20

x − 3x − 4 x−5 1 = lim =− 5 x→4 x + 1

lim ( z 2 − 5 z − 4)

=

19.

−9

x−3 1 1 = lim = x →3 ( x + 3)( x − 3) x →3 x + 3 6

= lim

lim h

x→4

18. lim

x−3

x →3 x 2

0

30.

02 + 1

x→4

x 4 − 81

lim

x →−3 x 2

+ 8 x + 15

lim

y →15

y+3 =

( x − 4)( x − 5) ( x − 4)( x + 1)

( x 2 + 9)( x 2 − 9) x →−3 ( x + 3)( x + 5)

= lim

( x 2 + 9)( x + 3)( x − 3) ( x + 3)( x + 5) x →−3

= lim

(

lim p 2 + p + 5

p →4

)

( x 2 + 9)( x − 3) x+5 x →−3 = −54 = lim

= 42 + 4 + 5 = 25 = 5

20.

t −3 3 =− 4 4 t − t →0

= lim

=5

27. lim

=

t →0 t 2 (t − 4)

x →3

x →−6

− 7h + 1

t 2 (t + 3)

x2 − x − 6 ( x − 3)( x + 2) = lim x −3 x−3 x →3 x →3 = lim ( x + 2)

42 7 =− −12 2

2

− 4t

2

= lim

25. lim

t →−3

h

x →−1

t 3 + 3t 2

t →0 t 3

lim (t − 2) t − 2 t →−3 −3 − 2 −5 5 15. lim = = = =− lim (t + 5) −3 + 5 2 2 t →−3 t + 5

h →0 h 2

= lim 1 = 1

x →2

4r − 3 4(9) − 3 36 − 3 33 = = = =3 14. lim 11 11 11 r →9 11

17. lim

x +1

x2 − x − 2 ( x − 2)( x + 1) = lim x−2 x−2 x→2 x →2 = lim ( x + 1) = 3

= 3(−2)3 − 4(−2) 2 + 2(−2) − 3 = −24 − 16 − 4 − 3 = −47

=

lim

x →−1 x + 1

23. lim

x →−2

x2 + 6 16. lim = t →−6 x − 6

x2 + 2 x x( x + 2) = lim = lim x = −2 x →−2 x + 2 x →−2 x + 2 x →−2

lim

lim ( y + 3) = 15 + 3 = 18

31. lim

y →15

3x 2 − x − 10

=3 2

358

(3x + 5)( x − 2) x → 2 ( x + 7)( x − 2)

= lim

+ 5 x − 14 3 x + 5 11 = lim = 9 x→2 x + 7 x→2 x 2


32.

x2 + 2 x − 8

lim

x →−4 x 2

+ 5x + 4

Section 10.1

x−2 ( x + 4)( x − 2) = lim =2 x →−4 x + 1 x →−4 ( x + 4)( x + 1)

= lim

⎡ 4 + 4h + h 2 ⎤ − 4 4h + h 2 h(4 + h) (2 + h) 2 − 22 ⎦ 33. lim = lim (4 + h) = 4 = lim = lim = lim ⎣ h h h h h →0 h →0 h→0 h →0 h →0 ( x + 2) 2 − 4 x2 + 4 x = lim ( x + 4) = 4 = lim x x x →0 x →0 x →0

34. lim

( x + h) 2 − x 2 2 xh + h 2 = lim (2 x + h) = 2 x = lim h h h →0 h→0 h→0

35. lim

3( x + h) 2 + 7( x + h) − 3x 2 − 7 x 3 x 2 + 6 xh + 3h 2 + 7 x + 7 h − 3 x 2 − 7 x = lim h h h →0 h →0 6 xh + 3h 2 + 7h h(6 x + 3h + 7) = lim = lim h h h →0 h →0 = lim (6 x + 3h + 7) = 6 x + 7

36. lim

h →0

f ( x + h) − f ( x ) [7 − 3( x + h)] − (7 − 3x) −3h = lim = lim = lim − 3 = −3 h h h →0 h →0 h →0 h h →0

37. lim

38. lim

h →0

f ( x + h) − f ( x ) [2( x + h) + 3] − (2 x + 3) 2h = lim = lim = lim 2 = 2 h h h →0 h →0 h h→0

(

⎡ ( x + h ) 2 − 3⎤ − x 2 − 3 f ( x + h) − f ( x ) ⎦ 39. lim = lim ⎣ h h h →0 h →0 = lim

h →0

(

x 2 + 2 xh + h 2 − 3 − x 2 − 3

) = lim 2 xh + h h →0

h

h

2

) = lim (2 x + h) = 2 x h →0

(

)

⎡ ( x + h)2 + ( x + h) + 1⎤ − x 2 + x + 1 f ( x + h) − f ( x ) 2 xh + h 2 + h ⎦ = lim = lim (2 x + h + 1) = 2 x + 1 = lim ⎣ h h h h →0 h →0 h →0 h →0

40. lim

f ( x + h) − f ( x ) [( x + h)3 − 4( x + h)2 ] − [ x3 − 4 x 2 ] = lim h h h →0 h →0 3 2 2 x + 3 x h + 3xh + h3 − 4 x 2 − 8 xh − 4h 2 − x3 + 4 x 2 = lim h h →0 2 2 3 3 x h + 3xh + h − 8 xh − 4h 2 = lim h h →0 h(3x 2 + 3xh + h 2 − 8 x − 4h) = lim h h →0 = lim (3 x 2 + 3 xh + h 2 − 8 x − 4h) = 3x 2 − 8 x

41. lim

h →0

359



f ( x + h) − f ( x ) [3 − ( x + h) + 4( x + h)2 ] − (3 − x + 4 x 2 ) = lim h h h →0 h →0 3 − x − h + 4 x 2 + 8 xh + 4h 2 − (3 − x + 4 x 2 ) = lim h h →0 − h + 8 xh + 4h 2 = lim h h →0 h(−1 + 8 x + 4h) = lim h h →0 = lim (−1 + 8 x + 4h)

42. lim

h →0

= −1 + 8 x x−2 −2 = lim x−6 x →6 x →6

43. lim

x →6 ( x − 6)

x →6

x−2 −2 ( x − 6)

( x − 2) − 4

= lim

= lim

(

(

x−2 +2

1 x−2 +2

44. For lim

=

x−2 +2

x−2 +2

= lim

x →6 ( x − 6)

)

)

x−6

(

x−2 +2

)

1 4

x2 + x + c

x →3 x 2

)

(

)(

− 5x + 6

x2 + x + c to exist, x – 3 must be a factor of the numerator x 2 + x + c : x →3 ( x − 3)( x − 2)

= lim

x + x + c = ( x − 3)( x + r ) = x 2 + (r − 3) x − 3r Thus r – 3 = 1, or r = 4. So c = –3r = –3(4) = –12. For c = –12, x2 + x + c x 2 + x − 12 ( x − 3)( x + 4) x+4 7 = lim = lim = lim = =7 lim 2 2 1 x →3 x − 5 x + 6 x →3 x − 5 x + 6 x →3 ( x − 3)( x − 2) x →3 x − 2 2

Th − Tc Th − 0 Th = = =1 Th Th Tc →0 Th

45. a.

lim

Th − Tc Th − Th 0 = = =0 Th Th Tc →Th Th

b.

46.

lim

lim

r →7.5×107

47.

E=

lim

r →7.5×107

−

7.0 7.0 × 1017 7.0 × 1017 =− =− × 1010 ≈ −9.33 × 109 ft-lb 7 r 7.5 7.5 × 10

15

0

0

5

11.00

360


Section 10.2

5

48.

lim P ( x) = 2343.056

x →53.2

lim P ( x) = 224(53.2) − 3.1(53.2)2 − 800

x →53.2

5

–5


–5

1

1. The graph of p(x) is shown. 10,000

10

49.

0 5

15

0

x →∞

quickly drops down toward zero. According to this function, a low price corresponds to a high demand and a high price corresponds to a low demand.

2

0

0

10

0

From the graph, it is apparent that lim p( x) = 0 . The graph starts out high and

4.00 50.

= 2343.056

2. The graph of y(x) is shown.

5

550

0.80 51. The graph of C(p) is shown. (Negative amounts of impurities and money are not reasonable, so only the first quadrant is shown.)

0

100,000

1000

0

From the graph, it is apparent that lim y ( x) = 500 . The greatest yearly sales that x →∞

0

0

the company can expect is $500,000, even with unlimited spending on advertising.

5

As p gets closer and closer to 0, the values of C(p) increase without bound, so lim C ( p) does

3. The graph of C(x) is shown. 1,000,000

p →0

not exist. 52. The graph of P(x) is shown with the value x = 53.2 indicated.

0

3500

1000

0

From the graph it is apparent that lim C ( x) = ∞ . This indicates that the cost x →∞

increases without bound the more units that you make. 0

0

75

4. lim f ( x) does not exist since x →1

lim f ( x) = lim 100 = 100 while

x →1−

361

x →1−



lim f ( x) = lim 175 = 175 .

x →1+

6.

x →1+

lim 19 = 19

x →−∞

lim f ( x) = 250 since

x → 2.5

lim

x →2.5−

f ( x) = lim

x →2.5+

7.

f ( x) = 250

6x

lim

x →0

−

x

4

= lim x →0

−

6 x3

= −∞ since x3 is negative

and close to 0 for x → 0− .

Problems 10.2 1. a.

lim 7 7 7 x→2 8. lim = = =7 lim ( x − 1) 1 x→2 x − 1

2

x →2

b. 3 c.

does not exist

d.

−∞

e.

∞

f.

∞

g.

∞

9. 10. 11.

1

j.

1

12.

13.

k. 1 2. a.

14.

0

b. –∞ c.

15.

does not exist

3.

2

f.

1

g.

1

h is close to 0 when h is

5−h = 0

lim

h →5−

lim

x →−2

−

−3 =∞ x+2

lim 21/ 2 = 21/ 2

x →0−

(

)

lim 4 x − 1 . As x → 1+ , then x – 1

x →1+

(

)

lim 4 x − 1 = 4 ⋅ lim

16.

lim ( x − 2)

17.

x →3+

As x → 3 , then x − 2 → 1 .

5.

h = 0 since

lim

h →0 +

x →1+

+

4.

lim (t − 1)3 = ∞

t →∞

approaches 0 through positive values. So x − 1 → 0 . Thus

d. ∞ e.

x →−∞

positive and close to 0.

h. 0 i.

lim x 2 = ∞ since x 2 is positive for x → −∞ .

x →1+

x −1 = 4 ⋅ 0 = 0 .

lim ⎛⎜ x x 2 − 4 ⎞⎟ = 0 ⎠ x → 2+ ⎝ lim

x →∞

x + 10

As x becomes very large, so does x + 10. Because square roots of very large numbers are very large, lim x + 10 = ∞ .

lim (1 − x 2 ) = 0

x →−1+

x →∞

18.

lim 5 x

x →−∞

As x becomes very negative, so does 5x. Thus lim 5 x = −∞ .

lim − 1 − 10 x

x →−∞

As x becomes very negative, 1 − 10x becomes very positive. Because square roots of very large numbers are very large, lim − 1 − 10 x = −∞.

x →−∞

x →−∞

362


19.

20.

21.

22.

23.

24.

3

lim

x →∞

x

= 3 lim

x →∞

−6

lim

x →∞ 5 x

3

=− x

1 x

Section 10.2

= 3⋅ 0 = 0

1 2

30.

6 1 6 lim = − ⋅0 = 0 4 / 3 5 x→∞ x 5

31.

x +8 x = lim = lim 1 = 1 x →∞ x − 3 x →∞ x x →∞ lim

lim

x →∞

2x − 4 2x = lim = lim (−1) = −1 3 − 2 x x →∞ −2 x x →∞ x2 − 1

lim

x →−∞ x3

r3

lim

r →∞ r 2

25. lim

+ 4x − 3

+1

= lim

= lim

= lim

x →−∞ x3

r3

r →∞ r 2

3t 3 + 2t 2 + 9t − 1 5t − 5 2

t →∞

x2

x →−∞

1 =0 x

33.

r →∞

3t 3

t →∞ 5t

7 − 5 x3 + 2 x 2 2 2 = lim = 5 x →−∞ 5 x →−∞

x+3

lim

−

x →3

34.

2

3t t →∞ 5 3 = lim t 5 t →∞ =∞

27.

28.

= lim

35.

36.

5x

−x +4 5 5 1 5 = lim = ⋅ lim = ⋅0 = 0 6 3 x →∞ x 6 3 x →∞ 3 x x →∞ 3 x 7

3

5w + 7 w − 1 2

w→∞

4 − 3 x3

lim

x3 − 1

x →∞

x →∞

4 + 5x − 7 x 3 x − x3

lim

x →−∞ x3

x →∞ 3 x 7

37.

lim

−3 x3 x3

x →∞

+ x +1

2

x →−3

lim

2

2

(4 x − 1)

⋅ lim

3

= lim

1

43 x →−∞ x3

lim

3 − 4x − 2x

3

5x − 8x + 1 −2 2 = lim =− 5 x→∞ 5 x →∞

3

2 43

= lim

x →∞

38. lim

x + 3x

t 2 − 4t + 3

= lim (−1) = −1 x →−∞

= lim

−

(t − 1)(t − 3)

t →3 (t + 1)(t − 3)

− 2t − 3 t −1 2 1 = lim = = 2 t →3 t + 1 4

t →3 t 2

−2 x

x3

x = −∞ x →∞ −7

= lim

2

(5 x − 1)( x + 3) x( x + 3) x →−3 5x − 1 = lim x x →−3− −16 = −3 16 = 3

2

⋅0 = 0

−7 x

= lim

2

x →−∞ 43 x 3

=

2 2 = 5 5

= lim (−3) = −3

− x3

x →−∞

7 7 7 1 7 = lim = ⋅ lim = ⋅ 0 = 0 2 x →∞ x 2 x →∞ 2 x + 1 x →∞ 2 x

x →−∞

w→∞

x3

x →∞

= lim

5w

= lim

2

x →∞

= lim

5 x 2 + 14 x − 3 −

2 w2

w→∞

= lim

6 − 4 x 2 + x3

lim

= lim

lim

=

29.

5x

lim

x+3 x →3 ( x + 3)( x − 3) −

2 w2 − 3w + 4

lim

= lim

26.

−5 x3

x →−∞

= lim

x −9 1 = lim = −∞ x →3− x − 3 2

−2 x3

= lim

32. As x → −3− , then 3x → −9 and 9 − x 2 → 0 through negative values. Thus 3x = ∞. lim − x →−3 9 − x 2

= lim r = ∞

= lim

3 − 2 x − 2 x3

lim

3

5 x3

39. lim

x →1

363

x 2 − 3x + 1 x2 + 1

=

(

) = −1 = − 1

lim x 2 − 3 x + 1

x →1

(

)

lim x 2 + 1

x →1

2

2


3 x3 − x 2 = 40. lim x →−1 2 x + 1

(

lim 3 x3 − x 2

x →−1

lim (2 x + 1)


) = −4 = 4

48.

−1

x →−1

1 → ∞ . Thus x −1

41. As x → 1+ , then

49. As x → 1+ , then 1 − x → 0 through negative −5 = ∞. values. Thus, lim + 1− x x →1

1 ⎤ ⎡ lim ⎢1 + =∞ + x − 1 ⎥⎦ x →1 ⎣

42.

lim −

x5 + 2 x3 − 1

x5 − 4 x 2 = lim (−1) = −1 x →−∞

= lim − x →−∞

x5 x5

50.

x →−∞

43.

lim

x →−7

x2 + 1 −

x − 49 2

51.

x 2 + 1 → 50 and x 2 − 49 approaches 0 through positive values. Thus x2 + 1 →∞. x 2 − 49

lim x = lim x = 0

x →0 +

x →0 +

lim x = lim (− x) = 0

x →0 −

x →0 −

Thus, lim x = 0. x →0

16 − x 4 → 0

52.

lim

1 1 = lim = ∞ + x x →0 x

lim

1 ⎛ 1⎞ = lim − =∞ x x→0− ⎜⎝ x ⎟⎠

x →0

x →0

45. As x → 0+ , x + x 2 approaches 0 through 5 positive values. Thus →∞. x + x2 46. As x → ∞ , then

7 ⎞ ⎛ lim ⎜ − = −∞ + − 3 ⎟⎠ x x →3 ⎝ 7 ⎞ ⎛ lim ⎜ − ⎟ = +∞ − x →3 ⎝ x − 3 ⎠ Answer: does not exist.

. As x → −7 − , then

44. As x → −2+ , then x → −2 and through positive values. Thus, x lim = −∞. + x →−2 16 − x 4

1 = −∞ 2x −1 1 lim =∞ + 2x −1 x →1/ 2 Answer: does not exist lim

x →1/ 2−

+

−

Thus, lim

x →0

53.

1 → 0 . Thus x

54.

1⎞ ⎛ lim ⎜ x + ⎟ = ∞ . x⎠ x →∞ ⎝

lim

x →−∞

x +1 x = lim = lim 1 = 1 x x →−∞ x x →−∞

⎡ 3 2 x2 ⎤ 3 x 2 + 3 − 2 x3 lim ⎢ − = lim ⎥ x →∞ ⎢ x x 2 + 1 ⎥ x →∞ x3 + x ⎣ ⎦ = lim

x →∞

x 47. lim = −∞ x →1− x − 1 x lim =∞ + x −1 x →1 Answer: does not exist

55.

−2 x3 x3

= lim (−2) = −2 x →∞

⎧2 if x ≤ 2 f ( x) = ⎨ ⎩1 if x > 2

a. b.

364

1 = ∞. x

lim f ( x) = lim 1 = 1

x →2+

x → 2+

lim f ( x) = lim 2 = 2

x →2−

x →2−


Section 10.2

lim f ( x) ≠ lim f ( x) , so lim f ( x) does

x →2+

x → 2−

lim g ( x) = lim x = 0

a.

x →2

x →0 +

x →0 +

not exist. d. e.

56.

x →∞

b. c.

x →−∞

d. e.

x →0 +

x →−∞

lim g ( x) = 0

if x ≤ 2

x →∞

x →2−

59.

x →−∞

b. c.

x →−∞

⎛ 5000 ⎞ lim c = lim ⎜ + 6⎟ = 0 + 6 = 6 q →−∞ q →∞ ⎝ q ⎠

lim f ( x) = lim f ( x) = lim f ( x) = 2

c

x →2−

lim f ( x) = lim (−2 + 4 x − x 2 ) = −∞

x →∞

x →∞

lim c = 6

q→∞

lim f ( x) = lim x = −∞

x →−∞

x →−∞

6 q

if x < 0 ⎧x 57. g ( x) = ⎨ ⎩− x if x > 0

a.

x →∞

lim g ( x) = lim x 2 = ∞

e.

lim f ( x) = lim x = 2

x → 2+

lim g ( x) = lim x = ∞

d.

if x > 2

x →2+

x →2−

x →0−

x →0

lim f ( x) = lim (−2 + 4 x − x 2 ) = 2

x → 2+

x →0 −

lim g ( x) = lim g ( x ) = 0 , so

c.

lim f ( x) = lim 2 = 2

x→2

x →0 −

x →∞

⎧⎪ x f ( x) = ⎨ 2 ⎪⎩−2 + 4 x − x

a.

lim g ( x) = lim x 2 = 0

b.

lim f ( x) = lim 1 = 1

5000

60.

lim g ( x) = lim (− x) = 0

x →0 +

x →0 +

q →∞

7q + 12, 000 q

⎛ 12, 000 ⎞ = lim ⎜ 7 + ⎟ = 7+0 = 7 q ⎠ q →∞ ⎝

lim g ( x) = lim x = 0

x →0 −

lim c = lim

q →∞

x →0 −

c

lim g ( x) = lim g ( x ) = 0 , so

x →0 +

x →0−

c=7+

lim g ( x) = 0

x →0

12,000 q

lim c = 7

q→∞

d.

lim g ( x) = lim (− x ) = −∞

x →∞

7

x →∞

q

e.

lim g ( x) = lim x = −∞

x →−∞

⎪⎧ x 2 58. g ( x) = ⎨ ⎪⎩ x

12,000

x →−∞

2000 ⎞ ⎛ 61. lim ⎜ 50, 000 − = 50, 000 − 0 = 50, 000 t + 1 ⎟⎠ t →∞ ⎝

if x < 0 if x > 0

365


62.

lim ⎛⎜ x 2 + x − x ⎞⎟ x →∞ ⎝ ⎠ ⎛ x2 + x − x ⎞ ⎛ x2 + x + x ⎞ ⎜ ⎟⎜ ⎟ ⎠⎝ ⎠ = lim ⎝ x →∞

(x = lim x →∞

= lim

x →∞

2

)

x2 + x + x

–5

= lim

x →∞

x 1+

1 +x x x

⎛ ⎞ 1 x ⎜⎜ 1 + + 1⎟⎟ x ⎝ ⎠ 1 1 = = 1+ 0 +1 2

–10

x

–∞

⎛ 1⎞ x 2 ⎜1 + ⎟ + x x⎠ ⎝

69.

= lim

20

x →∞

0

1 1+

1 +1 x

c.

1.

f ( x) = x3 − 5 x; x = 2

(i) f is defined at x = 2: f(2) = −2

⎧⎪ 2 − x if x < 2 f ( x) = ⎨ 3 ⎪⎩ x + k ( x + 1) if x ≥ 2

(ii) lim f ( x) = lim ( x3 − 5 x ) = 23 − 5(2) = −2, x→2

2− x = 0

x→2

which exists. (iii) lim f ( x) = −2 = f (2)

lim f ( x) = lim ⎡ x3 + k ( x + 1) ⎤ = 8 + 3k ⎦ x → 2+ ⎣

x→2

x →2+

Thus f is continuous at x = 2.

8 If lim f ( x) exists, then 8 + 3k = 0. So k = − . 3 x →2

2.

f ( x) =

65. 1, 0.5, 0.525, 0.631, 0.912, 0.986, 0.998; conclude limit is 1.

x −3 ; x = −3 5x

(i) f is defined at x = −3: f (−3) =

−44

66. 0.368, 0.135, 0.00674, 0.0000454, 3.72 × 10 , can’t do last two. Conclude that the limit is 0.

(ii)

4

67.

does not exist

Problems 10.3

x→∞

x → 2−

11

b. 9

900 x 900 x = lim x →∞ x →∞ 10 + 45 x x →∞ 45 x = lim 20 = 20

lim f ( x) = lim

5

0

a.

lim y = lim

x → 2−

0

x

x →∞

64.

10

68.

x2 + x + x

+ x − x2

= lim

63.


(iii) lim f ( x) = 0

x →−3

1

x−3 2 = , which exists 5 x →−3 5 x

lim f ( x ) = lim

x →−3

2 = f (−3) 5

Thus f is continuous at x = −3.

–1

0

366

−6 2 = −15 5


Section 10.3 8. Continuous at 2 and –2 because f is a polynomial function (which is continuous everywhere).

3. g ( x) = 2 − 3 x ; x = 0 (i) g is defined at x = 0; g (0) = 2 .

9. Discontinuous at 3 and −3 because at both points the denominator of this rational function is 0.

(ii) lim g ( x) = lim 2 − 3 x = 2 , which exists x →0

x →0

10. Continuous at 2 and –2 because f is a rational function and at neither point is the denominator zero.

(iii) lim g ( x) = 2 = g (0) x →0

Thus g is continuous at x = 0. 4.

f ( x) =

11.

x ;x=2 8

x → 2+

2 1 (i) f is defined at x = 2; f (2) = = . 8 4

x →2

lim f ( x) = lim x 2 = 0 . Since

x →0

x →0

lim f ( x) = 4 = f (2) and lim f ( x) = 0 = f (0) ,

x →2

x →0

f is continuous at both 2 and 0. Answer: Continuous at 2 and 0.

x−4 ;x=4 x+4

12.

(i) h is defined at x = 4, h (4) = 0.

1 ⎪⎧ f ( x) = ⎨ x ⎪⎩0

if x ≠ 0 if x = 0

1 = ∞ , lim f ( x) x →0 x →0 x →0 x does not exist. Thus f is discontinuous at x = 0. At x = –1, f is defined; f(–1) = –1. 1 lim f ( x) = lim = −1 . Since x →−1 x →−1 x lim f ( x) = −1 = f (−1) , f is continuous at

Because lim f ( x) = lim

x−4 0 (ii) lim h( x) = lim = = 0 , which exists. 8 x→4 x→4 x + 4

+

(iii) lim h( x ) = 0 = h(4) x→4

Thus h is continuous at x = 4. 6.

x →2−

lim f ( x) = 4 . In addition,

x 2 1 = = , which exists. 8 4 x→2 8

1 (iii) lim f ( x) = = f (2) . 4 x→2 Thus f is continuous at x = 2.

5. h( x) =

x →2+

lim f ( x) = lim x 2 = 4 , we have

x →2−

(ii) lim f ( x ) = lim x→2

⎧⎪ x + 2 if x ≥ 2 f ( x) = ⎨ 2 if x < 2 ⎪⎩ x f is defined at x = 2 and x = 0; f(2) = 4, f(0) = 0. Because lim f ( x) = lim ( x + 2) = 4 and

+

x →−1

f ( x) = 3 x ; x = –1

x = –1. Answer: Discontinuous at 0, continuous at –1.

(i) f is defined at x = –1; f(–1) = –1.

13. f is a polynomial function. (ii)

lim f ( x) = lim

x →−1

x →−1

3

x = 3 −1 = −1 , which

14. f is a polynomial function 2 3 1 2⎤ ⎡ ⎢ f ( x) = 5 + 5 x − 5 x ⎥ . ⎣ ⎦

exists. (iii) lim f ( x) = −1 = f (−1) x →−1

15. f is a rational function and the denominator is never zero.

Thus f is continuous at x = –1. 7. Continuous at –2 and 0 because f is a rational function and at neither point is the denominator zero.

16. f is a polynomial function ⎡ f ( x) = x − x 2 ⎤ . ⎣ ⎦ 17. None, because f is a polynomial function. 367



18. None, because h is a polynomial function.

30.

19. The denominator of this rational function is zero only when x = –4. Thus f is discontinuous only at x = –4. 20. The denominator of this rational function is zero only when x = ±2. Thus f is discontinuous only at x = ±2.

x →−1−

31.

23. x 2 + 2 x − 15 = 0 , (x + 5)(x – 3) = 0, x = –5 or 3. Discontinuous at –5 and 3. 24. x 2 + x = 0 , x(x + 1) = 0, x = 0 or –1. Discontinuous at 0 and –1.

(

lim f ( x) = lim ( x − 1) = 0 , then lim f ( x) = 0 . x →1

32.

27. x 2 + 1 = 0 has no real roots, so no discontinuity exists.

)

28. x 4 − 1 = 0 , x 2 + 1 x 2 − 1 = 0 ,

( x + 1) ( x + 1)( x − 1) = 0 , x = ±1. 2

33.

Discontinuous at ±1. ⎧1 if x ≥ 0 f ( x) = ⎨ ⎩−1 if x < 0 For x < 0, f(x) = –1, which is a polynomial and hence continuous. For x > 0, f(x) = 1, which is a polynomial and hence continuous. Because lim f ( x) = lim (−1) = −1 and x →0 −

x →0 +

⎧ x − 3 if x > 2 f ( x) = ⎨ ⎩3 − 2 x if x < 2 For x < 2, f(x) = 3 – 2x, which is a polynomial and hence continuous. For x > 2, f(x) = x – 3, which is a polynomial and hence continuous. Because f is not defined at x = 2, it is discontinuous there. ⎪⎧ x 2 + 1 if x > 2 f ( x) = ⎨ if x < 2 ⎪⎩8 x For x < 2, f(x) = 8x, which is a polynomial and

hence continuous. For x > 2, f ( x) = x 2 + 1, which is a polynomial and hence continuous. Because f is not defined at x = 2, it is discontinuous there.

34.

lim f ( x) = lim 1 = 1 , lim f ( x) does not

x →0 +

x →1

x = 1. f has no discontinuities.

3 26. Discontinuous at x = , for which the 2 denominator is zero.

x →0 −

x →1+

Since lim f ( x ) = 0 = f (0) , f is continuous at

x = 0, ±1. Discontinuous at 0, ±1.

29.

x →1−

x →1+

)

)(

x →−1

if x ≤ 1 ⎧0 f ( x) = ⎨ ⎩ x − 1 if x > 1 For x < 1, f(x) = 0, which is a polynomial and hence continuous. For x > 1, f(x) = x – 1, which is a polynomial and hence continuous. For x = 1, f is defined [f(1) = 0]. Because lim f ( x) = lim 0 = 0 and x →1−

25. x − x = 0 , x x − 1 = 0 , x(x + 1)(x – 1) = 0,

(

x →−1+

does not exist. Thus f is discontinuous at x = –1.

22. None, because f is a polynomial function.

2

x →−1−

lim f ( x ) = lim (2 x + 1) = −1 , lim f ( x)

x →−1+

21. None, because g is a polynomial function. 8 6 12 4 18 2 9 ⎤ ⎡ ⎢ g ( x) = 15 x − 5 x + 5 x − 5 ⎥ ⎣ ⎦

3

⎧2 x + 1 if x ≥ −1 f ( x) = ⎨ if x < −1 ⎩1 For x < –1, f(x) = 1, which is a polynomial and hence continuous. For x > –1, f(x) = 2x + 1, which is a polynomial and hence continuous. Because lim f ( x) = lim 1 = 1 and

x →0

exist. Thus f is discontinuous at x = 0.

⎧ 16 if ⎪ f ( x) = ⎨ x 2 ⎪3x − 2 if ⎩

x≥2 x<2

For x < 2, f(x) = 3x − 2, which is a polynomial 16 , and hence continuous. For x > 2, f ( x) = x2 which is continuous because x > 2 means that the denominator is never zero. 368


Section 10.4

For x = 2, f is defined [f(2) = 4]. Because lim f ( x) = lim (3 x − 2) = 4 and x →2−

38.

x →2−

lim f ( x) = lim

16

+

x →2

+

x→2

x = 2. f has no discontinuities. 35.

0.34

Answer: Yes

–2


y

1. We need to solve V(x) > 0. The zeros of V(x) occur when x = 0, 8 – 2x = 0, and 10 – 2x = 0, or x = 0, 4, and 5. These zeros determine the intervals (–∞, 0) (0, 4), (4, 5), and (5, ∞). Using x = –1, 1, 4.5, and 6 for test points, we find the sign of V(x): V(–1) = (–)(+)(+) = –, so V(x) < 0 on (–∞, 0); V(1) = (+)(+)(+) = +, so V(x) > 0 on (0, 4); V(4.5) = (+)(–)(+) = –, so V(x) < 0 on (4, 5); V(6) = (+)(–)(–) = +, so V(x) > 0 on (5, ∞). The volume is positive when 0 < x < 4 or 5 < x. However, x > 5 is unrealistic (as is x < 0) since the longest side of the piece of metal has length 2(5) = 10 inches. Thus, the volume is positive when 0 < x < 4.

0.28 0.22 0.16 0.10 x 1

2

3

4

5

Discontinuous at 1, 2, 3, 4. 36.

10

–10

= 4, then lim f ( x) = 4. x→2 x2 Since lim f ( x) = 4 = f (2), f is continuous at x→2

2

5

y

x 5

Problems 10.4 1. x 2 − 3 x − 4 > 0 f ( x) = x 2 − 3x − 4 = ( x + 1)( x − 4) has zeros –1

For –3.5 ≤ x ≤ 3.5, discontinuities at –3, –2, – 1, 0, 1, 2, 3. 37.

1000

and 4. By considering the intervals (–∞, –1), (–1, 4), and (4, ∞), we find f(x) > 0 on (–∞, –1) and (4, ∞). Answer: (–∞, –1), (4, ∞)

y

2. x 2 − 8 x + 15 > 0 f ( x) = x 2 − 8 x + 15 = ( x − 3)( x − 5) has zeros 3

and 5. By considering the intervals (–∞, 3), (3, 5), and (5, ∞), we find f(x) > 0 on (–∞, 3) and (5, ∞). Answer: (–∞, 3), (5, ∞)

x

f is continuous at 2. f is discontinuous at 5. f is discontinuous at 10.

20

3. x 2 − 3x − 10 ≤ 0 f(x) = (x + 2)(x − 5) has zeros −2 and 5. By considering the intervals (−∞, −2), (−2, 5), and (5, ∞), we find f(x) < 0 on (−2, 5). Answer: [−2, 5]

369


ISM: Introductory Mathematical Analysis 9. (x + 2)(x – 3)(x + 6) ≤ 0 f(x) = (x + 2)(x – 3)(x + 6) has zeros –2, 3, and –6. By considering the intervals (–∞, –6), (–6, –2), (–2, 3), and (3, ∞), we find f(x) < 0 on (–∞, –6) and (–2, 3). Answer: (–∞, –6], [–2, 3]

4. 14 − 5 x − x 2 ≤ 0 , or equivalently, x 2 + 5 x − 14 ≥ 0 f ( x) = x 2 + 5 x − 14 = ( x + 7)( x − 2) has zeros –7

and 2. By considering the intervals (–∞, –7), (–7, 2), and (2, ∞), we find f(x) ≥ 0 on (–∞, –7) and (2, ∞). Answer: (–∞, –7], [2, ∞)

10. (x + 5)(x + 2)(x − 7) ≤ 0 f(x) = (x + 5)(x + 2)(x − 7) has zeros −5, −2 and 7. By considering the intervals (−∞, −5), (−5, −2), (−2, 7) and (7, ∞), we find f(x) < 0 on (−∞, −5) and (−2, 7). Answer: (−∞, −5], [−2, 7]

5. 2 x 2 + 11x + 14 < 0 f ( x) = 2 x 2 + 11x + 14 = (2 x + 7)( x + 2) has zeros 7 and –2. By considering the intervals 2 7⎞ ⎛ 7 ⎛ ⎞ ⎜ −∞, − 2 ⎟ , ⎜ − 2 , − 2 ⎟ , and (–2, ∞), we find ⎝ ⎠ ⎝ ⎠ −

11. –x(x – 5)(x + 4) > 0, or equivalently, x(x – 5)(x + 4) < 0. f(x) = x(x – 5)(x + 4) has zeros, 0, 5, and –4. By considering the intervals (–∞, –4), (–4, 0), (0, 5), and (5, ∞), we find f(x) < 0 on (–∞, –4) and (0, 5). Answer: (–∞, –4), (0, 5)

⎛ 7 ⎞ f(x) < 0 on ⎜ − , − 2 ⎟ . ⎝ 2 ⎠ ⎛ 7 ⎞ Answer: ⎜ − , − 2 ⎟ 2 ⎝ ⎠

12. ( x + 2)2 > 0 f ( x) = ( x + 2) 2 has –2 as zero. By considering

6. x 2 − 4 < 0 . f ( x) = x 2 − 4 = ( x + 2)( x − 2) has

the intervals (–∞, –2) and (–2, ∞), we find f(x) > 0 on both intervals. Answer: (–∞, –2), (–2, ∞)

zeros ±2. By considering the intervals (–∞, –2), (–2, 2), and (2, ∞), we find f(x) < 0 on (–2, 2). Answer: (–2, 2)

13. x3 + 4 x ≥ 0

7. x 2 + 4 < 0 . Since x 2 + 4 is always positive, the

(

By considering the intervals (–∞, 0) and (0, ∞), we find f(x) > 0 on (0, ∞). Answer: [0, ∞)

8. 2 x 2 − x − 2 ≤ 0 . f ( x) = 2 x 2 − x − 2 has zeros

(

) f ( x) = ( x + 2) ( x − 1) = ( x + 2) ( x + 1)( x − 1)

1 ± 17 . By considering the intervals 4 ⎛ 1 − 17 ⎞ ⎛ 1 − 17 1 + 17 ⎞ , ⎜⎜ −∞, ⎟, ⎜ ⎟ , and 4 ⎟⎠ ⎜⎝ 4 4 ⎟⎠ ⎝

14. ( x + 2)2 x 2 − 1 < 0 2

2

2

has zeros –2, –1, and 1. By considering the intervals (–∞, –2), (–2, –1), (–1, 1), and (1, ∞), we find f(x) < 0 on (–1, 1). Answer: (–1, 1)

⎛ 1 + 17 ⎞ , ∞ ⎟ , we find f(x) < 0 on ⎜⎜ ⎟ ⎝ 4 ⎠ ⎛ 1 − 17 1 + 17 , ⎜⎜ 4 ⎝ 4

)

f ( x) = x x 2 + 4 has 0 as the only (real) zero.

inequality x 2 + 4 < 0 has no solution. Answer: no solution

⎞ ⎟⎟ . ⎠

15. x3 + 8 x 2 + 15 x ≤ 0 f(x) = x(x + 3)(x + 5) has zeros 0, –3, and −5. By considering the intervals (−∞, −5), (−5, −3), (−3, 0), and (0, ∞), we find f(x) < 0 on (−∞, −5) and (−3, 0). Answer: (−∞, −5], [−3, 0]

⎡ 1 − 17 1 + 17 ⎤ Answer: ⎢ , ⎥ 4 ⎦⎥ ⎣⎢ 4

370


Section 10.4 discontinuous at x = –5 and x = 1; f has zeros 3 and –2. By considering the intervals (–∞, –5), (–5, –2), (–2, 1), (1, 3), and (3, ∞), we find f(x) > 0 on (–∞, –5), (–2, 1), and (3, ∞). Answer: (–∞, –5), [–2, 1), [3, ∞)

16. x3 + 6 x 2 + 9 x < 0 f ( x) = x( x 2 + 6 x + 9) = x( x + 3)2 has zeros −3

and 0. By considering the intervals (∞, −3), (−3, 0) and (0, ∞), we find f(x) < 0 on (−∞, −3) and (−3, 0). Answer: (−∞, −3), (−3, 0) 17.

x x −9 2

22.

<0

x2 − 1 <0 x

23.

24.

4 ≥0 x −1

3 x − 5x + 6

x2 + 4 x − 5 f ( x) =

≤0

2x +1 x2

3

2

=

≤0 2x +1 x2

is discontinuous at x = 0 and f has

1⎤ ⎛ Answer: ⎜ −∞, − ⎥ 2⎦ ⎝

25. x 2 + 2 x ≥ 2 , or equivalently, x 2 + 2 x − 2 ≥ 0 . f ( x) = x 2 + 2 x − 2 has zeros −1 ± 3 . By

( ) ( −1 − 3, − 1 + 3 ) , and ( −1 + 3, ∞ ) , we find f(x) > 0 on ( −∞, − 1 − 3 ) and ( −1 + 3, ∞ ) . Answer: ( −∞, − 1 − 3 ⎤ , ⎡ –1 + 3, ∞ ) ⎦ ⎣ considering the intervals – ∞, – 1 – 3

≥0

x + 4x − 5 2

x + 6x + 5

1⎞ ⎛ f(x) < 0 on ⎜ −∞, − ⎟ . 2⎠ ⎝

>0

x2 − x − 6

( x + 5)( x − 1) is ( x + 2)( x + 1)

1 as a zero. By considering the intervals 2 1⎞ ⎛ 1 ⎞ ⎛ ⎜ −∞, − 2 ⎟ , ⎜ − 2 , 0 ⎟ , and (0, ∞), we find ⎝ ⎠ ⎝ ⎠

3 is never zero, but is ( x − 2)( x − 3) discontinuous when x = 2, 3. By considering the intervals (–∞, 2), (2, 3),and (3, ∞), we find f(x) > 0 on (–∞, 2) and (3, ∞). Answer: (–∞, 2), (3, ∞)

21.

=

−

f ( x) =

x2 − x − 6

3

f ( x) =

4 is discontinuous when x = 1, and x −1 f(x) = 0 has no root. By considering the intervals (–∞, 1) and (1, ∞), we find f(x) > 0 on (1, ∞). Note also that f(x) ≠ 0 for any x. Answer: (1, ∞)

2

2

3 is never zero, x + 6 x + 5 ( x + 5)( x + 1) but is discontinuous at x = –5 and x = –1. By considering the intervals (–∞, –5), (–5, –1), and (–1, ∞), we find that f(x) < 0 on (–5, –1). Answer: (–5, –1)

f ( x) =

20.

x2 + 4 x − 5

2

f ( x) =

x2 − 1 is discontinuous at x = 0; f has x zeros at ±1. By considering the intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞), we find f(x) < 0 on (–∞, –1) and (0, 1). Answer: (–∞, –1), (0, 1) f ( x) =

19.

≤0

x + 3x + 2 discontinuous at x = −1 and −2; f has zeros −5 and 1. By considering the intervals (−∞, −5), (−5, −2), (−2, −1), (−1, 1), and (1, ∞), we find f(x) < 0 on (−5, −2) and (−1, 1). Answer: [−5, −2), (−1, 1]

is discontinuous when x = ±3;

x2 − 9 f has 0 as a zero. By considering the intervals (–∞, –3), (–3, 0), (0, 3), and (3, ∞), we find f(x) < 0 on (–∞, –3) and (0, 3). Answer: (–∞, –3), (0, 3)

18.

x 2 + 3x + 2 f ( x) =

x

f ( x) =

x2 + 4 x − 5

=

( x − 3)( x + 2) is ( x + 5)( x − 1) 371


26. x 4 − 16 ≥ 0 .

(


4( x − 8)2 ≥ 324

)

f ( x) = x 2 + 4 ( x + 2)( x − 2) has (real) zeros –2

( x − 8)2 ≥ 81 x 2 − 16 x − 17 ≥ 0 (x – 17)(x + 1) ≥ 0 Solving gives x ≤ –1 or x ≥ 17. Since x must be positive, we have x ≥ 17. Answer: 17 in. by 17 in.

and 2. By considering the intervals (–∞, –2), (–2, 2), and (2, ∞), we find f(x) > 0 on (–∞, –2) and (2, ∞). Answer: (–∞, –2], [2, ∞) 27. Revenue = (no. of units)(price per unit). We q (28 − 0.2q ) ≥ 750 want

30. Let n = no. of persons over the 50 that attend. Then each of 50 + n persons will pay 50 – 0.50n. We want (50 + n)(50 – 0.50n) ≥ 50(50) 1 25n − n 2 ≥ 0 2 1 ⎞ ⎛ n ⎜ 25 − n ⎟ ≥ 0 2 ⎠ ⎝ Solving gives 0 ≤ n ≤ 50. Thus the size of the group is between 50 and 100 inclusive. Answer: 50 ≤ size of group ≤ 100

0.2q − 28q + 750 ≤ 0 2

q 2 − 140q + 3750 ≤ 0 Using the quadratic formula, q 2 − 140q + 3750 = 0 when q ≈ 36.09, 103.91.

Thus q 2 − 140q + 3750 ≤ 0 when 36.09 ≤ q ≤ 103.91, so sales revenue will be at least $750 when between 37 and 103 units, inclusive, are sold. x

28. x

x

–10

2

(2 − 2 x)(1 − 2 x) ≥

21 16

–10

64 x − 96 x + 11 ≥ 0 (8x – 11)(8x – 1) ≥ 0 1 11 Solving gives x ≤ or x ≥ . From the 8 8 1 diagram, clearly, x cannot exceed . Thus 2 1 x≤ . 8 1 mi Answer: 8

4

4

5

32. –5

5 –5

(–∞, 1.51) 5

33.

x–8 4

10

(–∞, –7.72]

2

29.

10

31.

x 1

–5

5

4

–5

x–8

(–∞, –0.5), (0.667, ∞) 4

4

4

4

If x is the length of a side of the piece of aluminum, then the box will be 4 by x – 8 by x – 8.

372


Chapter 10 Review

5

34.

9. As x → ∞, x + 1 → ∞ . Thus lim

x →∞

5

–5

10. –5

(−2, −1.62), (0.62, 1)

11.

x2 + 1

lim

x →∞

2x

2

x2

= lim

x →∞

2x

2

= lim

x →∞

2 =0. x +1

1 1 = 2 2

2x + 5 2x 2 2 = lim = lim = x →∞ 7 x − 4 x →∞ 7 x x →∞ 7 7 lim

Chapter 10 Review Problems 1.

(

)

12.

lim 2 x + 6 x − 1 = 2(−1) + 6(−1) − 1 = −5

x →−1

2. lim

2

2 x − 3x + 1 2

2 x2 − 2

x →0

=

2

(

)

13.

lim 2 x 2 − 3x + 1

= x →0 lim 2 x 2 − 2 x →0

(

)

1 1 =− −2 2

14.

x −9

( x + 3)( x − 3) 3. lim = lim 2 x( x − 3) x →3 x − 3 x x →3 x+3 6 = lim = =2 3 x →3 x 2

2x + 3

15.

2t − 3 2t − 3 = −∞ and lim = ∞ . Thus + t −3 3 − t t →3 t →3 2t − 3 lim does not exist. t →3 t − 3 lim

−

x6

lim

= lim x = −∞

x →−∞ x 5

x →−∞

x+3

lim

x →−∞ 1 − x

x→4

lim (2 x + 3)

17.

5. lim ( x + h) = x + 0 = x

x2 − 1

lim

x →∞

(3 x + 2)2 x

2

x →∞ 9 x

x −4 2

= lim

− 3x + 2 x+2 4 = lim = =4 1 x→2 x − 1 x →2 x 2

x→2

x3 + 4 x 2

lim

x →−4 x 2

+ 2x − 8

( x + 2)( x − 2) ( x − 2)( x − 1)

2

= lim

x →∞ 9 x 2

x2 − 1 + 12 x + 4

1 1 = 9 x →∞ 9

= lim

x2 + x − 2 ( x + 2)( x − 1) = lim x −1 x −1 x →1 x →1 = lim ( x + 2) = 3 x →1

x 2 ( x + 4) x →−4 ( x + 4)( x − 2)

19.

= lim

x2 16 8 = =− 3 −6 x →−4 x − 2

lim

−

x+3

x+3 x →3 ( x + 3)( x − 3)

= lim

x −9 1 = lim = −∞ − x−3 x →3 x →3

2

−

2− x ⎡ x − 2⎤ = lim ⎢ − ⎥ = lim (−1) = −1 x→2 x − 2 x→2 ⎣ x − 2 ⎦ x →2

( x − 1)( x + 2) x →1 ( x − 1)( x + 5)

20. lim

= lim

+ 4x − 5 x+2 3 1 = lim = = 6 2 x →1 x + 5 x →1 x 2

x = lim (−1) = −1 − x x →−∞

18. lim

lim

x2 + x − 2

x →−∞

x→4

= lim

h →0

6. lim

= lim

x →−4

x →−4

8. lim

=0

x →−∞ x 4

16. lim 3 64 = lim 4 = 4

−5 5 4. lim = = =− 2 12 12 x →−4 x 2 − 4 lim ( x − 4)

7.

1

lim

21. As x becomes large, so does 3x. Because the square roots of large numbers are also large, lim 3x = ∞ . x →∞

373



22. As y → 5+ , y – 5 approaches 0 through positive y → 5+

23.

lim

π− x

x →∞

= lim

x 97

x →∞

(

x3 x100 + 13

x →∞

x103 + 1

= lim

πx3 − x100

h →0

y −5 = 0.

values. Thus lim x100 + 13

f ( x + h) − f ( x ) h [8( x + h) − 2] − [8 x − 2] = lim h h→0 8h = lim = lim 8 = 8 h →0 h h →0

29. lim

)

)

f ( x + h) − f ( x ) h h →0

x103

= lim

30. lim

− x100

x →∞

( )

(

x3 π − x

x 97

⎡ 2( x + h)2 − 3⎤ − ⎡ 2 x 2 − 3⎤ ⎦ ⎣ ⎦ = lim ⎣ h h→0

= lim − x3 = −∞ x →∞

24.

ex 2 − x 4

lim

x →−∞ 31x − 2 x3

4 xh + 2h 2 h h→0 h(4 x + 2h) = lim = lim (4 x + 2h) = 4 x h h→0 h →0

− x4

= lim

x →−∞

= lim

−2 x3

x = −∞ x →−∞ 2

= lim

25.

1 ⎞ ⎛ 31. y = 23 ⎜ 1 − ⎟ ⎝ 1 + 2x ⎠ 1 Considering , we have 1 + 2x 1 1 1 1 lim = ⋅ lim = ⋅ 0 = 0 . Thus 2 x →∞ x 2 x →∞ 1 + 2 x

lim f ( x ) = lim x 2 = 1

x →1−

x →1−

lim f ( x) = lim x = 1

x →1+

x →1+

Thus lim f ( x) = 1 . x →1

26.

lim f ( x) = lim ( x + 5) = 8

x →3−

x →3−

⎡ ⎛ 1 ⎞⎤ lim y = lim ⎢ 23 ⎜ 1 − ⎥ = 23(1 − 0) = 23 1 + 2 x ⎟⎠ ⎦ x →∞ x →∞ ⎣ ⎝ Answer: 23

lim f ( x ) = lim 6 = 6

x →3+

x →3+

Because lim f ( x) ≠ lim f ( x) , x →3−

x →3+

lim f ( x ) does not exist.

32.

x →3

27.

x 2 − 16 = lim 4− x x → 4+

lim

x→4

+

= lim − x→4

+

33. f(x) = x + 5; x = 7 (i) f is defined at x = 7; f(7) = 12

x − 4 approaches 0 through

(ii) lim f ( x) = lim ( x + 5) = 7 + 5 = 12 , which

x + 4 → 8 . Thus

x →7

x+4

lim

x →5

+

x 2 − 3x − 10 x −5

10 x

x →∞ 0.1x

x →∞

x →7

exists

→ −∞ . x−4 Answer: –∞

28.

= lim

Answer: 100

x−4

positive values and −

10 x

x →∞ 1 + 0.1x

= lim 100 = 100

x−4 x+4 −( x − 4)

x+4

As x → 4+ ,

lim y = lim

x →∞

(iii) lim f ( x ) = 12 = f (7) x →7

= lim x →5

+

= lim

x →5+


( x − 5)( x + 2) x−5 x − 5( x + 2)

= 0⋅7 =0

374


34.

x −5 x +2 2

Chapter 10 Review

; x=5

43.

(i) f is defined at x = 5; f(5) = 0 (ii) lim f ( x) = lim

x −5

x →5 x 2

x →5

+2

=

0 = 0, which 27

⎧ x + 4 if x > −2 f ( x) = ⎨ ⎩3x + 6 if x ≤ −2 For x < –2, f(x) = 3x + 6, which is a polynomial and hence continuous. For x > –2, f(x) = x + 4, which is a polynomial and hence continuous. Because lim f ( x) = lim (3x + 6) = 0 and x →−2−

x →−2−

lim f ( x) = lim ( x + 4) = 2 , lim f ( x ) does

exists

x →−2+

x →−2+

x →−2

not exist. Thus f is discontinuous at x = –2.

(iii) lim f ( x ) = 0 = f (5) x →5


44.

1 35. Since f ( x) = x 2 is polynomial function, it is 5 continuous everywhere.

38.

x2 is a rational function and the x+3 denominator is zero at x = –3. Thus f is discontinuous at x = –3.

x →1+

x →1

45. x 2 + 4 x − 12 > 0 f ( x) = x 2 + 4 x − 12 = ( x + 6)( x − 2) has zeros –6

and 2. By considering the intervals (–∞, –6), (–6, 2), and (2, ∞), we find f(x) > 0 on (–∞, –6) and (2, ∞). Answer: (–∞, –6), (2, ∞)

is a rational function

2x + 3 whose denominator is never zero, f is continuous everywhere. 2

46. 3x 2 − 3 x − 6 ≤ 0

40. Since f ( x) = (2 − 3 x)3 is a polynomial function, it is continuous everywhere.

f ( x) = 3x 2 − 3 x − 6 = 3( x − 2)( x + 1) has zeros

−1 and 2. By considering the intervals (−∞, −1), (−1, 2), and (2, ∞), we find f(x) < 0 on (−1, 2). Answer: [−1, 2]

4 − x2

4 − x2 is a rational 41. f ( x) = = x 2 + 3 x − 4 ( x + 4)( x − 1) function and the denominator is zero only when x = –4 or x = 1, so f is discontinuous at x = –4, 1.

42.

f ( x) =

2x + 6 x +x 3

=

2x + 6

(

)

x x2 + 1

x →1

x = 1. f is discontinuous at x = 0.

is a rational function and the x3 denominator is zero at x = 0. Thus f is discontinuous at x = 0. x −1

x →1+

Since lim f ( x) = 1 = f (1) , f is continuous at

0

39. Since f ( x) =

if x ≥ 1

1 , which is a rational x function whose denominator is zero when x = 0. Thus f is discontinuous at x = 0. If x > 1, then f(x) = 1, which a polynomial function and hence continuous. At x = 1, f is defined [f(x) = 1]. 1 Because lim f ( x) = lim = 1 and x →1− x →1− x lim f ( x) = lim 1 = 1 , then lim f ( x) = 1 .

f ( x) =

f ( x) =

if x < 1

If x < 1, then f ( x) =

36. Since f ( x) = x 2 − 2 is a polynomial function, it is continuous everywhere. 37.

⎧⎪ 1 f ( x) = ⎨ x ⎪⎩1

47. x5 ≤ 7 x 4 , x5 − 7 x 4 ≤ 0 f ( x) = x5 − 7 x 4 = x 4 ( x − 7) has zeros 0 and 7.

By considering the intervals (−∞, 0), (0, 7), and (7, ∞), we find f(x) < 0 on (−∞, 0) and (0, 7). Answer: (−∞, 7]

is a rational function

and the denominator is zero only when x = 0, so f is discontinuous at x = 0.

375



f ( x) = x3 + 8 x 2 + 15 x = x( x + 5)( x + 3) has zeros 0, –5, and –3. By considering the intervals (–∞, –5), (–5, –3), (–3, 0), and (0, ∞), we find f(x) > 0 and (–5, –3) and (0, ∞). Answer: [–5, –3], [0, ∞)

49.

x+5 x2 − 1

5

53.

48. x3 + 8 x 2 + 15 x ≥ 0

0

3

–5

1.00 <0

1

54.

x+5 is discontinuous when ( x + 1)( x − 1) x = ±1, and f has –5 as a zero. By considering the intervals (–∞, –5), (–5, –1), (–1, 1), and (1, ∞), we find f(x) < 0 on (–∞, –5) and (–1, 1). Answer: (–∞, –5), (–1, 1) f ( x) =

50.

51.

0 –1

0.25

x( x + 5)( x + 8) <0 3 x( x + 5)( x + 8) f ( x) = has zeros 0, –5, and –8. 3 By considering the intervals (–∞, –8), (–8, –5), (–5, 0), and (0, ∞), we find f(x) < 0 on (–∞, –8) and (–5, 0). Answer: (–∞, –8), (–5, 0) x 2 + 3x x2 + 2 x − 8 f ( x) =

x2 − 9 x 2 − 16 f ( x) =

5

55. –5

0

≥0

2

=

x( x + 3) is ( x + 4)( x − 2)

0.50

=

5 –5 10

57. –5

≤0

[2.00, ∞)

( x + 3)( x − 3) is discontinuous ( x + 4)( x − 4)

5 –10 10

58.

x − 16 when x = ±4 and has zeros x = ±3. By considering the intervals (−∞, −4), (−4, −3), (−3, 3), (3, 4), and (4, ∞) we find f(x) < 0 on (−4, −3) and (3, 4). Answer: (−4, −3], [3, 4) 2

–5

–5

x2 + 3x

x2 − 9

5

5

56.

x + 2x − 8 discontinuous when x = −4, 2 and has zeros x = −3, 0. By considering the intervals (–∞, –4), (–4, –3), (–3, 0), (0, 2), and (2, ∞) we find f(x) > 0 on (–∞, –4), (–3, 0), and (2, ∞). Answer: (–∞, –4), [–3, 0], (2, ∞)

52.

5

–5

5

–10

(−1, 1.32]

376



Mathematical Snapshot Chapter 10 1. D = 8432e − rt A year from now, t = 1 and D = 8000. Thus 8000 = 8432e − r 8000 e−r = 8432 8000 −r = ln 8432 8000 r = − ln ≈ 0.053 8432 The rate is 5.3%.

2. D = 8432e −0.06t We want to find t when D =

8432 . 2

8432 = 8432e −0.06t 2 1 = e−0.06t 2 1 −0.06t = ln 2 1 ln 2 ln 2 t= = ≈ 12 −0.06 0.06 It would take about 12 years.

3. An exponential model assumes a fixed repayment rate. In reality, the repayment rate is constantly changing as a result of changing fiscal policy and other factors.

377

Chapter 11 Principles in Practice 11.1 1.

(

)

dH d = 6 + 40t − 16t 2 dt dt H (t + h) − H (t ) = lim h h →0 ⎡ 6 + 40(t + h) − 16(t + h)2 ⎤ − 6 + 40t − 16t 2 ⎦ = lim ⎣ h h→0

(

)

6 + 40t + 40h − 16t 2 − 32th − 16h 2 − 6 − 40t + 16t 2 h h →0

= lim

40h − 32th − 16h 2 = lim (40 − 32t − 16h) h h →0 h →0 = 40 – 32t dH = 40 − 32t dt = lim

Problems 11.1 1. a.

f ( x) = x3 + 3 , P = (−2, −5) [(−3)3 + 3] − (−5) [(−2.5)3 + 3] − (−5) = 19. If x = −2.5, then mPQ = = 15.25. −3 − (−2) −2.5 − (−2) Continuing in this manner, we complete the table:

To begin, if x = −3, then mPQ =

b. 2. a.

x-value of Q

−3

−2.5

−2.2

−2.1

−2.01

−2.001

mPQ

19

15.25

13.24

12.61

12.0601

12.0060

We estimate that mtan at P is 12. f ( x) = e 2 x , P = (0, 1) e 2(1) − 1 e 2(0.5) − 1 ≈ 6.3891 . If x = 0.5, then mPQ = ≈ 3.4366 . 1− 0 0.5 − 0 Continuing in this manner, we complete the table:

To begin, if x = 1, then mPQ =

x-value of Q mPQ

1

0.5

0.2

0.1

0.01

0.001

6.3891

3.4366

2.4591

2.2140

2.0201

2.0020

b. We estimate that mtan at P is 2. 3. f(x) = x f ( x + h) − f ( x ) ( x + h) − x h = lim = lim = lim 1 = 1 h h h →0 h →0 h → 0 h h →0

f ′( x) = lim

378


Section 11.1

4. f(x) = 4x – 1

10. f(x) = 7.01 f ( x + h) − f ( x ) h 7.01 − 7.01 = lim h h →0 0 = lim = lim 0 = 0 h → 0 h h →0

f ( x + h) − f ( x ) h [4( x + h) − 1] − [4 x − 1] = lim h h →0 4h = lim = lim 4 = 4 h →0 h h →0

f ′( x) = lim

f ′( x) = lim

h →0

h →0

5. y = 3x + 5. Let y = f(x). dy f ( x + h) − f ( x ) = lim dx h→0 h [3( x + h) + 5] − [3x + 5] = lim h h →0 3h = lim = lim 3 = 3 h →0 h h →0

11. Let f ( x) = x 2 + 4 x − 8.

(

6. y = –5x. Let y = f(x). dy f ( x + h) − f ( x ) = lim dx h→0 h [−5( x + h)] − [−5 x] = lim h h →0 −5h = lim = lim (−5) = −5 h →0 h h →0

x 2 + 2 xh + h 2 + 4 x + 4h − 8 − x 2 − 4 x + 8 h h →0

= lim

2 xh + h 2 + 4h h h →0 = lim (2 x + h + 4) = 2 x + 0 + 4 = 2 x + 4 = lim

h →0

12. y = f ( x) = x 2 + 5 x + 1

7. Let f(x) = 5 – 4x. d f ( x + h) − f ( x ) (5 − 4 x) = lim dx h h →0 [5 − 4( x + h)] − [5 − 4 x] = lim h h →0 −4h = lim = lim (−4) = −4 h →0 h h→0 8. Let f ( x) = 1 −

f ( x + h) − f ( x ) h [( x + h) 2 + 5( x + h) + 1] − [ x 2 + 5 x + 1] = lim h h →0 x 2 + 2 xh + h 2 + 5 x + 5h + 1 − x 2 − 5 x − 1 = lim h h →0 2 xh + h 2 + 5h = lim h h →0 = lim (2 x + h + 5) = 2 x + 0 + 5 = 2 x + 5

y ′ = lim

h →0

x 2

h →0

⎡1 − x + h ⎤ − ⎡1 − x ⎤ d ⎛ x⎞ 2 ⎦ ⎣ 2⎦ ⎣ − = 1 lim ⎜ ⎟ dx ⎝ 2 ⎠ h→0 h = lim

− h2

h →0

h

)

d 2 x + 4x − 8 dx f ( x + h) − f ( x ) = lim h h →0 ⎡ ( x + h)2 + 4( x + h) − 8⎤ − ⎡ x 2 + 4 x − 8⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

13. p = f (q) = 3q 2 + 2q + 1 dp f ( q + h) − f ( q ) = lim dq h→0 h

1 ⎛ 1⎞ = lim ⎜ − ⎟ = − 2 h→0 ⎝ 2 ⎠

⎡3(q + h)2 + 2(q + h) + 1⎤ − ⎡3q 2 + 2q + 1⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

9. f(x) = 3 f ( x + h) − f ( x ) h h →0 3−3 0 = lim = lim = lim 0 = 0 h →0 h h →0 h h → 0 f ′( x) = lim

6qh + 3h 2 + 2h h h →0 = lim (6q + 3h + 2) = 6q + 0 + 2 = 6q + 2 = lim

h →0

379

Chapter 11: Differentiation


14. Let f ( x) = x 2 − x − 3.

(

x+h+2 − x+2 h

)

d 2 x − x−3 dx f ( x + h) − f ( x ) = lim h ⎡ ( x + h ) 2 − ( x + h ) − 3⎤ − ⎡ x 2 − x − 3⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

x+h+2 − x+2 x+h+2 + x+2 ⋅ h x+h+2 + x+2 ( x + h + 2) − ( x + 2) 1 = = x+h+2 + x+2 h x+h+2 + x+2

=

(

Thus f ′( x) = lim

2 xh + h 2 − h = lim = lim (2 x + h − 1) = 2 x − 1 h h →0 h →0

15. y = f ( x) =

)

h →0

18. H ( x) =

6 x 6 x+h

=

3 x−2

H ( x + h) − H ( x ) h h →0 3 − 3 = lim x + h − 2 x − 2 h h →0 Multiplying the numerator and denominator by (x + h − 2)(x − 2) gives 3( x − 2) − 3( x + h − 2) H ′( x ) = lim h →0 h( x + h − 2)( x − 2) −3h = lim h →0 h( x + h − 2)( x − 2) −3 3 = lim =− h →0 ( x + h − 2)( x − 2) ( x − 2) 2

⎡ 6 ⎤ 6 6 = lim ⎢ − =− ⎥=− x( x + 0) h → 0 ⎣ x ( x + h) ⎦ x2

16. C = f (q) = 7 + 2q − 3q 2 dC f ( q + h) − f ( q ) = lim dq h→0 h

19. y = f ( x) = x 2 + 4

⎡ 7 + 2(q + h) − 3(q + h)2 ⎤ − ⎡ 7 + 2q − 3q 2 ⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

f ( x + h) − f ( x ) h 2 ⎡ ( x + h) + 4 ⎤ − ⎡ x 2 + 4 ⎤ ⎦ ⎣ ⎦ = lim ⎣ h h→0 y ′ = lim

h →0

2h − 6qh − 3h 2 = lim (2 − 6q − 3h) h h →0 h →0 = 2 − 6q

= lim

2 xh + h 2 = lim (2 x + h) = 2 x + 0 = 2 x h h→0 h →0 The slope at (–2, 8) is y ′(−2) = 2(−2) = −4 . = lim

17.

f ( x) = x + 2 f ′( x) = lim

h →0

1 2 x+2

H ′( x ) = lim

6 x

− f ( x + h) − f ( x ) = lim h h h →0 h →0 Multiplying the numerator and denominator by x(x + h) gives 6 x − 6( x + h) −6h y ′ = lim = lim h →0 hx( x + h) h→0 hx( x + h) y ′ = lim

1 x+h+2 + x+2

f ( x + h) − f ( x ) h

20. y = f ( x) = 1 − x 2 f ( x + h) − f ( x ) y ′ = lim h h →0 ⎡1 − ( x + h)2 ⎤ − ⎡1 − x 2 ⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

x+h+2 − x+2 h h →0 Rationalizing the numerator gives

= lim

−2 xh − h 2 h h →0 = lim (−2 x − h) = −2 x = lim

h →0

The slope at (1, 0) is y ′(1) = −2(1) = −2. 380


Section 11.1

26. y = ( x − 7) 2 = x 2 − 14 x + 49

21. y = f ( x) = 4 x 2 − 5 f ( x + h) − f ( x ) y ′ = lim h h →0 ⎡ 4( x + h) 2 − 5⎤ − ⎡ 4 x 2 − 5⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0

⎡ ( x + h)2 − 14( x + h) + 49 ⎤ − ⎡ x 2 − 14 x + 49 ⎤ ⎦ ⎣ ⎦ ′ = y lim ⎣ h h →0 2 xh + h 2 − 14h = lim (2 x + h − 14) = 2 x − 14 h h →0 h →0 If x = 6, then y ′ = 2(6) − 14 = −2 . The tangent line at (6, 1) is y – 1 = –2(x – 6), or y = –2x + 13. = lim

8 xh + 4h 2 = lim (8 x + 4h) = 8 x h h→0 h →0 The slope when x = 0 is y ′(0) = 8(0) = 0 . = lim

d 22. As shown in Example 5, dx

If x = 1, the slope is y ′(1) =

( x) = 2

27. y =

1 x

.

3 x −1

y ′ = lim

1 . 2

3 − 3 ( x + h ) −1 x −1

h h →0 3( x −1) −3( x + h −1) ( x + h −1)( x −1)

= lim

23. y = x + 4

h −3h −3 = lim = lim h→0 h( x + h − 1)( x − 1) h→0 ( x + h − 1)( x − 1) h →0

[( x + h) + 4] − [ x + 4] h = lim = 1 h h →0 h →0 h If x = 3, then y ′ = 1 . The tangent line at the point (3, 7) is y – 7 = 1(x – 3), or y = x + 4. y ′ = lim

=−

2

24. y = 3 x − 4

3 ( x − 1)2

3 If x = 2, then y ′ = − = −3 . The tangent line at 1 (2, 3) is y − 3 = −3( x − 2) , or y = −3 x + 9 .

[3( x + h) 2 − 4] − [3x 2 − 4] y ′ = lim h h →0 6 xh + 3h 2 = lim = lim (6 x + 3h) = 6 x h h →0 h →0 If x = 1, then y ′ = 6(1) = 6.

28. y =

The tangent line at (1, −1) is y + 1 = 6(x − 1) or y = 6x − 7.

5 1 − 3x

y ′ = lim

5 1−3( x + h )

− 1−53 x

h 5(1 − 3 x) − 5[1 − 3( x + h)] = lim h→0 h[1 − 3( x + h)](1 − 3 x) h →0

25. y = x 2 + 2 x + 3 ⎡ ( x + h)2 + 2( x + h) + 3⎤ − ⎡ x 2 + 2 x + 3⎤ ⎦ ⎣ ⎦ y ′ = lim ⎣ h h →0

15h h→0 h[1 − 3( x + h)](1 − 3 x)

= lim

2 xh + h 2 + 2h h h →0 = lim (2 x + h + 2) = 2 x + 2

15 h→0 [1 − 3( x + h)](1 − 3 x)

= lim

= lim

h →0

=

If x = 1, then y ′ = 2(1) + 2 = 4. The tangent line at the point (1, 6) is y − 6 = 4(x − 1), or y = 4x + 2.

15 (1 − 3 x)2

15 3 = . The tangent line at 25 5 3 3 11 (2, –1) is y + 1 = ( x − 2) , or y = x − . 5 5 5

If x = 2, then y ′ =

381



⎛ η ⎞⎛ dC ⎞ 29. r = ⎜ ⎟ ⎜ rL − ⎟ η dD + 1 ⎠ ⎝ ⎠⎝ dC ⎞ ⎛ (1 + η )r = η ⎜ rL − ⎟ dD ⎠ ⎝ dC ⎞ ⎛ r + η r = η ⎜ rL − ⎟ dD ⎠ ⎝ dC ⎞ ⎛ r = η ⎜ rL − ⎟ −η r dD ⎠ ⎝ dC ⎛ ⎞ −r⎟ r = η ⎜ rL − dD ⎝ ⎠ r η= dC rL − r − dD

30. 1.565, 1.470 31. –3.000, 13.445 32. 0.680, 1820.369 33. –5.120, 0.038 34. y = f ( x) = x 2 + x f ( x + h) − f ( x ) h 2 ⎡ ( x + h) + ( x + h) ⎤ − ⎡ x 2 + x ⎤ ⎦ ⎣ ⎦ = lim ⎣ h h →0 f ′( x) = lim

h →0

2 xh + h 2 + h = lim (2 x + h + 1) = 2 x + 1 h h →0 h →0 If x = –2, then f ′( x) = −3 . The tangent line at the point (–2, 2) is y – 2 = –3(x + 2), or y = –3x – 4. = lim

7

–5

5 –3

5

35.

–5

5

–5

382


Section 11.1

For the x-values of the points where the tangent to the graph of f is horizontal, the corresponding values of f ′( x) are 0. This is expected because the slope of a horizontal line is zero and the derivative gives the slope of the tangent line. 3

36. n = 4: ( z − x) ∑ xi z 3−i = ( z − x)( z 3 + xz 2 + x 2 z + x3 ) i =0

= z 4 − xz 3 + xz 3 − x 2 z 2 + x 2 z 2 − x3 z + x3 z − x 4 = z 4 − x4

2

n = 3: ( z − x) ∑ xi z 2−i = ( z − x)( z 2 + xz + x 2 ) i =0

= z 3 − xz 2 + xz 2 − x 2 z + x 2 z − x3 = z 3 − x3

1

n = 2: ( z − x) ∑ xi z1−i = ( z − x)( z + x) = z 2 − x 2 4

i =0 3

f ( x) = 2 x + x − 3 x 2 f ( z ) − f ( x) z−x z→x 2 z 4 + z 3 − 3z 2 − (2 x 4 + x3 − 3x 2 ) = lim z−x z→x 2( z 4 − x 4 ) + ( z 3 − x3 ) − 3( z 2 − x 2 ) = lim z−x z→x 2( z − x)( z 3 + xz 2 + x 2 z + x3 ) + ( z − x)( z 2 + xz + x 2 ) − 3( z − x)( z + x) = lim z−x z→x = lim [2( z 3 + xz 2 + x 2 z + x3 ) + ( z 2 + xz + x 2 ) − 3( z + x)]

f ′( x) = lim

z→x

= 2(4 x3 ) + (3 x 2 ) − 3(2 x) = 8 x3 + 3 x 2 − 6 x 4

37. n = 5: ( z − x) ∑ xi z 4−i = ( z − x)( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 ) i =0

= z 5 − xz 4 + xz 4 − x 2 z 3 + x 2 z 3 − x3 z 2 + x3 z 2 − x 4 z + x 4 z − x5 = z 5 − x5

2

n = 3: ( z − x) ∑ xi z 2−i = ( z − x)( z 2 + xz + x 2 ) i =0

= z 3 − xz 2 + xz 2 − x 2 z + x 2 z − x3 = z 3 − x3

f ( x ) = 4 x5 − 3 x3

383



f ( z ) − f ( x) z−x z→x 4 z 5 − 3z 3 − (4 x5 − 3x3 ) = lim z−x z→x 4( z 5 − x5 ) − 3( z 3 − x3 ) = lim z−x z→x 4( z − x)( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 ) − 3( z − x)( z 2 + xz + x 2 ) = lim z−x z→x = lim [4( z 4 + xz 3 + x 2 z 2 + x3 z + x 4 ) − 3( z 2 + xz + x 2 )]

f ′( x) = lim

z→x

= 4(5 x 4 ) − 3(3x 2 ) = 20 x 4 − 9 x 2

Principles in Practice 11.2 d (50q − 0.3q 2 ) dq d d = (50q ) − 0.3q 2 dq dq

1. r ′(q) =

(

) d d = 50 (q ) − 0.3 ( q 2 ) dq dq

= 50(1) – 0.3(2q) = 50 – 0.6q The marginal revenue is r ′(q) = 50 − 0.6q . Problems 11.2 1. f(x) = 5 is a constant function, so f ′( x) = 0 2.

⎛6⎞ f ( x) = ⎜ ⎟ ⎝7⎠

2/3

is a constant function, so f ′( x) = 0

3. y = x6 , y ′ = 6 x6−1 = 6 x5 4.

f ′( x) = 21x 21−1 = 21x 20

5. y = x80 ,

dy = 80 x80−1 = 80 x79 dx

6. y = x5.3 , y ′ = 5.3 x5.3−1 = 5.3 x 4.3 7.

(

)

f ( x) = 9 x 2 , f ′( x) = 9 2 x 2−1 = 18 x

(

)

8. y ′ = 4 3 x3−1 = 12 x 2 9. g ( w) = 8w7 , g ′( w) = 8(7 w7 −1 ) = 56w6

384


Section 11.2

10. v ′( x) = exe −1 11. y =

28.

(

)

2 4 2 8 x , y ′ = 4 x 4−1 = x3 3 3 3

(

f ′( p ) = 3 4 p 4−1 = 4 3 p3

13.

t7 1 7 f (t ) = , f ′(t ) = (7t 7 −1 ) = t 6 25 25 25

14. y ′ =

(

(

f ′( x) = 3(1) − 0 = 3

17.

f ′( x) = 4(2 x) − 2(1) + 0 = 8 x − 2

( ) ( )

32. p ′( x) =

)

f ′(t ) = −13 ( 2t ) + 14(1) + 0 = −26t + 14

1 ⎛ 1 1 −1 ⎞ 21. y ′ = 3 x3−1 − ⎜ x 2 ⎟ = 3 x 2 − 2 x ⎝2 ⎠

(

)

34.

28 − 19 ⎛ 14 ⎞ ( − 14 )−1 f ′( x) = 2 ⎜ − ⎟ x 5 =− x 5 5 ⎝ 5⎠

37.

2

= −39 x + 28 x − 2

(

f ′( x) = 2 0 − 4 x

)

) = −8x

38. y = x7 / 2 , y ′ =

3

26. φ ′(t ) = 5(3t 3−1 − 0) = 15t 2

( (

39.

)

1 13 − x 4 , 3 1 4 g ′( x) = 0 − 4 x 4−1 = − x3 3 3

27. g ( x) =

1

f ( x) = 11 x = 11x 2 , 11 ⎛ 1 ⎞ ( 1 )−1 11 − 1 f ′( x) = 11⎜ ⎟ x 2 = x 2 = 2 2 x ⎝2⎠

24. V ′(r ) = 8r 8−1 − 7 6r 6−1 + 3(2r ) + 0 = 8r 7 − 42r 5 + 6r 25.

⎛ 5 ( 5 )−1 ⎞ 3 − 1 10 2 3 ( 34 )−1 + 2⎜ x 3 ⎟ = x 4 + x3 x 4 3 ⎝3 ⎠ 4

2 −7 ⎛ 2 ⎞ −7 36. y ′ = 5(3x 2 ) − ⎜ − ⎟ x 5 = 15 x 2 + x 5 5 ⎝ 5⎠

)

4 −1

3 35 −1 3 −2 / 5 = x x 5 5

f ′( x) =

23. y ′ = −13 3x3−1 + 14(2 x) − 2(1) + 0

(

( )

1 2 2 7 x 6 + (1) = x 6 + 7 3 3

33.

35. y ′ =

22. y ′ = −8 4 x 4−1 + 0 = −32 x3

(

)

3 4 7 3 x + x 10 3 3 7 6 f ′( x) = 4 x3 + 3 x 2 = x3 + 7 x 2 10 3 5

19. g ′( p ) = 4 p 4−1 − 3 3 p3−1 − 0 = 4 p3 − 9 p 2 20.

9 2 x + 8x 2 9 + 3 x3−1 − (2 x) + 8(1) 2

f ( x) =

18. F ′( x) = 5(2 x) − 9(1) = 10 x − 9

(

)

5 5 30. k ′( x) = −2(2 x) + (1) + 0 = −4 x + 3 3

31.

16.

(

= 16 x3 + 3 x 2 − 9 x + 8

)

f ( x) = x + 3, f ′( x) = 1 + 0 = 1

)

h′( x) = 4 4 x 4−1

1 7 x 7 −1 = x 6 7

15.

(

5 4 5 x − 6 , f ′( x) = 4 x 4−1 − 0 = 10 x3 2 2

29. h( x) = 4 x 4 + x3 −

)

12.

f ( x) =

7 72 −1 7 5 / 2 = x x 2 2

1 −2 ⎛ 1 −2 ⎞ f (r ) = 6r 3 , f ′(r ) = 6 ⎜ r 3 ⎟ = 2r 3 ⎝3 ⎠

1 −3 ⎛ 1 −3 ⎞ 40. y = 4 x 4 , y ′ = 4 ⎜ x 4 ⎟ = x 4 ⎝4 ⎠

)

41.

385

f ( x) = x −4 , f ′( x) = −4 x −4−1 = −4 x −5



42.

f ′( s ) = 2(−3s −4 ) = −6 s −4

43.

f ( x) = x −3 + x −5 − 2 x −6 ,

(

53.

) (

f ′( x) = −3x −3−1 + −5 x −5−1 − 2 −6 x −6−1 = −3 x

44.

−4

− 5x

−6

+ 12 x

(

(

)

54. Φ ( x) =

)

− 12

55.

56.

f ( x) = 2 x −3 f ′( x) = 2(−3 x

47. y =

8

−4

) = −6 x

(

)

1 5

=

58.

4x 1 5 y ′ = −5 x −6 = − x −6 4 4

49. g ( x) =

)

4 3

=

50. y =

51.

1 x

2

1

=

2/3

=

1 −2 / 3 x 2

3 4 3

= 3x

− 34

2 1

= 2x

− 12

x2 −3 ⎛ 1 −3 ⎞ y′ = 2 ⎜ − x 2 ⎟ = − x 2 ⎝ 2 ⎠

4 −3 x 3

)

1 − 12 x 2 1 −3 y′ = − x 2 4

60. y =

= x −2 , y ′ = −2 x −3

5 2+( 1 ) ⎛ 1⎞ 61. y = x 2 x = x 2 ⎜ x 2 ⎟ = x 2 = x 2 ⎝ ⎠ 5 32 y′ = x 2

1 ⎛1⎞ 1 f (t ) = ⎜ ⎟ = t −1 2⎝t ⎠ 2 1 1 f ′(t ) = −1 ⋅ t −2 = − t −2 2 2

(

f ( x) =

59. y =

3x 4 g ′( x) = −3 x −4 = −4 x −4 3

(

3 2

x 9 −7 ⎛ 3 −7 ⎞ f ′( x) = 3 ⎜ − x 4 ⎟ = − x 4 4 ⎝ 4 ⎠

1 −5 x 4

(

1 3

2x 8 x 1⎛ 2 1 ⎞ q ′( x) = ⎜ − x −5 / 3 ⎟ = − x −5 / 3 2⎝ 3 3 ⎠

y ′ = 8 −5 x −6 = −40 x −6

48. y =

−7 ⎛ 1 −3 ⎞ ⎛ 3 −7 ⎞ 3 −3 f ′( z ) = 3 ⎜ z 4 ⎟ − 0 − 8 ⎜ − z 4 ⎟ = z 4 + 6 z 4 ⎝4 ⎠ ⎝ 4 ⎠ 4

57. q( x) =

−4

= 8 x −5

x5

f ( x) = −9 x1/ 3 + 5 x −2 / 5 , −2 −7 ⎛ 1 −2 ⎞ ⎛ 2 −7 ⎞ f ′( x) = −9 ⎜ x 3 ⎟ + 5 ⎜ − x 5 ⎟ = −3 x 3 − 2 x 5 ⎝3 ⎠ ⎝ 5 ⎠

1 45. y = = x −1 x dy 1 = −1 ⋅ x −1−1 = − x −2 = − dx x2

46.

)

1 3 x − 3 x −3 , 3 1 Φ ′( x) = (3x 2 ) − 3(−3x −4 ) = x 2 + 9 x −4 3

−7

⎛ 1 −1 ⎞ f ′( x) = 100 −3 x −4 + 10 ⎜ x 2 ⎟ ⎝2 ⎠ = −300 x −4 + 5 x

1 x + 7 x −1 7 1 1 f ′( x) = (1) + 7 −1x −2 = − 7 x −2 7 7

f ( x) =

)

7 −1 x 9 7 7 g ′( x) = (−1x −2 ) = − x −2 9 9

62.

52. g ( x) =

386

f ( x) = (8 x5 ), f ′( x) = 40 x 4


63.

(

Section 11.2

)

f ( x) = x 3 x 2 − 10 x + 7 = 3x3 − 10 x 2 + 7 x

72.

f ( x ) = 3 x9 − 5 x5 + 4 x3

73. w( x ) =

(

65.

3

2

3

)

( ) = 9x

f ( x) = x (3 x) = x 9 x

2

74.

5

f ′( x) = 45 x 4

66. s ( x) = 3 x

( 4 x − 6 x + 3)

= x1/ 3 ( x1/ 4 − 6 x + 3) = x 7 /12 − 6 x 4 / 3 + 3x1/ 3 7 s ′( x) = x −5 /12 − 8 x1/ 3 + x −2 / 3 12

67. v( x) = x v′( x) =

68.

− 23

1

( x + 5) = x 3 + 5 x

(

)

13

f (q) =

8

3

f ( w) =

w−5 w5

7 x3 + x

6 x 1 ⎛ 7 x3 x ⎞ = ⎜ + ⎟ 1/ 2 1/ ⎜ 6⎝ x x 2 ⎟⎠ 1 = (7 x5 / 2 + x1/ 2 ) 6 1 ⎛ 35 1 ⎞ f ′( x) = ⎜ x3 / 2 + x −1/ 2 ⎟ 6⎝ 2 2 ⎠ 1 1/ 2 −1 = x (35 x + x ) 12

39 2 ′ y x =−3 = −60 y ′ x =3 / 2 = −

77. y is a constant, so y ′ = 0 for all x.

−1

78. y ′ = 3 − 2 x −1/ 2 = 3 −

2 q2

y ′ x =4 = 2 7 y ′ x =9 = 3 13 y ′ x = 25 = 5

= w−4 − 5w−5

f ′( w) = −4w−5 + 25w−6 = − w−6 (4w − 25)

71.

= 1+ x

y ′ x =0 = −6

3q 2 + 4q − 2 3q 2 4q 2 = + − q q q q2

f ′(q ) = 3(1) + 0 − 2(− q −2 ) = 3 + 2q −2 = 3 +

70.

x2

76. y ′ = −6 − 6 x 2

)

= 3q + 4 − 2q

f ( x) =

x3

+

y ′ x = 2 = 16 y ′ x =−3 = −14

13 8 56 3 33 − 2 f ′( x) = x 5 + x 5 + x 5 5 5 5 1 − 52 = x 13x 2 + 56 x + 33 5

69.

x2

y ′ x =0 = 4

f ( x) = x 5 x 2 + 7 x + 11 = x 5 + 7 x 5 + 11x 5

(

=

75. y ′ = 6 x + 4

− 32

1 − 23 10 − 53 1 − 35 x − x = x ( x − 10) 3 3 3 3

x 2 + x3

x2 x2 w′( x) = 0 + 1 = 1

f ′( x) = 27 x8 − 25 x 4 + 12 x 2 = x 2 27 x6 − 25 x 2 + 12

(

f ′( x) = 4 x3 + 6 x 2 − 16 x = 2 x 2 x 2 + 3 x − 8

f ′( x) = 9 x 2 − 20 x + 7

64.

f ( x) = x 2 ( x − 2)( x + 4) = x 4 + 2 x3 − 8 x 2

f ( x) = ( x + 1)( x + 3) = x 2 + 4 x + 3 f ′( x) = 2 x + 4 = 2( x + 2)

387

2 x

)



79. y = 4 x 2 + 5 x + 6 y′ = 8x + 5

84. y =

y ′ x =1 = 13

(

1 x3

)

85. y =

( 2 − x2 ) = 2 x−

1 2

3

− x2 .

3 12 x 2

5 2 x − x3 2

5 x − 3x 2 = 0. Then x(5 − 3x) = 0, x = 0 or 5 x= . 3 5 125 . This If x = 0, then y = 0. If x = , y = 3 54 ⎛ 5 125 ⎞ gives the points (0, 0) and ⎜ , ⎟. ⎝ 3 54 ⎠

3 x4

3 16 An equation of the tangent line is 1 3 3 1 y − = − ( x − 2), or y = − x + . 8 16 16 2 y ′ x =2 = −

82. y = − 3 x = − x

−

1 2

y ′ = 5 x − 3x 2 A horizontal tangent line has slope 0, so we set

= x −3

y ′ = −3x −4 = −

− 32

) = x−

1 25 y ′ x =4 = − − 3 = − 8 8 When x = 4, then y = –7. The tangent line is 25 25 11 y + 7 = − ( x − 4) , or y = − x + . 8 8 2

1 1 − x2 5 1 y ′ = ( −2 x ) 5 8 y ′ x =4 = − 5 An equation of the tangent line is 8 8 17 y + 3 = − ( x − 4) , or y = − x + . 5 5 5

81. y =

x

y′ = − x

An equation of the tangent line is y – 15 = 13(x – 1), or y = 13x + 2. 80. y =

(

x 2 − x2

86. y =

x5 − x +1 5

y′ = x4 − 1 A horizontal tangent line has slope 0, so we set

1 3

x 4 − 1 = 0 . Then x 4 = 1 , so x = 1 or x = –1. If 1 9 x = 1, then y = ; if x = –1, then y = . This 5 5 9⎞ ⎛ 1⎞ ⎛ gives the points ⎜1, ⎟ and ⎜ −1, ⎟ . 5⎠ ⎝ 5⎠ ⎝

1 −2 1 y′ = − x 3 = − 2 3 3x 3 1 1 1 y ′ x =8 = − =− =− 3⋅ 4 12 ⎛ 23 ⎞ 3⎜ 8 ⎟ ⎝ ⎠ An equation of the tangent line is 1 4 1 y + 2 = − ( x − 8) , or y = − x − . 12 3 12

87. y = x 2 − 5 x + 3 y′ = 2 x − 5 Setting 2x – 5 = 1 gives 2x = 6, x = 3. When x = 3, then y = –3. This gives the point (3, –3).

83. y = 3 + x − 5 x 2 + x 4 y ′ = 1 − 10 x + 4 x3 .

88. y = x 4 − 31x + 11

When x = 0, then y = 3 and y ′ = 1 . Thus an equation of the tangent line is y – 3 = 1(x – 0), or y = x + 3.

y ′ = 4 x3 − 31

If 4 x3 − 31 = 1, then x3 = 8, x = 2. When x = 2, then y = −35. This gives the point (2, −35).

388


89.

1

f ( x) = x +

x

1

= x2 + x

Section 11.3 Principles in Practice 11.3

− 12

1. Here

1 − 32 1 1 x −1 − = x = 2 2 x 2x x 2x x x −1 x −1 − f ′( x) = − =0. Thus 2x x 2x x 2x x 1 x 2 x −1

f ′( x) =

− 12

−

dP = 5 and ∆p = 25.5 – 25 = 0.5. dp

dP ∆p = 5(0.5) = 2.5 dp The profit increases by 2.5 units when the price is changed from 25 to 25.5 per unit. ∆P ≈

90. z = (1 + b) w p − bwc dw p dz = (1 + b) −b dwc dwc Rewriting the right side and factoring out 1 + b dw p b(1 + b) dz , gives = (1 + b) − 1+ b dwc dwc

2.

(

)

dy d = 16t − 16t 2 = 16 − 16(2t ) = 16 − 32t dt dt dy = 16 − 32(0.5) = 16 − 16 = 0 dt t =0.5 The graph of y(t) is shown. 5

⎡ dw p dz b ⎤ = (1 + b) ⎢ − ⎥. dwc ⎣ dwc 1 + b ⎦

91. y = x3 − 3x y ′( x ) = 3x 2 − 3

0

( )−3 = 9

y ′ x =2 = 3 2

1

0

2

When t = 0.5, the object is at the peak of its flight.

The tangent line at (2, 2) is given by y – 2 = 9(x – 2), or y = 9x – 16.

3. V ′(r ) =

8

( )

4 π 3r 2 + 4π(2r ) = 4πr 2 + 8πr 3

When r = 2, V ′(r ) = 4π(2) 2 + 8π(2) = 32π and 4 32π 80 π(2)3 + 4π(2)2 = + 16π = π . 3 3 3 The relative rate of change of the volume when V ′(2) 32π 6 = = = 1.2 . Multiplying 1.2 r = 2 is V (2) 80 π 5 V (r ) =

5

–5 –2

92. y = 3 x = x1/ 3

3

by 100 gives the percentage rate of change: (1.2)(100) = 120%.

1 −2 / 3 1 x = 3 2 3 3 x 1 y ′ x =−8 = 12 The tangent line at (−8, −2) is given by 1 4 1 y + 2 = ( x + 8), or y = x − . 12 3 12 y ′( x ) =

Problems 11.3 1. s = f (t ) = 2t 2 + 3t If ∆t = 1, then over [1, 2] we have ∆s f (2) − f (1) 14 − 5 = = = 9. 2 −1 1 ∆t If ∆t = 0.5, then over [1, 1.5] we have ∆s f (1.5) − f (1) 9 − 5 = = = 8. 1.5 − 1 0.5 ∆t Continuing this way, we obtain the following table:

1 –15

5

–3

389


∆t ∆s ∆t

1

0.5

0.2

9

8

7.4


0.1 7.2

0.01

0.001

7.02

7.002

We estimate the velocity when t = 1 to be 7 m/s. With differentiation we get v = ds = 4(1) + 3 = 7 m/s. dt t =1

2. y = f ( x) = 2 x + 5 . If ∆x = 1, then over [3, 4] we have ∆y f (4) − f (3) 13 − 11 = = ≈ 0.2889 ∆x ∆x 1 If ∆x = 0.5, then over [3, 3.5] we have ∆y f (3.5) − f (3) 12 − 11 = = ≈ 0.2950 ∆x ∆x 0.5 Continuing in this way we obtain the following table: ∆x

1

0.5

0.2

0.1

0.01

0.001

∆y ∆x

0.2889

0.2950

0.2988

0.3002

0.3014

0.3015

We estimate the rate of change to be 0.3015. 1 ⎛ ⎞ ≈ 0.3015. ⎟ ⎜ Note: The actual rate of change is 11 ⎝ ⎠ 3. s = f (t ) = 2t 2 − 4t a.

When t = 7, then s = 2(7 2 ) − 4(7) = 70 m.

b.

∆s f (7.5) − f (7) [2(7.5)2 − 4(7.5)] − 70 = = = 25 m/s ∆t 0.5 0.5

c.

v=

ds = 4t − 4. If t = 7, then v = 4(7) − 4 = 24 m/s dt

1 4. s = f (t ) = t + 1 2

a.

When t = 2, s =

1 (2) + 1 = 2 m. 2

b.

1 ∆s f (2.1) − f (1) ⎡⎣ 2 (2.1) + 1⎤⎦ − 2 = = = 0.5 m/s 0.1 0.1 ∆t

c.

v=

ds 1 1 = . If t = 2, then v = m/s dt 2 2 390

ds = 4t + 3, dt


Section 11.3

5. s = f (t ) = 2t 3 + 6 a. b.

c.

b.

When t = 1, s = 2(1)3 + 6 = 8 m.

=

∆s f (1.02) − f (1) = ∆t 0.02 ⎡ 2(1.02)3 + 6 ⎤ − 8 ⎦ =⎣ 0.02 = 6.1208 m/s v=

c.

ds = 6t 2 . If t = 1, then dt

9.

v = 6(1)2 = 6 m/s

6. s = f (t ) = −3t 2 + 2t + 1 a. b.

c.

( )

When t = 1, s = −3 12 + 2(1) + 1 = 0 m. ∆s f (1.25) − f (1) = ∆t 0.25 ⎡ −3(1.25) 2 + 2(1.25) + 1⎤ − 0 ⎦ =⎣ = −4.75 m/s 0.25 v=

b.

c.

4

7/2⎤

⎥⎦ − 0

1 4

ds 7 = 12t 3 − t 5 / 2 . If t = 0, then dt 2 7 v = 12(0)3 − (0)5 / 2 = 0 m/s. 2 v=

dy 25 32 dy 25 = (27) = 337.50 . = x . If x = 9, dx 2 dx 2

dT = 0 + 0.27(1 − 0) = 0.27 dTe

12.

dV = 4πr 2 dr

13. c = 500 + 10q,

dc = 10 . When q = 100, dq

dc = 10 . dq

∆s f (2.1) − f (2) = ∆t 0.1 ⎡(2.1)4 − 2(2.1)3 + 2.1⎤ − 2 ⎦ =⎣ = 10.261 m/s 0.1

14. c = 5000 + 6q,

15.

ds = 4t 3 − 6t 2 + 1. If t = 2, then dt

dc dc = 6 . When q = 36, =6. dq dq

dc = 0.1(2q ) + 3 = 0.2q + 3 . When q = 5, dq dc = 0.2(5) + 3 = 4. dq

( ) ( )

v = 4 23 − 6 22 + 1 = 9 m/s

a.

1 m/s 64

( 14 ) − ( 14 )

11.

( )

8. s = f (t ) = 3t − t

1 4

⎡ ⎢3 ⋅ =⎣

dA dA = 2πr . If r = 3, = 2π(3) = 6π . dr dr

ds = −6t + 2. If t = 1, v = –4 m/s dt

4

− f (0)

10.

When t = 2, s = 24 − 2 23 + 2 = 2 m.

v=

( ) 1 4

When r = 6.3 × 10−4 , dV = 4π[6.3 × 10−4 ]2 = 158.76π×10−8 dr ≈ 4.988 × 10−6.

7. s = f (t ) = t 4 − 2t 3 + t a.

∆s = ∆t

f

16.

7/2

When t = 0, s = 3 ⋅ 04 = 07 / 2 = 0.

17.

391

dc dc = 0.2q + 3 . When q = 3, = 3.6 . dq dq dc = 2q + 50 . Evaluating when q = 15, 16 and dq 17 gives 80, 82 and 84, respectively.


18.


1 ⎞ 1 ⎛ 24. r = q ⎜ 15 − q ⎟ = 15q − q 2 30 30 ⎝ ⎠ dr 1 = 15 − q dq 15

dc = 0.12q 2 − q + 4.4 dq Evaluating when q = 5, 25, and 1000 gives 2.4, 54.4 and 119,004.4, respectively.

19. c = 0.01q + 5 +

500 q

For q = 5,

c = cq = 0.01q 2 + 5q + 500

q = 150,

dc = 0.02q + 5 dq

dr = 250 + 90q − 3q 2 . Evaluating when dq q = 5, 10 and 25 gives 625, 850 and 625, respectively.

dc =7 dq q =100

26. r = 60q − 0.2q 2

1000 20. c = 2 + q c = cq = 2q + 1000

dr = 60 − 0.4q dq Evaluating when q = 10 and 20 gives 56 and 52, respectively.

dc = 2 for all q dq 3

27.

2

21. c = cq = 0.00002q − 0.01q + 6q + 20, 000 dc = 0.00006q 2 − 0.02q + 6 dq

dc = 6.750 − 0.000328(2q) = 6.750 − 0.000656q dq dc = 6.750 − 0.000656(2000) = 5.438 dq q = 2000 c −10, 484.69 c= = + 6.750 − 0.000328q q q

dc = 4.6 . If q = 500, then dq

dc = 11 . dq

−10, 484.69 + 6.750 − 0.000328(2000) 2000 = 0.851655

c (2000) =

22. c = cq = 0.002q3 − 0.5q 2 + 60q + 7000 28.

dc = 0.006q 2 − q + 60 dq

If q = 15, then

dr =5. dq

25. r = 250q + 45q 2 − q3

dc =6 dq q =50

If q = 100, then

dr 44 dr = ; for q =15, = 14 ; for dq 3 dq

dc = −0.79 + 0.04284q − 0.0003q 2 dq dc = 0.7388 dq q =70

dc = 46.35. If q = 25, then dq

dc = 38.75. dq

29. PR 0.93 = 5, 000, 000 P = 5, 000, 000 R −0.93 dP = −4, 650, 000 R −1.93 dR

23. r = 0.8q dr = 0.8 for all q. dq

30.

392

dv = −10,500 for all t. dt


31. a.

Section 11.3 e.

dy = −1.5 − x dx dy = −1.5 − 6 = −7.5 dx x =6

36. a.

b. Setting –1.5 – x = –6 gives x = 4.5. 32. c = f (q ) = 0.4q 2 + 4q + 5 dc = 0.8q + 4 dq

If q = 2, then

dc = 5.6 . Over the interval [2, 3], dq

∆c f (3) − f (2) 20.6 − 14.6 = = =6. ∆q 3− 2 1

33. a. b.

y′ 1 = y x+4

c.

y ′(5) = 1

d.

1 1 = ≈ 0.111 5+ 4 9

e.

11.1%

34. a.

−9 x 2 y′ = y 5 − 3x3

c.

y ′(1) = −9

d.

−9 9 = − = −4.5 5−3 2

e.

−450% y′ = −3 x 2

b.

y′ −3 x 2 = y 8 − x3

c.

y ′(1) = −3

d.

−3 3 = − ≈ −0.429 8 −1 7

e.

–42.9%

38. a.

y ′ = −3

y ′ = −9 x 2

b.

37. a.

y′ = 1

63.2%

y′ = 2 x + 3

b.

y′ 2x + 3 = y x2 + 3x − 4

c.

y ′(−1) = 2(−1) + 3 = 1

d.

1 1 = − ≈ −0.167 1− 3 − 4 6

e.

–16.7%

b.

y′ −3 3 = = y 7 − 3x 3x − 7

c.

y ′(6) = −3

d.

3 3 = ≈ 0.2727 3(6) − 7 11

e.

27.27%

39. c = 0.3q 2 + 3.5q + 9

35. a.

y′ = 6 x

dc = 0.6q + 3.5 dq

b.

y′ 6x = y 3x 2 + 7

dc = 0.6(10) + 3.5 = 9.5. If dq q = 10, then c = 74 and

c.

y ′(2) = 6(2) = 12

d.

12 12 = ≈ 0.632 12 + 7 19

If q = 10, then dc dq

c

393

(100) =

9.5 (100) ≈ 12.8% . 74


40. y =


100 = 100 x −1 x

dy 100 = −100 x −2 = − dx x2 dy 100 y′ −1 =− = −1 and (100) = (100) = −10% . If x = 10, dx 100 y 10

41. a.

dr = 30 − 0.6q dq

b. If q = 10, c. 42. a.

9% dq = 10 − 0.4q dr

b. If q = 25, c. 43.

r′ 30 − 6 24 4 = = = ≈ 0.09 . r 300 − 30 270 45

r′ 10 − 0.4(25) = = 0. r 10(25) − 0.2(25) 2

0%

W ′ 0.864t −0.568 0.432 = = W t 2t 0.432 1.3 I 0.3

44. a.

R1′ 1855.24 1.3 = 1.3 = I R1 I 1855.24 1.3 I 0.3

R2′ 1101.29 1.3 = 1.3 = I R2 I 1101.29

b. They are equal. c.

f ′ x nC1 x n −1 n = = f ( x) x C1 x n g ′( x ) nC2 x n −1 n = = g ( x) x C2 x n The rates are equal.

45. The cost of q = 20 bikes is qc = 20(150) = $3000 . The marginal cost, $125, is the approximate cost of one additional bike. Thus the approximate cost of producing 21 bikes is $3000 + $125 = $3125.

394


Section 11.4

dc dq

dc

1 dq 1 dc c = = c , and the marginal cost 46. The relative rate of change of c is , which is given to be : = . Thus c dq q q c q ⎛ dc ⎞ function ⎜ ⎟ and the average cost function (c ) are equal. ⎝ dq ⎠

47. $5.07 per unit 48. 11,275 people per year Principles in Practice 11.4 1.

dR d d = (2 − 0.15 x) (225 + 20 x) + (225 + 20 x) (2 − 0.15 x) dx dx dx = (2 – 0.15x)(20) + (225 + 20x)(–0.15) = 40 – 3x – 33.75 – 3x = 6.25 – 6x dR = 6.25 − 6 x dx

1 2. T ( x) = x 2 − x3 3 T ′( x ) = 2x − x 2

When the dosage is 1 milligram the sensitivity is T ′(1) = 2(1) − 12 = 1 . Problems 11.4 1.

f ′( x) = (4 x + 1)(6) + (6 x + 3)(4) = 24x + 6 + 24x + 12 = 48x + 18 = 6(8x + 3)

2.

f ′( x) = (3x − 1)(7) + (7 x + 2)(3) = 42 x − 1

3. s ′(t ) = (5 − 3t )(3t 2 − 4t ) + (t 3 − 2t 2 )(−3) = 15t 2 − 20t − 9t 3 + 12t 2 − 3t 3 + 6t 2 = −12t 3 + 33t 2 − 20t 4. Q′( x) = (3 + x)(10 x) + (5 x 2 − 2)(1) = 15 x 2 + 30 x − 2 5.

(

)

(

(

)

(

)

f ′(r ) = 3r 2 − 4 (2r − 5) + r 2 − 5r + 1 (6r ) = 6r 3 − 15r 2 − 8r + 20 + 6r 3 − 30r 2 + 6r = 12r 3 − 45r 2 − 2r + 20

)

(

6. C ′( I ) = 2 I 2 − 3 (6 I − 4) + 3I 2 − 4 I + 1 (4 I ) = 12 I 3 − 8I 2 − 18I + 12 + 12 I 3 − 16 I 2 + 4 I = 2 12 I 3 − 12 I 2 − 7 I + 6 7. Without the product rule we have

(

)

f ( x) = x 2 2 x 2 − 5 = 2 x 4 − 5 x 2 f ′( x) = 8 x3 − 10 x

8. Without the product rule we have

(

)

f ( x ) = 3 x3 x 2 − 2 x + 2 = 3 x5 − 6 x 4 + 6 x3 f ′( x) = 15 x 4 − 24 x3 + 18 x 2

395

)


(

)


(

)

9. y′ = x 2 + 3x − 2 (4 x − 1) + 2 x 2 − x − 3 (2 x + 3)

(

) (

= 4 x3 + 12 x 2 − 8 x − x 2 − 3x + 2 + 4 x3 − 2 x 2 − 6 x + 6 x 2 − 3x − 9

)

= 8 x3 + 15 x 2 − 20 x − 7

10. φ ′( x) = (3 − 5 x + 2 x 2 )(1 − 8 x) + (2 + x − 4 x 2 )(−5 + 4 x) = 3 − 5 x + 2 x 2 − 24 x + 40 x 2 − 16 x3 − 10 − 5 x + 20 x 2 + 8 x + 4 x 2 − 16 x3 = −32 x3 + 66 x 2 − 26 x − 7

11.

f ′( w) = ( w2 + 3w − 7)(6 w2 ) + (2 w3 − 4)(2w + 3) = 6 w4 + 18w3 − 42 w2 + 4 w4 + 6 w3 − 8w − 12 = 10w4 + 24w3 − 42 w2 − 8w − 12

12.

(

)

(

)

f ′( x) = 3 x − x 2 (−1 − 2 x) + 3 − x − x 2 (3 − 2 x) = −3 x − 5 x 2 + 2 x3 + 9 − 3x − 3 x 2 − 6 x + 2 x 2 + 2 x3 = 4 x3 − 6 x 2 − 12 x + 9

(

)(

) (

)

13. y′ = x 2 − 1 9 x 2 − 6 + 3 x3 − 6 x + 5 (2 x) − 4(8 x + 2) = 9 x 4 − 15 x 2 + 6 + 6 x 4 − 12 x 2 + 10 x − 32 x − 8 = 15 x 4 − 27 x 2 − 22 x − 2

( ) (

)

14. h′( x) = 4 5 x 4 + 3 ⎡⎢ 8 x 2 − 5 (2) + (2 x + 2)(16 x) ⎤⎥ ⎣ ⎦ = 20 x 4 + 3(16 x 2 − 10 + 32 x 2 + 32 x) = 20 x 4 + 144 x 2 + 96 x − 30

3 ⎡ 1/ 2 ⎛ 1 ⎞⎤ (5 p − 2)(3) + (3 p − 1) ⎜ 5 ⋅ p −1/ 2 ⎟ ⎥ ⎢ 2⎣ ⎝ 2 ⎠⎦ 3⎡ 15 5 ⎤ = ⎢15 p1/ 2 − 6 + p1/ 2 − p −1/ 2 ⎥ 2⎣ 2 2 ⎦ 3 −1/ 2 1/ 2 = [45 p − 12 − 5 p ] 4

15. F ′( p ) =

3 ⎛1 ⎞ ⎛1 ⎞ 16. g ′( x) = ( x1/ 2 + 5 x − 2) ⎜ x −2 / 3 − x −1/ 2 ⎟ + ( x1/ 3 − 3 x1/ 2 ) ⎜ x −1/ 2 + 5 ⎟ 2 ⎝3 ⎠ ⎝2 ⎠ 1 −1/ 6 5 1/ 3 2 −2 / 3 3 15 1/ 2 1 3 = x + x − x − − x + 3 x −1/ 2 + x −1/ 6 + 5 x1/ 3 − − 15 x1/ 2 3 3 3 2 2 2 2 1 −1/ 6 −1/ 2 −2 / 3 1/ 2 1/ 3 = (−135 x + 40 x + 5 x + 18 x − 4x − 18) 6

17. y = 7 ⋅

2 is a constant function, so y′ = 0 . 3

18. y = x3 − 6 x 2 + 11x − 6 y′ = 3x 2 − 12 x + 11 396


Section 11.4

19. y = 6 x3 + 47 x 2 + 31x − 28 y′ = 18 x 2 + 94 x + 31

20.

21.

dy (4 x + 1)(2) − (2 x − 3)(4) 8 x + 2 − 8 x + 12 = = dx (4 x + 1)2 (4 x + 1)2 14 = (4 x + 1)2 ( x − 1)(5) − (5 x)(1)

f ′( x) = =−

( x − 1)

2

=

5x − 5 − 5x ( x − 1)2

5 ( x − 1)2 (5 − x)(−5) − (−5 x)(−1)

22. H ′( x) =

(5 − x)2 −25 + 5 x − 5 x 25 = =− 2 (5 − x) (5 − x) 2

23.

24.

−13

f ( x) =

5

=−

13 −5 x 3

3x 13 65 f ′( x ) = − (−5 x −6 ) = 3 3x6

(

)

5 2 x −2 7 5 10 f ′( x) = (2 x) = x 7 7 f ( x) =

25. y′ = =

( x − 1)(1) − ( x + 2)(1) ( x − 1)2

x −1− x − 2 ( x − 1)2

=−

3 ( x − 1)2

26. h′( w) = = =

(

)

( w − 3)(6w + 5) − 3w2 + 5w − 1 (1) ( w − 3)2

6w2 − 13w − 15 − 3w2 − 5w + 1 ( w − 3) 2 3w2 − 18w − 14 ( w − 3)2

397


27.


z 2 − 4 ) (−2) − (6 − 2 z )(2 z ) ( h′( z ) = 2 ( z2 − 4) =

−2 z 2 + 8 − 12 z + 4 z 2

( z2 − 4) 2 ( z2 − 6z + 4) = 2 ( z2 − 4)

2

2 z 2 − 12 z + 8

( z2 − 4)

2

(3 x 2 + 5 x + 3)(4 x + 5) − (2 x 2 + 5 x − 2)(6 x + 5)

28. z ′ = =

=

(3x 2 + 5 x + 3) 2 12 x3 + 35 x 2 + 37 x + 15 − (12 x3 + 40 x 2 + 13x − 10) (3x 2 + 5 x + 3)2 2

−5 x + 24 x + 25

=

(3x 2 + 5 x + 3) 2

( x2 − 5x ) (16 x − 2) − (8x2 − 2x + 1) (2 x − 5) 2 ( x2 − 5x ) 16 x3 − 82 x 2 + 10 x − (16 x3 − 44 x 2 + 12 x − 5 ) −38 x 2 − 2 x + 5 = = 2 2 2 x − 5 x ( ) ( x2 − 5x )

29. y′ =

30.

x 2 + 1)( 3x 2 − 2 x ) − ( x3 − x 2 + 1) (2 x) ( f ′( x) = 2 ( x2 + 1) =

=

31.

3 x 4 − 2 x3 + 3 x 2 − 2 x − 2 x 4 + 2 x3 − 2 x

(

x x3 + 3 x − 4

( x2 + 1)

)

( x 2 + 1) 2

2

2 x 2 − 3 x + 2 ) (2 x − 4) − ( x 2 − 4 x + 3) (4 x − 3) ( y′ = 2 ( 2 x 2 − 3x + 2 ) 4 x3 − 14 x 2 + 16 x − 8 − ( 4 x3 − 19 x 2 + 24 x − 9 ) = 2 ( 2 x 2 − 3x + 2 ) =

5x2 − 8x + 1

( 2 x 2 − 3x + 2 )

2

398


Section 11.4

32. The quotient rule can be used, or we can write z4 + 4 1 3 = z + 4 z −1 , F ( z) = 3z 3

(

so F ′( z ) =

33.

)

38.

1 3z 4 − 4 3 z 2 − 4 z −2 = . 3 3z 2

(

)

=

x100 + 7 ) (0) − (1) (100 x99 ) ( 100 x99 g ′( x) = =− 2 2 ( x100 + 7 ) ( x100 + 7 )

=

−9

9 = − x −5 5 2 2x 45 −6 y′ = x 2

34. y =

35. u (v) =

x

41. y′ =

3x2 − x − 1 1

5 3

⎞ x+5 ⎟= 3 ⎟ ⎠ 16 x 2 2 3

= 3x − x − x

=

− 13

=

x3

2 2 − 13 1 − 43 2 1 x + x = 5x 3 − 1 + 4 3 3 3x 3 3x 3 15 x 2 − 2 x + 1

3x

2

(3 x − 5)(5) − (5 x + 1)(3) (3 x − 5)2 28 (3 x − 5)

2

+ 6 x −4

+ 6 x −4

[( x + 2)( x − 4)](1) − ( x − 5)(2 x − 2) [( x + 2)( x − 4)]2

)

(

[( x + 2)( x − 4)]

− x 2 − 10 x + 18

)

[( x + 2)( x − 4)]2 (9 x − 1)(3 x + 2) 27 x 2 + 15 x − 2 = 4 − 5x 4 − 5x

(

)

(4 − 5 x)(54 x + 15) − 27 x 2 + 15 x − 2 (−5) (4 − 5 x)

2

−270 x 2 + 141x + 60 + 135 x 2 + 75 x − 10

=−

399

(3 x + 1)2

2

y′ = =

(3x + 1)(2) − (2 x)(3)

(

42. y =

4 3

+

x 2 − 2 x − 8 − 2 x 2 − 12 x + 10

2

y′ = 5 x 3 − =

)

)

2

( x − 8)(0) − (4)(1)

= 6 x2 −

1⎛ 1 −1 ⎞ = ⎜ x 2 − 5x 2 ⎟ ⎠ 8 x 8⎝

3

(

x0.7 2 x 2.1 + 1

40. q ′( x) = 6 x 2 +

x−5

=

0.3 1 + 28 x1.8 − 12 x 2.1

⎞ 2(v3 + 4) ⎟= v2 ⎠

1 ⎛ 1 −1 5 −3 ⎞ 1 ⎛ 1 5 y′ = ⎜ x 2 + x 2 ⎟ = ⎜ 1 + 3 ⎜ 8⎝ 2 2 ⎠ 16 ⎝ x 2 x 2

37. y =

(

(2 x 2.1 + 1) 2

( x − 8) 4 2 = + 2 ( x − 8) (3 x + 1) 2

v3 − 8 v3 8 = − = v 2 − 8v −1 v v v

3x 2 − x − 1

0.6 x1.4 + 0.3x −0.7 − 4.2 x1.4 + 8.4 x1.1

39. y′ = −

4 ⎛ u′(v) = 2v + 8v −2 = 2 ⎜ v + v2 ⎝

36. y =

2 x 2.1 + 1)( 0.3 x −0.7 ) − ( x 0.3 − 2 )( 4.2 x1.1 ) ( y′ = 2 ( 2x2.1 + 1)

(4 − 5 x) 2 135 x − 216 x − 50 (4 − 5 x)2


(

)(


)

(

)(

⎡ t 2 − 1 t 3 + 7 ⎤ (2t + 3) − t 2 + 3t 5t 4 − 3t 2 + 14t ⎢ ⎦⎥ 43. s′(t ) = ⎣ 2 ⎡ t 2 −1 t3 + 7 ⎤ ⎢⎣ ⎥⎦

(

=

44.

)(

)

)

−3t 6 − 12t 5 + t 4 + 6t 3 − 21t 2 − 14t − 21

(

)(

)

⎡ t 2 −1 t3 + 7 ⎤ ⎣⎢ ⎦⎥

2

17

f ( s) =

3

5s − 10 s 2 + 4s 0 − 17 ⎡15s 2 − 20s + 4 ⎤ 17 15s 2 − 20s + 4 ⎣ ⎦ =− f ′( s ) = 2 2 5s3 − 10s 2 + 4 s 5s3 − 10s 2 + 4 s

(

45. y = 3x −

2 x

(

(

)

)

)

2( x −1) −3 x x ( x −1)

− x3−1

= 3x − x−2 x−2 x+2 x+2 = 3x + = 3x + 3 x( x − 1)( x − 2) x − 3x2 + 2 x y′ = 3 + = 3−

( x3 − 3 x 2 + 2 x)(1) − ( x + 2)(3 x 2 − 6 x + 2) [ x( x − 1)( x − 2)]2

2 x3 + 3 x 2 − 12 x + 4 [ x( x − 1)( x − 2)]2

3

46. y = 3 − 12 x + y ′ = −36 x 2 +

47.

f ′( x ) =

1 − 25

x +2

x2 + 5

3

= 3 − 12 x +

x 2 + 2 −5 x2 + 2 2

x +5

= 3 − 12 x3 +

( x 4 + 7 x 2 + 10)(2 x) − ( x 2 − 3)(4 x3 + 14 x) ( x 4 + 7 x 2 + 10) 2

(a − x)(1) − (a + x)(−1) (a − x)

2

=

x2 − 3 x 4 + 7 x 2 + 10

= −36 x 2 +

2a (a − x)2

x −1 + a −1 ax a + x ⋅ = x −1 − a −1 ax a − x (a − x)(1) − (a + x)(−1) 2a f ′( x ) = = 2 (a − x) (a − x)2

48. Simplifying, f ( x) =

( )( ) y′ = ( 4 x 2 + 2 x − 5 )( 3 x 2 + 7 ) + ( x3 + 7 x + 4 ) (8 x + 2)

49. y = 4 x 2 + 2 x − 5 x3 + 7 x + 4

y′(−1) = (−3)(10) + (−4)(−6) = −6

400

−2 x5 + 12 x3 + 62 x [( x 2 + 2)( x 2 + 5)]2


50. y = y′ =

x3 x4 + 1 ( x 4 + 1)(3 x 2 ) − ( x3 )(4 x3 ) ( x 4 + 1)2

y′(−1) =

51. y = y′ =

Section 11.4

(2)(3) − (−1)(−4) (2)

6 x −1 ( x − 1)(0) − (6)(1) ( x − 1)

y′(3) = −

6 2

2

2

=−

=

2

1 2

6

=−

( x − 1) 2

3 2

3 3 15 The tangent line is y − 3 = − ( x − 3) , or y = − x + . 2 2 2

52. y =

x+5 x2

= x −1 + 5 x −2

y ′ = − x −2 − 10 x −3 = − y ′(1) = −1 − 10 = −11

1 x

2

−

10 x3

The tangent line is y − 6 = −11(x − 1) or y = −11x + 17.

( ) y′ = (2 x + 3) ⎡⎢ 2 ( 4 x3 − 10 x ) ⎤⎥ + ⎡⎢ 2 ( x 4 − 5 x 2 + 4 ) ⎤⎥ (2) ⎣ ⎦ ⎣ ⎦

53. y = (2 x + 3) ⎡⎢ 2 x 4 − 5 x 2 + 4 ⎤⎥ ⎣ ⎦

y′(0) = (3)(0) + [2(4)](2) = 16 The tangent line is y – 24 = 16(x – 0), or y = 16x + 24.

54. y =

x +1 2

x ( x − 4)

=

x +1 3

x − 4 x2

x3 − 4 x 2 ) (1) − ( x + 1) ( 3 x 2 − 8 x ) ( y′ = 2 ( x3 − 4 x 2 ) y′(2) =

(−8)(1) − (3)(–4) (−8)

2

The tangent line is y +

=

4 1 = 64 16

3 1 1 1 = ( x − 2) , or y = x − . 8 16 16 2

401


55. y = y′ =

x 2x − 6 (2 x − 6)(1) − x(2) (2 x − 6)

2


500 q r = pq = 500

60. p = =

−6 (2 x − 6) 2

dr =0 dq

1 1 If x = 1, then y = = − and 2−6 4 −6 −6 3 y′ = = =− . 8 (−4)2 16

61. p =

3 y′ − 8 3 Thus = = = 1.5 . y −1 2 4

56. y = y′ =

r = pq =

1− x 1+ x (1 + x)(−1) − (1 − x)(1) (1 + x) 2

v=

2

= =−

2

t +1

58. s = v=

=

6 = −1.5 m/s. 4

ds (t 2 + 7)(1) − (t + 3)(2t ) = dt (t 2 + 7) 2 2

(t + 7)

2

=

(7 + t )(1 − t ) 2

(t + 7)

q 2 + 750q q + 50

(

)

2 dr (q + 50)(2q + 750) − q + 750q (1) = dq (q + 50)2

q 2 + 100q + 37,500 (q + 50)2

63.

dC = 0.672 dI

64.

dC = 0.712 dI

t2 + 7

7 − 6t − t 2

−3

q + 750 q + 50

r = pq =

t +3

=

(q + 2)2

62. p =

ds (t 3 + 1)(0) − 2(3t 2 ) 6t 2 = =− dt (t 3 + 1)2 (t 3 + 1)2

If t = 1, then v = −

216

(1 + x )2

. When t = 1, then s = 1 m.

3

108q − 3q q+2

dr (q + 2)(108) − (108q)(1) = −3 dq (q + 2) 2

1 y′ − 18 1 When x = 5, then . = = y − 2 12 3

57. s =

108 −3 q+2

65. C = 3 + I 1/ 2 + 2 I 1/ 3 dC 1 2 1 2 = 0 + I −1/ 2 + I −2 / 3 = + dI 2 3 2 I 33 I 2 dC 1 2 7 = + = . When I = 1, then dI 2 3 6 dS dC 1 2 = 1− = 1− − dI dI 2 I 33 I 2

2

v = 0 when t = −7 or t = 1. Since t is positive, we choose t = 1. 59. p = 50 − 0.01q r = pq = 50q − 0.01q 2 dr = 50 − 0.02q dq

When I = 1, then 1 −

402

dC 7 1 = 1− = − . dI 6 6


66.

Section 11.4

dC 3 1 = − dI 4 6 I dC 43 dS 43 17 = , so = 1− = dI I = 25 60 dI I = 25 60 60

dC 67. = dI

(

I +4

( )

)

) ( 8I + 1.2

I − 0.2 − ⎛⎜16 I + 0.8 I 3 − 0.2 I ⎞⎟ 1 ⎝ ⎠ 2 I

(

I +4

)

2

dS dC ≈ 0.615 , so ≈ 1 − 0.615 = 0.385 when I = 36. dI dI I =36 dC 68. = dI

(

I +5

) ( 10I + 0.75

( )

)

I − 0.4 − ⎛⎜ 20 I + 0.5 I 3 − 0.4 I ⎞⎟ 1 ⎝ ⎠ 2 I

(

I +5

)

2

dS dC ≈ 0.393 , so ≈ 1 − 0.393 = 0.607 when I = 100. dI dI I =100 1

69. Simplifying gives C = 10 + 0.7 I − 0.2 I 2 a.

b.

dC 0.1 −1 = 0.7 − 0.1I 2 = 0.7 − dI I dS dC 0.1 = 1− = 0.3 + dI dI I dS 0.1 = 0.3 + = 0.32 dI I = 25 5 dC dI

C

when I = 25 is

0.7 − 0.1 5 10 + 0.7(25) − 0.2(5)

70. Simplifying S gives S=

I − 2 I −8

(

I +2

)(

I −4

)=

I −4 I +2 I +2 dS 1 −1/ 2 1 Thus . = I = dI 2 2 I dS 1 dC = ≈ 0.04082 and ≈ 1 − 0.04082 ≈ 0.9592. dI I =150 2 ⋅ 150 dI I =150

71.

=

≈ 0.026

dc (q + 2)(2q ) − q 2 (1) q 2 + 4q 6q(q + 4) = 6⋅ = 6⋅ = dq (q + 2) 2 (q + 2) 2 (q + 2) 2

403



dc dc d ⎛ c ⎞ q ⋅ dq − c (1) d = 72. We assume that (c ) = 0 . Thus 0 = . ⎜ ⎟= dq dq ⎝ q ⎠ dq q2

This implies that q ⋅

dc dc dc c −c = 0 , q⋅ = = c , so the marginal cost function =c, dq dq dq q

cost function ( c ) are equal.

⎛ dc ⎞ ⎜ ⎟ and the average ⎝ dq ⎠

900 x 10 + 45 x dy (10 + 45 x)(900) − (900 x)(45) = dx (10 + 45 x )2

73. y =

dy (100)(900) − (1800)(45) 9 = = dx x = 2 10 (100)2 0.05V A + xV d ( A + xV )(0.05) − (0.05V )( x) (RT) = dV ( A + xV ) 2

74. RT =

=

0.05 A ( A + xV ) 2

Both numerator and denominator are always positive, so increases by one unit, RT increases. 0.7355 x 1 + 0.02744 x dy (1 + 0.02744 x)(0.7355) − (0.7355 x)(0.02744) = dx (1 + 0.02744 x )2

75. y =

=

76.

0.7355 (1 + 0.02744 x)2

a (1 + x) − b(2 + n) x a (2 + n)(1 + x ) − b(2 + n) x For convenience let c = 2 + n. a (1 + x) − bcx 1 a (1 + x) − bcx = ⋅ . Then f ( x) = ac(1 + x) − bcx c a(1 + x) − bx f ( x) =

1 [a (1 + x) − bx](a − bc) − [a (1 + x) − bcx](a − b) f ′( x) = ⋅ c [a (1 + x) − bx]2 1 1 (−c + 1)ab −abc + ab = ⋅ = ⋅ c [a (1 + x) − bx]2 c [a(1 + x) − bx]2 =

1 [−1(2 + n) + 1]ab −(1 + n)ab ⋅ = 2 2 + n [a(1 + x) − bx] [a(1 + x) − bx]2 (2 + n)

g ( x) =

A + Bx C + Dx 404

d (RT) > 0 . This rate of change means that if V dV


g ′( x) = = =

Section 11.5

(C + Dx )( B ) − ( A + Bx)( D) (C + Dx) 2 CB + BDx + AD − BDx (C + Dx)2 BC − AD (C + Dx )2

C⎞ ⎛ Thus, g ′( x) has the form given. When g ′( x) is defined ⎜ for x ≠ ⎟ , its sign is constant. D⎠ ⎝

dc d ⎛c⎞ 77. = ⎜ ⎟= dq dq ⎝ q ⎠

78.

dc − c(1) q ⋅ dq

q

2

. When q = 20 we have

dc dq

c

dc − c q⋅ dq

=

q2

=

c

20(125) − 20(150) (20)2

150

=−

1 120

dy = (3)(2 x − 1)( x − 4) + (3 x + 1)(2)( x − 4) + (3 x + 1)(2 x − 1)(1) dx = 18 x 2 − 50 x + 3

Principles in Practice 11.5 1. By the chain rule, dy dy dx d d = ⋅ = 4 x 2 ⋅ (6t ) = (8 x)(6) = 48 x . dt dx dt dx dt dy Since x = 6t, = 48(6t ) = 288t . dt

( )

Problems 11.5

(

)

(

)

1.

dy dy du = ⋅ = (2u − 2)(2 x − 1) = ⎡⎢ 2 x 2 − x − 2 ⎤⎥ (2 x − 1) = 2 x 2 − 2 x − 2 (2 x − 1) = 4 x3 − 6 x 2 − 2 x + 2 ⎣ ⎦ dx du dx

2.

dy dy du = ⋅ = 6u 2 − 8 7 − 3x 2 = 2 3x6 − 42 x 4 + 147 x 2 − 4 7 − 3 x 2 dx du dx

3.

dy dy dw ⎛ 2 ⎞ 2 2 = ⋅ = ⎜− ⎟ (−1) = 3 = 3 dx dw dx ⎝ w ⎠ w (2 − x)3

4.

5.

(

)(

) (

)(

)

dy dy dz 1 −3 / 4 4 5 x 4 − 4 x3 = ⋅ = z (5 x − 4 x3 ) = dx dz dx 4 4 ⎛⎜ 4 ( x5 − x 4 + 3)3 ⎞⎟ ⎝ ⎠ ⎡ (t + 1) − (t − 1) ⎤ dw dw du 2 ⎤ 1 −1 dw ⎡2⎤ 2⎡ = 0, so = 3(0) 2 ⎢ ⎥ = 0 . = ⋅ = (3u 2 ) ⎢ ⎥ = 3u ⎢ ⎥ . If t = 1, then u = 2 2 dt du dt + 1 1 dt ⎣4⎦ t =1 ⎣⎢ (t + 1) ⎦⎥ ⎣⎢ (t + 1) ⎦⎥

405


6.


dz dz du ⎛ 1 ⎞ = ⋅ = ⎜ 2u + ⎟ (4 s ) . If s = –1, then ds du ds ⎝ 2 u⎠ dz ⎛5⎞ u = 1, so = ⎜ ⎟ (−4) = −10 ds s =−1 ⎝ 2 ⎠

14.

)

3

(

(

= −3 x 2 − 2

(

(

)

(

3

2

)

2

)

(

=

= 4(2 x + 1)( x 2 + x)3

)(

)

− 52

(

d 3 x − 8x2 + x dx

(

1 2

)

(

)

−1 1 (10 x − 1) 5 x 2 − x 2 2

(

)

)

−1 1 5 x 2 − x 2 (10 x − 1) 2

20. y = 3x 2 − 7 = 3 x 2 − 7 y′ =

( ) (3x − 16 x + 1) 99 = 200 ( 3 x 2 − 16 x + 1)( x3 − 8 x 2 + x )

= 200 x − 8 x + x

99

−4

− 52

(

d 2 ( x + x) dx = 4( x 2 + x)3 (2 x + 1)

2

)

) ( 7 − 4 x3 )

19. y = 5 x 2 − x = 5 x 2 − x

12. y ′ = 4( x 2 + x)3

3

)

(

= −6 7 − 4 x3 7 x − x 4

y′ =

⋅

(

d 2 x −2 dx

(2 x) = −6 x x 2 − 2

(

= 30 x 2 (3 + 2 x3 )

99

⋅

⎛ 3⎞ 18. y′ = 4 ⎜ − ⎟ 7 x − x 4 ⎝ 2⎠

3

d (3 + 2 x3 ) dx = 5(3 + 2 x3 ) 4 (6 x 2 )

)

−4

−4

d (2 x3 − 8 x) dx = −12(6 x 2 − 8)(2 x3 − 8 x) −13

11. y ′ = 5(3 + 2 x3 ) 4 ⋅

(

)

)

3

d ⎛ 5⎞ 17. y′ = 2 ⎜ − ⎟ ( x 2 + 5 x − 2)−12 / 7 ⋅ ( x 2 + 5 x − 2) dx ⎝ 7⎠ 10 2 −12 / 7 = − (2 x + 5)( x + 5 x − 2) 7

= 4 x − 4 (2 x) = 8 x x − 4

13. y′ = 2 ⋅100 x3 − 8 x 2 + x

)

16. y ′ = −12(2 x3 − 8 x) −13 ⋅

d (3 x + 2) dx

d 2 x −4 dx

)

15. y′ = −3 x 2 − 2

= 6(3x + 2)5 (3) = 18(3x + 2)5

(

)

(

)

10. y′ = 4 x 2 − 4 ⋅

(

4 1 2x2 + 1 2 2 3 d 1 2 y′ = ⋅ 4 2 x + 1 (2 x 2 + 1) 2 dx

=

= 2(2 x 2 + 1)3 (4 x) = 8 x 2 x 2 + 1

dy dy du 8. = ⋅ = 9u 2 − 2u + 7 (5) . If x = 1, then dx du dx dy u = 3, so = (82)(5) = 410 dx x =1

9. y′ = 6(3 x + 2)5 ⋅

4

(

dy dy dw 7. = ⋅ = (6w − 8)(4 x) . If x = 0, then dx dw dx dy = 0. dx

(

2 x 2 + 1) ( y=

2

(

)

)

1 2

(

)

−1 −1 1 3 x 2 − 7 2 (6 x) = 3 x 3 x 2 − 7 2 2 1

21. y = 4 2 x − 1 = (2 x − 1) 4 y′ =

1 1 −3 −3 (2 x − 1) 4 (2) = (2 x − 1) 4 4 2

(

3

)

22. y = 8 x 2 − 1 = 8 x 2 − 1 y′ =

406

(

)

1 3

(

)

−2 −2 1 16 8 x 2 − 1 3 (16 x) = x 8 x 2 − 1 3 3 3


(

)

( ) − − 12 ⎛2⎞ y′ = 2 ⎜ ⎟ ( x3 + 1) ( 3 x 2 ) = x 2 ( x3 + 1) 5 ⎝5⎠

23. y = 2 5 x3 + 1

2

= 2 x3 + 1

Section 11.5

2 5

3 5

6 2

2x − x +1

(

(

)

= 6 2x2 − x + 1

)

2

y′ = 6(−1) 2 x − x + 1

−2

(

)

= −6(4 x − 1) 2 x 2 − x + 1

26. y =

(

3

= 3 x4 + 2

4

x +2

(

y′ = 3(−1) x 4 + 2

27. y =

1

(

x 2 − 3x

)

(

2

28. y =

1 (2 + x)

4

32. y = 2 x +

)

= (2 x)

4 2

1

1

= (2 x) 2 + (2 x)

− 12

− 12

− (2 x)

− 32

33. y′ = x 2 ⎡5( x − 4) 4 (1) ⎤ + ( x − 4)5 (2 x) ⎣ ⎦

−2

= x( x − 4)4 [5 x + 2( x − 4)] = x( x − 4)4 (7 x − 8)

−2

34. y ′ = x ⎡ 4( x + 4)3 (1) ⎤ + ( x + 4)4 (1) ⎣ ⎦ = ( x + 4)3 (4 x + x + 4) = ( x + 4)3 (5 x + 4)

−3

1

35. y = 4 x 2 5 x + 1 = 4 x 2 (5 x + 1) 2 −1 ⎛1 ⎞ y ′ = 4 x 2 ⎜ (5 x + 1) 2 (5) ⎟ + 5 x + 1(8 x) 2 ⎝ ⎠

= (2 + x) −4

= 10 x 2 (5 x + 1)

y ′ = −4(2 + x)−5 (1) = −4(2 + x)−5

29. y =

– 53

2x 1 −1 −3 ⎛ ⎞ ⎛ 1⎞ y′ = ⎜ ⎟ (2 x) 2 (2) + ⎜ − ⎟ (2 x) 2 (2) ⎝2⎠ ⎝ 2⎠

(2 x − 3)

= −2(2 x − 3) x 2 − 3x

(6 x − 1)

1 7 −2 −2 y′ = (7 x) 3 (7) + 3 7(1) = (7 x) 3 + 3 7 3 3

−1

) ( 4x3 ) = −12 x3 ( x4 + 2)

−3

− 23

1

−2

)

)

5 3

)

31. y = 3 7 x + 3 7 x = (7 x) 3 + 3 7 x

−2

(

)

(

−1

2

2 3

= −2(6 x − 1) 3 x 2 − x

(4 x − 1)

= x − 3x

y′ = −2 x 2 − 3x

(

)

(

= 3 3x 2 − x

( 3x2 − x ) − ⎛ 2⎞ y′ = 3 ⎜ − ⎟ ( 3 x 2 − x ) ⎝ 3⎠

3 5

24. y = 7 3 ( x5 − 3)5 = 7( x5 − 3)5 / 3 5 y ′ = 7 ⋅ ( x5 − 3) 2 / 3 (5 x 4 ) 3 175 4 5 x ( x − 3) 2 / 3 = 3 25. y =

3

30. y =

= 4(9 x 2 + 1)−1/ 2

9x +1 ⎛ 1⎞ y′ = 4 ⎜ − ⎟ (9 x 2 + 1) −3 / 2 (18 x ) ⎝ 2⎠ = −36 x(9 x 2 + 1)−3 / 2

407

− 12

+ 8x 5x + 1



1

36. y = 4 x3 1 − x 2 = 4 x3 (1 − x 2 ) 2 ⎡⎛ 1 ⎞ ⎤ y′ = 4 x3 ⎢⎜ ⎟ (1 − x 2 ) −1/ 2 (−2 x) ⎥ + 1 − x 2 (12 x 2 ) ⎣⎝ 2 ⎠ ⎦ 4 4x =− + 12 x 2 1 − x 2 2 1− x

(

)

(

)

3 2 ⎡ ⎤ 37. y′ = x 2 + 2 x − 1 (5) + (5 x) ⎢3 x 2 + 2 x − 1 (2 x + 2) ⎥ ⎣ ⎦

( ) ( ) 2 = 5 ( x 2 + 2 x − 1) ( 7 x 2 + 8 x − 1) 2

= 5 x 2 + 2 x − 1 ⎡⎢ x 2 + 2 x − 1 + 3 x(2 x + 2) ⎤⎥ ⎣ ⎦

(

)( ) (

)

3 4 ⎡ ⎤ 38. y′ = x 2 ⎢ 4 x3 − 1 3 x 2 ⎥ + x3 − 1 (2 x) ⎣ ⎦

(

)

(

3

)⎤⎥⎦ = 2 x ( 7 x3 − 1)( x3 − 1)

3

= 2 x x3 − 1 ⎡⎢ 6 x3 + x3 − 1 ⎣

39. y′ = (8 x − 1)3 ⎡ 4(2 x + 1)3 (2) ⎤ + (2 x + 1) 4 ⎡3(8 x − 1)2 (8) ⎤ ⎣ ⎦ ⎣ ⎦ = 8(8 x − 1) 2 (2 x + 1)3 [(8 x − 1) + 3(2 x + 1)] = 8(8 x − 1) 2 (2 x + 1)3 (14 x + 2) = 16(8 x − 1)2 (2 x + 1)3 (7 x + 1)

40. y ′ = (3x + 2)5 [2(4 x − 5)(4)] + (4 x − 5)2 [5(3 x + 2) 4 (3)] = (3x + 2) 4 (4 x − 5)[8(3 x + 2) + 15(4 x − 5)] = (3x + 2) 4 (4 x − 5)(84 x − 59) 11 ⎛ x − 3 ⎞ ⎡ ( x + 2)(1) − ( x − 3)(1) ⎤ 41. y′ = 12 ⎜ ⎥ ⎟ ⎢ ⎝ x + 2 ⎠ ⎣⎢ ( x + 2)2 ⎦⎥ 11 ⎡ 5 ⎤ ⎛ x −3 ⎞ = 12 ⎜ ⎥ ⎟ ⎢ ⎝ x + 2 ⎠ ⎢⎣ ( x + 2) 2 ⎥⎦ 60( x − 3)11 = ( x + 2)13

⎛ 2x ⎞ 42. y′ = 4 ⎜ ⎟ ⎝ x+2⎠

3

1⎛ x−2⎞ 43. y′ = ⎜ ⎟ 2⎝ x+3⎠

⎡ ( x + 2)(2) − 2 x(1) ⎤ 128 x3 ⎢ ⎥= ( x + 2)2 ⎢⎣ ⎥⎦ ( x + 2)5

− 12

⎡ ( x + 3)(1) − ( x − 2)(1) ⎤ ⎢ ⎥ ( x + 3)2 ⎣⎢ ⎦⎥

⎛ x−2⎞ = 2 ⎜⎝ x + 3 ⎟⎠ 2( x + 3) 5

− 12

=

5 2( x + 3)

2

x+3 x−2 408


1 ⎛ 8x2 − 3 ⎞ 44. y′ = ⎜ ⎟ 3 ⎜⎝ x 2 + 2 ⎟⎠ 1 ⎛ 8x2 − 3 ⎞ = ⎜ ⎟ 3 ⎜⎝ x 2 + 2 ⎟⎠ =

(

)

(

Section 11.5

)

⎡ 2 ⎤ 2 ⎢ x + 2 (16 x) − 8 x − 3 (2 x) ⎥ ⎢ ⎥ 2 ⎢ ⎥ x2 + 2 ⎣ ⎦

− 23

− 23

(

)

38 x

( x2 + 2)

2

38 x

(

3 8x2 − 3

) ( x2 + 2) 2 3

( x2 + 4)

3

45. y′ =

4 3

(

⎡ (2) − (2 x − 5) ⎢3 x 2 + 4 ⎣

)

2

⎤ (2 x ) ⎥ ⎦

( x2 + 4) 2 x 2 + 4 ) {( x 2 + 4 ) (2) − (2 x − 5)[3(2 x)]} ( = 6 ( x2 + 4)

=

=

2 x 2 + 8 − 12 x 2 + 30 x

(

= =

47. y′ =

=

( x2 + 4)

4

−2 5 x 2 − 15 x − 4

46. y ′ =

=

6

( x2 + 4)

=

−10 x 2 + 30 x + 8

)

( x2 + 4)

4

4

(3x 2 + 7)[4(4 x − 2)3 (4)] − (4 x − 2)4 (6 x) (3x 2 + 7) 2 3

(4 x − 2) [16(3x 2 + 7) − 6 x (4 x − 2)] (3 x 2 + 7)2 (4 x − 2)3 (24 x 2 + 12 x + 112) (3 x 2 + 7)2 (3 x − 1)3 ⎡5(8 x − 1)4 (8) ⎤ − (8 x − 1)5 ⎡3(3x − 1)2 (3) ⎤ ⎣ ⎦ ⎣ ⎦ 6 (3 x − 1)

(3 x − 1)2 (8 x − 1) 4 [(3x − 1)(40) − (8 x − 1)(9)] (3 x − 1)6 (8 x − 1)4 (48 x − 31) (3x − 1) 4

409



48. y = 3 ( x − 2) 2 ( x + 2) = [( x − 2) 2 ( x + 2)]1/ 3 1 y′ = [( x − 2) 2 ( x + 2)]−2 / 3 [(1)( x − 2)2 + 2( x − 2)( x + 2)] 3 1 = [( x − 2) 2 ( x + 2)]−2 / 3 ( x − 2)[ x − 2 + 2( x + 2)] 3 1 = [( x − 2) 2 ( x + 2)]−2 / 3 ( x − 2)(3 x + 2) 3 1 = ( x − 2)−1/ 3 ( x + 2)−2 / 3 (3 x + 2) 3

(

49. y = 6 5 x 2 + 2

1 ⎡ ⎤ x4 + 5 = 6 ⎢ 5x2 + 2 x4 + 5 2 ⎥ ⎣ ⎦

)

(

)(

)

1 −1 ⎡ ⎤ 1 y′ = 6 ⎢ 5 x 2 + 2 ⋅ x 4 + 5 2 4 x3 + x 4 + 5 2 (10 x) ⎥ 2 ⎣ ⎦ 1 1 − ⎡ ⎤ = 6 ⎢ 5 x 2 + 2 x 4 + 5 2 2 x3 + x 4 + 5 2 (10 x) ⎥ ⎣ ⎦ 1 1 − ⎡ 2 ⎤ = 12 x ⎢ 5 x + 2 x 4 + 5 2 x 2 + x 4 + 5 2 (5) ⎥ ⎣ ⎦

(

) (

(

)(

(

) ( ) (

) ( ) (

)(

)

) ( ) (

(

Factoring out x 4 + 5

(

)

)

− 12

)

gives

) ⎡⎢⎣(5x2 + 2)( x2 ) + ( x4 + 5) (5)⎤⎥⎦ − = 12 x ( x 4 + 5 ) (10 x 4 + 2 x 2 + 25 ) y′ = 12 x x 4 + 5

− 12

1 2

50. y′ = 3 − 4 ⎡ x(2)(7 x + 1)(7) + (7 x + 1)2 (1) ⎤ ⎣ ⎦ = 3 − 4 ⎡147 x 2 + 28 x + 1⎤ = −588 x 2 − 112 x − 1 ⎣ ⎦

51. y′ = 8 + = 8+

(t + 4)(1) − (t − 1)(1) (t + 4) 5

(t + 4)

2

2

⎛ 8t − 7 ⎞⎛ 1 ⎞ − 2⎜ ⎟⎜ ⋅ 8 ⎟ ⎝ 4 ⎠⎝ 4 ⎠

− (8t − 7) = 15 − 8t +

5 (t + 4)2

410


52. y = y′ = = = =

53. y′ =

(2 x3 + 6)(7 x − 5) (2 x + 4) 2

=

Section 11.5

14 x 4 − 10 x3 + 42 x − 30 (2 x + 4)2

(2 x + 4)2 (56 x3 − 30 x 2 + 42) − (14 x 4 − 10 x3 + 42 x − 30)[2(2 x + 4)(2)] (2 x + 4) 4 (2 x + 4)[(2 x + 4)(56 x3 − 30 x 2 + 42) − 4(14 x 4 − 10 x3 + 42 x − 30)] (2 x + 4) 4 112 x 4 − 60 x3 + 84 x + 224 x3 − 120 x 2 + 168 − 56 x 4 − 40 x3 − 168 x + 120 (2 x + 4)3 4(14 x 4 + 51x3 − 30 x 2 − 21x + 72) (2 x + 4)3 ( x3 − 5)5 [(2 x + 1)3 (2)( x + 3)(1) + ( x + 3)2 (3)(2 x + 1)2 (2)] − (2 x + 1)3 ( x + 3)2 [5( x3 − 5)4 (3 x 2 )] ( x3 − 5)10

(

)

(

) ( 12 ) ( x + 2)−

2 ⎡ (9 x − 3) ⎢ x + 2(2) 4 x 2 − 1 (8 x) + 4 x 2 − 1 ⎣ 54. y′ = (9 x − 3)2

55.

(

)

dy = 0. dx

dz dz dy dx = ⋅ ⋅ = (4 y − 4)(6)(2) dt dy dx dt

When t = 1, then x = 2 and y = 7. Thus

(

57. y′ = 3 x 2 − 7 x − 8

)

2

dz = (24)(6)(2) = 288 . dt t =1

(2 x − 7)

If x = 8, then slope = y′ = 3(64 − 56 − 8) 2 (16 − 7) = 0 . 1

58. y = ( x + 1) 2 1 −1 ( x + 1) 2 2 1 If x = 8, y′ = . 6 y′ =

411

(

)

2 ⎤ 2 ⎥ − x + 2 4 x − 1 (9) ⎦

3 dy dy du ⎡ ⎡ ⎤ = ⋅ = 3(5u + 6)2 (5) ⎤ ⎢ 4 x 2 + 1 (2 x) ⎥ ⎣ ⎦ dx du dx ⎣ ⎦

When x = 0, then

56.

1 2


(

)

59. y = x 2 − 8 y′ =


2 3

(

2 2 x −8 3

64. y =

)

− 13

4x

(2 x) =

(

3 x2 − 8

)

If x = −1, y ′ = 3(2)2 = 12.

=

( x + 1)

( ) (7 x + 2)

m = 6, then q = 30, so (7) − 7 x + 2(1)

Also,

( x + 1) 2

( 72 )

1 − 7 x+2 2

( x + 1)

If x = 1, then y′ =

7x + 2

( )( ) − 3 = − 1 . The

2

1 3

4 6 3 1 tangent line is y − = − ( x − 1) , or 2 6 1 5 y = − x+ . 6 3

(

)

62. y = −3 3 x 2 + 1

(

)

−4

(

)

3

(

)

2

(

)

dr = −0.2q + 70 . If dq

dq dq 1 =6. = (200 − 2m) . When m = 40, dm dm 20 dr Thus = (6)(6) = 36 . dm m = 40

(6 x)

and y′ = 6 x x 2 + 9

dq dr = 5. Thus = (26)(5) = 130. dm dm m =6

m = 40, then q = 320, so dr = −64 + 70 = 6 . dq m = 40

If x = 0, then y′ = 0 . The tangent line is y + 3 = 0(x – 0), or y = –3. 63. y = x 2 + 9

= −24 + 50 = 26. m =6

r = pq = −0.1q 2 + 70q , so

−3

y′ = −3(−3) 3x 2 + 1

dr dq

dr = −0.8q + 50, . For dq

1 200m − m 2 20 p = –0.1q + 70; m = 40 dr dr dq = ⋅ dm dq dm

66. q = 7 2

( x − 1)4

1 12 4 and y ′ = − = − , so 4 27 27 3

r = pq = −0.4q 2 + 50q,

The tangent line is y − 8 = 12(x + 1) or y = 12x + 20. ( x + 1)

6x 2

65. q = 5m, p = –0.4q + 50; m = 6 dr dr dq = ⋅ dm dq dm

60. y ′ = 3( x + 3)2 (1) = 3( x + 3) 2

61. y′ =

and y ′ = −

⎛ y′ ⎞ 4 ⎜ ⎟ (100) = − ⋅ 27(100) = −400% y 27 ⎝ ⎠

12 = 4 . Thus the tangent line 3(1) is y – 1 = 4(x – 3), or y = 4x – 11.

1 2

( x − 1)

3

When x = 2, y =

1 3

If x = 3, then y′ =

− 12

1 2

. When

67. q =

10m 2

m2 + 9 525 ;m=4 p= q+3

x = 4, then y = (25)3 and y′ = 6(4)(25) 2 , so y′ 6(4)(25)2 24 (100) = (100) = (100) = 96% 3 y 25 (25)

dr dr dq = ⋅ dm dq dm r = pq =

525q , so q+3

dr (q + 3)(1) − q(1) 1575 . = 525 ⋅ = 2 dq (q + 3) (q + 3)2

412


dr dq

If m = 4, then q = 32, so

(

2

m +9 dq = dm m2 + 9 ) ( = =

)

1 2

= m=4

(20m) − 10m2 ⋅ 12 2

Section 11.5

1575 9 = . 1225 7

(m

2

+9

)

− 12

b.

(

=−

)

⎡ 20m m 2 + 9 − 10m3 ⎤ ⎢⎣ ⎥⎦ 2 m +9 3 10m + 180m

(

m2 + 9

)

=−

c.

3 2

70. p =

q q 2 + 20 ⎛⎜ 100 − q 2 + 20 ⎞⎟ ⎝ ⎠ q 100 q 2 + 20 − q 2 − 20

r = pq = 100q − q q 2 + 20

dr 45, 000 4500q , so . = dq (q + 10) 2 q + 10 dr dq

If m = 9, then q = 90, so

( m2 + 19)

3 2

= m =9

71.

9 . 2

. When m = 9, then

dq 19 . = dm 10

72.

dr 9 19 = ⋅ = 8.55 . dm m =9 2 10

(

dp 1 = 0 − q 2 + 20 dq 2

2

q + 20

− 12

⎤ (2q) + q 2 + 20(1) ⎥ ⎦

− q 2 + 20

k ; q = f(m) q

r = pq = k, so

1900

q2

)

dr dr dq = ⋅ dm dq dm

dr dr dq = ⋅ dm dq dm

69. a.

100 − q 2 + 20

= 100 −

m 2 + 19 4500 ;m=9 p= q + 10

Thus

q 2 + 20

(

100m

dq = dm

=

⎡ 1 = 100 − ⎢ q ⋅ q 2 + 20 ⎣ 2

dr 9 272 = ⋅ ≈ 13.99 . dm m = 4 7 25

r = pq =

−q

dr dq

When m = 4, then dq 10(64) + 180(4) 1360 272 . Thus = = = 3 125 25 dm (25) 2

68. q =

p

(2m)

m +9

− 12

dp dq

)

− 12

(2q ) =

dr dq dr = 0 . Thus = 0⋅ =0. dm dm dq

dc dc dq = ⋅ = (12 + 0.4q )(−1.5) dp dq dp When p = 85, then q = 772.5, so dc = −481.5. dp p =85 ⎛ 250 ⎞ f (t ) = 1 − ⎜ ⎟ ⎝ 250 + t ⎠

3

2 250 ⎤ ⎛ 250 ⎞ ⎡ f ′(t ) = −3 ⎜ ⎥ ⎟ ⎢− ⎝ 250 + t ⎠ ⎣⎢ (250 + t )2 ⎦⎥

−q

2

⎛ 250 ⎞ ⎡ 250 ⎤ f ′(100) = −3 ⎜ ⎟ ⎢− ⎥ ⎝ 350 ⎠ ⎣ 3502 ⎦ ⎛ 25 ⎞ ⎛ 1 ⎞ = −3 ⎜ ⎟ ⎜ − ⎟ ⎝ 49 ⎠ ⎝ 490 ⎠ 15 . = 4802 Thus when t increases from 100 to 101, the proportion discharged increases by 15 . approximately 4802

2

q + 20

413


73.

dc = dq

( q 2 + 3)

1 2


⎡ (10q ) − 5q 2 ⎢ 12 q 2 + 3 ⎣ q2 + 3

( ) (

)

− 12

⎤ (2q ) ⎥ ⎦

(

Multiplying numerator and denominator by q 2 + 3

(

) (

(

)

1 2

gives

)

q 2 + 3 (10q) − 5q 2 (q ) 5q3 + 30q 5q q 2 + 6 dc . = = = 3 3 3 dq 2 2 2 2 2 2 q +3 q +3 q +3

74. a.

(

)

)

(

)

dS dS = 680 E − 4360 . If E = 16, = 6520 . dE dE

b. Solving 680 E − 4360 = 5000 gives 680 E = 9360, E ≈ 13.8. 75.

(

)

( )

dV dV dr = ⋅ = 4πr 2 ⎡10−8 (2t ) + 10−7 ⎤ . When t = 10, then r = 10−8 102 + 10−7 (10) = 10−6 + 10−6 = 2(10)−6 . ⎣ ⎦ dt dr dt Thus 2 dV = 4π ⎡ 2(10) −6 ⎤ ⎡10−8 (2)(10) + 10−7 ⎤ = 4π ⎡ 4(10)−12 ⎤ ⎡⎢3 10−7 ⎤⎥ = 48π(10)−19 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎣ ⎦ dt t =10

(

76. a.

b.

77. a.

)

dp 1 −1 −1 = (2 ρVI ) 2 (2 ρV ) = ρV (2 ρVI ) 2 dI 2 dp dI

p

=

ρV (2 ρVI ) (2 ρVI )

− 12

1 2

=

1 2I

d ( I x ) = −0.001416 x3 + 0.01356 x3 + 1.696 x − 34.9 dx d ( I x ) = −256.238. If x = 65, dx

b. If x = 65,

d (I ) dx x

Ix

≈

−256.238 ≈ −0.01578 16, 236.484

If x = 65, the percentage rate of change is

d (I ) dx x

Ix

⋅=

78. (P + a)(v + b) = k k v+b = P+a k v= −b P+a v = k ( P + a ) −1 − b dv k = k (−1)( P + a )−2 = − dP ( P + a)2 414

−25, 623.8 = −1.578%. 16, 236.484


79. By the chain rule,

Section 11.5

dq −100 dc dc dq 100 . We are given that q = = 100 p −1 , so . Thus = ⋅ = −100 p −2 = dp dp dq dp p p2

100 1 dc dc dc ⎡ −100 ⎤ = and we are given that = 0.01 . Therefore = ⎢ 2 ⎥ . When q = 200, then p = dp dq ⎢⎣ p ⎥⎦ 200 2 dq ⎡ ⎤ dc −100 ⎥ ⎢ = 0.01 ⎢ = −4 . 2 ⎥ dp ⎢⎣ 12 ⎥⎦

( )

80. a.

b.

When m = 12, then q = 3000, so r = 1500. r 1500 1 Thus p = = = = $0.50 . q 3000 2

( )

1000 + 3q (50) − 50q 12 (1000 + 3q) dr = 1000 + 3q dq dr dq

c.

= q =3000

− 12

(3)

2750 11 = 10, 000 40

dr dr dq dr . From part (b) we know . Now, = ⋅ dm dq dm dq 3 1 dq dq ⎛3⎞ = 610 . = (2m) ⎜ ⎟ (2m + 1) 2 (2) + (2m + 1) 2 (2) , so dm dm m =12 ⎝2⎠

Thus

81.

dr 11 671 = ⋅ 610 = . dm m =12 40 4

dy dy dx = ⋅ = f ′( x) g ′(t ) . We are given that g(2) = 3, so x = 3 when t = 2. Thus dt dx dt dy dy dx = ⋅ = f ′(3) g ′(2) = 10(4) = 40 . dt t = 2 dx x = g (2) dt t = 2

82. a.

b.

⎛ 324 5 19 ⎞ 19 19 + + ⎟ = 0+0+ = lim c = lim ⎜ ⎜ ⎟ 2 q 18 18 18 q →∞ q →∞ ⎝ q + 35 ⎠ c = cq =

dc = dq dc dq

c.

324q q 2 + 35

+5+

19 q 18

q 2 + 35(324) − 324q

( 12 ) ( q2 + 35)

q 2 + 35

− 12

(2q ) 19 + 18

=3 q =17

From part (b) the increase in cost of the additional unit is approximately $300. Since the corresponding revenue increases by $275, the move should not be made. 415



83. 86,111.37 84. 5.25 Chapter 11 Review Problems 1.

f ( x) = 2 − x 2

(

⎡ 2 − ( x + h) 2 ⎤ − 2 − x 2 f ( x + h) − f ( x ) ⎦ f ′( x) = lim = lim ⎣ h h h →0 h →0 ⎡ 2 − x 2 − 2hx − h 2 ⎤ − 2 − x 2 −2hx − h 2 ⎦ = lim ⎣ = lim h h h→0 h →0 −h(2 x + h) = lim = lim − (2 x + h) = −2 x h h→0 h →0

(

2.

)

f ( x) = 2 x 2 − 3 x + 1 f ′( x) = lim

h →0

f ( x + h) − f ( x ) h

(

)

⎡ 2( x + h) 2 − 3( x + h) + 1⎤ − 2 x 2 − 3x + 1 ⎦ = lim ⎣ h h→0 ⎡ 2 x 2 + 4hx + 2h 2 − 3 x − 3h + 1⎤ − 2 x 2 − 3x + 1 ⎦ = lim ⎣ h h→0

(

)

4hx + 2h 2 − 3h h(4 x + 2h − 3) = lim h h h→0 h→0 = lim (4 x + 2h − 3) = 4 x − 3

= lim

h→0

3.

f ( x) = 3x f ′( x) = lim

h →0

3( x + h) − 3x 3( x + h) + 3 x ⋅ h 3( x + h) + 3 x

= lim

h→0

= lim

h→0 h

= lim

h→0

=

3( x + h) − 3x f ( x + h) − f ( x ) = lim h h h →0

(

3( x + h) − 3 x 3( x + h) + 3x

)

= lim

h→0 h

(

3h 3( x + h) + 3x

)

3 3( x + h) + 3x 3

3x + 3x

=

3 2 3x

=

3 2 x

416

)


4.

f ( x) =

Chapter 11 Review

2 1 + 4x 2

2

− f ( x + h) − f ( x ) 1+ 4( x + h ) 1+ 4 x = lim f ′( x) = lim h h h →0 h →0 2(1 + 4 x) − 2[1 + 4( x + h)] = lim h→0 h[1 + 4( x + h)](1 + 4 x) −8h h→0 h[1 + 4( x + h)](1 + 4 x)

= lim

−8 −8 = x h x x [1 + 4( + )](1 + 4 ) [1 + 4( )](1 + 4 x) h→0

= lim =−

8 (1 + 4 x)2

5. y is a constant function, so y′ = 0 . 6. y′ = e(1) x1−1 = ex0 = e

( ) ( )

7. y′ = 7 4 x3 − 6 3 x 2 + 5(2 x) + 0

(

= 28 x3 − 18 x 2 + 10 x = 2 x 14 x 2 − 9 x + 5

)

8. y′ = 4(2 x + 0) − 7(1) = 8 x − 7 9.

(

)

f ( s ) = s 2 s 2 + 2 = s 4 + 2s 2

(

)

f ′( s ) = 4s 3 + 2(2 s ) = 4 s3 + 4 s = 4 s s 2 + 1 1

10. y = ( x + 3) 2 y′ =

1 1 −1 −1 ( x + 3) 2 (1) = ( x + 3) 2 2 2

(

)

1 2 x +1 5 1 2x y′ = (2 x) = 5 5

11. y =

12. y = −

2 2x

2

= − x −2 , so y ′ = −1(−2) x −3 = 2 x −3 .

13. y ′ = ( x3 + 7 x 2 )(3x 2 − 2 x) + ( x3 − x 2 + 5)(3 x 2 + 14 x) = 3 x5 + 19 x 4 − 14 x3 + 3x5 + 11x 4 − 14 x3 + 15 x 2 + 70 x = 6 x5 + 30 x 4 − 28 x3 + 15 x 2 + 70 x

(

)

14. y′ = x 2 + 1

100

(

)

(1) + ( x − 6)(100) x 2 + 1

99

(

) ( 201x2 − 1200 x + 1)

(2 x) = ( x 2 + 1)99 [ x 2 + 1 + 200 x( x − 6)] = x 2 + 1 417

99


(


)

99

(4 x + 4) = 400( x + 1)[(2 x)( x + 2)]99

15.

f ′( x) = 100 2 x 2 + 4 x

16.

f ( w) = w w + w2 = w 2 + w2 f ′( w) =

3

3 12 w + 2w 2

17. y = 3(2 x + 1)−1 y′ = 3(−1)(2 x + 1) −2 (2) = −

6 (2 x + 1)2

5x2 − 8x 5 = x−4 2x 2 5 y′ = 2

18. y =

(

)

(

)

3 4 ⎡ ⎤ 19. y′ = (8 + 2 x) ⎢ (4) x 2 + 1 (2 x) ⎥ + x 2 + 1 (2) ⎣ ⎦

( ) ( ) 3 = 2 ( x 2 + 1) ( 32 x + 8 x 2 + x 2 + 1) 3 = 2 ( x 2 + 1) ( 9 x 2 + 32 x + 1) 3

= 2 x 2 + 1 ⎡⎢ 4 x(8 + 2 x) + x 2 + 1 ⎤⎥ ⎣ ⎦

6 −2 −2 ⎛3⎞ 20. g ′( z ) = ⎜ ⎟ (2 z ) 5 (2) + 0 = (2 z ) 5 5 ⎝5⎠

21.

z 2 + 4 ) (2 z ) − ( z 2 − 1) (2 z ) ( 10 z f ′( z ) = = 2 2 ( z2 + 4) ( z2 + 4)

22. y′ =

( x + 2)2 (1) − ( x − 5)(2)( x + 2) ( x + 2)

4

=

12 − x ( x + 2)3

1

23. y = (4 x − 1) 3 1 4 −2 −2 y′ = (4 x − 1) 3 (4) = (4 x − 1) 3 3 3

24. f is a constant function, so f ′( x) = 0. 25. y = (1 − x 2 )

− 12

−3 −3 ⎛ 1⎞ y′ = ⎜ − ⎟ (1 − x 2 ) 2 (−2 x) = x(1 − x 2 ) 2 ⎝ 2⎠

418


26. y =

Chapter 11 Review

x2 + x 2 x2 + 3

2 x 2 + 3) (2 x + 1) − ( x 2 + x ) (4 x) −2 x 2 + 6 x + 3 ( y′ = = 2 2 2 2 x + 3 ( ) ( 2 x 2 + 3) 2 27. h′( x) = ( x − 6) 4 ⎢⎡3 ( x + 5 ) ⎥⎤ + ( x + 5)3 ⎡ 4( x − 6)3 ⎤ ⎣ ⎦ ⎣ ⎦

= ( x − 6)3 ( x + 5)2 [3( x − 6) + 4( x + 5)] = ( x − 6)3 ( x + 5)2 (7 x + 2)

28. y′ =

29. y′ =

30.

x(5)( x + 3) 4 − ( x + 3)5 (1)

=

x2 ( x + 6)(5) − (5 x − 4)(1) ( x + 6)

2

=

( x + 3)4 (4 x − 3) x2 34

( x + 6)2

f ( x) = 5 x3 3 + 2 x 4 = 5 x3 (3 + 2 x 4 )1/ 2 ⎡1 ⎤ f ′( x) = (3 + 2 x 4 )1/ 2 (15 x 2 ) + 5 x3 ⎢ (3 + 2 x 4 ) −1/ 2 (8 x3 ) ⎥ 2 ⎣ ⎦ = 15 x 2 (3 + 2 x 4 )1/ 2 + 20 x6 (3 + 2 x 4 )−1/ 2

3 − 11 3 ⎛ − 11 ⎞ − 11 − 11 ⎛ 3 ⎞ − 11 ⎛ 3 ⎞ 31. y′ = 2 ⎜ − ⎟ x 8 + ⎜ − ⎟ (2 x) 8 (2) = − x 8 − ⎜ 2 8 ⎟ x 8 4 4⎝ ⎠ ⎝ 8⎠ ⎝ 8⎠ 11 11 11 11 3 − ⎛ 3⎛ − ⎞ − ⎞ − = − x 8 ⎜1 + 2 8 ⎟ = − ⎜1 + 2 8 ⎟ x 8 4 4⎝ ⎝ ⎠ ⎠ 1⎛ x⎞ 32. y′ = ⎜ ⎟ 2⎝2⎠

− 12

⎛1⎞ 1⎛2⎞ ⎜ ⎟+ ⎜ ⎟ ⎝2⎠ 2⎝ x⎠

− 12

(

1

)

1 ⎛ 2 ⎞2 1 ⎛ 2 ⎞ −2 x −2 = ⎜ ⎟ − ⎜ ⎟ 4⎝ x⎠ x2 ⎝ x ⎠

− 12

⎛2⎞ =⎜ ⎟ ⎝ x⎠

1

1 ⎤ x−2 x ⎛ x ⎞2 ⎡ 1 =⎜ ⎟ ⎢ − ⎥= ⎝ 2 ⎠ ⎣ 2 x x2 ⎦ 2 x2 2

33.

x2 + 5) ( y′ =

1 2

(

(2 x) − x 2 + 6

) ( 12 ) ( x2 + 5)

− 12

(2 x)

x2 + 5

(

Multiplying the numerator and denominator by x 2 + 5 x 2 + 5 ) (2 x) − x ( x 2 + 6 ) x ( x2 + 4 ) ( x3 + 4 x = = y′ = 2 2 ( x + 5) ( x + 5) ( x 2 + 5) 3 2

3 2

)

3 2

419

1 2

gives

− 12

⎡1 ⎛ 2 ⎞ 1 ⎤ ⎢4 ⎜ x ⎟ − 2 ⎥ ⎣ ⎝ ⎠ x ⎦


(

34. y = 7 − 3 x 2

)

2 3

(

2 y′ = 7 − 3 x 2 3

35. y′ =


1

41. y = x 3

)

− 13

(

(

( − 6 x ) = −4 x 7 − 3 x

3 3 x + 6 x2 + 9 5

(

)

)

y′ =

1 2 −3

1 . An 12 1 equation of the tangent line is y − 2 = ( x − 8) , 12 1 4 or y = x + . 12 3

When x = 8, then y = 2 and y′ =

) (3x2 + 12 x ) − 52

2

− 3 3 x + 6 x 2 + 9 5 (3x)( x + 4) 5 −2 9 = x( x + 4) x3 + 6 x 2 + 9 5 5

=

(

)

x2 x − 12

42. y =

36. z ′ = 0.4[ x 2 (−3)( x + 1) −4 (1) + ( x + 1) −3 (2 x)] + 0 = 0.4( x + 1)−4 [−3x 2 + ( x + 1)(2 x)]

y′ =

37. g ( z ) = − z ( z − 1)2 = − z 3 + 2 z 2 − z g ′( z ) = −3z 2 + 4 z − 1

(

=

(

) 5 ( z5 + 2 z − 5)

43.

)

−4 3 5 z + 2z − 5 4 3 g ′( z ) = − (−4) z 5 + 2 z − 5 4

(

( x − 12)(2 x) − x 2 (1)

=

x 2 − 24 x

( x − 12) 2 ( x − 12)2 When x = 13, then y = 169 and y ′ = −143. An equation of the tangent line is y − 169 = −143(x − 13) or y = −143x + 2028.

= 0.4( x + 1)−4 (− x 2 + 2 x)

38. g ( z ) = −

1 − 23 x 3

) (5z 4 + 2) −5

f ( x) = 4 x 2 + 2 x + 8 f ′( x) = 8 x + 2 f(1) = 14 and f ′(1) = 10 . The relative rate of f ′(1) 10 5 = = ≈ 0.714 , so the f (1) 14 7 percentage rate of change is 71.4%.

change is

3 5z 4 + 2

44.

39. y = x 2 − 6 x + 4 y′ = 2 x − 6 When x = 1, then y = –1 and y′ = −4 . An equation of the tangent line is y – (–1) = –4(x – 1), or y = –4x + 3.

f ( x) = f ′( x) =

x x+4 ( x + 4)(1) − x(1) ( x + 4)

2

=

4 ( x + 4)2

1 4 and f ′(1) = . The relative rate of 5 25 f ′(1) 4 change is = = 0.8 , so the percentage rate f (1) 5 of change is 80%. f (1) =

40. y = −2 x3 + 6 x + 1 y′ = −6 x 2 + 6 When x = 2, then y = –3 and y′ = −18 . An equation of the tangent line is y – (–3) = –18(x – 2), or y = –18x + 33.

45. r = q (20 − 0.1q) = 20q − 0.1q 2 dr = 20 − 0.2q dq

46.

dc = 0.0003q 2 − 0.04q + 3 dq dc dq

420

=2 q =100


47.

Chapter 11 Review

1 dC ⎛ 1 ⎞ −1 = 0.6 − 0.25 ⎜ ⎟ I 2 = 0.6 − 2 dI I 8 ⎝ ⎠ dC ≈ 0.569 dI I =16 Thus the marginal propensity to consume is 0.569, so the marginal propensity to save is 1 – 0.569 = 0.431.

54. y = 12 −

dy 36 = −12(−1)(1 + 3 x)−2 (3) = dx (1 + 3x)2 36 1 Setting = gives (1 + 3x) 2 = 108, 2 3 (1 + 3 x) 1 + 3 x = ±6 3, x =

dp (q + 5)(1) − (q + 12)(1) 7 = =− 48. 2 dq (q + 5) (q + 5)2

dr = 500 − 0.2q. dq

dt dT d dT

55. a.

3 50. Since c = 0.03q + 1.2 + , then q c = qc = 0.03q 2 + 1.2q + 3 . Thus dc dc = 0.06q + 1.2 , so dq dq

51.

= 7.2 . q =100

when T = 38 is 175 ⎤ 4 4 ⎡4 = = . ⎢3T − 4 ⎥ ⎣ ⎦ T =38 3 T =38 3

when T = 35 is 11 ⎤ 1 1 ⎡1 = = . ⎢ 24 T + 4 ⎥ ⎣ ⎦ T =35 24 T =35 24

(

56. s = 9 2t 2 + 3

= 0.7396

v=

q = 70

52. q = 50m − m2 p = –0.01q + 9; m = 10 dr dr dq = ⋅ dm dq dm

)

−1

(

ds = −9 2t 2 + 3 dt

If t = 1, then v = −

57. V ′ =

)

−2

(4t ) =

−36t

( 2t 2 + 3)

2

36 m/s. 25

ft 3 1 2 πd . If d = 4 ft, then V ′ = 8π . ft 2

r = pq = −0.01q 2 + 9q , so

dr = −0.02q + 9 . dq

58. v = 128 – 32t. Set 128 – 32t = 64 to get t = 2.

If m = 10, then q = 400, so

dr dq

59. c = cq = 2q 2 +

= −8 + 9 = 1 . m =10

dq dq = 50 − 2m . When m = 10, = 30 . dm dm dr Thus = (1)(30) = 30 . dm m =10

53.

dt dT d dT

b.

dc = 0.125 + 0.00878q dq dc dq

−1 ± 6 3 , x ≈ 3.13 or 3

x ≈ −3.80. Because we must have x ≥ 0, then x ≈ 3.13.

49. Since p = −0.1q + 500, then r = pq = −0.1q 2 + 500q. Thus

12 1 + 3x

10, 000 = 2q 2 + 10, 000q −1 q

dc 10, 000 = 4q − 10, 000q −2 = 4q − dq q2

dy = 42 x 2 − 34 x − 16 dx dy = 84 eggs/mm dx x = 2

421


60. y =

( x3 + 2) x + 1 x4 + 2 x

=


( x3 + 2) x + 1 x( x3 + 2)

=

65. –0.32

x +1 x


−1 ⎛ ⎞ x ⎜ 12 ( x + 1) 2 (1) ⎟ − x + 1(1) dy ⎠ = ⎝ dx x2 dy 3 =− 2 and y = 2 when x = 1. An dx x =1 4 equation of the tangent line is 3 3 7 y− 2 =− 2( x − 1) or y = − 2x + 2. 4 4 4

61. a.

1. In Problems 63 and 64 of Sec. 11.4, the slope is ≈ 0.7 . In Fig. 11.15 the slope is above 0.9. More is spent; less is saved. 2. In the lowest quintile, the average family spends more than it earns, thus accumulating debt. 3. The slope of the family consumption curve is 112, 040 , which for 1.9667 × 1010 + 224, 080x x = 25,000 equals about 0.705. You would expect the family to spend $705 and save $295.

q = 10 m 2 + 4900 − 700 p = 19,300 − 8q ; m = 240 dr dr dq . = ⋅ dm dq dm

4. For x = 90,000, the slope of the consumption curve is 0.561. You would expect the family to spend $561 and save $439.

r = pq = q 19,300 − 8q , so dr −1 ⎛1⎞ = q ⎜ ⎟ (19,300 − 8q ) 2 (−8) + 19,300 − 8q (1). dq ⎝2⎠ If m = 240, then q = 1800, so dr 230 =− ≈ −32.86. dq m = 240 7

(

dq 1 = 10 ⋅ m2 + 4900 dm 2 dq = 9.6. Thus dm m = 240

)

− 12

(2m) .

dr ≈ (−32.86)(9.6) = −315.456 dm m = 240

b.

dr dm

r

= m = 240

=

−315.456 r q =1800 −315.456

1800 4900 = −0.0025

c.

dr < 0, there would be no dm additional revenue generated to offset the cost of $400.

No. Since

62. 21.094 63. 0.305 64. $5.05

422


Chapter 12 Principles in Practice 12.1 1.

(

)

dq d ⎡ = 25 + 2 ln 3 p 2 + 4 ⎤ ⎦⎥ dp dp ⎣⎢ = 0+2

(

)

d ⎡ ln 3 p 2 + 4 ⎤ ⎦⎥ dp ⎣⎢

⎛ 1 ⎞ d 2 3 p2 + 4 = (6 p) = 2⎜ ⎟ 2 ⎜ 3 p 2 + 4 ⎟ dp 3p + 4 ⎝ ⎠ 12 p = 3 p2 + 4

(

)

2. With I 0 = 1 , R(I) = log I. dR d d ⎡ ln I ⎤ = [log I ] = dI dI dI ⎢⎣ ln10 ⎥⎦ 1 1 1 = ⋅ = ln10 I I ln10

Problems 12.1 1.

dy d 1 4 = 4 ⋅ (ln x) = 4 ⋅ = dx dx x x

2.

dy 5 ⎛ 1 ⎞ 5 = = dx 9 ⎜⎝ x ⎟⎠ 9 x

3.

dy 1 3 = (3) = dx 3x − 7 3x − 7

4.

dy 1 5 = (5) = dx 5 x − 6 5x − 6

5. y = ln x 2 = 2 ln x dy 1 2 = 2⋅ = dx x x

6.

dy 1 6x + 2 (6 x + 2) = = 2 dx 3x 2 + 2 x + 1 3x + 2 x + 1

7.

dy 1 2x ( −2 x ) = − = dx 1 − x 2 1 − x2

423

Chapter 12: Additional Differentiation Topics

8.

9.

dy 1 −2 x + 6 (−2 x + 6) = = dx − x 2 + 6 x − x2 + 6 x −2( x − 3) 2( x − 3) = = − x( x − 6) x( x − 6)

f ′( X ) = = = =

10.


f ′(r ) = =

=

1 6

3

(24 X 5 + 6 X 2 )

4X + 2X 24 X 5 + 6 X 2

4X 6 + 2X 3 6 X 2 (4 X 3 + 1) 2 X 3 (2 X 3 + 1) 3(4 X 3 + 1) X (2 X 3 + 1)

( 8r + 2r + 1

1 4

2r − 3r

2

3

− 6r + 2

)

8r 3 − 6r + 2 2r 4 − 3r 2 + 2r + 1

(

)

2 4r 3 − 3r + 1

2r 4 − 3r 2 + 2r + 1

11.

⎛1⎞ f ′(t ) = t ⎜ ⎟ + (ln t )(1) = 1 + ln t ⎝t ⎠

12.

dy ⎛1⎞ = x 2 ⎜ ⎟ + (ln x)(2 x) = x + 2 x ln x dx ⎝x⎠ = x(1 + 2 ln x)

13.

dy ⎡ 1 ⎤ = x3 ⎢ (2) ⎥ + ln(2 x + 5) ⋅ 3 x 2 dx ⎣ 2x + 5 ⎦ =

14.

2 x3 + 3x 2 ln(2 x + 5) 2x + 5

⎡ 1 ⎤ dy = (ax + b)3 ⎢ (a ) ⎥ + [ln(ax + b)]3(ax + b) 2 (a ) dx ⎣ (ax + b) ⎦ = a(ax + b)2 + 3a (ax + b)2 ln(ax + b) = a(ax + b)2 [1 + 3ln(ax + b)]

424


15. y = log3 (8 x − 1) =

Section 12.1

ln(8 x − 1) ln 3

21.

dy 1 d = ⋅ [ln(8 x − 1)] dx ln 3 dx 1 1 8 = ⋅ (8) = ln 3 8 x − 1 (8 x − 1)(ln 3)

16.

(

)

=

(

f ( w) = log w2 + w = log10 w2 + w =

(

ln w2 + w

)

)

(

23. y = ln x 2 + 4 x + 5

(

(

)

2

2

(

ln x 2 + 4

)

(

)

ln 2

25. y = 9 ln 1 + x 2 =

ln x 1 = ( x 2 ln x) ln 2 ln 2 ⎤ dy 1 ⎡ 2⎛1⎞ = + ln x(2 x) ⎥ ⎢x dx ln 2 ⎣ ⎜⎝ x ⎟⎠ ⎦ x = (1 + 2 ln x) ln 2

20.

26.

( 1z ) − (ln z)(1) = 1 − ln z z

2

z

( )

2 x ln x − x 2

ln x

=

)

(

27.

x[2 ln x − 1] ln 2 x

425

)

⎛ ⎞ t5 f (t ) = ln ⎜ ⎟ = 5ln t − ln(1 + 3t 2 + t 4 ) ⎜ 1 + 3t 2 + t 4 ⎟ ⎝ ⎠ 1 1 ⎛ ⎞ f ′(t ) = 5 ⎜ ⎟ − (6t + 4t 3 ) ⎝ t ⎠ 1 + 3t 2 + t 4 5(1 + 3t 2 + t 4 ) − t (6t + 4t 3 ) = t (1 + 3t 2 + t 4 )

=

2

2 1 dy (ln x)(2 x) − x x = dx (ln x)2

=

(

= 3ln x 2 + 4 x + 5

9 ln 1 + x 2 2 dy 9 1 9x = ⋅ (2 x) = dx 2 1 + x 2 1 + x2

⎤ ⎥ ⎥ ⎦⎥

18. y = x 2 log 2 x = x 2 ⋅

z

3

1 24. y = 6 ln 3 x = 6 ⋅ ln x = 2 ln x 3 dy 1 2 = 2⋅ = dx x x

⎤ 1 ⎡ 1 dy (2 x) ⎥ = 2x + dx ln 2 ⎢⎣ x 2 + 4 ⎦

⎡ 1 = 2 x ⎢1 + ⎢ (ln 2) x 2 + 4 ⎣⎢

)

dy 1 (2 x + 4) = 3⋅ 2 dx x + 4x + 5 3(2 x + 4) 6( x + 2) = = 2 2 x + 4x + 5 x + 4x + 5

)

17. y = x + log 2 x + 4 = x +

f ′( z ) =

x(ln x )3

dy 1 100 = 100 ⋅ = dx x x

1 1 ⋅ f ′( w) = (2 w + 1) 2 ln10 w + w 2w + 1 = (ln10) w2 + w

19.

2 x 2 ln x − 2( x 2 + 3)

22. y = ln x100 = 100 ln x

ln10

2

2 2 1 dy (ln x) (2 x) − ( x + 3)2(ln x) x = dx (ln x) 4

t 4 + 9t 2 + 5 t (1 + 3t 2 + t 4 )

⎛ 1+ l ⎞ f (l ) = ln ⎜ ⎟ = ln(1 + l ) − ln(1 − l ) ⎝ 1− l ⎠ 1 1 f ′(l ) = (−1) − 1+ l 1− l (1 − l ) + (1 + l ) 2 = = (1 + l )(1 − l ) 1− l2



(

⎛ 2x + 3 ⎞ 28. y = ln ⎜ ⎟ = ln(2 x + 3) − ln(3 x − 4) ⎝ 3x − 4 ⎠ dy 2 3 = − dx 2 x + 3 3x − 4 2(3x − 4) − 3(2 x + 3) 17 = =− (2 x + 3)(3 x − 4) (2 x + 3)(3 x − 4)

29. y = ln 4

1 + x2

=

(

)

(

2 2 ⎡ 1 ⎢ 2x 1 − x + 2x 1 + x = 4⎢ 1 + x2 1 − x2 ⎢⎣

(

)(

)

) ⎤⎥ = ⎥ ⎥⎦

= 13ln x + 13ln(5 x + 2)1/ 3 13 = 26 ln x + ln(5 x + 2) 3 dy 1 1 26 65 ⎛ ⎞ 13 = 26 ⎜ ⎟ + ⋅ (5) = + dx x 3(5 x + 2) ⎝ x ⎠ 3 5x + 2

)

1

= 6 ln x − 6 ln(2 x + 1) 2

2x +1 = 6 ln x − 3ln(2 x + 1)

dy 6 1 6 6 = − 3⋅ (2) = − dx x 2x + 1 x 2x +1 x 1 − x4

35.

1 = [ln( x3 − 1) − ln( x3 + 1)] 30. y = ln 3 3 x +1 3

(

36.

(

) + 2 x ln(2 x + 1)

2 x2 + 1 2x + 1

dy ⎡1 ⎤ = (ax + b) ⎢ (a ) ⎥ + ln(ax ) ⋅ (a ) dx ⎣ ax ⎦ =

2 x2

)

dy ⎡ 1 ⎤ (2) ⎥ + ln(2 x + 1) ⋅ (2 x) = x2 + 1 ⎢ dx ⎣ 2x + 1 ⎦ =

dy 1 ⎡ 3x 2 3x 2 ⎤ = ⎢ − ⎥ dx 3 ⎢⎣ x3 − 1 x3 + 1 ⎦⎥ 1 ⎡ 3x 2 ( x3 + 1) − 3 x 2 ( x3 − 1) ⎤ = ⎢ ⎥ 3 ⎢⎣ ( x3 − 1)( x3 + 1) ⎥⎦

ax + b + a ln(ax) x

37. y = ln x3 + ln 3 x = 3ln x + (ln x)3

x6 − 1

(

dy 1 1 3 3(ln x) 2 = 3 ⋅ + 3(ln x )2 ⋅ = + dx x x x x

) ( x + x − 1)⎤⎥⎦ = 2 ln ( x + 2 ) + ln ( x + x − 1)

⎡ 31. y = ln ⎢ x 2 + 2 ⎣

2

3

2

=

3

(

)

dy 1 1 = 2⋅ (2 x) + 3x 2 + 1 2 3 dx x +2 x + x −1 =

x

34. y = 6 ln

x3 − 1

=

)

2

1⎡ ln 1 + x 2 − ln 1 − x 2 ⎤ ⎦⎥ 4 ⎣⎢

1 − x2 dy 1 ⎡ 2 x −2 x ⎤ = ⎢ − ⎥ 2 dx 4 ⎣ 1 + x 1 − x2 ⎦

(

) (

33. y = 13ln x 2 3 5 x + 2

4x x2 + 2

+

38.

3x2 + 1

(

3 1 + ln 2 x

)

x

dy = (ln 2) x (ln 2) −1 dx

39. y = ln 4 (ax) = [ln(ax)]4

x3 + x − 1

3 dy ⎛ 1 ⎞ 4 ln (ax) = 4[ln(ax )]3 ⎜ ⋅ a ⎟ = dx x ⎝ ax ⎠

32. y = ln ⎡ (5 x + 2)4 (8 x − 3)6 ⎤ ⎣ ⎦ = 4 ln(5 x + 2) + 6 ln(8 x − 3)

40. y = ln 2 (2 x + 11) = [ln(2 x + 11)]2

dy 1 1 = 4⋅ (5) + 6 ⋅ (8) dx 5x + 2 8x − 3 20 48 = + 5x + 2 8x − 3

dy 1 4 ln(2 x + 11) = 2[ln(2 x + 11)] ⋅ (2) = dx 2 x + 11 2 x + 11

426


41. y = x ln x − 1 =

Section 12.1

1 x ln( x − 1) 2

47. y =

⎤ dy 1 ⎡ ⎛ 1 ⎞ = ⎢x + ln( x − 1) ⋅ (1) ⎥ dx 2 ⎣ ⎜⎝ x − 1 ⎟⎠ ⎦ x = + ln x − 1 2( x − 1)

(

y′ =

)

1 ln(2 x + 1) 4 dy 1 1 1 3 1 = 3⋅ + ⋅ (2) = + dx x 4 2x + 1 x 2(2 x + 1)

x 1+ x 2

=

=

x + 1 + x2 1

=

(ln 3) − 1 ln 2 3

.

( )

= 25 ⋅

(q + 2) ln(q + 2) − q (q + 2) ln 2 (q + 2)

1 q+2

.

49. c = 25 ln(q + 1) + 12 dc 25 dc 25 , so . = = dq q + 1 dq q =6 7

−1 ⎡ 1 ⎤ 2 2 x 1 1 (2 x) ⎥ + + ⎢ ⎥⎦ ⎣⎢ 2

)

1 + x2 + x 1 + x 2 ⎜⎛ x + 1 + x 2 ⎟⎞ ⎝ ⎠

50. c =

500 ln(q + 20)

c = cq =

1 + x2

500q ln(q + 20)

( ) 1

[ln(q + 20)](1) − q q + 20 dc = 500 ⋅ dq [ln(q + 20)]2

45. y = ln( x 2 − 3 x − 3) y′ =

ln 2 x

ln(q + 2)(1) − q dr = 25 ⋅ dq ln 2 (q + 2)

3 dy 1 −1 3 = (4 + 3ln x) 2 ⋅ = dx 2 x 2 x 4 + 3ln x

1+

ln x

25 25q , so r = pq = . Thus the ln(q + 2) ln(q + 2) marginal revenue is

1

(

2

48. p =

43. y = 4 + 3ln x = (4 + 3ln x ) 2

dy 1 = dx x + 1 + x 2

( ) = ln x − 1

(ln x)(1) − x 1x

When x = 3 the slope is y′(3) =

42. y = ln x3 4 2 x + 1 = 3ln x +

44.

x ln x

2x − 3 2

x − 3x − 3 The slope of the tangent line at x = 4 is 8−3 y′(4) = = 5. Also, if x = 4, then 16 − 12 − 3 y = ln(16 − 12 − 3) = ln 1 = 0. Thus an equation of the tangent line is y – 0 = 5(x – 4), or y = 5x − 20.

50

ln 70 − 70 dc = 500 ⋅ ≈ $97.90 dq q =50 (ln 70)2

51.

dq d = [25 + 10 ln(2 p + 1)] dp dp = 0 + 10

46. y = x[ln(x) – 1] ⎛1⎞ y′ = x ⎜ ⎟ + [ln( x) − 1](1) = ln x ⎝x⎠ When x = e, y = 0 and y′ = 1 . The equation of the tangent line is y – 0 = 1(x – e), or y = x – e.

=

427

⎛ 1 ⎞ d d [ln(2 p + 1)] = 10 ⎜ ⎟ [2 p + 1] dp ⎝ 2 p + 1 ⎠ dp

10 20 (2) = 2 p +1 2 p +1


52. With I 0 = 17, L( I ) = 10 log

ISM: Introductory Mathematical Analysis Principles in Practice 12.2

I . 17

1. The rate of change of temperature with respect to dT . T(t) has the form Ceu where C is a time is dt constant and u = kt. dT d ⎡ kt ⎤ d Ce = C ⎡ ekt ⎤ = ⎦ dt dt ⎣ dt ⎣ ⎦

dL d ⎡ I ⎤ d = 10 log ⎥ = 10 [log I − log17] dI dI ⎢⎣ 17 ⎦ dI = 10 =

d dI

⎡ ln I ⎤ ⎡ 1 1 ⎤ ⎢ ln10 − log17 ⎥ = 10 ⎢ ln10 ⋅ I − 0 ⎥ ⎣ ⎦ ⎣ ⎦

10 I ln10

( ) dtd [kt ] = Ce

= C ekt

⎛ T ⎞ 53. A = 6 ln ⎜ − a ⎟ . Rate of change of A with ⎝ a −T ⎠ respect to T: ⎡ (a − T )(1) − T (−1) ⎤ dA 1 = 6⋅ ⎢ ⎥ T −a dT (a − T )2 ⎢⎣ ⎥⎦ a −T = 6⋅ = 6⋅ =

1 T − a ( a −T ) a −T

⋅

1. y ′ = 5 ⋅

2. y′ =

a

T − a 2 + aT (a − T ) 2 6a

(T − a

2

)

d x (e ) = 5e x dx

2e x 5

3. y′ = e 2 x

2

+3

(4 x) = 4 xe2 x

2

+3

4. y′ = e 2 x

2

+5

(4 x) = 4 xe2 x

2

+5

+ aT (a − T )

dy 1 f ′( x) = f ′( x) = , dx f ( x) f ( x) which is the relative rate of change of y = f(x) with respect to x.

6.

7.

1 d 1 ⎛ 1 du ⎞ ⋅ (ln u ) = ⋅ ln b dx ln b ⎜⎝ u dx ⎟⎠

x4 f ′( x) = 0 for x ≈ −1.65, 1.65

2

2

+ 4r + 4

+ 6 x3 +1

( −3q

2

+6

)

+ 6 q −1

(6r + 4) = 2(3r + 2)e3r

2

+ 4r + 4

(2 x + 18 x 2 ) 2

+ 6 x3 +1

( )

9. y′ = x e x + e x (1) = e x ( x + 1) 10. y′ = 3x 4 ⎡ e− x (−1) ⎤ + e− x (12 x3 ) = 3 x3e− x (4 − x) ⎣ ⎦

57. Note that f(x) is defined for all x ≠ 0. x

f ′(r ) = e3r

3

= 2 x(1 + 9 x)e x

f ′( x) = x (1 + 3ln x) f ′( x) = 0 for x ≈ 0.72

f ′( x) =

+ 6 q −1

)

8. y ′ = e x

2

x 2 ⋅ 12 (2 x) − ln( x 2 ) ⋅ 2 x

3

(

du ⎛ 1 du ⎞ 1 = ( logb e ) ⎜ ⋅ ⎟ = ( logb e ) dx ⎝ u dx ⎠ u

56.

f ′(q) = e− q

= −3 q 2 − 2 e − q

d d ln u ( logb u ) = ⎛⎜ ⎞⎟ dx dx ⎝ ln b ⎠ =

d ( 9 − 5 x ) = e9−5 x (−5) = −5e9−5 x dx

5. y′ = e9−5 x ⋅

54. If y = ln f(x), then

55.

(k ) = Ckekt

Problems 12.2

⎡ ⎤ a ⎢ ⎥ 2 ⎢⎣ (a − T ) ⎥⎦

a −T

kt

=

2 2 11. y′ = x 2 ⎡ e − x (−2 x) ⎤ + e− x (2 x) ⎣⎢ ⎦⎥

2 − 2 ln( x 2 ) x3

2

(

= 2 xe− x 1 − x 2

)

12. y′ = x ⎡ e3 x (3) ⎤ + e3 x (1) = e3 x (3 x + 1) ⎣ ⎦ 428


(

1 x e + e− x 3

13. y = y′ =

14.

)

1⎡ x e x − e− x e + e − x (−1) ⎤ = ⎦ 3⎣ 3

dy (e x + e − x )[e x − e− x (−1)] − (e x − e− x )[e x + e− x (−1)] = dx (e x + e − x ) 2 =

15.

Section 12.2

(e x + e − x ) 2 − (e x − e − x ) 2 x

(e + e

−x 2

)

=

4 (e + e − x ) 2 x

( )

d 2 x3 d ⎡ (ln 5)2 x3 ⎤ = e 5 ⎥⎦ dx dx ⎢⎣ 3

= e(ln 5)2 x [(ln 5)6 x 2 ] 3

= (6 x 2 )52 x ln 5

16. y = 2 x x 2 = e(ln 2) x x 2 y ′ = e(ln 2) x (2 x) + x 2 ⎡ e(ln 2) x (ln 2) ⎤ ⎣ ⎦

( )

( )

( )

= 2 x 2 x + x 2 2 x (ln 2) = x 2 x (2 + x ln 2)

17.

w2 ⎡ e2 w (2) ⎤ − e2 w [2 w] ⎣ ⎦ f ′( w) = w4 =

2e 2 w ( w − 1) w3

18. y′ = e x −

x

19. y′ = e1+

x

1 ⎞ ⎛ 1 − 12 ⎞ x− x ⎛ ⎜1 − ⎟ ⎜1 − 2 x ⎟ = e ⎝ ⎠ ⎝ 2 x⎠ 1+ x ⎛ 1 − 12 ⎞ e ⎜2x ⎟= ⎝ ⎠ 2 x

20. y ′ = 3(e2 x + 1) 2 (e2 x (2) + 0) = 6e2 x (e2 x + 1)2 21. y = x5 − 5 x = x5 − e(ln 5) x y′ = 5 x 4 − e(ln 5) x (ln 5) = 5 x 4 − 5 x ln 5

22.

2

f ( z ) = e−1/ z = e− z 2

−2

f ′( z ) = e−1/ z [−(−2 z −3 )] =

2 −1/ z 2 e z3

429


(

)

(


)

e x + 1 ⎡e x ⎤ − e x − 1 ⎡e x ⎤ dy ⎣ ⎦ ⎣ ⎦ 23. = 2 dx ex +1 =

2e

(

)

33.

dp = −0.015e−0.5 dq q =500

x

( e + 1) x

dp = 15e−0.001q (−0.001) = −0.015e−0.001q dq

2

34.

24. y′ = e2 x [1] + ( x + 6) ⎡ e2 x (2) ⎤ = e2 x (2 x + 13) ⎣ ⎦

dp ⎛ 5 ⎞ −5q / 750 = 9e−5q / 750 ⎜ − ⎟ = −0.06e dq ⎝ 750 ⎠ dp = −0.06e−2 dq q =300

25. y = ln e x = x so y ′ = 1.

q

(

q 7000e 700 35. c = , so c = cq = 7000e 700 . The q

)

1 ⎛1 ⎞ 26. y′ = e− x ⋅ + (ln x) −e− x = e− x ⎜ − ln x ⎟ x ⎝x ⎠

27. y ′ = e x

2

ln x 2

= 2 xe

⎡ 2 1 ⎤ 2 ⎢ x ⋅ 2 (2 x) + (ln x )(2 x) ⎥ x ⎣ ⎦

x 2 ln x 2

q

= 10e 700 . Thus

2

(1 + ln x )

28. y = ln e4 x +1 = 4 x + 1 , so 29.

marginal cost function is

2

f ( x) = ee x e x = e1+ x + x

dc = 10e0.5 and dq q =350

dc = 10e . dq q =700

dy = 4. dx

2

2q+6

2

f ′( x) = e1+ x + x (1 + 2 x) = (1 + 2 x)e1+ x + x

850 e 800 36. c = + 4000 q q

2

2

f ′(−1) = [1 + 2(−1)]e1+ ( −1) + ( −1) = −e

30.

f ( x) = 5

x 2 ln x

( )

= e

ln 5

x 2 ln x

=e

c = cq = 850 + 4000e

37. w = e x

f ′(1) = e0 (ln 5)[1 + 0] = ln 5

y′ = e−2 . Thus an equation of the tangent line is y−e

3

−4 x

q +3 dc = 10e 400 . dq

+ x ln( x − 1) and x =

By the chain rule,

31. y = e x , y′ = e x . When x = –2, then y = e−2 and −2

q +3

= 850 + 4000e 400

dc dc = 10e0.25 and = 10e0.5 . dq q =97 dq q =197

2 = e(ln 5) x ln x (ln 5)[ x + 2 x ln x]

−2

2q+6 800

The marginal cost function is

(ln 5) x 2 ln x

2 ⎧ 1 ⎡ ⎤⎫ f ′( x) = e(ln 5) x ln x ⎨(ln 5) ⎢ x 2 ⋅ + (ln x)(2 x) ⎥ ⎬ x ⎣ ⎦⎭ ⎩

−2

q dc ⎛ 1 ⎞ = 7000e 700 ⎜ ⎟ dq ⎝ 700 ⎠

(

t +1 t −1

dw dw dx = ⋅ dt dx dt

)

⎡ 3 ⎤ ⎛ 1 ⎞ = ⎢e x − 4 x 3 x 2 − 4 + x ⎜ + [ln( x − 1)(1)]⎥ ⎟ ⎝ x −1 ⎠ ⎣ ⎦

−2

= e ( x + 2) , or y = e x + 3e .

⎡ (t − 1)(1) − (t + 1)(1) ⎤ · ⎢ ⎥ (t − 1) 2 ⎢⎣ ⎥⎦

32. y ′ = e x When x = 1, y = e and y ′ = e. Thus an equation

3 x ⎡ ⎤ ⎡ −2 ⎤ = ⎢ 3x2 − 4 e x −4 x + + ln( x − 1) ⎥ ⎢ ⎥. x −1 ⎣ ⎦ ⎣⎢ (t − 1) 2 ⎦⎥

(

of the tangent line is y − e = e(x − 1) or y = ex.

430

)


When t = 3, then x =

Section 12.2

3 +1 4 = = 2 and 3 −1 2

42.

dw ⎡ 1⎤ = [8 + 2 + 0] ⎢ − ⎥ = −5 . dt ⎣ 2⎦

38.

(

)

( )

f ( x) = 10

t

t

t

= −(ln10)10

(

45. Since S = Pert , then dS dt

x−2

S

(

rPert Pert

(

dS = Pert r = rPert . Thus dt

= r.

)

dy = K ⎡ −e− ax (−a ) ⎤ = aKe− ax ⎣ ⎦ dx

Solving the original equation for e− ax gives y e− ax = − + 1 . Thus substitution, K dy ⎛ y ⎞ = aK ⎜ − + 1⎟ = a(− y + K ) = a (K – y), as dx ⎝ K ⎠ was to be shown.

1 + 10

dq = 500 −e−0.2t dt dq Thus = 100e−2 . dt t =10

=

46. y = K 1 − e− ax

1 + 0.01e x − 2 8+ x

) ) (−0.2) = 100e

d (ln β )t [e ] dt

t

+ 0.01 f ′(2) −(ln10)10 = ≈ 0.0374 2 − f (2) 10 + ln(10) + 0.01

41. q = 500 1 − e−0.2t

d t [β ] dt

= kα β ( β t ln α ) ln β

1 + + 0.01e x − 2 8+ x −2

t

= kα β (ln α )e(ln β )t (ln β )

= e(ln10)( − x ) + ln(8 + x) + 0.01e x − 2

−x

/2

Y ′ = ke(ln α ) β (ln α ) = kα β (ln α )

+ ln(8 + x) + 0.01e

2

t

−x ⎥ ⎦

f ′( x) = e(ln10)( − x ) (− ln10) +

e− x

/2

44. Y = kα β = ke(ln α ) β

c⎤

d x (c − x c ) = (ln c)c − c dx x =1 If this is zero, (ln c)c – c = 0, or c[ln(c) – 1] = 0. Since c > 0, we must have ln(c) – 1 = 0, ln c = 1, or c = e.

40.

2

43. P = 1.92e0.0176t dP = 1.92e0.0176t (0.0176) = P(0.0176) dt = 0.0176P = kP for k = 0.0176.

= (ln c)e(ln c ) x − cx c −1 = (ln c)c x − cxc −1

−x

e− x

(− x) 2π 1 −1/ 2 e (−1) ≈ −0.242 2π

f ′(1) =

d dy [ f (u )] = and by the chain rule dx dx d dy dy du du [ f (u )] = = ⋅ = f ′(u ) = u3 ⋅ e x dx dx du dx dx = ( e x )3 ⋅ e x = e 3 x ⋅ e x = e 4 x x

2π 1

f ′( x) =

f ′( x) = x3 and u = e x . Let y = f(u). Then

d x d ⎡ ln c 39. c − xc = e dx dx ⎢⎣ d ⎡ (ln c ) x = e − xc ⎤ ⎦ dx ⎣

1

f ( x) =

−0.2t

47. N = 10 A10−bM = 10 A−bM = e(ln10)( A−bM ) dN = e(ln10)( A−bM ) (ln10)(−b) , so dM dN = 10 A−bM (ln10)(−b) = −b 10 A−bM ln10 dM

(

431

)



48. p = 0.89 ⎡ 0.01 + 0.99(0.85)t ⎤ ⎣ ⎦ a.

52. S = ln

dP = 0.89 ⎡0.99(0.85)t ln(0.85) ⎤ ⎣ ⎦ dt

a.

= 0.8811(0.85)t ln(0.85) This represents the rate of change of proportion of correct recalls with respect to length of recall interval.

−

b. If

50. C (t ) =

a.

b.

⎞ ⎟ C (t ) ⎠

=

7 ≈ $0.847 billion 3 = $(0.847)(1000) million = $847 million

R⎡ R 1 − e0 ⎤ = [1 − 1] = 0 ⎣ ⎦ r r

53.

3 2 f ′( x) = (6 x 2 + 2 x − 3)e2 x + x −3 x

f ′( x) = 0 for x ≈ −0.89, 0.56

54.

f ′( x) = 1 − e− x f ′( x) = 0 gives e x = 1 or x = 0.

− ( r )t ⎞ ⎤ R⎡ r R⎛ ⎢1 − ⋅ ⎜ 1 − e V ⎟ ⎥ V ⎢⎣ R r ⎝ ⎠ ⎥⎦

Problems 12.3

R⎡ r ⎤ R r = ⎢1 − C (t ) ⎥ = − C (t ) V⎣ R ⎦ V V

1. η =

p q dp dq

=

p q

−2

.

When q = 5 then p = 40 – 2(5) = 30, so

f (t ) = 1 − e−0.008t f ′(t ) = 0.008e

−I

I = ln

− ( r )t ⎞ ⎤ R⎡ ⎛ = ⎢1 − ⎜ 1 − e V ⎟ ⎥ V ⎣⎢ ⎝ ⎠ ⎦⎥

51.

dS 1 e− I 1 = , then = . −I 8 dI 8 3+ e

1 (e )(3 + e ) 8 1 1 = I 3e + 1 8 3e I + 1 = 8 7 eI = 3

dC R ⎡ r −( Vr )t ⎤ R −( Vr )t = ⎢ e ⎥= e dt r ⎣V ⎦ V

=

dC dS e− I 3 . = 1− = 1− = −I dI dI 3+ e 3 + e− I

I

−( Vr )t ⎤ R⎡ ⎢1 − e ⎥ r ⎣ ⎦

C (0) =

dC dS = 1− . dI dI

(e I ) e − I

− ( r )t ⎛ r ⎞ dC = C0 e V ⎜ − ⎟ dt ⎝ V⎠

⎞ ⎛r ⎟ = −⎜ V ⎠ ⎝

Recall that

Thus

( Vr )t

⎛ r = [C (t ) ] ⎜ − ⎝ V

3+e

= ln 5 − ln(3 + e− I )

−I

dS 1 e− I (e− I )(−1) = =− dI 3 + e− I 3 + e− I

b. If t = 2, then dp = 0.8811(0.85)2 ln(0.85) ≈ −0.10 dt 49. C (t ) = C0 e

5

η=

30 5

= −3 −2 Because η > 1 , demand is elastic.

−0.008t

f ′(100) = 0.008e−0.8 ≈ 0.0036

p q

6

= 100 = −1.5 −0.04 −0.04 Because η > 1, demand is elastic.

2. η =

432


3. p =

Section 12.3 9. q = 1200 − 150p

3500 = 3500q −1 q

η=

dp 3500 = −3500q −2 = − dq q2

η=

p q dp dq

=

p q − 3500 q2

=

(3500 / q ) q − 3500 q2

4. η =

5. η =

p q dp dq

(500 / q )

= −1

=

=

p q − 500 2 ( q + 2)

q 1000 − 3 q

=

1 = − , inelastic 2

[500 /( q + 2)] q − 500 2 ( q + 2)

=

=−

η=

6. η =

=

800 /(2 q +1) q 1600 − (2 q +1) 2

When q = 24, η = −

7. η =

p q dp dq

=

=−

q+2 q

η=

2q + 1 2q

η=

49 , elastic 48

8. η = −e

−

q 200

p q dp dq

=

p dq ⋅ q dp

p⎛ 1 ⎞ p ⎜− ⎟ = − 2 q ⎝ 2q ⎠ 2q

If p = 400, then q = 500 − 400 = 10, so

η=−

p q

400 = −2. η > 1 , so demand is elastic. 200

q

12. q = 2500 − p 2

⎛ 150 ⎞ = −⎜ − 1⎟ . Because η > 1 , ⎝ e ⎠

η=

−

=

100e q

2

−e

−

=−

200 q

=

2

When q = 200, η = −

=

p dq ⋅ q dp

)

− 12 dq 1 = 2500 − p 2 (−2 p) dp 2

q 200

q 200

p q dp dq

(

demand is elastic. p q

p dq ⋅ q dp

−1 1 dq 1 −1 = (500 − p ) 2 (−1) = =− 2q dp 2 2 500 − p

When q = 100, then p = 150 – e and

η=

=

11. q = 500 − p

−e100 100

150 −e 100 e − 100

p q dp dq

50 dq = −1 , so η = (−1) = −1 , unit elasticity. 50 dp

106 53 When q = 104, then η = − = − . Because 104 52 η > 1 , demand is elastic. p q − 1600 2 (2 q +1)

p dq p ⋅ = (−150) q dp q

10. q = 100 – p When p = 50, then q = 50.

2

p q −1000 q3

=

If p = 4, then q = 1200 − 150(4) = 600, so 4 η= (−150) = −1. Since η = 1, demand has 600 unit elasticity.

Because η = 1 , demand has unit elasticity. p q dp dq

p q dp dq

η=

200 = −1, so demand has 200

−p 2500 − p 2

=−

p⎛ p⎞ ⎜− ⎟ = − q⎝ q⎠

p q

p2 q2

If p = 20, then q = 2100 , so we have

unit elasticity.

η=−

433

400 4 = − , inelastic. 2100 21


13. q =


( p − 100) 2 2 p q dp dq

η=

=

η=

p dq ⋅ q dp

η=

(20 − 100) 2 = 3200 . Thus 2 20 1 η= (20 − 100) = − . Demand is inelastic. 3200 2 q=

=

p dq ⋅ q dp

=−

18. η =

η

demand

10

60

− 10 3

elastic

3 − 10

inelastic

−1

unit elasticity

130

η=

η=

Setting

=

=

p q dp dq

=

p dq ⋅ q dp

−p 2500 − p

p ⎛ −p ⎞ ⎜ ⎟=− q⎝ q ⎠

2

=

p2 q2

−p , so q

.

Now, if p = 30, then q = 2500 − 302 = 40, so (30) 2 9 = − = − . If the price of 30 p =30 2 16 (40) decreases to 28.5, that is, it changes by −1.5 = −5%, then demand would change by 30 ⎛ 9 ⎞ approximately −5 ⎜ − ⎟ %, or 2.8%. (That is, ⎝ 16 ⎠ demand increases by 2.8%.)

p = 36 − 0.25q p q dp dq

p =15

dq = dp

η

16. a.

p dq ⋅ q dp

q = 2500 − p 2

q

6.50

=

1 %, 2 then the change in demand is approximately ⎛1 ⎞ ⎜ 2 % ⎟ (−1.2) = −0.6%. Thus demand decreases ⎝ ⎠ approximately 0.6%.

p 0.05q

200

p q dp dq

demand. Thus if the price of 15 increases

p

3

= −1 yields q = ±10.

15 6 (30 − 40) = − = −1.2. Now, 125 5 (% change in price) · (η ) = % change in

so η

15. p = 13 – 0.05q

η=

−2q 2

When p = 15, then q = 500 − 40(15) + 152 = 125,

dq p = 2 p − 50, so η = (2 p − 50). dp q If p = 20, then q = 250, and 20 200 4 η= = − , inelastic. (40 − 50) = − 250 250 5 p q dp dq

300 − q 2

dq p = −40 + 2 p , so η = (2 p − 40). dp q

14. q = p 2 − 50 p + 850

η=

=

17. q = 500 − 40 p + p 2

p ( p − 100) . If p = 20, then q

p q dp dq

p q dp dq

Since q > 0, we must have q = 10.

dq 1 = (2)( p − 100)(1) = p − 100 , so dp 2

η=

p = 300 − q 2

b.

36 − 0.25q −0.25q

36 − 0.25q = −1 yields q = 72. −0.25q

434


Section 12.3 22. p = mq + b

19. p = 500 – 2q

η=

p q dp dq

=

500 − 2 q q

−2

=

Note: q =

q − 250 q

q − 250 < −1. For q q > 0, we have q – 250 < –q, 2q < 250, so q < 125. Thus, if 0 < q < 125, demand is elastic. q − 250 > −1. If demand is inelastic, then η = q For q > 0, the inequality implies q > 125. Thus if 125 < q < 250, then demand is inelastic.

If demand is elastic, then η =

a.

b.

23. a.

3q − 50 = = −3 3q

b.

c.

1000 q2

b + cq

2

= a (b + cq 2 ) −1/ 2

p q dp dq

=

a(b + cq 2 ) −1/ 2 −acq 2 (b + cq 2 ) −3 / 2

=−

b + cq 2 cq 2

⎛ b ⎞ = −⎜ + 1⎟ 2 ⎜ ⎟ cq ⎝ cq ⎠ b If b, c > 0, then + 1 > 1 so |η| > 1 and cq 2 demand is elastic.

η=−

b + cq 2 2

If |η| = 1, then

b cq 2

+ 1 = 1, which can only

occur if b = 0.

1000 r = pq = q

24. η =

dr 1000 = −1000q −2 = − dq q2 =

a

p=

Thus η does not depend on a.

⎛ 1⎞ ⎛ 3q ⎞ p ⎜ 1 + ⎟ = (50 − 3q) ⎜ 1 + ⎟ η 3 − 50 ⎠ q ⎝ ⎠ ⎝ ⎛ 3q − 50 + 3q ⎞ = (50 − 3q) ⎜ ⎟ ⎝ 3q − 50 ⎠ dr = 50 − 6q = dq

η=

p p −b Thus if p = 0, then η = 0.

η=

η=

50 −3q q

1000 q3 − 2000 q3

−

= −acq (b + cq 2 )−3 / 2

dr = 50 − 6q dq

p q dp dq

m

p →b −

dp 1 = − a (b + cq 2 )−3 / 2 (2cq) dq 2

r = pq = 50q − 3q 2

21. p =

p →b −

p ( p −b ) / m

= lim

p = −∞ −b p p →b

20. p = 50 − 3q

η=

p q p →b − dp dq

lim η = lim

= lim

Since Total Revenue = r = pq = 500q − 2q 2 , then r ′ = 500 − 4q = 4(125 − q ). If 0 < q < 125, then r ′ > 0, so r is increasing. If 125 < q < 250, then r ′ < 0, so r is decreasing.

p q dp dq

p−b m

p q dp dq

=

p dq ⋅ q dp

We differentiate implicitly for

1 =− 2

dq . dp

q 2 (1 + p) 2 = p

(

)

⎛ dq ⎞ q 2 ⋅ 2(1 + p)(1) + 1 + p 2 ⎜ 2q ⎟ = 1 ⎝ dp ⎠ dq 2q 2 (1 + p) + 2q (1 + p) 2 =1 dp

⎛ 1 ⎞ 1000 1000 dr p ⎜1 + ⎟ = (1 − 2) = − = 2 η dq q q2 ⎝ ⎠

435


Thus


dq 1 − 2q 2 (1 + p) = dp 2q(1 + p) 2

q 2 (1 + p) 2 1 − 2q 2 (1 + p) 1 − 2q 2 (1 + p) ⋅ = q 2 2q(1 + p) 2 If p = 9, we find q from the given equation:

Hence η =

q 2 (1 + 9)2 = 9 q2 =

9 100

( )

2

3 1 − 2 10 (1 + 9) 3 since q > 0. Thus η p =9 = = −0.4 q= 10 2

25. a.

(

q=

60 + ln 65 − p3 p

η=

p q dp dq

=

)

p dq p ⎡ 60 3 p2 ⎤ = ⎢− − ⎥ q dp q ⎢⎣ p 2 65 − p3 ⎦⎥

If p = 4, then q =

60 4 ⎡ 60 3(16) ⎤ 207 =− ≈ −13.8, and demand is elastic. + ln1 = 15 , so η = ⎢ − − 4 15 ⎣ 16 65 − 64 ⎥⎦ 15

b. The percentage change in q is (–2)(–13.8) = 27.6%, so q increases by approximately 27.6%. c. 26. a.

Lowering the price increases revenue because demand is elastic. 0.02 q +19 ⎤ p = 50 ⎡⎢ (151 − q ) ⎥⎦ ⎣ ln p = ln 50 + 0.02 q + 19 ln(151 − q)

⎡ q + 19 ⎤ 1 dp 1 (−1) + ln(151 − q) ⋅ = 0 + 0.02 ⎢ ⎥ p dq 2 q + 19 ⎦⎥ ⎣⎢ 151 − q

When q = 150, then p = 50, so

b.

η q =150 =

p q dp dq q =150

dp ⎡ 13 0 ⎤ = 0.02(50) ⎢ − + ⎥ = −13 dq q =150 ⎣ 1 26 ⎦

50

= 150 ≈ −0.0256 −13

Thus demand is inelastic. c.

(elasticity)(% change in price) = % change in demand −10 (−0.0256)(% change in price) = ⋅100 150 100 ⎛ 1 ⎞ % change in price = − = 260% 15 ⎜⎝ −0.0256 ⎟⎠ Thus price per unit of $50 changes by 2.6(50) = $130, so it is approximately 50 + 130 = $180.

d. The manufacturer should increase the price because demand is inelastic. 436


Section 12.4

−5 25 50 ⋅100 = − % and the percentage change in quantity is ⋅100 = 10%. 80 4 500 Thus, since (elasticity)(% change in price) ≈ % change in demand, ⎛ 25 ⎞ (elasticity) ⎜ − ⎟ ≈ 10. ⎝ 4 ⎠

27. The percentage change in price is

40 8 = − = −1.6 25 5 dr To estimate when p = 80, we have dq

elasticity ≈ −

⎛ ⎛ 1⎞ dr 1 ⎞ ⎟ = 30. = p ⎜ 1 + ⎟ = 80 ⎜1 + ⎜ −8 ⎟ dq ⎝ η⎠ 5⎠ ⎝

28. η =

p q dp dq

=

2000 − q 2 −2q

2

=

For 5 ≤ q ≤ 40, η =

1 1000 − 2 q2 1000 2

−

1 2000 and η ′ = − . Since η ′ < 0 , η is decreasing on [5, 40] and thus η is 2 q3

q maximum at q = 5 and a minimum at q = 40.

29.

dp −200 = 200(−1)(q + 5) −2 = dq (q + 5)2

Thus η =

p q dp dq

=

200 q ( q + 5) − 200 2 ( q + 5)

For 5 ≤ q ≤ 95, η =

=−

q+5 . q

5 q+5 5 = 1 + and η ′ = − . q q q2

Since η ′ < 0, η is decreasing on [5, 95], and thus η is maximum at q = 5 and minimum at q = 95. Principles in Practice 12.4 ⎛ P ⎞ 1. Assume that P is a function of t and differentiate both sides of ln ⎜ ⎟ = 0.5t with respect to t. ⎝ 1− P ⎠ d dt

⎡ ⎛ P ⎞⎤ d ⎢ ln ⎜ ⎟ ⎥ = [0.5t ] ⎣ ⎝ 1 − P ⎠ ⎦ dt

⎛ 1 ⎞d ⎡ P ⎤ ⎜ ⎟ = 0.5 ⎜ P ⎟ dt ⎢⎣ 1 − P ⎥⎦ ⎝ 1− P ⎠ 1 − P (1)(1 − P) − P (−1) dP ⋅ ⋅ = 0.5 P dt (1 − P )2 1 − P + P dP ⋅ = 0.5 P (1 − P ) dt dP = 0.5P (1 − P ) dt 437


2.


( )

3. 6 y 2 y ′ − 14 x = 0 14 x 7x y′ = = 2 6y 3y2

dV d ⎡4 dr dr ⎤ 4 = ⎢ πr 3 ⎥ = π 3r 2 = 4πr 2 dt dt ⎣ 3 3 dt dt ⎦ dr = 5 and r = 12, When dt dV = 4π(12) 2 (5) = 2880π . The balloon is dt increasing at the rate of 2880π cubic inches/minute.

4. 4 x − 6 yy′ = 0 y′ =

5.

3. The hypotenuse is the length of the ladder, so x 2 + y 2 = 100 . Differentiate both sides of the equation with respect to t. d ⎡ 2 d x + y 2 ⎤ = [100] ⎦ dt dt ⎣ dx dy 2x + 2 y =0 dt dt When y = 8, we can find x by using the Pythagorean theorem.

2x 3y

x1/ 3 + y1/ 3 = 3 1 −2 / 3 1 −2 / 3 x + y y′ = 0 3 3 y −2 / 3 y ′ = − x − 2 / 3 y′ = − =−

x −2 / 3 y −2 / 3 y2 / 3 x2 / 3 3

x + 8 = 100

=−

x 2 = 100 − 64 = 36 x=6

= −3

2

2

When x = 6, y = 8, and 2(6)(3) + 2(8)

dx = 3 , we have dt

y2 x2

⎛ 1 ⎞ −4 ⎛ 1 ⎞ −4 6. ⎜ ⎟ x 5 + ⎜ ⎟ y 5 y′ = 0 ⎝5⎠ ⎝5⎠

dy =0 dt

dy =0 dt dy 36 9 =− =− dt 16 4 dy 9 = − , thus the top of the ladder is sliding dt 4 9 down the wall at the rate of feet/sec. 4

4

y′ = −

y5 4

x5

4

⎛ y ⎞5 = −⎜ ⎟ ⎝x⎠

⎛ 3 ⎞ −1 ⎛ 3 ⎞ −1 7. ⎜ ⎟ x 4 + ⎜ ⎟ y 4 y′ = 0 ⎝4⎠ ⎝4⎠ 1

y′ = −

y4 1

x4

Problems 12.4

8. 3 y 2 y′ = 4

1. 2 x + 8 yy′ = 0 x + 4 yy′ = 0 4 yy′ = − x

y′ =

4 3y2

9. By the product rule xy′ + y (1) = 0 , xy′ = − y ,

x 4y

y′ = −

2. 6 x + 12 yy′ = 0 y′ = −

3 2

x

36 + 16

y′ = −

y2

x 2y

438

y x


Section 12.4

10. 2 x + xy ′ + y (1) − 4 yy ′ = 0 xy ′ − 4 yy ′ = −2 x − y −2 x − y 2x + y = y′ = −x + 4 y x − 4y 11. xy′ + y (1) − y′ − 11 = 0 y′( x − 1) = 11 − y y′ =

11 − y x −1 3x 2 − 3 y 2 y ′ = 3x 2 y ′ + 6 xy − 3x(2 yy ′) − 3 y 2

12.

y ′(−3 y 2 − 3x 2 + 6 xy ) = 6 xy − 3 y 2 − 3 x 2 y′ = 1

13. 6 x 2 + 3 y 2 y′ − 12 ( xy′ + y ) = 0 3 y 2 y′ − 12 xy′ = 12 y − 6 x 2

( y′ ( y

)

y′ 3 y 2 − 12 x = 12 y − 6 x 2

y′ =

)

2

− 4 x = 4 y − 2 x2

4 y − 2x2 y2 − 4x

14. 6 x 2 + (3x) y′ + y (3) + 3 y 2 y′ = 0

( ) y′ ( x + y ) = −2 x

y ′ 3 x + 3 y 2 = −6 x 2 − 3 y 2

y′ = −

15. x =

2

−y

2x 2 + y x + y2

y + 4 y = y1/ 2 + y1/ 4

1 −1/ 2 1 y y ′ + y −3 / 4 y ′ 2 4 ⎛ 2 y1/ 4 + 1 ⎞ ⎛ 1 1 ⎞ = y′ ⎜ + = y′ ⎜ ⎟ ⎟ ⎜ 2 y1/ 2 4 y 3 / 4 ⎟ ⎜ 4 y3 / 4 ⎟ ⎝ ⎠ ⎝ ⎠

1=

y′ =

4 y3 / 4 2 y1/ 4 + 1

(

)

( )

16. x3 3 y 2 y′ + y 3 3 x 2 + 1 = 0 y′ = −

1 + 3x2 y3 3 x3 y 2

439



(

)( ) x +1 y (1 + y′) 6e (1 + e ) ( x + y ) = 1 + y′ y′ = 6e (1 + e ) ( x + y ) − 1

17. 5 x3 (4 y 3 y ′) + 15 x 2 y 4 − 1 + 2 yy ′ = 0 y ′(20 x3 y 3 + 2 y ) = 1 − 15 x 2 y 4 y′ =

23. 2 1 + e3 x 3e3 x =

1 − 15 x 2 y 4

3x

20 x3 y 3 + 2 y

3x

1 x 1 (2 y + 1) y′ = x 1 y′ = x(2 y + 1)

18. 2 yy′ + y′ =

24.

( )

⎛1⎞ 19. y ⎜ ⎟ + (ln x ) y′ = x e y y′ + e y (1) ⎝x⎠ ⎡ ln( x ) − xe y ⎤ y′ = e y − y ⎣ ⎦ x

20.

1 (1 + y ′) x+ y 1 y′ e x + y + y ′e x + y = + x+ y x+ y ⎛ 1 ⎞ 1 y′ ⎜ e x+ y − − e x+ y ⎟= x y x y + + ⎝ ⎠ y ′ = −1 e x + y (1 + y ′) =

y′ = −

xe y − y

xy′ = − y ( x + 1) y ( x + 1) x

( )

21. ⎡ x e y y′ + e y (1) ⎤ + y′ = 0 ⎣⎢ ⎦⎥

⎛ ⎞ x y − x + 1 ⎟ y′ = − y +1 ⎜ ⎜ 2 y +1 ⎟ 2 x +1 ⎝ ⎠

xe y y′ + e y + y′ = 0 y

y′ = −

1+ 2 3 =− . 1+ 4 5

⎛ ⎞ 1 ⋅ y′ ⎟ + y + 1(1) 26. x ⎜ ⎜ 2 y +1 ⎟ ⎝ ⎠ ⎛ 1 ⎞ = y⎜ ⎟ + x + 1( y′) ⎝ 2 x +1 ⎠ x y ⋅ y′ − x + 1 ⋅ y′ = − y +1 2 y +1 2 x +1

xy′ + y (1) +1 = 0 xy xy′ + y + xy = 0

( xe + 1) y′ = −e

1+ y x + 2y

At the point (1, 2), y′ = −

x ⎡ ln( x) − xe y ⎤ ⎣ ⎦

y′ = −

3x

25. 1 + [ xy′ + y (1)] + 2 yy′ = 0 xy′ + 2 yy′ = −1 − y ( x + 2 y ) y′ = −(1 + y )

y ⎡ ln( x ) − xe y ⎤ y′ = xe − y ⎣ ⎦ x

y′ =

3x

y

y′ =

ey xe y + 1

y − 2 x +1 x − 2 y +1

y +1 x +1 3

At (3, 3),

22. 8 x + 18 yy ′ = 0 8 x = −18 yy ′ 8x 4x y′ = − =− 18 y 9y

dy 4 − 2 = = 1. dx 3 − 2 4

440


Section 12.4

27. 8 x + 18 yy′ = 0

32. p = 400 − q

8x 4x y′ = − =− 18 y 9y

(

d d ( p) = 400 − q dp dp

⎛ 1⎞ Thus at ⎜ 0, ⎟ , y′ = 0; at ( x0 , y0 ) , ⎝ 3⎠ 4 x0 y′ = − . 9 y0

28.

1= −

2

2

33. p =

3

x + x yy ′ + xy + y y ′ = 2 yy ′ ( x 2 y + y 3 − 2 y ) y ′ = − x3 − xy 2 y′ =

− x( x 2 + y 2 ) y ( x 2 + y 2 − 2)

40 (q + 5)

3

⋅

dq dp

dq (q + 5)3 =− dp 40

3x2 + y y′ = − x + 2y At (−1, 1), y ′ = −4 and the tangent line is given

34.

by y − 1 = −4[x − (−1)], or y = −4x − 3. 30. 2 yy′ + [ xy′ + y (1)] − 2 x = 0 2x − y 2y + x

1 and the tangent line is given by 2 1 1 y − 3 = ( x − 4), or y = x + 1. 2 2

At (4, 3), y′ =

p=

10 2

q +3 d d ⎡ 10 ⎤ ( p) = ⎢ ⎥ dp dp ⎢⎣ q 2 + 3 ⎥⎦ 20q dq 1= − ⋅ 2 2 dp (q + 3) dq (q 2 + 3)2 =− dp 20q From the original equation, we have 10 dq q 2 + 3 = . Thus we can write as p dp

31. p = 100 − q 2

(

(q + 5)2

1= −

29. 3x 2 + xy ′ + y + 2 y ′ = 0

d d ( p) = 100 − q 2 dp dp

20

d d ⎡ 20 ⎤ ( p) = ⎢ ⎥ dp dp ⎢⎣ (q + 5)2 ⎥⎦ d d ⎡ ( p) = 20(q + 5)−2 ⎤ ⎦ dp dp ⎣

At (0, 2), y ′ = 0.

y′ =

1 dq ⋅ 2 q dp

dq = −2 q dp

2( x 2 + y 2 )(2 x + 2 yy ′) = 8 yy ′ ( x 2 + y 2 )( x + yy ′) = 2 yy ′ 3

)

)

( ) dq =− 10 p

dq 1 = −2q ⋅ dp

dp

dq 1 =− dp 2q

35. ln

2

20q

=−

I = −λ t I0

ln I − ln I 0 = −λ t 1 dI = −λ I dt dI = −λ I dt

441

5 qp 2

.


⎛ ⎞ E 36. 1.5M = log ⎜ 11 ⎟ ⎝ 2.5 × 10 ⎠

(

1.5M = log E − log 2.5 × 1011 d (1.5M ) = dM d (1.5M ) = dM 1.5 =

d dM d dM


1 2 I = SI + I . Differentiating implicitly 4 with respect to I: dS 1 dS ⎤ ⎡ 2S + I = ⎢ S (1) + I + 1, dI 2 dI ⎥⎦ ⎣

39. S 2 +

)

( ) ⎡ ln E ⎤ ⎢ ln10 − log ( 2.5 × 10 ) ⎥ ⎣ ⎦

⎡log E − log 2.5 × 1011 ⎤ ⎣⎢ ⎦⎥

dS dS I −I = S +1− , dI dI 2 dS 2S + 2 − I dS 2S + 2 − I (2 S − I ) = , = . dI 2 dI 2(2S − I ) 2S

11

1 ⎛ 1 dE ⎞ ⋅ ln10 ⎜⎝ E dM ⎟⎠

Marginal propensity to consume =

dE = 1.5E ln10 dM d d ⎡ ln E ⎤ (1.5M ) = − log(2.5 × 1011 ) ⎥ dE dE ⎢⎣ ln10 ⎦ dM 1 1 1.5 = ⋅ dE ln10 E dM 1 = dE 1.5E ln10

Thus

df df f , =− . dλ dλ λ

Solving v = f λ for f and differentiating: f =

λ

f (t ) 1 +σ = C1 + C2t. Thus 1 − f (t ) 1 − f (t )

f ′(t ) f ′(t ) σ f ′(t ) + + = C2 f (t ) 1 − f (t ) [1 − f (t )]2

to λ:

v

dC 24 + 2 − 16 10 6 3 = 1− = 1− = = . dI 2(24 − 16) 16 16 8

ln f (t ) − ln[1 − f (t )] + σ [1 − f (t )]−1 = C1 + C2t ,

37. v = f λ. Differentiating implicitly with respect 0 = f (1) + λ

dC 2S + 2 − I = 1− . When I = 16 and dI 2(2 S − I )

S = 12,

40. ln

dC dS . = 1− dI dI

⎡ 1 ⎤ 1 σ f ′(t ) ⎢ + + ⎥ = C2 2 ⎣⎢ f (t ) 1 − f (t ) [1 − f (t )] ⎦⎥

,

⎡ [1 − f (t )]2 + f (t )[1 − f (t )] + σ f (t ) ⎤ f ′(t ) ⎢ ⎥ = C2 f (t )[1 − f (t )]2 ⎢⎣ ⎥⎦ ⎡ [1 − f (t )][1 − f (t ) + f (t )] + σ f (t ) ⎤ f ′(t ) ⎢ ⎥ = C2 f (t )[1 − f (t )]2 ⎢⎣ ⎥⎦ ⎡ [1 − f (t )] + σ f (t ) ⎤ f ′(t ) ⎢ ⎥ = C2 2 ⎣⎢ f (t )[1 − f (t )] ⎦⎥

df v fλ f =− =− = − , which is the same 2 2 λ dλ λ λ as before.

so

38. (P + a)(v + b) = k d d [( P + a )(v + b)] = (k ) dP dP dv ( P + a) + (v + b)(1) = 0 dP dv v+b =− . From the original equation, dP P+a dv k v+b = . Thus we can write as dP ( P + a)

Thus f ′(t ) =

dv k =− . dP ( P + a)2

442

C2 f (t )[1 − f (t )]2 σ f (t ) + [1 − f (t )]


Section 12.5

Problems 12.5

(

)

1. y = ( x + 1)2 ( x − 2) x 2 + 3 . Take natural logarithms of both sides,

(

)

ln y = ln ⎡ ( x + 1) 2 ( x − 2) x 2 + 3 ⎤ . ⎢⎣ ⎥⎦ Using properties of logarithms on the right side gives

(

)

ln y = 2 ln( x + 1) + ln( x − 2) + ln x 2 + 3 .

Differentiating both sides with respect to x, y′ 2 1 2x = + + . 2 y x +1 x − 2 x + 3 Solving for y′ , ⎡ 2 1 2x ⎤ + + y′ = y ⎢ ⎥. 2 x + x − 1 2 x + 3⎦ ⎣ Expressing y′ in terms of x,

(

)

⎡ 2 1 2x ⎤ + + y′ = ( x + 1)2 ( x − 2) x 2 + 3 ⎢ ⎥ 2 1 2 x + x − x + 3⎦ ⎣

(

)

4⎤ ⎡ 2. ln y = ln ⎢ (3 x + 4)(8 x − 1) 2 3x 2 + 1 ⎥ ⎣ ⎦

(

)

= ln(3x + 4) + 2 ln(8 x − 1) + 4 ln 3 x 2 + 1 y′ 3 8 6x = + 2⋅ + 4⋅ y 3x + 4 8x −1 3x2 + 1

⎡ 3 16 24 x ⎤ + + y′ = y ⎢ ⎥ ⎣ 3x + 4 8 x − 1 3x2 + 1 ⎦ 4 ⎡ 3 16 24 x ⎤ + + = (3 x + 4)(8 x − 1)2 3 x 2 + 1 ⎢ ⎥ x + x − 3 4 8 1 3x2 + 1 ⎦ ⎣

(

(

)

)

2 ⎡ ⎤ 3. ln y = ln ⎢ 3x3 − 1 (2 x + 5)3 ⎥ ⎣ ⎦

(

)

= 2 ln 3 x3 − 1 + 3ln(2 x + 5) y′ 9 x2 2 = 2⋅ + 3⋅ 3 +5 y 2 x 3x − 1 ⎡ 18 x 2 6 ⎤ y′ = y ⎢ + ⎥ 3 ⎢⎣ 3 x − 1 2 x + 5 ⎥⎦ ⎡ 18 x 2 2 6 ⎤ y′ = 3 x3 − 1 (2 x + 5)3 ⎢ + ⎥ 3 ⎢⎣ 3x − 1 2 x + 5 ⎥⎦

(

)

443


4.


y = (2 x 2 + 1) 8 x 2 − 1 ln y = ln ⎡⎢ (2 x 2 + 1) 8 x 2 − 1 ⎤⎥ ⎣ ⎦ 1 = ln(2 x 2 + 1) + ln(8 x 2 − 1) 2 y′ 4x 1 16 x = + ⋅ y 2 x2 + 1 2 8x2 − 1 ⎡ 4x 8x ⎤ y′ = y ⎢ + ⎥ 2 ⎣ 2 x + 1 8x2 − 1 ⎦ ⎡ 4x 8x ⎤ = (2 x 2 + 1) 8 x 2 − 1 ⎢ + ⎥ 2 ⎣ 2 x + 1 8x2 − 1 ⎦

5. y = x + 1 x 2 − 2 x + 4 ln y = ln ⎛⎜ x + 1 x 2 − 2 x + 4 ⎞⎟ ⎝ ⎠ 1 1 1 2 ln y = ln( x + 1) + ln x − 2 + ln( x + 4) 2 2 2 y′ 1 ⎡ 1 2x 1 ⎤ = + + y 2 ⎢⎣ x + 1 x 2 − 2 x + 4 ⎥⎦

(

y′ = =

)

y⎡ 1 2x 1 ⎤ + + 2 ⎢⎣ x + 1 x 2 − 2 x + 4 ⎥⎦ x + 1 x2 − 2 x + 4 ⎡ 1 2x 1 ⎤ + + ⎢ ⎥ 2 2 ⎣ x +1 x − 2 x + 4 ⎦

6. ln y = ln ⎡ (2 x + 1) x3 + 2 3 2 x + 5 ⎤ ⎢⎣ ⎥⎦ 1 1 = ln(2 x + 1) + ln( x3 + 2) + ln(2 x + 5) 2 3 2 1 3x 2 1 2 y′ = + ⋅ + ⋅ 3 y 2x + 1 2 x + 2 3 2x + 5 ⎡ 2 ⎤ 3x2 2 + + y′ = y ⎢ ⎥ 3 ⎣⎢ 2 x + 1 2( x + 2) 3(2 x + 5) ⎦⎥ ⎡ 2 ⎤ 3x 2 2 = (2 x + 1) x3 + 2 3 2 x + 5 ⎢ + + ⎥ 3 ⎢⎣ 2 x + 1 2( x + 2) 3(2 x + 5) ⎥⎦

444


Section 12.5

1 − x2 1 7. ln y = ln = ln 1 − x 2 − ln(1 − 2 x ) 1− 2x 2 y′ 1 −2 x −2 = ⋅ − y 2 1 − x2 1 − 2 x ⎡ x 2 ⎤ + y′ = y ⎢ − 2 1 − 2x ⎥ ⎣ 1− x ⎦

(

y′ =

)

10. ln y = ln

(

)

y′ =

9. y =

11.

x2 + 5 ⎡ 2 x 1 ⎤ − x + 9 ⎢⎣ x 2 + 5 x + 9 ⎥⎦

1 2

( 2x

2

+2

)

2

( x + 1)2 (3x + 2)

(

)

2 ⎤ ⎡ 2 ⎢ 2x + 2 ⎥ ln y = ln ⎢ ⎥ 2 ⎢ ( x + 1) (3x + 2) ⎥ ⎢⎣ ⎥⎦

(

)

2

= 2 ln 2 x + 2 − 2 ln( x + 1) − ln(3 x + 2)

+2

)

(

)

)

2

⎡1 4x x ⎤ + − ⎢ ⎥ 2 2 + x2 ⎦ 2 + x2 ⎣ x 1 + x

( x + 3)( x − 2) 2x −1 ( x + 3)( x − 2) ln y = ln 2x −1 1 1 1 = ln( x + 3) + ln( x − 2) − ln(2 x − 1) 2 2 2 1 1 1 2 y′ 1 1 = ⋅ + ⋅ − ⋅ y 2 x + 3 2 x − 2 2 2x −1 1 2 ⎤ y⎡ 1 + − y′ = ⎢ 2 ⎣ x + 3 x − 2 2 x − 1 ⎥⎦ 1 ( x + 3)( x − 2) ⎡ 1 1 2 ⎤ = + − ⎢ ⎥ 2 2x −1 ⎣ x + 3 x − 2 2x −1⎦ y=

3

(

)

6 x3 + 1

2

x 6 e−4 x

( (

) )

1⎡ ln(6) + 2 ln x3 + 1 − 6 ln( x) − (−4 x ) ln e ⎤ ⎦⎥ 3 ⎣⎢ 1 = ⎡ ln(6) + 2 ln x3 + 1 − 6 ln( x) + 4 x ⎤ ⎦⎥ 3 ⎣⎢

=

⎡ 8x 2 3 ⎤ y′ = y ⎢ − − 2 x 1 3 x + + 2 ⎦⎥ ⎣ 2x + 2 2

(

x 1 + x2

12. ln y = ln

y′ 4x 1 3 = 2⋅ − 2⋅ − 2 y x + 1 3x + 2 2x + 2

( 2x

)

⎡1 4x x ⎤ y′ = y ⎢ + − ⎥ 2 x 1+ x 2 + x2 ⎦ ⎣

y′ 1 ⎡ 2 x 1 ⎤ = ⎢ − 2 y 2 ⎣ x + 5 x + 9 ⎥⎦ y ⎡ 2x 1 ⎤ y′ = ⎢ − 2 ⎣ x 2 + 5 x + 9 ⎥⎦ y′ =

2

2 + x2

(

x +5 1 ⎡ = ln x 2 + 5 − ln( x + 9) ⎤ ⎦⎥ x+9 2 ⎣⎢

8. ln y = ln

)

1 = ln x + 2 ln 1 + x 2 − ln 2 + x 2 2 y′ 1 2x 1 2x = + 2⋅ − ⋅ y x 1 + x2 2 2 + x2

1 − x2 ⎡ x 2 ⎤ + ⎢ 2 1 − 2 x ⎣ x − 1 1 − 2 x ⎥⎦ 2

(

x 1 + x2

2

⎤ y′ 1 ⎡ 3 x 2 6 = ⎢2 ⋅ − + 4⎥ y 3 ⎢⎣ x3 + 1 x ⎦⎥ 2 ⎤ y ⎡ 6x 6 − + 4⎥ y′ = ⎢ 3 3 ⎣⎢ x + 1 x ⎦⎥

⎡ 4x 2 3 ⎤ = − − ⎢ 2 ⎥ 2 ( x + 1) (3x + 2) ⎣ x + 1 x + 1 3 x + 2 ⎦

(

)

3 1 3 6 x +1 y= 3 x 6 e−4 x

445

2

⎡ 6 x2 ⎤ 6 − + 4⎥ ⎢ 3 ⎣⎢ x + 1 x ⎦⎥


2


17. y = (3 x + 1) 2 x . Thus

2

13. y = x x +1 , thus ln y = ln x x +1 = ( x 2 + 1) ln x. y′ 1 = ( x 2 + 1) ⋅ + (ln x)(2 x) y x 2 ⎛ x +1 ⎞ + 2 x ln x ⎟ y′ = y ⎜ ⎜ x ⎟ ⎝ ⎠ 2 ⎛ ⎞ 2 + x 1 = x x +1 ⎜ + 2 x ln x ⎟ ⎜ x ⎟ ⎝ ⎠ 14. y = (2 x)

x

ln y = ln ⎡( 3x + 1) ⎣⎢

⎡ 3x ⎤ = 2(3 x + 1) 2 x ⎢ + ln(3x + 1) ⎥ + 3 x 1 ⎣ ⎦

18. y = ( x 2 + 1) x +1 , thus ln y = ln( x 2 + 1) x +1 = ( x + 1) ln( x 2 + 1). y′ 2x = x + 1⋅ + ln( x 2 + 1) ⋅1 2 y x +1 ⎡ 2 x( x + 1) ⎤ + ln( x 2 + 1) ⎥ y′ = y ⎢ 2 ⎣ x +1 ⎦ + 2 x ( x 1) ⎡ ⎤ = ( x 2 + 1) x +1 ⎢ + ln( x 2 + 1) ⎥ 2 ⎣ x +1 ⎦

y′ 1 ⎡1⎤ = x ⎢ ⎥ + [ln 2 + ln x] ⋅ y 2 x ⎣x⎦ ⎡ 1 ln(2 x) ⎤ y′ = y ⎢ + ⎥ 2 x ⎦ ⎣ x ⎡ 2 + ln(2 x) ⎤ y′ = (2 x) x ⎢ ⎥ ⎣ 2 x ⎦

( )

1 ln x ln x = . x x

19. y = 4e x x3 x . Thus

(

= ln 4 + x + 3 x ln x. ⎡ ⎛1⎞ ⎤ y′ = 1 + 3 ⎢ x ⎜ ⎟ + (ln x)(1) ⎥ y x ⎣ ⎝ ⎠ ⎦ y′ = y (4 + 3ln x)

1

x x (1 − ln x)

y′ = 4e x x3 x (4 + 3ln x)

x2

x

20. y = (ln x)e . Thus ln y = e x ln(ln x).

x

⎛ 3 ⎞ 16. y = ⎜ ⎟ . Thus ⎝ x2 ⎠ ⎛ 3 ⎞ ln y = x ln ⎜ ⎟ = x[ln 3 − 2 ln x]. ⎝ x2 ⎠ y′ ⎛ 2⎞ = x ⎜ − ⎟ + (ln 3 − 2 ln x )(1) y ⎝ x⎠ ⎛ 3 ⎞ = −2 + ln ⎜ ⎟ ⎝ x2 ⎠ ⎡ ⎛ 3 y ′ = y ⎢ −2 + ln ⎜ ⎝ x2 ⎣

)

ln y = ln 4 + ln e x x3 x = ln 4 + ln e x + ln x3 x

1 y′ x x − (ln x)(1) = y x2 ⎡ 1 − ln x ⎤ y′ = y ⎢ ⎥ ⎣ x2 ⎦

y′ =

= 2 x ln(3 x + 1)

⎡ 3x ⎤ + ln(3 x + 1) ⎥ y′ = 2 y ⎢ + 3 x 1 ⎣ ⎦

ln y = ln(2 x) x = x [ln 2 + ln x ].

1

⎦⎥

⎧ ⎛ 3 ⎞ ⎫ y′ = 2 ⎨x ⎜ ⎟ + [ln(3x + 1)](1) ⎬ + y 3 x 1 ⎠ ⎩ ⎝ ⎭

. Thus

15. y = x x . Thus ln y =

2x ⎤

x

⎞⎤ ⎛ 3 ⎞ ⎡ ⎛ 3 ⎟ ⎥ = ⎜ 2 ⎟ ⎢ −2 + ln ⎜ 2 ⎠⎦ ⎝ x ⎠ ⎣ ⎝x

y′ ⎡ 1 ⎤ x = ex ⎢ ⎥ + [ln(ln x)]e y ⎣ x ln x ⎦ ⎡ 1 ⎤ + ln(ln x) ⎥ e x y′ = y ⎢ x x ln ⎣ ⎦ x ⎡ 1 ⎤ = (ln x)e ⎢ + ln(ln x ) ⎥ e x x x ln ⎣ ⎦

⎞⎤ ⎟⎥ ⎠⎦

446


Section 12.5

21. y = (4 x − 3) 2 x +1

25. y = e x ( x 2 + 1) x

ln y = ln(4 x − 3) 2 x +1 = (2 x + 1) ln(4 x − 3)

ln y = ln e x + ln( x 2 + 1) x

(

y′ ⎡ 4 ⎤ = (2 x + 1) ⎢ ⎥ + [ln(4 x − 3)](2) y ⎣ 4x − 3 ⎦

(

)

⎡ ⎛ 2 x ⎞⎤ ⎡ y′ 2 ⎤ = 1+ ⎢x ⎜ ⎟ ⎥ + ⎣⎢ln x + 1 (1) ⎥⎦ 2 y ⎣ ⎝ x + 1 ⎠⎦

⎡ 4(2 x + 1) ⎤ + 2 ln(4 x − 3) ⎥ y′ = y ⎢ − 4 x 3 ⎣ ⎦

⎡ ⎤ 2 x2 + ln x 2 + 1 ⎥ y′ = y ⎢1 + 2 ⎣⎢ x + 1 ⎦⎥ When x = 1, then y = 2e and y′ = 2e[1 + 1 + ln(2)] = 2e(2 + ln 2). Thus an equation of the tangent line is y − 2e = 2e(2 + ln 2)(x − 1), or y = (4e + 2e ln 2)x − 2e − 2e ln 2.

(

dy ⎡ 12 ⎤ When x = 1, then = 1 ⎢ + 2 ln(1) ⎥ = 12. dx ⎣1 ⎦

22. y = (ln x)ln x ln y = ln(ln x)ln x = (ln x) ln(ln x) y′ ⎡ 1 1⎤ ⎛1⎞ = (ln x ) ⎢ ⋅ ⎥ + [ln(ln x)] ⎜ ⎟ y ⎣ ln x x ⎦ ⎝x⎠

26.

⎡ 1 ln(ln x) ⎤ y′ = y ⎢ + x ⎥⎦ ⎣x ⎡ 1 + ln(ln x) ⎤ y′ = (ln x)ln x ⎢ ⎥ x ⎣ ⎦

)

y = xx ln y = x ln x y′ 1 = x ⋅ + (ln x )(1) = 1 + ln x y x

When x = 1,

dy 1 ⎡ 1 + ln(1) ⎤ −1 =1 ⎢ When x = e, ⎥=e . dx ⎣ e ⎦

y′ = 1 + ln1 = 1 + 0 = 1. y

27. y = (3x )−2 x ln y = –2x ln(3x) ⎧ ⎡1 ⎫ y′ ⎤ = −2 ⎨ x ⎢ (3) ⎥ + [ln(3 x)](1) ⎬ y ⎩ ⎣ 3x ⎦ ⎭

23. y = ( x + 1)( x + 2) 2 ( x + 3) 2 ln y = ln(x + 1) + 2 ln(x + 2) + 2 ln(x + 3) y′ 1 2 2 = + + y x +1 x + 2 x + 3

= –2[1 + ln(3x)] y′ ⋅100 gives the percentage rate of change. y Thus –2[1 + ln(3x)](100) = 60 1 + ln(3x) = –0.3 ln(3x) = –1.3

2 2 ⎤ ⎡ 1 y′ = y ⎢ + + ⎥ ⎣ x +1 x + 2 x + 3⎦ When x = 0, then y = 36 and y′ = 96. Thus an equation of the tangent line is y – 36 = 96(x – 0), or y = 96x + 36.

24.

)

= x + x ln x 2 + 1

3x = e −1.3 1 x= 3e1.3

y = xx ln y = x ln x y′ 1 = x ⋅ + (ln x )(1) = 1 + ln x y x y ′ = y (1 + ln x) = x x (1 + ln x) When x = 1, then y = 1 and y ′ = 11 (1 + ln1) = 1(1 + 0) = 1. An equation of the

28. y = [ f ( x)]g ( x ) ln y = g(x) ln[f(x)] ⎛ 1 ⎞ y′ = g ( x) ⎜ ⋅ f ′( x) ⎟ + ln[ f ( x)]g ′( x) y ⎝ f ( x) ⎠

tangent line is y − 1 = 1(x − 1) or y = x.

⎛ ⎞ g ( x) y′ = y ⎜ f ′( x) + g ′( x) ln[ f ( x)] ⎟ f ( x) ⎝ ⎠ ⎛ ⎞ g ( x) y′ = [ f ( x)]g ( x ) ⎜ f ′( x) + g ′( x) ln[ f ( x)] ⎟ f ( x) ⎝ ⎠ 447


29.


r′ p′ q′ ⋅100% = ⋅100% + ⋅100% r p q p′ = (1 + η ) 100% p

where η =

η=

p q dp dq

=

p dq ⋅ . q dp

p 500 − 40 p + p 2

⋅ (−40 + 2 p)

1 ⎛1 ⎞ % increase in price will result in a (1 − 1.2) ⎜ % ⎟ = −0.1% change in 2 ⎝2 ⎠ revenue, which is a 0.1% decrease in revenue.

When p = 15, then η = −1.2 and a

30.

r′ p′ q′ ⋅100% = ⋅100% + ⋅100% r p q p′ = (1 + η ) 100% p

where η =

p q dp dq

=

p dq ⋅ . q dp

p

η=

⋅ (−40 + 2 p) 500 − 40 p + p 2 When p = 15, then η = −1.2 and a 10% decrease in price will result in a (1 − 1.2)(−10%) = 2% change in revenue, which is a 2% increase in revenue.

Principles in Practice 12.6 3 1. Let f ( x) = 20 x − 0.01x 2 − 850 + 3ln x, then f ′( x) = 20 − 0.02 x + . f(10) ≈ –644 and f(50) ≈ 137, x so we use 50 to be the first approximation, x1 , to find the break-even quantity between 10 and 50. xn +1 = xn − = xn −

= =

f ( xn )

f ′ ( xn )

= xn −

20 xn − 0.01xn2 − 850 + 3ln xn 20 − 0.02 xn + 3 xn−1

20 xn2 − 0.01xn3 − 850 xn + 3xn ln xn 20 xn − 0.02 xn2 + 3

(

20 xn2 − 0.02 xn3 + 3xn − 20 xn2 − 0.01xn3 − 850 xn + 3 xn ln xn 20 xn − 0.02 xn2

+3

−0.01xn3 + 853 xn − 3xn ln xn 20 xn − 0.02 xn2 + 3

x2 = 50 −

f (50) ≈ 42.82602 f ′(50)

x3 = 42.82602 −

f (42.82602) ≈ 42.85459 f ′(42.82602)

448

)


x4 = 42.85459 −

Section 12.6

f (42.85459) ≈ 42.85459 f ′(42.85459)

Since the values of x3 and x4 differ by less than 0.0001, we take the first break-even quantity to be x ≈ 42.85459 or 43 televisions. f(1900) ≈ 1073 and f(2000) ≈ –827, so we use 2000 to be the first approximation, x1 , for the break-even quantity between 1900 and 2000. x2 = 2000 −

f (2000) ≈ 1958.63703 f ′(2000)

x3 = 1958.63703 −

f (1958.63703) ≈ 1957.74457 f ′(1958.63703)

x4 = 1957.74457 −

f (1957.74457) ≈ 1957.74415 f ′(1957.74457)

x5 = 1957.74415 −

f (1957.74415) ≈ 1957.74415 f ′(1957.74415)

Since the values of x4 and x5 differ by less than 0.0001, we take the second break-even quantity to be x ≈ 1957.74415 or 1958 televisions. Problems 12.6 1. We want a root of f ( x) = x3 − 4 x + 1 = 0. We see that f(0) = 1 and f(1) = –2 have opposite signs, so there must be a root between 0 and 1. Moreover, f(0) is closer to 0 than is f(1), so we select x1 = 0 as our initial estimate. Since f ′( x) = 3x 2 − 4, the recursion formula is xn +1 = xn −

f ( xn )

x3 − 4 xn + 1 = xn − n . f ′ ( xn ) 3 xn2 − 4

Simplifying gives xn +1 =

2 xn3 − 1 3 xn2 − 4

. Thus we obtain:

n

xn

xn +1

1

0.00000

0.25000

2

0.25000

0.25410

3

0.25410

0.25410

Because x4 − x3 < 0.0001, the root is approximately x4 = 0.25410. 3 ⎛1⎞ 2. Let f ( x) = x3 + 2 x 2 − 1. f ⎜ ⎟ = − and 2 8 ⎝ ⎠ 1 ⎛1⎞ f(1) = 2 (note the sign change). Since f ⎜ ⎟ is closer to 0 than is f(1), we select x1 = . We have 2 ⎝2⎠ x3 + 2 xn2 − 1 f ′( x) = 3x 2 + 4 x, so the recursion formula is xn +1 = xn − n 3 xn2 + 4 xn

449



n

xn

xn +1

xn +1

1

2.50000

2.58974

0.50000

0.63636

2

2.58974

2.58425

2

0.63636

0.61838

3

2.58425

2.58423

3

0.61838

0.61803

4

2.58423

2.58423

4

0.61803

0.61803

n

xn

1

Since x5 − x4 < 0.0001, the root is

Because x5 − x4 < 0.0001, the root is

approximately x5 = 2.58423.

approximately x5 = 0.61803. (Note that f ′(0) = 0, so we cannot use 0 for x1.)

5. Let f ( x) = x3 + x + 1. We have f(−1) = −1 and f(0) = 1 (note the sign change). Choose x1 = −1. Since f ′( x) = 3 x 2 + 1, the recursion formula is

3

3. Let f ( x) = x − x − 1. We have f(1) = –1 and f(2) = 5 (note the sign change). Since f(1) is closer to 0 than is f(2), we choose x1 = 1. We

xn +1 = xn −

have f ′( x) = 3 x 2 − 1, so the recursion formula is xn +1 = xn − =

f ( xn )

f ′ ( xn )

2 xn3 + 1 3xn2

= xn −

=

xn3 − xn − 1 3xn2 − 1

−1

n

xn

xn +1

1

1.00000

1.50000

2

1.50000

1.34783

3

1.34783

1.32520

4

1.32520

1.32472

5

1.32472

1.32472

3xn2 + 1

n

xn

xn +1

1

−1

−0.75000

2

−0.75000

−0.68605

3

−0.68605

−0.68234

4

−0.68234

−0.68233

approximately x5 = −0.68233. 6. x3 = 2 x + 5, so use f ( x) = x3 − 2 x − 5 = 0. We have f(2) = –1 and f(3) = 16, so f(2) is closer to 0 than is f(3). We choose x1 = 2. Since f ′( x) = 3x 2 − 2, the recursion formula is


xn +1 = xn −

4. Let f ( x) = x3 − 9 x + 6. We have f(2.5) = –0.875 and f(3) = 6. Since f(2.5) is closer to 0 than is f(3), we choose x1 = 2.5. We have f ′( x) = 3x − 9 , so xn +1 = xn −

2 xn3 − 1


Since x6 − x5 < 0.0001, the root is

2

xn3 − 9 xn + 6 3xn2 − 9

f ( xn ) x3 + xn + 1 = xn − n f ′( xn ) 3 xn2 + 1

.

xn3 − 2 xn − 5 3 xn2 − 2

=

2 xn3 + 5 3 xn2 − 2

n

xn

xn +1

1

2.00000

2.10000

2

2.10000

2.09457

3

2.09457

2.09455

Because x4 − x3 < 0.0001, the root is approximately x4 = 2.09455. 450


Section 12.6

7. x 4 = 3x − 1 , so use f ( x) = x 4 − 3x + 1 = 0 . Since f(0) = 1 and f(1) = –1 (note the sign change), f(0) and f(1) are equally close to 0. We shall choose

9. Let f ( x) = x 4 − 2 x3 + x 2 − 3. f(1) = –3 and f(2) = 1 (note the sign change), so f(2) is closer to 0 than is f(1). We choose x1 = 2. Since

x1 = 0. Since f ′( x) = 4 x3 − 3, the recursion formula is f ( xn ) x 4 − 3 xn + 1 = xn − n xn +1 = xn − f ′ ( xn ) 4 xn3 − 3 =

f ′( x) = 4 x3 − 6 x 2 + 2 x, the recursion formula is xn +1 = xn −

x 4 − 2 xn3 + xn2 − 3 f ( x) = xn − n f ′ ( xn ) 4 xn3 − 6 xn2 + 2 xn

3xn4 − 1

n

xn

xn +1

4 xn3 − 3

1

2.00000

1.91667

2

1.91667

1.90794

3

1.90794

1.90785

n

xn

xn +1

1

0.00000

0.33333

2

0.33333

0.33766

3

0.33766

0.33767

Because x4 − x3 < 0.0001, the root is approximately x4 = 1.90785. 10. Let f ( x) = x 4 − x3 + x − 2. f(1) = –1 and f(2) = 8, so f(1) is closer to 0 than is f(2). We

Because x4 − x3 < 0.0001, the root is approximately x4 = 0.33767.

choose x1 = 1. Since f ′( x) = 4 x3 − 3x 2 + 1, the recursion formula is

8. Let f ( x) = x 4 + 4 x − 1. Since f(–2) = 7 and f(–1) = –4, f(–1) is closer to 0 than is f(–2). However, f ′(−1) = 0, so we shall choose

xn +1 = xn −

x1 = −2. Since f ′( x) = 4 x3 + 4, the recursion formula is xn +1 = xn −

xn4 + 4 xn − 1 4 xn3 + 4

=

3xn4 + 1 4 xn3 + 4

xn4 − xn3 + xn − 2 4 xn3 − 3 xn2 + 1

n

xn

xn +1

1

1.00000

1.50000

2

1.50000

1.34677

n

3

1.34677

1.31040

xn

xn +1

4

1.31040

1.30858

1

−2.00000

−1.75000

5

1.30858

1.30857

2

−1.75000

−1.67092

3

−1.67092

−1.66332

4

−1.66332

−1.66325

Because x6 − x5 < 0.0001, the root is approximately x6 = 1.30857. 11. The desired number is x, where x3 = 71, or


x3 − 71 = 0. Thus we want to find a root of

approximately x5 = −1.66325.

f ( x) = x3 − 71 = 0. Since 43 = 64 , the solution should be close to 4, so we choose x1 = 4 as our

initial estimate. We have f ′( x) = 3x 2 , so the recursion formula is f ( xn ) x3 − 71 2 xn3 + 71 = xn − n = xn +1 = xn − f ′ ( xn ) 3xn2 3xn2

451



n

xn

xn +1

n

xn

xn +1

1

4

4.146

1

3

2.37

2

4.146

4.141

2

2.37

2.03

4

4.141

4.141

3

2.03

1.94

4

1.94

1.94

Thus to three decimal places,

3

71 = 4.141.

Thus the solutions are –4.99 and 1.94.

12. The desired number is x, where x 4 = 19, or

14. We must solve ln x = 5 – x. That is, we must determine all roots of f(x) = ln(x) + x – 5 = 0. A rough sketch shows that the graph of the logarithmic function y = ln x intersects the line y = 5 – x at one point, where x is between 3 and 1 4. We choose x1 = 3. Since f ′( x) = + 1, the x recursion formula is f ( xn ) ln( xn ) + xn − 5 = xn − xn +1 = xn − 1 +1 f ′ ( xn )

x 4 − 19 = 0. Thus we want to find a root of f ( x) = x 4 − 19. Since 24 = 16, the solution should be close to 2, so we choose x1 = 2 as our

initial estimate. We have f ′( x) = 4 x3 , so the recursion formula is xn +1 = xn − =

f ( xn ) x 4 − 19 = xn − n f ′( xn ) 4 xn3

3 xn4 + 19

xn

4 xn3

n

xn

xn +1

n

xn

xn +1

1

3

3.676

1

2

2.09

2

2.676

3.693

2

2.09

2.09

3

3.693

3.693

Thus to two decimal places,

4 19

Thus the solution is approximately 3.693.

= 2.09.

15. The break-even quantity is the value of q when total revenue and total cost are equal: r = c, or r – c = 0. Thus we must find a root of

13. We want real solutions to e x = x + 5. Thus we want to find roots of f ( x) = e x − x − 5 = 0. A

(

f (q) = q − 250 + 0.1q3 = 0, so f ′(q) = 1 + 0.3q 2 . The recursion formula is f ( qn ) q − 250 + 0.1qn3 = qn − n qn +1 = qn − f ′ ( qn ) 1 + 0.3qn 2

two roots. Since f ′( x) = e x − 1, the recursion formula is f ( xn ) e xn − xn − 5 = xn − xn +1 = xn − f ′ ( xn ) e xn − 1

We choose q1 = 13, as suggested.

If x1 = −5, we obtain n

xn

xn +1

1

−5

−4.99

2

−4.99

−4.99

)

3q − 250 + 2q − 0.1q3 = 0, or

rough sketch of the exponential function y = e x and the line y = x + 5 shows that there are two intersection points: one when x is near –5, and the other when x is near 3. Thus we must find

n

qn

qn +1

1

13

13.33

2

13.33

13.33

Thus q ≈ 13.33.

If x1 = 3, we obtain: 452


Section 12.6

The break-even quantity is the value of q when total cost = total revenue: c = r, c – r = 0. Thus we solve q2 1 40 + 3q + + = 7 q. Multiplying both 1000 q sides by q and simplifying, we see that the problem is equivalent to solving q3 f (q) = − 4q 2 + 40q + 1 = 0. 1000

= qn −

qn

qn +1

1

3

2.875

2

2.875

2.880

3

2.880

2.880

Thus q ≈ 2.880. 18. In the same manner as problem 17, we must find a root of f (q) = 0.2q3 + 1.5q − 8 = 0, so

3q 2 − 8q + 40, the recursion b. Since f ′(q) = 1000 formula is f ( qn ) qn +1 = qn − f ′ ( qn ) qn3 − 4qn 2 + 40qn 1000 3qn2 − 8qn + 40 1000

n

f ′(q ) = 0.6q 2 + 1.5. The recursion formula is qn +1 = qn −

f (qn ) 0.2qn3 + 1.5qn − 8 = qn − f ′(qn ) 0.6qn2 + 1.5

We select q1 = 5 as suggested.

+1

We select q1 = 10 as suggested.

n

qn

qn +1

1

5

3.54

2

3.54

2.85

n

qn

qn +1

3

2.85

2.71

1

10

10.05

4

2.71

2.70

2

10.05

10.05

5

2.70

2.70

Thus q = 2.70, so p = 10 − 2.70 = 7.30 (from the demand equation).

Thus q ≈ 10.05. 17. The equilibrium quantity is the value of q for which supply and demand are equal, that is, it is 100 , or of a root of 2q + 5 = q2 + 1 100 f ( q ) = 2q + 5 − = 0. Since q2 + 1 200q f ′(q) = 2 + , the recursion formula is 2 2 q +1

(

qn +1 = qn −

19. For a critical value of f ( x) =

we want a root of f ′( x) = x 2 − 2 x − 5 = 0. Since d [ f ′( x)] = 2 x − 2, the recursion formula is dx xn 2 − 2 xn − 5 . 2 xn − 2 For the given interval [3, 4], note that f ′(3) = −2 and f ′(4) = 3 have opposite signs. Thus there is a root x between 3 and 4. Since 3 is closer to 0, we shall select x1 = 3. xn +1 = xn −

)

f ( qn )

f ′ ( qn )

= qn −

100 qn 2 +1 200 qn

2qn + 5 − 2+

( q +1) n

2

x3 − x 2 − 5 x + 1, 3

2

A rough sketch shows that the graph of the supply equation intersects the graph of the demand equation when q is near 3. Thus we select q1 = 3.

n

xn

xn +1

1

3.0

3.5

2

3.5

3.45

3

3.45

3.45

Thus x ≈ 3.45. 453



Principles in Practice 12.7 1.

6.

dh = 0 − 16(2t ) = −32t ft/sec dt

d 2F dq

d 2h

d = [−32t ] = −32 feet/sec2 2 dt dt The acceleration of the rock at time t is –32 feet/sec2 or 32 feet/sec2 downward.

dq

7.

2 (q + 1)3

1 = x −1 x

y′ = − x −2 y ′′ = 2 x −3 y ′′′ = −6 x −4 = −

2. y′ = 5 x 4 − 8 x3 + 14 x

9.

y ′′ = 20 x3 − 24 x 2 + 14 2

y ′′′ = 60 x − 48 x

f (q) =

1 2q

4

=

6 x4

1 −4 q 2

f ′(q) = −2q −5 f ′′(q ) = 10q −6

dy 3. = −1 dx

f ′′′(q ) = −60q −7 = −

=0

10.

dy = −1 − 2 x dx dx 2

(q + 1) 2

f ( x) = x 2 ln x

8. y =

1. y′ = 12 x 2 − 24 x + 6 y ′′ = 24 x − 24 y ′′′ = 24

d2y

=

1

⎛2⎞ f ′′( x) = x ⎜ ⎟ + (1 + 2 ln x)(1) = 3 + 2 ln x ⎝x⎠

Problems 12.7

4.

3

=−

⎛1⎞ f ′( x) = x 2 ⎜ ⎟ + (ln x)(2 x) = x(1 + 2 ln x) ⎝ x⎠

c′(q) = 14q + 11 c′′ = 14 When x = 3, the rate of change of the marginal cost function is 14 dollars/unit2.

dx 2

2

d 3F

2. The rate of change of the marginal cost function with respect to x is c′′(q) .

d2y

dF 1 = dq q + 1

60 q7

1

f ( x) = x = x 2 1 − 12 x 2 1 −3 1 f ′′( x) = − x 2 = − 3 4 4x 2 f ′( x) =

= −2

5. y′ = 3x 2 + e x

11.

y ′′ = 6 x + e x

1

f (r ) = 9 − r = (9 − r ) 2 1 −1 f ′(r ) = − (9 − r ) 2 2 1 1 −3 f ′′(r ) = − (9 − r ) 2 = − 3 4 4(9 − r ) 2

y ′′′ = 6 + e x y (4) = e x

454


12. y = e−4 x

(2 x + 5)(5 x − 2) x +1 = ln(2 x + 5) + ln(5 x − 2) − ln( x + 1)

2

18. y = ln

y′ = −8 xe−4 x

2

2 ⎡ y ′′ = −8 ⎢ x ⎜⎛ −8 xe−4 x ⎣ ⎝

= 8e−4 x

13. y =

2

Section 12.7

(8x − 1)

⎞ −4 x 2 (1) ⎤ ⎟+e ⎥⎦ ⎠

2 5 1 + − 2 x + 5 5x − 2 x + 1 4 25 1 y ′′ = − − + 2 2 (2 x + 5) (5 x − 2) ( x + 1) 2

y′ =

2

1 = (2 x + 3)−1 2x + 3

19.

dx 2

= 8(2 x + 3)

−3

z

=

8

20. y =

4

y′ = =−

( x − 1)

dx

= −2( x − 1)−2

y ′′ = 4( x − 1) −3 = 1 2

16. y = 2 x + (2 x)

( )

2

=

e x (−1) − (1 − x)e x

(e ) x

2

=

x−2 ex

21. y = e2 x + e3 x dy = 2e2 x + 3e3 x dx

4 ( x − 1)3

d2y dx 2

1 2

1 −1 −1 −1 y′ = x + (2 x) 2 (2) = x 2 + (2 x) 2 2 ⎡ 1 −3 1 1 1 −3 y ′′ = − x 2 − (2 x) 2 (2) = − ⎢ 3 + 3 ⎢ 2 2 2 (2 x) 2 ⎣ 2x

ex

d2y

( x − 1) 2 2

x

( )

x +1 x −1 ( x − 1)(1) − ( x + 1)(1) 2

z

x x dy e (1) − x e 1− x = = 2 dx ex ex

y ′′ = 180(3x + 7)3

15. y =

z

2

z

(2 x + 3)3

14. y = (3 x + 7)5 y′ = 15(3 x + 7)

( ) ( ) f ′′( z ) = ( ze ) (1) + ( z + 2) ⎡ ze + e (1) ⎤ ⎣ ⎦ = e ( z + 4z + 2)

f ′( z ) = z 2 e z + e z (2 z ) = ze z ( z + 2)

dy = −2(2 x + 3)−2 dx d2y

f ( z) = z 2e z

d3y

− 12

dx3 d4y

⎤ ⎥ ⎥ ⎦

dx

4

d5 y dx

17. y = ln[ x( x + 6)] = ln( x) + ln( x + 6)

5

= 4e2 x + 9e3 x = 8e2 x + 27e3 x = 16e2 x + 81e3 x = 32e2 x + 243e3 x

d5 y

1 1 + = x −1 + ( x + 6)−1 x x+6 ⎡ 1 1 ⎤ y ′′ = − x −2 + (−1)( x + 6) −2 = − ⎢ + ⎥ 2 ( x + 6)2 ⎥⎦ ⎢⎣ x

y′ =

dx

455

5

= 32e0 + 243e0 = 32 + 243 = 275 x =0


22. y = e

(

) = eln( x +1)

2 ln x3 +1

3

(

2

(

)

= x3 + 1

)


26. 9 x 2 + 16 y 2 = 25 18 x + 32 yy ′ = 0

2

y′ = 6 x 2 x 3 + 1 = 6 x 5 + 6 x 2

y′ = −

y ′′ = 30 x 4 + 12 x When x = 1, then y ′′ = 30 + 12 = 42.

27.

( ) = − 4y

16 y 2 =−

2

+ x2

16 y3

−1

y3

y ′′ = =

x y y (1) − x( y′) y

y 2 − x2 y

3

2

=

−16 y

3

=

y−x

=−

y

1 ⎡ 12 ⎛ 1 − 12 ⎞ ⎛ −1 ⎞ ⎤ x ⎜ 2 y y′ ⎟ − y 2 ⎜ 12 x 2 ⎟ ⎥ ⎢ 1 ⎠ ⎝ ⎠⎥ y ′′ = − ⎢ ⎝ 4⎢ x ⎥ ⎢ ⎥ ⎣ ⎦ 1 1 ⎤ ⎡ 1⎛ 1 2 ⎞ ⎡ ⎢ x2 ⎜ − y ⎟ − y2 ⎥ 1 − y2 1 1 ⎢ 1 − ⎢ ⎜ ⎥ ⎟ 1 4x 2 ⎠ x 2 ⎥ 1 y2 1 ⎢ 4 x2 =− ⎢ ⎝ =− ⎢ ⎥ 8⎢ 8⎢ x x ⎢ ⎥ ⎢ ⎢ ⎥ ⎣⎢ ⎢⎣ ⎥⎦ 1 ⎡ 2 ⎤ ⎢ 1 + y1 ⎥ 1 ⎡ 12 4 2 ⎤ 1⎢ x2 ⎥ 1 ⎢ x + 4y ⎥ = ⎢ = ⎥ 8 ⎢ x ⎥ 8 ⎢ 4 x 32 ⎥ ⎣ ⎦ ⎢ ⎥ ⎢⎣ ⎥⎦ 1⎡ 4 ⎤ 1 = ⎢ 3⎥= 3 8 ⎢ 4 x 2 ⎥ 8x 2 ⎣ ⎦

() x y

2

16 y3

25. y 2 = 4 x 2 yy′ = 4 y′ =

2 = 2 y −1 y

(

)

y ′′ = −2 y −2 y′ = −2 y −2 2 y −1 = −

1

1 x 2 1 y2 y′ = − ⋅ =− ⋅ 1 1 2 2 y− 2 4 x2

1

24. x 2 − y 2 = 16 2 x − 2 yy′ = 0 y′ =

1

1 − 12 −1 x + 2 y 2 y′ = 0 2 1 −1 −1 2 y 2 y′ = − x 2 2

16 y 2

16 y 3

9 16 y 2 + 9 x 2 225 ⋅ =− 3 16 16 y 256 y 3

x + 4y 2 = 4

4 y (1) − x(4 y′)

16

)

x +4 y = 4 1 2

4 y − 4 x − 4xy

=− =−

=−

x 4y

y ′′ = −

(

9x 9 y (1) − xy ′ 9 y − x − 16 y y ′′ = − ⋅ =− ⋅ 16 16 y2 y2

23. x 2 + 4 y 2 − 16 = 0 2 x + 8 yy′ = 0 8 yy′ = −2 x y′ = −

9x 16 y

4 y3

456

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦⎥


Section 12.7 31. y = e x + y

28. y 2 − 6 xy = 4 2 yy′ − 6[ xy′ + y (1)] = 0 2 yy′ − 6 xy′ = 6 y (2 y − 6 x) y′ = 6 y

y′ = e x + y (1 + y′) y ′ − e x + y y′ = e x + y

(

y ′′ = 3 ⋅

( y − 3 x) y′ − y ( y′ − 3) ( y − 3x) 2

= 9⋅

= 9⋅

y − 6 xy ( y − 3 x)

3

= 9⋅

4 ( y − 3x)

3

=

y − xy′

y ′′ =

36 3

=

29. xy + y – x = 4 xy′ + y (1) + y′ − 1 = 0 xy′ + y′ = 1 − y ( x + 1) y′ = 1 − y

=

(1 + x)2

=

(1 − y )

y 1− y

(1 − y )

2

=

2

=

y′ (1 − y ) 2

y (1 − y )3

e x − e y y′ = 2 x + 2 yy′ y′ =

−2(1 − y ) (1 + x) 2

ex − 2x ey + 2y

(e y ′′ =

y

) (

)(

x

x

y

2( y − 1) (1 + x) 2

y

30. x 2 + 2 xy + y 2 = 1 2 x + 2 y + 2 xy ′ + 2 yy ′ = 0 ( x + y ) y ′ = −( x + y ) y ′ = −1 y ′′ = 0

2

y

y

=

)(

+ 2 y e x − 2 − e x − 2 x e y y ′ + 2 y′

(e + 2 y ) ( e + 2 y )( e − 2) − ( e − 2 x )( e + 2) y′ = (e + 2 y ) ( e + 2 y ) ( e − 2) − ( e − 2x ) ( e + 2) = (e + 2 y )

(1− y ) (1 + x) ⎡⎢ − (1+ x ) ⎤⎥ − (1 − y ) ⎣ ⎦

(1 + x)2 −(1 − y ) − (1 − y )

(1 − y ) y′ − y (− y′)

32. e x − e y = x 2 + y 2

1− y y′ = 1+ x (1 + x)(− y′) − (1 − y )(1) y ′′ = (1 + x) 2 =

x+ y

1 − e x+ y y y′ = 1− y

( y − 3x) 2

( y − 3x)

e

y′ =

3y y − x ⎡⎢ y −3 x ⎤⎥ ⎣ ⎦ = 9 ⋅ y ( y − 3 x) − 3 xy = 9⋅ 2 ( y − 3 x) ( y − 3 x )3 2

)

y′ 1 − e x + y = e x + y

6y 3y y′ = = 2 y − 6 x y − 3x

2

x

y

2

x

y

2

33. x 2 + 3x + y 2 = 4 y 2 x + 3 + 2 yy ′ = 4 y ′ 2 yy ′ − 4 y ′ = −2 x − 3 2x + 3 2x + 3 y′ = − = 2y − 4 4 − 2y

457

2

y

)


y ′′ =

= =

(4 − 2 y )(2) − (2 x + 3)(−2 y ′) (4 − 2 y ) 2 2(4 − 2 y ) + 2(2 x + 3)

( ) 2 x +3 4− 2 y

(4 − 2 y )2 2(4 − 2 y ) 2 + 2(2 x + 3)2 (4 − 2 y )3

When x = 0 and y = 0, then 34.


d2y dx

2

=

2(4)2 + 2(3) 2 3

4

=

25 . 32

f ( x) = (3 x − 5)e−2 x f ′( x) = (3x − 5) ⎡ −2e−2 x ⎤ + e−2 x [3] . Thus, ⎣ ⎦ f ′( x) = e−2 x [−2(3x − 5) + 3] = (13 − 6 x)e−2 x f ′′( x) = (13 − 6 x) ⎡ −2e−2 x ⎤ + e−2 x [−6] ⎣ ⎦ = 2e−2 x [−(13 − 6 x) − 3] = 4(3 x − 8)e−2 x f ′′( x) + 4 f ′( x) + 4 f ( x) = 4(3 x − 8)e−2 x + 4 ⎡ (13 − 6 x)e−2 x ⎤ + 4 ⎡ (3x − 5)e−2 x ⎤ = [4(3 x − 8) + 4(13 − 6 x) + 4(3 x − 5)]e−2 x ⎣ ⎦ ⎣ ⎦ = [0]e−2 x = 0, as was to be shown.

35.

f ( x) = (5 x − 3)4 f ′( x) = 20(5 x − 3)3 f ′′( x) = 300(5 x − 3)2

36.

1 2

f ( x) = 6 x + f ′( x) = 3x

− 12

x

− 12

6 −3

x 2 − 12 −5

3 −3 x 2 f ′′( x) = − x 2 + 2 8 −7

9 − 5 5x 2 f ′′′( x) = x 2 − 4 16

458


37.

Chapter 12 Review

dc = 0.6q + 2 dq d 2c dq 2

= 0.6

d 2c dq 2

42.

= 0.6 q =100

2

39.

2

2

= 3e x + 2 xe x + e2 xe

d r dq 2

= −104.

f ( x) = x 4 − 6 x 2 + 5 x − 6 f ′( x) = 4 x − 12 x + 5 f ′′( x) = 12 x 2 − 12 = 12( x + 1)( x − 1) Clearly f ′′( x) = 0 when x = ±1.

a.

y

=

b.

y ′′ =

=

41.

e y − 2ey

−2

y

( y − 2)2

( ) =− y y −2 2

( y − 2) x

2y ( y − 2)3

=

=

2

3r + 7r + 1

+ 4 x +5

2

+ 4 x +5

(2 x + 4) = 2( x + 2)e x

f (t ) = log 6 t 2 + 1 =

)

(

3

8. y = 35 x = e(ln 3)5 x

(

2

+ 4 x +5

)

1 log 6 t 2 + 1 2

)

(

)

)

3

3 3 y ′ = e(ln 3)5 x (ln 3)(15 x 2 ) = 15 x 2 ln 3 ⎛⎜ 35 x ⎞⎟ ⎝ ⎠

−2 y′ ( y − 2)2

9. y = ( x − 6)( x + 5)(9 − x)

2y

ln y = ln ( x − 6)( x + 5)(9 − x)

(2 − y )3

1 = [ln( x − 6) + ln( x + 5) + ln(9 − x)] 2 y′ 1 ⎡ 1 1 −1 ⎤ = + + y 2 ⎢⎣ x − 6 x + 5 9 − x ⎥⎦

f ′( x) = 6e − 3x − 30 x

(

6r + 7 2

7. y′ = e x (2 x) + x 2 + 2 e x = e x x 2 + 2 x + 2

2

f ′′( x) = 6 e x − x − 5

3r + 7 r + 1

(6r + 7) =

(

1 y = = 2 − y 2 1− y

( y − 2)( y′) − y ( y′)

1 2

(

2 x

⎛ y⎞ y2 ⎜ e 2 ⎟ y e ⎝y ⎠ y′ = = y x ⎛ y⎞ e − 2 ye ey − 2y ⎜ e2 ⎟ ⎝y ⎠ e

f ′(r ) =

2 1 ln t + 1 = ⋅ . Thus 2 ln 6 ⎞ 1⎛ 1 1 t . f ′(t ) = ⎜ ⋅ ⋅ [2t ] ⎟ = 2 ⎝ ln 6 t 2 + 1 ⎠ (ln 6) t 2 + 1

2 x

y

)

3.

6.

x

x

−1

f ′( w) = we w + e w + 2 w = we w + e w + 2 w

y′ = e x

( e ) + e (2 yy′) ( e − 2 ye ) y′ = y e e y′ = y

x

−1

2.

5. y = e x

40. e y = y 2 e x 2

(

2

2

4. y = eln x = x. Thus y′ = 1.

3

y

x 4 x3 5 x 2 + + +x 4 3 2

1. y ′ = 3e x + 0 + e x (2 x) + (e2 ) xe

= −80 − 6q

When q = 4,

f ′( x) =


dr = 400 − 80q − 3q 2 dq dq 2

x5 x 4 5 x3 x 2 + + + 20 12 6 2

f ′′( x) = x3 + x 2 + 5 x + 1 f ′′( x) = 0 when x ≈ −0.21.

38. r = pq = 400q − 40q 2 − q3

d 2r

f ( x) =

)

f ′′( x) = 0 when x ≈ –4.99 or 1.94. 459


y′ = =

10.

=

ex

e1/ t t

2

18. y′ = =

e2 x xe

2x

(

=

1 − x ln x xe

)

x 2 e x − e− x − e x + e− x (2 x) x4

x 2 e x − x 2 e− x − 2 xe x − 2 xe− x

x3

f (q) = ln ⎡(q + 1) 2 (q + 2)3 ⎤ ⎣ ⎦ = 2 ln(q + 1) + 3ln(q + 2)

21.

14. y = ( x + 2)3 ( x + 1)4 ( x − 2)2 ln y = 3 ln(x + 2) + 4 ln(x + 1) + 2 ln(x − 2) y′ 3 4 2 = + + y x + 2 x +1 x − 2 4 2 ⎤ ⎡ 3 + + y′ = y ⎢ ⎥ ⎣ x + 2 x +1 x − 2 ⎦ 4 2 ⎤ ⎡ 3 = ( x + 2)3 ( x + 1)4 ( x − 2)2 ⎢ + + x + x + x − 2 1 2 ⎥⎦ ⎣

ln 2 x

e x ( x ln x − 1) x ln 2 x

(

)

1

⎡1 + 2l + 3l 2 ⎤ ⎦ 1 + l + l2 + l3 ⎣

1 + 2l + 3l 2 1 + l + l 2 + l3

22. y = ( x 2 ) x

2

ln y = x 2 ln x 2 = 2 x 2 ln x y′ ⎛1⎞ = 2 x 2 ⎜ ⎟ + (ln x) ( 4 x ) y ⎝x⎠ y′ = 2 xy (1 + 2 ln x) 2

y′ = 2 x( x 2 ) x (1 + 2 ln x)

+ 2 x −5)(ln 2)

= (4 x + 2)(ln 2)2

( 1x ) = e x ( ln x − 1x )

f (l ) = ln 1 + l + l 2 + l 3

=

y′ = e

(ln x) 2

f ′(l ) =

2 3 f ′(q) = + q +1 q + 2

(2 x 2 + 2 x −5)(ln 2)

(ln x)e x − e x

⎞ ⎟ = ln 5 − 2 ln x ⎠ 1 2 y′ = 0 − 2 ⋅ = − x x

e x ( x − 2) − e− x ( x + 2)

2

4e2 x +1 x

⎛ 5 20. y = ln ⎜ ⎝ x2

x4

15. y = e(2 x

=

19. y = log 2 (8 x + 5) 2 = 2 log 2 (8 x + 5) ln(8 x + 5) = 2⋅ ln 2 1 8 16 y′ = 2 ⋅ ⋅ = ln 2 8 x + 5 (8 x + 5) ln 2

x

) (

4e3 x

xe x −1 x ⎡ e2 x +1 (2) ⎤ − e2 x +1[1] 4e2 x +1 (2 x − 1) ⎦ = y′ = 4 ⋅ ⎣ x2 x2

( 1x ) − (ln x) ( e x )

e x − xe x ln x

12. y′ = =

17. y =

( x − 6)( x + 5)(9 − x) ⎡ 1 1 1 ⎤ + + ⎢ ⎥ 2 ⎣ x −6 x +5 x −9⎦

11. y′ =

13.

y⎡ 1 1 1 ⎤ + − 2 ⎣⎢ x − 6 x + 5 9 − x ⎥⎦

f ′(t ) = e1/ t (−1 ⋅ t −2 ) = −

=


23. y = ( x + 1) x +1 ln y = ( x + 1) ln( x + 1)

(4 x + 2)(ln 2)

2 x 2 + 2 x −5

y′ 1 = ( x + 1) + ln( x + 1)[1] = 1 + ln( x + 1) y x +1

16. y is a constant, so y′ = 0.

y′ = y[1 + ln( x + 1)] = ( x + 1) x +1[1 + ln( x + 1)]

460


(1 − e ) e − (1 + e )( −e ) = 2e y′ = (1 − e ) (1 − e ) x

24.

Chapter 12 Review

x

x

x

x

x

2

x

2

1 25. φ (t ) = ln ⎛⎜ t 4 − t 2 ⎞⎟ = ln t + ln(4 − t 2 ) 2 ⎝ ⎠ 1 1 1 1 t φ ′(t ) = + ⋅ ⋅ (−2t ) = − 2 t 2 4−t t 4 − t2

26. y = ( x + 3)ln x ln y = [ln x] ln(x + 3) 1 1 y′ = (ln x) + ln( x + 3) ⋅ y x+3 x ⎡ ln x ln( x + 3) ⎤ y′ = y ⎢ + ⎥ x ⎣x+3 ⎦ ln( x + 3) ⎤ ln x ⎡ ln x = ( x + 3) ⎢ + ⎥ x ⎣x+3 ⎦

27.

y= ln y = y′ = y y′ = =

28. y′ =

( x 2 + 1)1/ 2 ( x 2 + 2)1/ 3 (2 x3 + 6 x) 2 / 5 1 1 2 ln( x 2 + 1) + ln( x 2 + 2) − ln(2 x3 + 6 x) 2 3 5 1⎛ 1 ⎞ 1⎛ 1 ⎞ 2⎛ 1 ⎞ 2 ⎜ ⎟ (2 x) + ⎜ 2 ⎟ (2 x) − ⎜ 3 ⎟ (6 x + 6) 2 ⎝ x2 + 1 ⎠ 3⎝ x + 2 ⎠ 5 ⎝ 2x + 6x ⎠ ⎡ x 2x 6( x 2 + 1) ⎤ y⎢ + − ⎥ 2 2 3 ⎢⎣ x + 1 3( x + 2) 5( x + 3 x) ⎥⎦ 2x 6( x 2 + 1) ⎤ ( x 2 + 1)1/ 2 ( x 2 + 2)1/ 3 ⎡ x + − ⎢ ⎥ 2 2 3 (2 x3 + 6 x) 2 / 5 ⎣⎢ x + 1 3( x + 2) 5( x + 3x) ⎦⎥ x

( 1x ) − ln x ( 12 ) x−

1 2

=

x

( )

29. y = x x

x

= xx

2 − ln x 3

2x 2

2

2

ln y = ln x x = x 2 ln x y′ ⎛1⎞ = x 2 ⎜ ⎟ + (ln x)(2 x) = x + 2 x ln x y ⎝x⎠

( )

y′ = y ( x + 2 x ln x) = x x

x

( x + 2 x ln x)

461


30. y = x


(x ) x

ln y = ln x

( x ) = x x ln x x

( )

y′ d x ⎛1⎞ = x x ⎜ ⎟ + (ln x ) x y x dx ⎝ ⎠

Note: If v = x x , then ln v = ln x x = x ln x; v′ ⎛1⎞ = x ⎜ ⎟ + (ln x)(1) = 1 + ln x v ⎝x⎠ d x v′ = x = v(1 + ln x) = x x (1 + ln x) dx y′ ⎛1⎞ Thus = x x ⎜ ⎟ + (ln x) ⎡ x x (1 + ln x) ⎤ ⎣ ⎦ y ⎝x⎠

( )

⎡1 ⎤ = x x ⎢ + (1 + ln x) ln x ⎥ ⎣x ⎦ ⎡1 ⎤ y′ = yx x ⎢ + (1 + ln x) ln x ⎥ ⎣x ⎦ =x

( x ) x x ⎡ 1 + (1 + ln x) ln x ⎤ x

⎢x ⎣

⎥ ⎦

31. y = ( x + 1) ln x 2 = 2( x + 1) ln x ⎡ ⎤ ⎛1⎞ ⎡ x +1 ⎤ y′ = 2 ⎢( x + 1) ⎜ ⎟ + (ln x)(1) ⎥ = 2 ⎢ + ln x ⎥ ⎝x⎠ ⎣ x ⎦ ⎣ ⎦ ⎡2 ⎤ When x = 1, then y′ = 2 ⎢ + ln1⎥ = 4. ⎣1 ⎦

32. y =

ex

2

+1

x2 + 1

2 1 1 ln y = ln ⎛⎜ e x +1 ⎞⎟ − ln( x 2 + 1) = x 2 + 1 − ln( x 2 + 1) 2 ⎝ ⎠ 2 ⎡ y′ 1 1 1 ⎤ = 2x − ⋅ (2 x) = x ⎢ 2 − ⎥ 2 y 2 x2 + 1 x + 1⎦ ⎣

⎡ 1 ⎤ y ′ = yx ⎢ 2 − ⎥ 2 x + 1⎦ ⎣ y′ =

ex

2

+1

⎡ 1 ⎤ x ⎢2 − ⎥ 2 x + 1⎦ x +1 ⎣ 2

When x = 1, then y ′ =

e1+1

1 ⎤ 3e2 2 ⎡ (1) ⎢ 2 − ⎥= 4 . 1 + 1 ⎣ 1 + 1⎦

462


33. y = e

e + x ln

Chapter 12 Review

( 1x ) = ee− x ln x

⎛ ⎡ ⎛1⎞ ⎤⎞ y′ = ee− x ln x ⎜⎜ − ⎢ x ⎜ ⎟ + (ln x)(1) ⎥ ⎟⎟ ⎦⎠ ⎝ ⎣ ⎝x⎠ = −(1 + ln x)ee− x ln x

When x = e, then y′ = −(1 + ln e)ee −e ln e = −(2)e0 = −2. ⎡ 25 x ( x 2 − 3x + 5)1/ 3 ⎤ 34. y = ⎢ ⎥ 2 3 ⎣⎢ ( x − 3 x + 7) ⎦⎥

−1

1 ⎡ ⎤ ln y = −1 ⎢5 x ln 2 + ln( x 2 − 3x + 5) − 3ln( x 2 − 3 x + 7) ⎥ 3 ⎣ ⎦ ⎡ y′ 1 2x − 3 2x − 3 ⎤ = − ⎢5ln 2 + ⋅ − 3⋅ ⎥ 2 2 y 3 x − 3x + 5 x − 3x + 7 ⎦ ⎣ ⎡ 2x − 3 3(2 x − 3) ⎤ y′ = − y ⎢5ln 2 + − ⎥ 3( x 2 − 3 x + 5) x 2 − 3 x + 7 ⎦⎥ ⎣⎢ ⎡ 25 x ( x 2 − 3x + 5)1/ 3 ⎤ y′ = (−1) ⎢ ⎥ 2 3 ⎣⎢ ( x − 3 x + 7) ⎦⎥

When x = 0, then y′ = −

−1

⎡ 2x − 3 3(2 x − 3) ⎤ ⎢5ln 2 + ⎥ 2 3( x − 3x + 5) x 2 − 3x + 7 ⎦⎥ ⎣⎢

343 ⎡ 1 9⎤ 1862 5ln 2 − + ⎥ = −343(ln 2)52 / 3 − . 1/ 3 ⎢ 5 7 ⎦ 5 ⎣ 54 / 3

35. y = 3e x y′ = 3e x

If x = ln 2, then y = 3eln 2 = 6 and y′ = 3eln 2 = 6. An equation of the tangent line is y – 6 = 6(x – ln 2), y = 6x + 6 – 6 ln 2, y = 6x + 6(1 – ln 2). Alternatively, since 6 ln 2 = ln 26 = ln 64, the tangent line can be written as y = 6x + 6 – ln 64.

36. y = x + x 2 ln x ⎡ ⎛1⎞ ⎤ y′ = 1 + ⎢ x 2 ⎜ ⎟ + (ln x)(2 x) ⎥ = 1 + x + 2 x ln x x ⎣ ⎝ ⎠ ⎦ When x = 1, then y = 1 + 1(0) = 1 and y′ = 1 + 1 + 2(0) = 2 . Thus an equation of the tangent line is y – 1 = 2(x – 1), or y = 2x – 1. 2 37. y = x ⎛⎜ 22− x ⎞⎟ . To find y′ we shall use logarithmic differentiation. ⎝ ⎠ ⎡ ⎛ 2− x 2 ⎞ ⎤ 2 ln y = ln ⎢ x ⎜ 2 ⎟ ⎥ = ln x + 2 − x ln 2 ⎠⎦ ⎣ ⎝ y′ 1 = + (−2 x) ln 2 y x

(

)

⎡1 ⎤ y′ = y ⎢ − 2(ln 2) x ⎥ ⎣x ⎦ 463



When x = 1, then y = 2 and y′ = 2(1 − 2 ln 2). The equation of the tangent line is y – 2 = 2(1 – 2 ln 2)(x – 1). The y-intercept of the tangent line corresponds to the point where x = 0: y – 2 = 2(1 – 2 ln 2)(–1) = –2 + 4 ln 2 Thus y = 4 ln 2 and the y-intercept is (0, 4 ln 2). 38. w = 2 x +1 + ln(1 + x 2 ) = e(ln 2)( x +1) + ln(1 + x 2 ) 2

x = log 2 (t 2 + 1) − e(t −1) = dw dw dx = ⋅ dt dx dt ⎛ 2x = ⎜ 2 x +1 (ln 2) + 1 + x2 ⎝

ln(t 2 + 1) (t −1)2 −e ln 2

⎞ 2 ⎞⎛ 2t − e(t −1) [2(t − 1)] ⎟ ⎟ ⎜⎜ 2 ⎟ ⎠ ⎝ (ln 2)(t + 1) ⎠ 2

When t = 1, x = log 2 (1 + 1) − e(1−1) = 1 − 1 = 0, w = 21 + ln1 = 2 + 0 = 2, and 39. y = e x

2

y′ = e x

− 2 x +1

2

− 2 x +1

[2 x − 2] = (2 x − 2)e x

y ′′ = 2( x − 1)e x = 2e x

2

2

− 2 x +1

− 2 x +1

2

− 2 x +1

(2 x − 2) + 2e x

2

− 2 x +1

(2( x − 1)2 + 1)

At (1, 1), y ′′ = 2e0 (2(0) + 1) = 2. 40. y = x 2 e x

(

y′ = x 2 e x + e x (2 x) = e x x 2 + 2 x

( y ′′′ = e (2 x + 4) + ( x

)

)

(

y ′′ = e x (2 x + 2) + x 2 + 2 x e x = e x x 2 + 4 x + 2 x

2

)

(

)

+ 4 x + 2 e x = e x x2 + 6 x + 6

)

At (1, e), y ′′′ = e(1 + 6 + 6) = 13e 41. y = ln(2x) 1 y′ = (2) = x −1 2x y ′′ = −1 ⋅ x −2 = − x −2 y ′′′ = −(−2) x −3 =

At (1, ln 2), y ′′′ =

2 x3 2 13

=2

464

dw ⎛ 2 ⎞ = (21 ln 2 + 0) ⎜ − 1(0) ⎟ = 2. dt ⎝ 2 ln 2 ⎠


Chapter 12 Review

42. y = x ln x 1 y′ = x ⋅ + (ln x )(1) = 1 + ln x x 1 1 y ′′ = 0 + = x x 1 At (1, 0), y ′′ = = 1 1

47. x + xy + y = 5 1 + xy′ + y (1) + y′ = 0 ( x + 1) y′ = −1 − y 1+ y x +1 ( x + 1) y′ − (1 + y ) y ′′ = − ( x + 1) 2 1+1 2 At (2, 1), y′ = − = − and 2 +1 3 y′ = −

43. 2 xy + y 2 = 10 2 ( xy′ + y ) + 2 yy′ = 0 2 xy′ + 2 yy′ = −2 y ( x + y ) y′ = − y y′ = −

y ′′ = −

y x+ y

y′ =

−3 x 2 y 3 3 x3 y 2

y′ = − =−

y ′′ = −

y x

( )

ln x + 2 ln y = xy 1 2 + y′ = xy′ + y x y

y ′′ = −

2

)

(e

2

xy − y 2 x − x2 y

y ′[ y (2 + y ln x)] = − y2 x(2 + y ln x)

4 27

9

)

− e x y′ = ( y + 1)e x

( y + 1)e x

2

⎡ ⎛ 1 ⎞⎤ y 2 ⎢e y ln x ⎜ y ⋅ + (ln x) y′ ⎟ ⎥ + e y ln x [ 2 yy′] = 0 ⎝ x ⎠⎦ ⎣

y′ = −

y

y′ =

y 2 (ln x) y′ + 2 yy′ = −

( ) ( )=

3 − 23 − 2 − 13

e y y′ − e x y′ = ( y + 1)e x

2x − x 2 y y′ = xy 2 − y

=e

( x + 2 y ) y′ − y (1 + 2 y′)

e y y′ = ( y + 1)e x + e x ( y′)

2

2xy′ − x yy′ = xy − y

46. y e

y x + 2y

49. e y = ( y + 1)e x

y + 2 xy′ = x 2 yy′ + xy 2

2 y ln x

4 9

( x + 2 y )2 2 At (–1, 2), y′ = − and 3

45. ln xy 2 = xy

y′ =

9

=

48. xy + y 2 = 2 x( y′) + y (1) + 2 yy′ = 0

44. 3x 2 y3 + 3 x3 y 2 y ′ = 0 y ′(3x3 y 2 ) = −3x 2 y3

(

( )

3 − 23 − 2

=

y3 x

1 1−

y ′′ =

3

y x

465

e y − ex

1 y +1

=

=

( )= −( ) e

( y + 1) ey

ey y +1

ey y +1

ey y

−

ey y +1

y +1 y

y ( y′) − ( y + 1)( y′) y

2

=

− y′ y

2

=−

y +1 y 2

y

=−

y +1 y3



50. x1/ 2 + y1/ 2 = 1

54.

1 −1/ 2 1 −1/ 2 dy x + y ⋅ =0 2 2 dx y dy =− dx x

( x ) ⎛⎜⎝ 2 1 y ⋅ dydx ⎞⎟⎠ − ( y ) ( 2 1 x ) =− 2 dx 2 ( x) =

51.

+

y 2 x

x

=

x+ y 2x x

=

x5 x 4 2 x3 + + + x2 + 1 10 6 3

f ′( x) =

x 4 2 x3 + + 2 x2 + 2 x 2 3

f ′′( x) = 2 x3 + 2 x 2 + 4 x + 2 f ′′( x) = 0 when x ≈ −0.57.

d2y

1 2

f ( x) =

55. p =

1

η=

2x x

f ′(t ) = ⎡ −0.8e−0.01t (−0.01) − 0.2e−0.0002t (−0.0002) ⎤ ⎣ ⎦

=

500 / q q − 500 q2

= −1

56. p = 900 − q 2

52. log N = A – bM d d (log N ) = ( A − bM ) dM dM d ⎛ ln N ⎞ d ( A − bM ) = ⎜ ⎟ dM ⎝ ln10 ⎠ dM

η=

p q dp dq

=

900 − q 2 q

−2q

=−

900 − q 2 2q 2

When q = 10, then η = −4. Since η > 1, demand is elastic.

1 1 dN ⋅ = −b ln10 N dM 1 dN (log e) = −b N dM dN bN − = dM log e

57. p = 18 – 0.02q

η=

=

18−0.02 q q

−0.02

=−

18 − 0.02q 0.02q

58. p = 20 − 2 q

η=

p q dp dq

a.

When p = 8, then η =

b.

η=

⎛b⎞ ⎛b⎞ = log ⎜ ⎟ + ( A − bM ) = A + log ⎜ ⎟ − bM q ⎝ ⎠ ⎝q⎠ 4

p q dp dq

When q = 600, then η = −0.5. Because η < 1, demand is inelastic.

⎛ b ⎞ ⎛ dN ⎞ log ⎜ − = log ⎜ ⋅N⎟ ⎟ ⎝ dM ⎠ ⎝ log e ⎠ ⎛ b ⎞ = log ⎜ ⎟ + log N ⎝ log e ⎠

f ′( x) = (12 x3 + 6 x 2 − 25)e3 x f ′( x) = 0 when x ≈ 1.13.

p q dp dq

Since η = 1, demand has unit elasticity when q = 200.

= 0.008e−0.01t + 0.00004e−0.0002t

53.

500 q

3

+ 2 x − 25 x

=

p q − 1 q

=

−p q

=

−p 10 −

p 2

=

2p p − 20

2(8) 4 =− . 8 − 20 3

2p p − 20

20 , then η < −1, so η > 1 and 3 demand is elastic.

If p >

466


59. η =

p q dp dq

=

Chapter 12 Review

40 1 =− 80 − 200 3 % change in q ≈ (% change in price) (η )

η

b.

p dq ⋅ q dp

η=

−p 2500 − p

2

=

−p , so q

61. We want a root of f ( x) = x3 − 2 x − 2 = 0. We have f(1) = –3 and f(2) = 2 (note the sign change). Since f(2) is closer to 0 than is f(1), we

p ⎛ −p ⎞ p2 ⎜ ⎟ = − 2 . Now, if p = 30, then q⎝ q ⎠ q

q = 2500 − 302 = 40, so

η

p =30

=−

(30) 2 (40)

2

=−

choose x1 = 2. We have f ′( x) = 3x 2 − 2, so the recursion formula is f ( xn ) x3 − 2 xn − 2 = xn − n xn +1 = xn − f ′ ( xn ) 3xn2 − 2

9 16

2 %, then demand 3 would change by approximately 3 ⎛ 2 ⎞⎛ 9 ⎞ ⎜ − 3 ⎟ ⎜ − 16 ⎟ %, or 8 %. (That is, demand ⎝ ⎠⎝ ⎠

If the price of 30 decreases

=

3 increases by approximately %.) 8

60. a.

η=

p q dp dq

=

5 ⎛ 1⎞ = 5 ⎜ − ⎟ % = − % = −1.67%. Thus 3 3 ⎝ ⎠ demand decreases by approximately 1.67%.

q = 2500 − p 2 dq = dp

p = 40

p dq = ⋅ q dp

2 xn3 + 2 3xn2 − 2

n

xn

xn +1

1

2.00000

1.80000

2

1.80000

1.76995

3

1.76995

1.76929

4

1.76929

1.76929

q = 100 − p , where 0 < p < 100.


dq −1 = . Thus dp 2 100 − p


η=

p

62. We want real solutions of e x = 3 x. Thus we

−1

⋅ 100 − p 2 100 − p

want to find roots of f ( x) = e x − 3 x = 0. A rough sketch of the exponential function y = e x and the line y = 3x shows that there are two intersection points: one when x is near 0.5, and the other when x is near 1.5. Thus we must find

−p p = = 2(100 − p) 2 p − 200 p < −1. 2 p − 200 Noting that the denominator is negative for 0 < p < 100, we multiply both sides of the inequality by 2p – 200 and reverse the direction of the inequality 200 p > −2 p + 200, 3 p > 200, p > 3 200 < p < 100 for elastic demand. Thus 3

For elastic demand we want

two roots. Since f ′( x) = e x − 3, the recursion formula is xn +1 = xn −

f ( xn )

f ′ ( xn )

= xn −

If x1 = 0.5, we obtain

467

n

xn

xn +1

1

0.5

0.610

2

0.610

0.619

3

0.619

0.619

e xn − 3 xn e xn − 3



If x1 = 1.5, we obtain n

xn

xn +1

1

1.5

1.512

2

1.512

1.512

Thus the solutions are 0.619 and 1.512. Mathematical Snapshot Chapter 12 1. F = 25, D = 3400, V = 36.5, R = 0.05. 2 FD 2(25)(3400) q= = ≈ 305.2 RV (0.05)(36.5) The economic order quantity is 305 units. 2. If the number of units maintained as a safety margin is denoted by m, then the amount in stock at any time is increased by m units. The average inventory level is thus increased by m q units, to m + units. The carrying cost is then 2 FD q⎞ ⎛ C (q) = + RV ⎜ m + ⎟ q 2⎠ ⎝ FD RVq = + + RVm q 2 d ( RVm) = 0, the maintenance of a dq safetly margin does not affect the calculation of the economic order quantity.

Since


468

Chapter 13 maximum, since C ′(t ) changes from + to –. The drug is at its greatest concentration 2 hours after the injection.

Principles in Practice 13.1 1. The graph of c(q ) = 2q3 − 21q 2 + 60q + 500 is shown.

Problems 13.1

C(q)

1. Decreasing on (–∞, –1) and (3, ∞); increasing on (–1, 3); relative minimum (–1, –1); relative maximum (3, 4). 2. Decreasing on (–∞, –1) and (0, 1); increasing on (–1, 0) and (1, ∞); relative minima (–1, –1) and (1, –1); relative maximum (0, 0).

q 2 5

There looks to be a relative maximum at q = 2 and a relative minimum at q = 5. c ′(q) = 6q 2 − 42q + 60 = 6(q 2 − 7q + 10) = 6(q – 5)(q – 2) c ′(q) = 0 when q = 2 or q = 5. If q < 2, then c ′(q) = 6(−)(−) = + , so c(q) is increasing. If 2 < q < 5, then c′(q) = 6(−)(+ ) = − , so c(q) is decreasing. If 5 < q, then c′(q) = 6(+)(+) = + , so c(q) is increasing. When q = 2, there is a relative maximum, since c′(q) changes from + to –. The relative maximum value is

3. Decreasing on (–∞, –2) and (0, 2); increasing on (–2, 0) and (2, ∞); relative minima (–2, 1) and (2, 1); no relative maximum. 4. Decreasing on (–∞, 0) and (0, ∞); never increasing; no relative maximum; no relative minimum. In the following problems, we denote the critical value by CV. 5.

2(2)3 − 21(2) 2 + 60(2) + 500 = 552 . When q = 5, there is a relative minimum, since c′(q) changes from – to +. The relative minimum value is

CV: x = −3, 1, 2 −

C ′(t ) =

0.14t (t + 2) 2

−

+

−3

2(5)3 − 21(5)2 + 60(5) + 500 = 525.

2. First, find C ′(t ) , with C (t ) =

f ′( x) = ( x + 3)( x − 1)( x − 2) f ′( x) = 0 when x = −3, 1, 2

1

+ 2

Increasing on (−3, 1) and (2, ∞); decreasing on (−∞, −3) and (1, 2); relative maximum when x = 1; relative minima when x = −3, 2.

. 6.

0.14(t + 2)2 − 0.14t (2)(t + 2)

(t + 2)4 0.14(t + 2) − 0.28t 0.28 − 0.14t = = (t + 2)3 (t + 2)3 0.14(2 − t ) = (t + 2)3 C ′(t ) = 0 when t = 2 and is undefined when t = –2. However, since t denotes the number of hours after an injection, negative values of t are + not reasonable. If 0 ≤ t < 2, C ′(t ) = = + , so + − C(t) is increasing. If 2 < t, C ′(t ) = = − , so C(t) + is decreasing. When t = 2, there is a relative

f ′( x) = 2 x( x − 1)3 CV: x = 0, 1 +

–

0

+

1

Increasing on (–∞, 0) and (1, ∞); decreasing on (0, 1); relative maximum when x = 0; relative minimum when x = 1. 7.

f ′( x) = ( x + 1)( x − 3)2 CV: x = –1, 3 –

–1

+

+

3

Decreasing on (–∞, –1); increasing on (–1, 3) and (3, ∞); relative minimum when x = –1.

469

Chapter 13: Curve Sketching

8.

f ′( x ) =


x( x + 2)

1 ⎛ 1 ⎞ on ⎜ − , 2 ⎟ ; relative maximum when x = − ; 3 3 ⎝ ⎠ relative minimum when x = 2.

2

x +1 CV: x = 0, –2 +

–

–2

+

0

13. y = −

Increasing on (–∞, –2) and (0, ∞); decreasing on (–2, 0); relative maximum when x = –2; relative minimum when x = 0.

(

y′ = − x 2 − 4 x + 5 = − x 2 + 4 x − 5

–

y′ = 6 x2 CV: x = 0

–5

+ 0

14. y =

10. y = x 2 + 4 x + 3 y′ = 2 x + 4 = 2( x + 2) CV: x = –2

–

+

(

)

y′ = 4 x3 − 4 x = 4 x x 2 − 1 = 4 x( x + 1)( x − 1)

CV: x = 0, ±1 –

–

+

–1

1 2

–

0

+

1

Decreasing on (–∞, –1) and (0, 1); increasing on (–1, 0) and (1, ∞); relative maximum when x = 0; relative minima when x = ±1.

1⎞ ⎛ ⎛1 ⎞ Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ; 2⎠ ⎝ ⎝2 ⎠ 1 relative maximum when x = . 2

16. y = −3 + 12 x − x3

(

)

y′ = 12 − 3 x 2 = 3 4 − x 2 = 3(2 + x)(2 − x)

5 2 x − 2x + 6 2

CV: x = ±2 –

y′ = 3x 2 − 5 x − 2 = (3x + 1)( x − 2)

–2

+

–

2

Decreasing on (–∞, –2) and (2, ∞); increasing on (–2, 2); relative minimum when x = –2; relative maximum when x = 2.

1 CV: x = − , 2 3 −

0

15. y = x 4 − 2 x 2

1 CV: x = 2

12. y = x3 −

+

Decreasing on (−∞, −3); increasing on (−3, 0) and (0, ∞); relative minimum at x = −3.

11. y = x − x 2 + 2 y′ = 1 − 2 x

3

x4 + x3 4

–3

–2

−1

1

CV: x = −3, 0

Decreasing on (–∞, –2); increasing on (–2, ∞); relative minimum when x = –2.

+

–

y ′ = x3 + 3x 2 = x 2 ( x + 3)

+

+

+


Increasing on (−∞, 0); increasing on (0, ∞); no relative maximum or minimum

–

)

= −( x + 5)( x − 1) CV: x = –5, 1

9. y = 2 x3 + 1

+

x3 − 2 x2 + 5x − 2 3

+

17. y = x3 −

2

1⎞ ⎛ Increasing on ⎜ −∞, − ⎟ and (2, ∞); decreasing 3⎠ ⎝

7 2 x + 2x − 5 2

y′ = 3x 2 − 7 x + 2 = (3 x − 1)( x − 2) 1 CV: x = , 2 3 470

ISM: Introductory Mathematical Analysis −

+ 1 3

Section 13.1

21. y =

2

y ′ = x 2 − 10 x + 22

1⎞ ⎛ Increasing on ⎜ −∞, ⎟ and (2, ∞); decreasing 3⎠ ⎝ 1 ⎛1 ⎞ on ⎜ , 2 ⎟ ; relative maximum when x = , 3 3 ⎝ ⎠ relative minimum when x = 2.

18. y = x3 − 6 x 2 + 12 x − 6

(

By the quadratic formula, y ′ = 0 when x=

)

(5 −

Increasing on (–∞, 2) and (2, ∞); no relative maximum or relative minimum. 11 2 x − 10 x + 2 2

22. y =

+

)(

–

– 5

(

)

(

)(

–

2 ,± 5 3

+ 2 3

)( x − 5 )( x + 5 )

)

–

+ 5

2 3

⎛ 2 2⎞ Increasing on −∞, − 5 , ⎜⎜ − , ⎟ , and 3 ⎟⎠ ⎝ 3 ⎛ 2⎞ 5, ∞ ; decreasing on ⎜⎜ − 5, − ⎟ and 3 ⎟⎠ ⎝ ⎛ 2 ⎞ , 5 ⎟⎟ ; relative maxima when x = − 5 , ⎜⎜ 3 ⎝ ⎠

(

(

2

y′ = −15 x 2 + 2 x + 1 = −(5 x + 1)(3 x − 1) 1 1 CV: − , 5 3 5

9 5 47 3 x − x + 10 x 5 3

CV: x = ±

20. y = −5 x + x + x − 7

–1

)

3, 5 + 3 ; increasing on 5 + 3, ∞ ;

(

5 2

+

)

= 3x − 2 3x + 2

+

3

(

y′ = 9 x 4 − 47 x 2 + 10 = 9 x 2 − 2 x 2 − 5

2⎞ ⎛ ⎛5 ⎞ Increasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 3⎠ ⎝ ⎝2 ⎠ 2 5 ⎛ ⎞ decreasing on ⎜ − , ⎟ ; relative maximum ⎝ 3 2⎠ 2 5 when x = − ; relative minimum when x = . 3 2

–

5+ 3

minimum at x = 5 + 3.

2 5 CV: x = − , 3 2 3

+

relative maximum at x = 5 − 3; relative

y′ = 6 x 2 − 11x − 10 = (2 x − 5)(3 x + 2)

–2

or x = 5 ± 3.

Increasing on −∞, 5 − 3 ; decreasing on

2

–

–

5– 3

+

19. y = 2 x3 −

2(1)

+

CV: x = 2 +

10 ± (−10) 2 − 4(1)(22)

CV: x = 5 ± 3

y′ = 3x 2 − 12 x + 12 = 3 x 2 − 4 x + 4 = 3( x − 2) 2

+

x3 − 5 x 2 + 22 x + 1 3

+

)

)

2 2 ; relative minima when x = − , 5. 3 3

– 1 3

23. y = 3x5 − 5 x3

1⎞ ⎛ ⎛1 ⎞ Decreasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 5 3 ⎝ ⎠ ⎝ ⎠ ⎛ 1 1⎞ increasing on ⎜ − , ⎟ ; relative minimum when ⎝ 5 3⎠ 1 1 x = − ; relative maximum when x = . 5 3

y′ = 15 x 4 − 15 x 2 = 15 x 2 ( x + 1)( x − 1) CV: x = 0, ±1 +

–

–1

–

0

+

1

Increasing on (–∞, –1) and (1, ∞); decreasing on (–1, 0) and (0, 1); relative maximum when x = –1; relative minimum when x = 1.

471


24. y = 3x −


x6 2

28. y =

y′ = 3 − 3x5 = 3(1 − x5 ) 4

3

(

2

1

+

0

−

(

)

+

0

1

30. y = 3 x ( x − 2)

27. y = 8 x 4 − x8

( = 8 x3 ( 2 + x 2 )( 2 − x 2 ) = 8 x3 ( 2 + x 2 ) ( 2 − x )(

y′ = 32 x3 − 8 x7 = 8 x3 4 − x 4

)

2+x

y′ =

( decreasing on ( −

)

–

– 0

)

(

)

(

)

+

1 2

1 2

⎛ 1⎞ Decreasing on (–∞, 0) and ⎜ 0, ⎟ ; increasing ⎝ 2⎠ 1 ⎛1 ⎞ on ⎜ , ∞ ⎟ ; relative minimum when x = ; no 2 ⎝2 ⎠ relative maximum.

)

Increasing on −∞, – 2 and 0, 2 ; 2, 0 and

2

CV: x = 0,

– 2

2(2 x − 1) 3x 3

CV: x = 0, ± 2 0

)

Increasing on (−1, 0) and (1, ∞); decreasing on (−∞, −1) and (0, 1); relative maximum when x = 0; relative minima when x = ±1.

2

Decreasing on (−∞, 0) and (0, 2); increasing on (2, ∞); relative minimum at x = 2.

– 2

−

+

−1

+

3

y′ = 8 x( x 2 − 1)3 = 8 x( x + 1)3 ( x − 1)3 CV: 0, –1, 1

+

–

1 2

+

29. y = ( x 2 − 1) 4

3x4 − 4 x3 + 17 26. y = 2 y ′ = 6 x3 − 12 x 2 = 6 x 2 ( x − 2) CV: x = 0, 2

+

–

(


0

)

⎛ 1 1⎞ Increasing on −∞, − 3 , ⎜ − , ⎟ , 3, ∞ ; ⎝ 2 2⎠ 1 ⎛ ⎞ ⎛1 ⎞ decreasing on ⎜ − 3, − ⎟ and ⎜ , 3 ⎟ ; 2⎠ ⎝ ⎝2 ⎠ 1 relative maxima when x = − 3, ; relative 2 1 minima when x = − , 3 . 2

–

–

+

2

y′ = −5 x 4 − 20 x3 = −5 x3 ( x + 4) CV: x = 0, –4

–

–

– 3 –1

25. y = − x5 − 5 x 4 + 200

–4

)(

)

1 CV: x = ± , ± 3 2

–

+

)(

= (2 x − 1)(2 x + 1) x + 3 x − 3

Increasing on (–∞, 1); decreasing on (1, ∞); relative maximum when x = 1.

–

(

y′ = 4 x 4 − 13x 2 + 3 = 4 x 2 − 1 x 2 − 3

= 3(1 − x)( x + x + x + x + 1) CV: x = 1 +

4 5 13 3 x − x + 3x + 4 5 3

2, ∞ ; relative

maxima when x = ± 2 , relative minimum when x = 0.

472


Section 13.1 sign chart because it is a point of discontinuity of y.

5 = 5( x − 1)−1 x −1

31. y =

y′ = −5( x − 1)

−2

=−

–

5

1

y′ = 8 x −

3

x2 CV: None, but x = 0 must be included in the sign chart because it is a point of discontinuity of y.

−

x

= 10 x

y′ = −5 x

− 32

− 12

. [Note: x > 0] 5

y′ =

3x 2x + 5 3(2 x + 5) − (3 x)(2) (2 x + 5)2

(2 x + 5) 2

5 must be included in the 2 sign chart because it is a point of discontinuity of y.

5⎞ ⎛ ⎛ 5 ⎞ Increasing on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; no 2⎠ ⎝ ⎝ 2 ⎠ relative extremum

x + 4x + 3 2

–

38. y = y′ =

=

=

( x + 1)( x + 3)

–

+

–1

2x2 4 x 2 − 25 (4 x 2 − 25)(4 x) − (2 x 2 )(8 x)

(4 x 2 − 25)2 100 x 100 x =− =− 2 2 (4 x − 25) (2 x − 5)2 (2 x + 5)2

x2 2− x 2

( x + 2)

–3 –2

2

(2 − x)(2 x) − x 2 (−1)

)

2

Increasing on (–∞, –3) and (–1, ∞); decreasing on (–3, –2) and (–2, –1); relative maximum when x = –3; relative minimum when x = –1.

–5

y′ =

(

( x + 2)(2 x ) − x 2 − 3 (1)

2

+

+

35. y =

+

( x + 2) ( x + 2)2 CV: x = –3, –1, but x = –2 must be included in the sign chart because it is a point of discontinuity of y.

15

CV: None but x = −

+

(2 x − 1)(4 x 2 + 2 x + 1)

1 2

y′ = =

=

=

x2 − 3 x+2

37. y =

< 0 for x > 0. x3 Decreasing on (0, ∞); no relative extremum.

34. y =

=−

1

⎛1 ⎞ Increasing on ⎜ , ∞ ⎟ ; decreasing on (−∞, 0) 2 ⎝ ⎠ 1 ⎛ 1⎞ and ⎜ 0, ⎟ ; relative minimum when x = . 2 ⎝ 2⎠

0

Decreasing on (–∞, 0) and (0, ∞); no relative extremum. 10

−

0

–

33. y =

1 x

x2 x2 1 CV: x = , but x = 0 must be included in the 2 sign chart because it is a point of discontinuity of y.

3 = 3 x −1 x

–

4

36. y = 4 x 2 +

Decreasing on (–∞, 1) and (1, ∞); no relative extremum.

y′ = −3x −2 = −

–

Decreasing on (–∞, 0) and (4, ∞); increasing on (0, 2) and (2, 4); relative minimum when x = 0; relative maximum when x = 4.

–

32. y =

+ 2

0

( x − 1)2 CV: None, but x = 1 must be included in the sign chart because it is a point of discontinuity of y. –

+

x(4 − x)

(2 − x) (2 − x)2 CV: x = 0, 4, but x = 2 must be included in the

CV: x = 0, but x = ± 473

5 must be included in the 2



(

sign chart because they are points of discontinuity of y. +

–

+ –5 2

–

5⎞ ⎛ ⎛ 5 Increasing on ⎜ −∞, − ⎟ and ⎜ − , 2⎠ ⎝ ⎝ 2

⎞ 0⎟; ⎠

y′ =

2 2 ( x − 1)−1/ 3 = 3 3 3 x −1 CV: x = 1 y′ = −

5x + 2

1

( x2 + 1) (5) − (5x + 2)(2 x) = −5x2 − 4 x + 5 2 2 ( x2 + 1) ( x2 + 1)

42. y = x 2 ( x + 3) 4 y′ = x 2 (4)( x + 3)3 + ( x + 3) 4 (2 x)

y′ = 0 when −5 x 2 − 4 x + 5 = 0 ; by the quadratic

= 2 x( x + 3)3 [2 x + ( x + 3)]

−2 ± 29 5

formula, x =

–

= 2 x( x + 3)3 (3x + 3) = 6 x( x + 3)3 ( x + 1) CV: x = 0, –3, –1 –

−2 ± 29 5

+ 5

⎛ −2 − 29 ⎞ Decreasing on ⎜⎜ −∞, ⎟⎟ and 5 ⎝ ⎠ ⎛ −2 + 29 ⎞ , ∞ ⎟⎟ ; increasing on ⎜⎜ 5 ⎝ ⎠ ⎛ −2 − 29 −2 + 29 ⎞ , ⎜⎜ ⎟⎟ ; relative minimum 5 5 ⎝ ⎠

) ( 3x 2 − 9 ) = − 23

= x 2 ( x − 6)3 (7 x − 18)

CV: x = 0, 6, +

+

–

–3 – 3 0

–

+ 3

( x + 3 )( x − 3 ) 2

[ x( x + 3)( x − 3)] 3

(

)(

Increasing on (–∞, –3), −3, − 3 ,

– 18 7

18 7

+ 6

⎛ 18 ⎞ Increasing on (–∞, 0), ⎜ 0, ⎟ , and (6, ∞); ⎝ 7⎠ ⎛ 18 ⎞ decreasing on ⎜ , 6 ⎟ ; relative maximum when ⎝7 ⎠ 18 ; relative minimum when x = 6. x= 7

+ 3

+ 0

CV: x = ± 3, 0, ± 3 +

( )

= x 2 ( x − 6)3 [4 x + 3( x − 6)]

3

40. y = x3 − 9 x

(

0

y′ = x3 ⎡ 4( x − 6)3 ⎤ + ( x − 6) 4 3 x 2 ⎣ ⎦

−2 − 29 ; relative maximum when 5

1 3 x − 9x 3

+

43. y = x3 ( x − 6) 4

−2 + 29 x= . 5

y′ =

–

–3 –1

–2 – 29 –2 + 29

when x =

+

Increasing on (–3, –1) and (0, ∞); decreasing on (–∞, –3) and (–1, 0); relative maximum when x = –1; relative minima when x = –3 and x = 0.

–

5

+

Increasing on (1, ∞); decreasing on (−∞, 1); relative minimum when x = 1.

x2 + 1

CV: x =

)

41. y = ( x − 1) 2 / 3

⎛ 5⎞ ⎛5 ⎞ decreasing on ⎜ 0, ⎟ and ⎜ , ∞ ⎟ ; relative ⎝ 2⎠ ⎝2 ⎠ maximum at x = 0.

39. y =

(

relative maximum when x = − 3 ; relative minimum when x = 3 .

5 2

0

)

(3, ∞); decreasing on − 3, 0 and 0, 3 ;

)

3, 3 , and 474


Section 13.1

⎛ 3 2⎞ Decreasing on ⎜⎜ 0, ⎟ ; increasing on 2 ⎟⎠ ⎝ ⎛3 2 ⎞ 3 2 . , ∞⎟⎟ ; relative minimum when x = ⎜⎜ 2 ⎝ 2 ⎠

2

44. y = x(1 − x) 5 2 −3 ⎤ ⎡ 2 y′ = x ⎢ − (1 − x) 5 ⎥ + (1 − x) 5 (1) ⎣ 5 ⎦ − 35 ⎡ 2 −3 ⎛ 7 ⎤ ⎞ = (1 − x) ⎢ − x + (1 − x) ⎥ = −(1 − x) 5 ⎜ x − 1⎟ ⎣ 5 ⎦ ⎝5 ⎠ 5 CV: x = , 1 7

+

–

48. y = x −1e x ⎛1 1 ⎞ ⎛ x −1 ⎞ y ′ = x −1e x − x −2 e x = e x ⎜ − ⎟ = e x ⎜ ⎟ 2 x x ⎠ ⎝ ⎝ x2 ⎠ CV: x = 1, but x = 0 must also be included in the sign chart because it is a point of discontinuity of y.

+ 1

5 7

5⎞ ⎛ Increasing on ⎜ −∞, ⎟ and (1, ∞); decreasing 7⎠ ⎝ 5 5 ⎛ ⎞ on ⎜ , 1⎟ ; relative maximum when x = ; 7 ⎝7 ⎠ relative minimum when x = 1. −πx

+π

y ′ = −πe

−πx

45. y = e

−

49. y = e x + e− x y′ = e x − e − x

< 0 for all x. Thus decreasing on

Setting y′ = 0 gives e x − e− x = 0 , e x = e− x , x = –x, x = 0 CV: x = 0 –

Decreasing on (–∞, 0); increasing on (0, ∞); relative minimum when x = 0.

1 x=e = e 1 CV: x = e

50. y = e− x

2

1 e

+

⎛ 1⎞ Decreasing on ⎜ 0, ⎟ ; increasing on ⎝ e⎠ 1 relative minimum when x = . e

⎛1 ⎞ ⎜ , ∞⎟ ; ⎝e ⎠

2

/2

–

0

Increasing on (−∞, 0); decreasing on (0, ∞); relative maximum at x = 0 51. y = x ln x – x. [Note: x > 0.] ⎡ 1 ⎤ y′ = ⎢ x ⋅ + (ln x)(1) ⎥ − 1 = ln x x ⎣ ⎦ CV: x = 1

47. y = x 2 − 9 ln x . [Note: x > 0.] y′ = 2 x −

/2

y ′ = − xe− x CV: x = 0

+

–

+

0

−1

CV: x =

1

0

46. y = x ln x. (Note: x > 0.) y′ = 1 + ln x y′ = 0 when 1 + ln x = 0, ln x = –1, or

0

+

Increasing on (1, ∞); decreasing on (−∞, 0) and (0, 1); relative minimum when x = 1.

(−∞, ∞); no relative extremum.

–

−

9 2 x2 − 9 = x x

–

0

3 2 2

+

1

Decreasing on (0, 1); increasing on (1, ∞); relative minimum when x = 1; no relative maximum.

+

0 3 2 5

475



( ) y′ = ( x 2 + 1)( −e− x ) + e− x (2 x) = −e− x ⎡⎢( x 2 + 1) − 2 x ⎤⎥ = −e− x ( x − 1) 2 ⎣ ⎦

52. y = x 2 + 1 e− x

4

x 5

CV: x = 1 –

y

–12

–

1

Decreasing on (–∞, 1) and (1, ∞); never increasing; no relative extremum.

55. y = 3x − x3 = x

(

)(

3−x

)

)

Symmetric about origin.

Intercepts (−2, 0), (5, 0), (0, −10) y′ = 2 x − 3

y′ = 3 − 3 x 2 = 3(1 + x )(1 − x) CV: x = ±1 Decreasing on (–∞, –1) and (1, ∞); increasing on (–1, 1); relative minimum when x = –1; relative maximum when x = 1.

3 2

3⎞ ⎛ Decreasing on ⎜ −∞, ⎟ ; increasing on 2⎠ ⎝ 3 ⎛3 ⎞ ⎜ , ∞ ⎟ ; relative minimum when x = . 2 ⎝2 ⎠ y

3+x

Intercepts: (0, 0), ± 3, 0

53. y = x 2 − 3x − 10 = ( x + 2)( x − 5)

CV: x =

(

5

y

x

5

5

x 6

(

)

56. y = x 4 − 16 = x 2 + 4 ( x + 2)( x − 2) Intercepts (±2, 0), (0, –16) Symmetric about y-axis. 54. y = 2 x 2 − 5 x − 12 = (2 x + 3)( x − 4)

y′ = 4 x 3 CV: x = 0 Decreasing on (–∞, 0); increasing on (0, ∞); relative minimum when x = 0.

⎛ 3 ⎞ Intercepts ⎜ − , 0 ⎟ , (4, 0), (0, –12) ⎝ 2 ⎠ 5⎞ ⎛ y′ = 4 x − 5 = 4 ⎜ x − ⎟ 4⎠ ⎝ 5 CV: x = 4 5⎞ ⎛ ⎛5 ⎞ Decreasing on ⎜ −∞, ⎟ ; increasing on ⎜ , ∞ ⎟ ; 4 4 ⎝ ⎠ ⎝ ⎠ 5 relative minimum when x = . 4

4

y

x –2

2

–16

476

5


(

57. y = 2 x3 − 9 x 2 + 12 x = x 2 x 2 − 9 x + 12

Section 13.1

)

y

8

Note that 2 x 2 − 9 x + 12 = 0 has no real roots. The only intercept is (0, 0).

(

y′ = 6 x 2 − 18 x + 12 = 6 x 2 − 3 x + 2

)

x

= 6(x – 2)(x – 1) CV: x = 1, 2 Increasing on (–∞, 1) and (2, ∞); decreasing on (1, 2); relative maximum when x = 1; relative minimum when x = 2. 8

–2 –1

6 6⎞ ⎛ 60. y = x 6 − x5 = x5 ⎜ x − ⎟ 5 5⎠ ⎝ ⎛6 ⎞ Intercepts (0, 0), ⎜ , 0 ⎟ ⎝5 ⎠

y

5 4

y′ = 6 x5 − 6 x 4 = 6 x 4 ( x − 1) CV: x = 0, 1 Increasing on (1, ∞); decreasing on (−∞, 0) and (0, 1); relative minimum when x = 1.

x 1 2

2

5

y

2

58. y = 2 x3 − x 2 − 4 x + 4 The x-intercept is not convenient to find. y-intercept is (0, 4). y ′ = 6 x 2 − 2 x − 4 = 2(3 x + 2)( x − 1)

x 3

2 CV: x = − , 1 3 2⎞ ⎛ Increasing on ⎜ −∞, − ⎟ and (1, ∞); decreasing 3⎠ ⎝

61. y = ( x − 1) 2 ( x + 2)2 Intercepts: (1, 0), (–2, 0), (0, 4) y′ = ( x − 1) 2 ⋅ 2( x + 2) + ( x + 2) 2 ⋅ 2( x − 1) = 2(x – 1)(x + 2)[(x – 1) + (x + 2)] = 2(x – 1)(x + 2)(2x + 1) 1 CV: x = 1, –2, − 2 ⎛ 1 ⎞ Decreasing on (–∞, –2) and ⎜ − , 1⎟ ; increasing ⎝ 2 ⎠ 1 ⎛ ⎞ on ⎜ −2, − ⎟ and (1, ∞); relative minima when 2⎠ ⎝ x = –2 or x = 1; relative maximum when 1 x=− . 2

2 ⎛ 2 ⎞ on ⎜ − , 1⎟ ; relative maximum when x = − ; 3 3 ⎝ ⎠ relative minimum when x = 1. 8

y

x 5

59. y = x 4 + 4 x3 + 4 x 2 = x 2 ( x + 2)2 Intercepts (0, 0), (–2, 0)

8

y′ = 4 x3 + 12 x 2 + 8 x = 4 x( x + 1)( x + 2) CV: x = 0, –1, –2 Increasing on (–2, –1) and (0, ∞); decreasing on (–∞, –2) and (–1, 0); relative maximum when x = –1; relative minima when x = –2 or x = 0.

y

4 x –2

477

1

5



62. y = x ( x 2 − x − 2) = x ( x − 2)( x + 1) [Note x ≥ 0.] Intercepts (0, 0), (2, 0)

10

y = x5 / 2 − x3 / 2 − 2 x1/ 2

y

x 10

–2

5 3 / 2 3 1/ 2 1 − x − 2 ⋅ x −1/ 2 x 2 2 2 1 2 = (5 x − 3 x − 2) 2 x 1 = (5 x + 2)( x − 1) 2 x CV: x = 0, 1 (x ≥ 0) Decreasing on (0, 1); increasing on (1, ∞); relative minimum when x = 1. y′ =

5

65.

y 10

y

x 10

66.

x

5

y

5

2

(

)

x

63. y = 2 x − x = x 2 − x . [Note: x ≥ 0.]

1

Intercepts (0, 0), (4, 0) 1 1− x y′ = −1 = x x CV: x = 0, 1 Increasing on (0, 1); decreasing on (1, ∞); relative maximum when x = 1. 5

4

67. c f = 25, 000 cf =

cf q

=

25, 000 q

d 25, 000 < 0 for q > 0, so c f is a cf = − dq q2

( )

y

decreasing function for q > 0. 1

68. c = 3q − 3q 2 + q3

x 1

4

8

Marginal cost is given by

5

2

dc = 3 − 6q + 3q 2 . dq

dc ⎤ d ⎡⎢ dq dc ⎣ ⎦⎥ < 0, that is, Thus is increasing when dq dq

2

64. y = x 3 + 5 x 3 = x 3 ( x + 5) Intercepts (0, 0), (–5, 0) 5 2 10 5( x + 2) y′ = x 3 + 1 = 1 3 3x 3 3x 3

when −6 + 6q > 0. Hence q > 1.

CV: x = 0, –2 Increasing on (–∞, –2) and (0, ∞); decreasing on (–2, 0); relative maximum when x = –2; relative minimum when x = 0.

478


Section 13.1

69. p = 400 – 2q Revenue is given by r = pq = (400 − 2q )q

⎛ T ⎞ 73. E = 0.71⎜1 − c ⎟ ⎝ Th ⎠ ⎛T dE = 0.71⎜ c ⎜T2 dTh ⎝ h

= 400q − 2q 2 Marginal revenue is r ′ = 400 − 4q . Marginal revenue is increasing when its derivative is positive. But (r′)′ = −4 < 0 . Thus marginal revenue is never increasing.

increases. a2 ⎛ a⎞ 74. r = 2 F + ⎜ 1 − ⎟ p − p 2 + b ⎝ b⎠ dr ⎛ a ⎞ b−a ⎛b−a = ⎜1 − ⎟ − 2 p = − 2 p = 2⎜ − dp ⎝ b ⎠ b ⎝ 2b dr Setting = 0 gives the critical value dp

70. c = q dc 1 Marginal cost = . Since = dq 2 q dc ⎤ d ⎡⎢ dq ⎣ ⎥⎦ = − 1 < 0 for q > 0, marginal cost is dq 4 q3 decreasing for q > 0.

Average cost = c =

p=

c 1 . Since = q q

144 ⎤ ⎡ 75. C (k ) = 100 ⎢100 + 9k + , 1 ≤ k ≤ 100 k ⎥⎦ ⎣

r ′ = 240 + 114q – 3q 2 = 3(40 − q )(2 + q ) Since q ≥ 0, we have q = 40 as the only CV. Since r is increasing on (0, 40) and decreasing on (40, ∞), r is a maximum when output is 40.

a.

C(1) = 25,300

b.

⎡ 9k 2 − 144 ⎤ ⎡ 144 ⎤ C ′(k ) = 100 ⎢9 − = 100 ⎢ ⎥ ⎥ k2 ⎦ k2 ⎣ ⎢⎣ ⎥⎦

72. z = (1 + b) w p − bwc , w p is function of wc , and

⎡ 9(k + 4)(k − 4) ⎤ = 100 ⎢ ⎥ k2 ⎣ ⎦ Since k ≥ 1, the only critical value is k = 4. If 1 ≤ k < 4, then C ′(k ) < 0 and C is decreasing. If 4 < k ≤ 100, then C ′(k ) > 0 and C is increasing. Thus C has an absolute minimum for k = 4.

b > 0. dw p dz = (1 + b) − b(1) dwc dwc ⎡ dw p b ⎤ = (1 + b) ⎢ − ⎥ (factoring) ⎣ dwc 1 + b ⎦ dw p

b−a b−a dr . If p < , then > 0 and r is 2b 2b dp

b−a dr , then < 0 and r is 2b dp decreasing. Thus revenue is maximum for b−a p= . 2b

71. r = 240q + 57 q 2 − q3

c.

dw p b b , then − <0. dwc b + 1 dwc b + 1 Because b > 0, then 1 + b > 0. Thus from dz part (a), < 0 so z is a decreasing dwc

b. If

⎞ p⎟ ⎠

increasing. If p >

dc 1 =− < 0 for q > 0, average cost is dq 2 q3 decreasing for q > 0.

a.

⎞ ⎟ > 0 , so as Th increases, E ⎟ ⎠

<

function of wc .

479

C(4) = 17,200


76. P =


100 1 + 100, 000e−0.36h

(

dP d ⎡ 100 1 + 100, 000e−0.36 h = dh dh ⎢⎣ 3, 600, 000 = 2 e0.36h 1 + 100, 000e−0.36 h

(

Since

)

−1 ⎤

–5

⎥ ⎦

)

dP > 0 , P is an increasing function of h. dh

84.

Problems 13.2

79. Relative maximum: (2.74, 3.74); relative minimum: (–2.74, –3.74)

1.

f ( x) = x 2 − 2 x + 3 and f is continuous over [0, 3]. f ′( x) = 2 x − 2 = 2( x − 1) The only critical value on (0, 3) is x = 1. We evaluate f at this point and at the endpoints: f(0) = 3, f(1) = 2, and f(3) = 6. Absolute maximum: f(3) = 6; absolute minimum: f(1) = 2

2.

f ( x) = −2 x 2 − 6 x + 5 and f is continuous over [–3, 2]. 3⎞ ⎛ f ′( x) = −4 x − 6 = −4 ⎜ x + ⎟ 2⎠ ⎝ 3 The only critical value on (–3, 2) is x = − . We 2 3 19 ⎛ ⎞ have f(–3) = 5, f ⎜ − ⎟ = , and f(2) = –15. ⎝ 2⎠ 2 ⎛ 3 ⎞ 19 Absolute maximum: f ⎜ − ⎟ = ; ⎝ 2⎠ 2 absolute minimum: f(2) = –15

80. Relative maximum: (0.05, 3.05) 81. Relative minima: 0, 1.50, 2.00; relative maxima: 0.57, 1.77 82. f has relative extrema when x ≈ 0.38, 1.18; f ′( x) = 0 when x ≈ 0.38, 1.18. 4

2.5

–2.5

f ′( x) = 4 − 6 x − 3 x 2 10

b.

3. 5

–5

–5

c.

f ′( x) = 4 x3 − 2 x − 2( x + 2) = 4 x3 − 4 x − 4 CV: x ≈ 1.32

78. Relative minimum: (1.26, −5.74)

–1

5

–10

77. Relative minimum: (−3.83, 0.69)

83. a.

10

d.

1 3 1 2 x + x − 2 x + 1 and f is continuous 3 2 over [−1, 0]. f ′( x) = x 2 + x − 2 = ( x + 2)( x − 1) f ( x) =

There are no critical values on (−1, 0), so we only have to evaluate f at the endpoints: 19 and f(0) = 1. f (−1) = 6 19 Absolute maximum: f (−1) = 6 Absolute minimum: f(0) = 1

f ′( x) > 0 on (–2.53, 0.53); f ′( x) < 0 on

(–∞, –2.53), (0.53, ∞), f is inc. on (–2.53, 0.53); f is dec. on (–∞, –2.53), (0.53, ∞).

480


4.

f ( x) =

[0, 1].

Section 13.2

1 4 3 2 x − x and f is continuous over 4 2

(

)(

f ′( x) = x3 − 3 x = x x + 3 x − 3

8.

)

f ′( x) = 7 x 2 + 4 x − 3 = (7 x − 3)( x + 1)

There are no critical values on (0, 1), so we only have to evaluate f at the end points: f(0) = 0 and 5 f (1) = − 4 Absolute maximum: f(0) = 0; 5 absolute minimum: f (1) = − 4 5.

2

(

2

9.

)

f ( x) = 3 x 4 − x6 and f is continuous over [–1, 2].

= 6x

3

(

(

2−x

)(

2+x

)

)

= 6(2 x + 3)( x − 1)

The only critical values on (–1, 2) are x = 0,

⎛1 ⎞ The only critical value on ⎜ , 3 ⎟ is x = 1. We ⎝2 ⎠ evaluate f at this point and the endpoints: 19 ⎛1⎞ f ⎜ ⎟ = − ; f(1) = –8, f(3) = 84. 4 ⎝2⎠ Absolute maximum: f(3) = 84; absolute minimum: f(1) = –8

We have f(–1) = 2, f(0) = 0, f f(2) = –16. Absolute maximum: f

10.

2 − 13 x . 3 The only critical value on (–8, 8) is x = 0. We have f(–8) = 4, f(0) = 0, and f(8) = 4. Thus there is an absolute maximum when x = –8 or x = 8, and an absolute minimum when x = 0. Absolute maximum: f(–8) = f(8) = 4; absolute minimum: f(0) = 0

f ( x) =

1 4 1 2 x − x + 3 and f is continuous over 4 2

The critical values of f on (−2, 3) are x = −1, 0, 11 1. We have f(−2) = 5, f (−1) = , f(0) = 3, 4 11 75 and f (3) = . f (1) = 4 4 75 Absolute maximum: f (3) = 4 11 Absolute minimum: f (−1) = f (1) = 4

f ( x) = −3x5 + 5 x3 and f is continuous over [–2, 0].

(

( 2) = 4 ;

[−2, 3]. f ′( x) = x3 − x = x( x 2 − 1) = x( x − 1)( x + 1)

2

f ( x) = x 3 and f is continuous over [–8, 8].

f ′( x) = −15 x 4 + 15 x 2 = 15 x 2 1 − x 2

( 2 ) = 4 , and

2.

absolute minimum: f(2) = –16

f ′( x) =

7.

3 . We 7

⎛ 3 ⎞ 13 have f(0) = 1, f ⎜ ⎟ = , and f(3) = 73. ⎝ 7 ⎠ 49 Absolute maximum: f(3) = 73; ⎛ 3 ⎞ 13 absolute minimum: f ⎜ ⎟ = ⎝ 7 ⎠ 49

f ′( x) = 12 x3 − 6 x5 = 6 x3 2 − x 2

f ′( x) = 12 x + 6 x − 18 = 6 2 x + x − 3

6.

The only critical value on (0, 3) is x =

f ( x) = 4 x3 + 3 x 2 − 18 x + 3 and f is continuous ⎡1 ⎤ over ⎢ , 3⎥ . ⎣2 ⎦

7 3 x + 2 x 2 − 3x + 1 and f is continuous 3 over [0, 3]. f ( x) =

)

11.

= 15 x 2 (1 + x)(1 − x) The only critical value on (–2, 0) is x = –1. We have f(–2) = 56, f(–1) = –2, and f(0) = 0. Absolute maximum: f(–2) = 56; absolute minimum: f(–1) = –2.

f ( x) = x 4 − 9 x 2 + 2 and f is continuous over [–1, 3].

(

f ′( x) = 4 x3 − 18 x = 2 x 2 x 2 − 9 = 2x

(

2x − 3

)(

2x + 3

)

)

The only critical values on (–1, 3) are x = 0 and 3 3 2 . We have f(–1) = –6, f(0) = 2, x= = 2 2 481



⎛3 2 ⎞ 73 f ⎜⎜ ⎟⎟ = − , and f(3) = 2. 4 ⎝ 2 ⎠ Absolute maximum: f(0) = f(3) = 2; ⎛3 2 ⎞ 73 absolute minimum: f ⎜⎜ ⎟⎟ = − 4 ⎝ 2 ⎠

12.

f ( x) =

–4

x

and f is continuous over [0, 2].

2

x +1

a.

c.

d. 14,283 Problems 13.3 1.

3 −4

+ 0

3⎞ ⎛ Concave up on ⎜ −∞, − ⎟ and (0, ∞); concave 4⎠ ⎝ ⎛ 3 ⎞ down on ⎜ − , 0 ⎟ . Inflection points when ⎝ 4 ⎠ 3 x = − , 0. . 4

10

28 –1

2.

Absolute maximum: f(−26) = f(28) = 9; absolute minimum: f(1) = 0

f ( x) =

x5 x 4 + − 2 x2 20 4

f ′′( x) = ( x − 1)( x + 2) 2 f ′′( x) is 0 when x = 1, –2. Sign chart for f ′′ :

f ( x) = 0.2 x3 − 3.6 x 2 + 2 x + 1 and f is continuous

–

over [−1, 2]. 5

–1

−

+

2

–26

f ( x ) = 2 x 4 + 3 x3 + 2 x − 3 f ′′( x) = 6 x(4 x + 3) 3 f ′′( x) is 0 when x = 0, − . Sign chart for f ′′ : 4

f ( x) = ( x − 1) 3 and f is continuous over

[−26, 28].

14.

9

2

The only critical value on (0, 2) is x = 1. We 2 1 have f(0) = 0, f (1) = ,and f (2) = . 5 2 1 Absolute maximum: f (1) = ; 2 absolute minimum: f(0) = 0 13.

–3.22, –0.78

b. 2.75

(1 + x)(1 − x)

( x2 + 1)

9 0

x 2 + 1) − x(2 x) ( 1 − x2 f ′( x) = = 2 2 ( x2 + 1) ( x2 + 1) =

10

15.

–

–2

+ 1

Concave down on (–∞, –2) and (–2, 1); concave up on (1, ∞). Inflection point when x = 1.

2

3. –10

f ( x) = f ′′( x) =

Absolute maximum f(0.28) ≈ 1.28; absolute minimum f(2) = −7.8

2 + x − x2 x2 − 2 x + 1 2(7 − x)

( x − 1) 4 f ′′( x) is 0 when x = 7. Although f ′′ is not defined when x = 1, f is not continuous at x = 1. So there is no inflection point when x = 1, but x = 1 must be considered in concavity analysis. 482


Section 13.3

Sign chart for f ′′ : +

+

6.

–

1

7

f ′′( x) =

Concave up on (–∞, 1) and (1, 7); concave down on (7, ∞). Inflection point when x = 7. 4.

( x − 1)2 2(2 x + 1) f ′′( x) = ( x − 1) 4

+

1 f ′′( x) = 0 when x = − . Although f ′′ is not 2 defined when x = 1, f is not continuous at x = 1. So there is no inflection point when x = 1, but x = 1 must be considered in concavity analysis. Sign chart of f ′′ : –

+

–1

–2

+

8. y = −74 x 2 + 19 x − 37 y′ = −148 x + 19 y ′′ = −148 < 0 for all x. Thus the graph is concave down on (–∞, ∞).

1 2

9. y = 4 x3 + 12 x 2 − 12 x

2

x +1

f ( x) =

y′ = 12 x 2 + 24 x − 12 y ′′ = 24 x + 24 = 24( x + 1) Possible inflection point when x = –1. Concave down on (–∞, –1): concave up on (–1, ∞); inflection point when x = –1.

2

x −2

f ′′( x) =

( ) = 6 ( 3x 2 + 2 ) 3 3 ( x2 − 2) ⎡⎣( x − 2 )( x + 2 )⎤⎦

6 3x2 + 2

f ′′( x) is never 0. Although f ′′ is not defined

10. y = x3 − 6 x 2 + 9 x + 1

when x = ± 2 , f is not continuous at x = ± 2 . So there is no inflection point when x = ± 2 , but x = ± 2 must be considered in concavity analysis. Sign chart of f ′′ : +

–

– 2

y′ = 3x 2 − 12 x + 9 y ′′ = 6 x − 12 = 6( x − 2) Possible inflection point when x = 2. Concave down on (–∞, 2); concave up on (2, ∞); inflection point when x = 2.

+ 2

(

)

Concave up on −∞, − 2 and

(

2

down for all x, that is, on (–∞, ∞).

1⎞ ⎛ down on ⎜ −∞, − ⎟ . 2⎠ ⎝

Inflection point when x =

– 0

7. y = −2 x 2 + 4 x y′ = −4 x + 4 y ′′ = −4 < 0 for all x, so the graph is concave

⎛ 1 ⎞ Concave up on ⎜ − , 1⎟ and (1, ∞); concave ⎝ 2 ⎠

5.

( 4 − x2 )

)

3 2

Concave up on (–2, 0); concave down on (0, 2). Inflection point when x = 0.

1

2

(

2 x x2 − 6

Note that the domain of f is [–2, 2]. f ′′( x) is 0 only when x = 0; f ′′ is not defined when x = ±2, which are the endpoints of the domain of f. The only possible point of inflection occurs when x = 0. Sign chart for f ′′ :

x2

f ( x) =

f ( x) = x 4 − x 2

)

(

11. y = 2 x3 − 5 x 2 + 5 x − 2

)

2, ∞ ;

y ′ = 6 x 2 − 10 x + 5

concave down on − 2, 2 . No inflection

5⎞ ⎛ y ′′ = 12 x − 10 = 12 ⎜ x − ⎟ 6⎠ ⎝

point.

5 Possible inflection point when x = . Concave 6 483



1

5⎞ ⎛ ⎛5 ⎞ down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ ; 6 6 ⎝ ⎠ ⎝ ⎠ 5 inflection point when x = 6

15. y = 2 x 5 2 − 54 x 5 8 −9 8 y ′′ = − x 5 = − 9 25 25 x 5

y′ =

12. y = x 4 − 8 x 2 − 6

y ′′ is not defined when x = 0 and y is continuous there. Thus there is a possible inflection point when x = 0. Concave up on (–∞, 0); concave down on (0, ∞); inflection point when x = 0.

y′ = 4 x3 − 16 x 4⎞ ⎛ y ′′ = 12 x 2 − 16 = 12 ⎜ x 2 − ⎟ 3⎠ ⎝ ⎛ 2 3 ⎞⎛ 2 3⎞ = 12 ⎜⎜ x − ⎟⎜ x + ⎟ ⎟⎜ 3 ⎠⎝ 3 ⎟⎠ ⎝

16. y =

x 4 19 x3 7 x 2 + − + x+5 2 6 2 19 y′ = 2 x 3 + x 2 − 7 x + 1 2 2 ′′ y = 6 x + 19 x − 7 = (3 x − 1)(2 x + 7)

17. y =

y ′ = 8 x3 − 96 x + 7 y ′′ = 24 x 2 − 96 = 24( x 2 − 4) = 24( x + 2)( x − 2)

Possible inflection points when x = ±2. Concave up on (−∞, −2) and (2, ∞); concave down on (−2, 2); inflection points when x = ±2.

7 1 Possible inflection points when x = − , . 2 3 7⎞ ⎛ ⎛1 ⎞ Concave up on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 2⎠ ⎝ ⎝3 ⎠ ⎛ 7 1⎞ concave down on ⎜ − , ⎟ ; inflection points ⎝ 2 3⎠ 7 1 when x = − , . 2 3

x4 9 x2 + + 2x 4 2

y′ = − x 3 + 9 x + 2 y ′′ = −3x 2 + 9 = −3( x 2 − 3)

(

)(

)

Possible inflection points when x = ± 3 .

(

)

Concave down on – ∞, − 3 and

(

84

x5 Although y ′′ is not defined when x = 0, y is not continuous there. Thus there is no possible inflection point. However, x = 0 must be considered in concavity analysis. Concave down on (−∞, 0); concave up on (0, ∞); no inflection point

2 3 . 3

= −3 x + 3 x − 3

= 7 x −3

y ′′ = 84 x −5 =

13. y = 2 x 4 − 48 x 2 + 7 x + 3

14. y = −

x

3

y ′ = −21x −4

2 3 Possible inflection points x = ± . Concave 3 ⎛ ⎛2 3 ⎞ 2 3⎞ up on ⎜⎜ −∞, – , ∞ ⎟⎟ ; concave ⎟⎟ and ⎜⎜ 3 ⎠ ⎝ ⎝ 3 ⎠ ⎛ 2 3 2 3⎞ down on ⎜⎜ − , ⎟ ; inflection points when 3 3 ⎟⎠ ⎝ x=±

7

)

(

)

5 1 1 1 2 18. y = − x 4 − x3 + x 2 + x − 2 6 2 3 5 1 1 y′ = −10 x3 − x 2 + x + 2 3

3, ∞ ;

concave up on − 3, 3 ; inflection points when x = ± 3 .

y ′′ = −30 x 2 − x + 1 = −(5 x + 1)(6 x − 1) 1 1 Possible inflection points when x = − , . 5 6 1⎞ ⎛ ⎛1 ⎞ Concave down on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 5⎠ ⎝ ⎝6 ⎠ 484


Section 13.3

( (

⎛ 1 1⎞ concave up on ⎜ − , ⎟ ; inflection points when ⎝ 5 6⎠ 1 1 x=− , . 5 6 1 5 1 4 1 3 1 2 x − x + x − x− 20 4 6 2 3 1 4 1 1 y′ = x − x 3 + x 2 − 4 2 2

6⎞ ⎛ y ′′ = 30 x 4 − 36 x 2 = 30 x 2 ⎜ x 2 − ⎟ 5⎠ ⎝ ⎛ 6 ⎞⎛ 6⎞ = 30 x 2 ⎜⎜ x − ⎟⎟ ⎜⎜ x + ⎟ 5 ⎠⎝ 5 ⎟⎠ ⎝

)

y ′′ is 0 when x = 0 or x − 3x + 1 = 0 . Using the

quadratic formula to solve x 2 − 3x + 1 = 0 gives

6 . 5 ⎛ ⎛ 6 ⎞ 6⎞ Concave up on ⎜⎜ −∞, − , ∞ ⎟⎟ ; ⎟⎟ and ⎜⎜ 5 5 ⎝ ⎠ ⎝ ⎠ ⎛ 6 ⎞ ⎛ 6⎞ concave down on ⎜⎜ − , 0 ⎟⎟ and ⎜⎜ 0, ⎟. 5 ⎟⎠ ⎝ 5 ⎠ ⎝

Possible inflection points when x = 0, ±

3± 5 . Thus possible inflection points 2 3± 5 occur when x = 0, . Concave down on 2 ⎛ 3− 5 3+ 5 ⎞ (–∞, 0) and ⎜⎜ , ⎟ ; concave up on 2 ⎟⎠ ⎝ 2 ⎛ 3− 5 ⎞ ⎛ 3+ 5 ⎞ ⎜⎜ 0, 2 ⎟⎟ and ⎜⎜ 2 , ∞ ⎟⎟ ; inflection points ⎝ ⎠ ⎝ ⎠ x=

Inflection points when x = ±

3± 5 . 2

23. y = y′ =

1 5 x − 3x3 + 17 x + 43 10 1 y′ = x 4 − 9 x 2 + 17 2 y ′′ = 2 x3 − 18 x = 2 x( x 2 − 9)

20. y =

1 6 7 4 x − x + 5x2 + 2 x − 1 21. y = 30 12 1 7 y′ = x5 − x3 + 10 x + 2 5 3

)(

y ′′ = x 4 − 7 x 2 + 10 = x 2 − 2 x 2 − 5

4

24. y = 1 − y′ =

x2

x3 6

x4 No possible inflection point, but we must consider x = 0 in the concavity analysis. Concave down on (−∞, 0) and (0, ∞).

)

Possible inflection points when x = ± 2, ± 5 .

)(

1

2

y ′′ = −

)( x − 2 )( x + 5 )( x − 5 ) (

( x − 1)2

( x − 1)3 No possible inflection point, but we consider x = 1 in the concavity analysis. Concave down on (–∞, 1); concave up on (1, ∞).

Possible inflection points when x = 0, ±3. Concave down on (−∞, −3) and (0, 3); concave up on (−3, 0) and (3, ∞); inflection points when x = 0, ±3.

(

6 . 5

x +1 x −1 −2

y ′′ =

= 2 x( x + 3)( x − 3)

= x+ 2

)

2, 5 ; inflection points when

y′ = 6 x5 − 12 x3

2

(

)

22. y = x 6 − 3 x 4

y ′′ = x3 − 3x 2 + x = x x 2 − 3 x + 1

when x = 0,

(

x = ± 5, ± 2 .

19. y =

(

)

5, ∞ ; concave down on − 5, − 2 and

)

Concave up on −∞, − 5 , − 2, 2 , and 485


25. y =


x2

27. y =

2

x +1

x 2 + 1) (2 x) − x 2 (2 x) ( 2x y′ = = 2 2 ( x2 + 1) ( x2 + 1) 2 x 2 + 1) (2) − 2 x(2) ( x 2 + 1) (2 x) ( y ′′ = 4 ( x2 + 1) x 2 + 1) (2) − 8 x 2 ( = 3 ( x2 + 1) 2 (1 − 3x 2 ) 2 (1 + 3x )(1 − 3 x ) = = 3 3 2 x 1 + ( ) ( x2 + 1)

Possible inflection points when x = ±

y′ =

4 x2 x+3

y′ = y ′′ = =

( x + 3)(8 x) − 4 x 2 (1) ( x + 3) 2

=

(

4 x2 + 6 x

6( x + 3)2 1 ( x + 3) 2 (21) − (21x + 40)[2( x + 3)] ⋅ 6 ( x + 3)4

=

1 ( x + 3)(21) − (21x + 40)(2) ⋅ 6 ( x + 3)3

=

1 −21x − 17 1 21x + 17 ⋅ =− ⋅ 3 6 ( x + 3) 6 ( x + 3)3

3 2⎤ ⎡ 1 ( x + 3) (21) − (21x + 17) ⎣3( x + 3) ⎦ ′′ y =− ⋅ 6 ( x + 3)6 1 ( x + 3)(21) − (21x + 17)(3) =− ⋅ 6 ( x + 3) 4

1 −42 x + 12 7x − 2 =− ⋅ = 4 6 ( x + 3) ( x + 3)4

1

. 3 ⎛ ⎛ 1 ⎞ 1 ⎞ , ∞⎟ ; Concave down on ⎜ −∞, − ⎟ and ⎜ 3 3 ⎝ ⎠ ⎝ ⎠ ⎛ 1 1 ⎞ , concave up on ⎜ − ⎟ ; inflection points 3 3⎠ ⎝ 1 when x = ± . 3

26. y =

21x + 40

2 (x = –3 7 must be considered in concavity analysis). 2⎞ ⎛ Concave down on (–∞, –3) and ⎜ −3, ⎟ ; 7⎠ ⎝ ⎛2 ⎞ concave up on ⎜ , ∞ ⎟ ; inflection point when ⎝7 ⎠ 2 x= . 7

Possible inflection point when x =

28. y = 3( x 2 − 2) 2

)

y ′ = 12 x( x 2 − 2) = 12( x3 − 2 x) 2⎞ ⎛ y ′′ = 12(3x 2 − 2) = 36 ⎜ x 2 − ⎟ 3⎠ ⎝ ⎛ 6 ⎞⎛ 6⎞ = 36 ⎜ x − ⎟⎜ x + ⎟ ⎜ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎝

( x + 3) 2

( x + 3)2 (4)(2 x + 6) − 4( x 2 + 6 x)(2)( x + 3) ( x + 3)4 72

( x + 3)3 No possible inflection point, but we must include x = –3 in the concavity analysis. Concave down on (–∞, –3); concave up on (–3, ∞).


6 . 3

⎛ ⎛ 6 ⎞ 6⎞ Concave up on ⎜ −∞, − , 0⎟; ⎟⎟ and ⎜⎜ ⎜ ⎟ 3 ⎠ ⎝ ⎝ 3 ⎠ ⎛ 6 6⎞ , concave down on ⎜ − ⎟ ; inflection ⎜ 3 3 ⎟⎠ ⎝

points when x = ±

486

6 . 3


Section 13.3

29. y = 5e x

3 ⎛ ⎞ Concave down on ⎜ 0, e 2 ⎟ ; concave up on ⎝ ⎠ 3 3 ⎛ 2 ⎞ 2 ⎜ e , ∞ ⎟ ; inflection point when x = e . ⎝ ⎠

y′ = 5e x y ′′ = 5e x Thus y ′′ > 0 for all x. Concave up on (–∞, ∞).

34. y =

x

−x

x

−x

x

−x

30. y = e − e

y′ = e + e

y ′′ = e − e

y′ =

Setting y ′′ = 0 gives e x = e − x or, equivalently,

=

x = 0. Concave down on (–∞, 0); concave up on (0, ∞); inflection point when x = 0.

y′ = 3xe x + 3e x = 3e x ( x + 1)

=

y ′′ = 3e x (1) + 3( x + 1)e x = 3e x ( x + 2) y ′′ = 0 if x = –2. Concave down on (–∞, –2);

=

2 2

2

2

y′ = 2 x 2 e x + e x = e x (2 x 2 + 1) x2

2

y ′′ = e (4 x) + 2 x(2 x + 1)e

x2

x2

(

)

3e x (2 x) − x 2 + 1 3e x 9e

2x

=

(

3e

3e x

(

)

3e x (2 − 2 x) − 2 x − x 2 − 1 3e x 9e 2 x

(

)

(2 − 2 x) − 2 x − x 2 − 1 3e x

x2 − 4 x + 3 x

=

( x − 1)( x − 3)

35. y = x 2 − x − 6 = ( x − 3)( x + 2)

3

= e (4 x + 6 x)

Intercepts: (0, −6), (3, 0) and (−2, 0) 1⎞ ⎛ y′ = 2x − 1 = 2 ⎜ x − ⎟ 2⎠ ⎝

concave up on (0, ∞); inflection point when x = 0.

CV: x =

33. y =

1 2

1⎞ ⎛ Decreasing on ⎜ −∞, ⎟ ; increasing on 2⎠ ⎝

2 x ⋅ 1x − (ln x)(2)

25 ⎞ ⎛1 ⎞ ⎛1 ⎜ 2 , ∞ ⎟ ; relative minimum at ⎜ 2 , − 4 ⎟ . ⎝ ⎠ ⎝ ⎠ y ′′ = 2 No possible inflection point. Concave up on (−∞, ∞).

y ′′ =

=

2

ln x . (Note: x > 0.) 2x

y′ =

=

x

2x − x −1

= 2 xe (2 x + 3) y ′′ = 0 when x = 0. Concave down on (–∞, 0);

x2

)

2 x − x2 + 1

3e 3e x Possible inflection points when x = 1, 3. Concave up on (–∞, 1) and (3, ∞); concave down on (1, 3); inflection point when x = 1, 3.

concave up on (–2, ∞); inflection point when x = –2. 32. y = xe x

3e x

2

y ′′ =

31. y = 3xe x

x2 + 1

4x2

( )

=

1 − ln x 2x2

2 x 2 − 1x − (1 − ln x)(4 x)

4 x4 −2 x − (1 − ln x)(4 x)

4

4

4x −1 − (1 − ln x)(2) 2 x3

=

x 10

2 ln( x) − 3 2 x3

y ′′ is 0 if 2ln(x) – 3 = 0, ln x =

y

3 3 , x = e2 . 2

487



⎛1 9⎞ relative maximum at ⎜ , ⎟ ⎝2 4⎠ y ′′ = −2 No possible inflection point. Concave down on (–∞, ∞).

36. y = x 2 + 2 Intercept (0, 2) y′ = 2 x CV: x = 0 Decreasing on (–∞, 0); increasing on (0, ∞); relative minimum at (0, 2). y ′′ = 2 No possible inflection point. Concave up on (–∞, ∞). Symmetric about the y-axis. 8

5

x

y

5

39. y = x3 − 9 x 2 + 24 x − 19 The x-intercepts are not convenient to find; the y-intercept is (0, –19).

x 5

y′ = 3x 2 − 18 x + 24 = 3( x − 2)( x − 4) CV: x = 2, x = 4 Increasing on (–∞, 2) and (4, ∞); decreasing on (2, 4); relative maximum at (2, 1); relative minimum at (4, –3). y ′′ = 6 x − 18 = 6( x − 3) Possible inflection point when x = 3. Concave down on (–∞, 3); concave up on (3, ∞); inflection point at (3, –1).

37. y = 5 x − 2 x 2 = x(5 − 2 x) ⎛5 ⎞ Intercepts (0, 0) and ⎜ , 0 ⎟ ⎝2 ⎠ y′ = 5 − 4 x

CV: x =

5 4

5⎞ ⎛ ⎛5 ⎞ Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ; 4 4 ⎝ ⎠ ⎝ ⎠ ⎛ 5 25 ⎞ relative maximum at ⎜ , ⎟. ⎝4 8 ⎠ y ′′ = −4 No possible inflection point. Concave down on (–∞, ∞). y

8

y

x 2

8

5

40. y = x3 − 25 x 2 = x 2 ( x − 25) Intercepts: (0, 0) and (25, 0) 50 ⎞ ⎛ y ′ = 3 x 2 − 50 x = 3 x ⎜ x − ⎟ 3 ⎠ ⎝

x 5

CV: x = 0, 38. y = x − x 2 + 2 = −( x − 2)( x + 1) Intercepts (2, 0), (–1, 0), and (0, 2) y′ = 1 − 2 x CV: x =

y

50 3

⎛ 50 ⎞ Increasing on (−∞, 0) and ⎜ , ∞ ⎟ ; decreasing ⎝ 3 ⎠ ⎛ 50 ⎞ on ⎜ 0, ⎟ ; relative maximum at (0, 0); relative ⎝ 3 ⎠

1 2

62,500 ⎞ ⎛ 50 . minimum at ⎜ , − 3 27 ⎟⎠ ⎝

1⎞ ⎛ ⎛1 ⎞ Increasing on ⎜ −∞, ⎟ ; decreasing on ⎜ , ∞ ⎟ ; 2⎠ ⎝ ⎝2 ⎠ 488


Section 13.3

42. y = x3 − 6 x 2 + 9 x = x( x − 3) 2 Intercepts (0, 0) and (3, 0)

25 ⎞ ⎛ y ′′ = 6 x − 50 = 6 ⎜ x − ⎟ 3 ⎠ ⎝

y′ = 3x 2 − 12 x + 9 = 3( x − 1)( x − 3) CV: x = 1 and x = 3 Increasing on (–∞, 1) and (3, ∞); decreasing on (1, 3); relative maximum at (1, 4); relative minimum at (3, 0). y ′′ = 6 x − 12 = 6( x − 2) Possible inflection point when x = 2. Concave down on (–∞, 2); concave up on (2, ∞); inflection point at (2, 2).

25 . Concave 3 ⎛ 25 ⎞ concave up on ⎜ , ∞ ⎟ ; 3 ⎝ ⎠

Possible inflection point when x = 25 ⎞ ⎛ ; down on ⎜ −∞, 3 ⎟⎠ ⎝

31, 250 ⎞ ⎛ 25 . inflection point at ⎜ , − 27 ⎟⎠ ⎝ 3 1200

y x 48

5

y

x 8

41. y =

x3 x3 − 12 x − 4x = 3 3

(

)(

1 = x x+2 3 x−2 3 3

(

43. y = x3 − 3 x 2 + 3 x − 3 Intercept (0, –3)

)

Intercepts (0, 0) and ±2 3, 0

y′ = 3x 2 − 6 x + 3 = 3( x − 1) 2 CV: x = 1 Increasing on (–∞, 1) and (1, ∞); no relative maximum or minimum y ′′ = 6( x − 1) Possible inflection point when x = 1. Concave down on (–∞, 1); concave up on (1, ∞); inflection point at (1, –2).

)

y′ = x 2 − 4 = ( x + 2)( x − 2) CV: x = ±2 Increasing on (–∞, –2) and (2, ∞); decreasing on 16 ⎞ ⎛ (–2, 2); relative maximum at ⎜ −2, ⎟ ; relative 3⎠ ⎝ 16 ⎞ ⎛ minimum at ⎜ 2, − ⎟ . 3⎠ ⎝ y ′′ = 2 x Possible inflection point when x = 0. Concave down on (–∞, 0); concave up on (0, ∞); inflection point at (0, 0). Symmetric about the origin. 10

5

y

x 5

y

5 2 5 ⎛ ⎞ x + 2 x = x ⎜ 2x2 + x + 2 ⎟ 2 2 ⎝ ⎠ Intercept (0, 0)

44. y = 2 x3 +

x 10

y′ = 6 x 2 + 5 x + 2 CV: none Increasing on (–∞, ∞). 5⎤ ⎡ y ′′ = 12 x + 5 = 12 ⎢ x + ⎥ ⎣ 12 ⎦

Possible inflection point at x = − 489

5 . Concave 12



5⎞ ⎛ down on ⎜ −∞, − ⎟ ; concave up on 12 ⎝ ⎠ 235 ⎞ ⎛ 5 ⎞ ⎛ 5 ⎜ − , ∞ ⎟ ; inflection point at ⎜ − , − ⎟. 12 12 432 ⎝ ⎠ ⎝ ⎠ y

⎛1 ⎞ on ⎜ , 1⎟ ; relative minimum at ⎝3 ⎠ relative maximum at (1, 4) 2⎞ ⎛ y ′′ = −6 x + 4 = −6 ⎜ x − ⎟ 3⎠ ⎝

3

⎛ 1 104 ⎞ ⎜ 3 , 27 ⎟ ; ⎝ ⎠

2 . Concave 3 ⎛2 ⎞ concave down on ⎜ , ∞ ⎟ ; ⎝3 ⎠

Possible inflection point when x = x

2⎞ ⎛ up on ⎜ −∞, ⎟ ; 3⎠ ⎝

3

⎛ 2 106 ⎞ inflection point at ⎜ , ⎟ ⎝ 3 27 ⎠

45. y = 4 x3 − 3 x 4 = x3 (4 − 3 x)

8

y

⎛4 ⎞ Intercepts (0, 0), ⎜ , 0 ⎟ ⎝3 ⎠ y′ = 12 x 2 − 12 x3 = 12 x 2 (1 − x) CV: x = 0 and x = 1 Increasing on (–∞, 0) and (0, 1); decreasing on (1, ∞); relative maximum at (1, 1). y ′′ = 24 x − 36 x 2 = 12 x(2 − 3x)

x 5

47. y = −2 + 12 x − x3 Intercept (0, –2)

2 Possible inflection points at x = 0 and x = . 3 ⎛2 ⎞ Concave down on (–∞, 0) and ⎜ , ∞ ⎟ ; concave ⎝3 ⎠ ⎛ 2⎞ up on ⎜ 0, ⎟ ; inflection points at (0, 0) and ⎝ 3⎠ ⎛ 2 16 ⎞ ⎜ , ⎟ ⎝ 3 27 ⎠ 3

y′ = 12 − 3x 2 = 3(2 + x)(2 − x) CV: x = ±2 Decreasing on (–∞, –2) and (2, ∞); increasing on (–2, 2); relative minimum at (–2, –18); relative maximum at (2, 14). y ′′ = −6 x Possible inflection point when x = 0. Concave up on (–∞, 0); concave down on (0, ∞); inflection point at (0, –2).

y

y

20 x 3

x 5

46. y = − x3 + 2 x 2 − x + 4 Intercept (0, 4) y ′ = −3x 2 + 4 x − 1 = −(3 x − 1)( x − 1)

48. y = (3 + 2 x)3

1 CV: x = , 1 3

⎛ 3 ⎞ Intercepts (0, 27), ⎜ − , 0 ⎟ ⎝ 2 ⎠

1⎞ ⎛ Decreasing on ⎜ −∞, ⎟ and (1, ∞); increasing 3⎠ ⎝

y′ = 6(3 + 2 x)2

CV: x = − 490

3 2


Section 13.3

x3 3 x 2 x 2 ( x − 3) − = 5 5 5 Possible inflection points when x = 0 and x = 3. Concave down on (–∞, 0) and (0, 3); concave up on (3, ∞); inflection point at (3, –1.62).

3⎞ ⎛ ⎛ 3 ⎞ Increasing on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; no 2 2 ⎝ ⎠ ⎝ ⎠ relative maximum or minimum. y ′′ = 24(3 + 2 x)

y ′′ =

3 .Concave 2 3⎞ ⎛ ⎛ 3 ⎞ down on ⎜ −∞, − ⎟ ; concave up on ⎜ − , ∞ ⎟ ; 2⎠ ⎝ ⎝ 2 ⎠ 3 ⎛ ⎞ inflection point at ⎜ − , 0 ⎟ . ⎝ 2 ⎠

Possible inflection point at x = −

40

10

y

x 10

y

(

) = x ( 5 + x 2 )( 5 − x 2 ) = x ( 5 + x 2 ) ( 4 5 + x )( 4 5 − x ) Intercepts (0, 0) and ( ± 4 5, 0 )

51. y = 5 x − x5 = x 5 − x 4

x 5

49. y = 2 x3 − 6 x 2 + 6 x − 2 = 2( x − 1)3 Intercepts (0, –2), (1, 0)

Symmetric about the origin.

( ) ( = 5(1 − x)(1 + x) (1 + x 2 )

)(

y′ = 5 − 5 x 4 = 5 1 − x 4 = 5 1 − x 2 1 + x 2

2

y′ = 6( x − 1) CV: x = 1 Increasing on (–∞, 1) and (1, ∞); no relative maximum or minimum. y ′′ = 12( x − 1) Possible inflection point when x = 1. Concave down on (–∞, 1); concave up on (1, ∞); inflection point at (1, 0).

)

CV: x = ±1 Decreasing on (–∞, –1) and (1, ∞); increasing on (–1, 1); relative minimum at (–1, –4); relative maximum at (1, 4). y ′′ = −20 x3 Possible inflection point when x = 0. Concave up on (–∞, 0); concave down on (0, ∞); inflection point at (0, 0).

y 3

5

x

y

3 x 5

x5 x 4 x4 − = ( x − 5) 100 20 100 Intercepts (0, 0), (5, 0) x 4 x3 x3 y′ = − = ( x − 4) 20 5 20 CV: x = 0 and x = 4 Increasing on (–∞, 0) and (4, ∞); decreasing on (0, 4); relative maximum at (0, 0); relative minimum at (4, –2.56).

50. y =

52. y = x 2 ( x − 1)2 Intercepts: (0, 0), (1, 0) y ′ = x 2 [2( x − 1)(1)] + 2 x( x − 1) 2 = 2 x( x − 1)(2 x − 1) = 4 x3 − 6 x 2 + 2 x

CV: x = 0, 1 and x = 491

1 2



⎛1 ⎞ Decreasing on (−∞, 0) and ⎜ , 1⎟ ; increasing ⎝2 ⎠

3

y

⎛ 1⎞ on ⎜ 0, ⎟ and (1, ∞); relative minima at (0, 0) ⎝ 2⎠

x 3

⎛1 1 ⎞ and (1, 0); relative maximum at ⎜ , ⎟ ⎝ 2 16 ⎠ y ′′ = 12 x 2 − 12 x + 2 = 2(6 x 2 − 6 x + 1) From the quadratic formula, there are possible 3± 3 . Concave up inflection points when x = 6 ⎛ ⎛ 3+ 3 ⎞ 3− 3 ⎞ on ⎜⎜ −∞, , ∞ ⎟⎟ ; concave ⎟⎟ and ⎜⎜ 6 ⎠ ⎝ ⎝ 6 ⎠ ⎛ 3− 3 3+ 3 ⎞ down on ⎜⎜ , ⎟ ; inflection points at 6 ⎟⎠ ⎝ 6 ⎛ 3− 3 1 ⎞ ⎛ 3+ 3 1 ⎞ , , ⎜⎜ ⎟⎟ and ⎜⎜ ⎟. 36 ⎠ 36 ⎟⎠ ⎝ 6 ⎝ 6 5

5⎤ ⎡ 54. y = 3x5 − 5 x3 = 3x3 ⎢ x 2 − ⎥ 3⎦ ⎣ ⎛ 5 ⎞⎛ 5⎞ = 3x3 ⎜⎜ x + ⎟⎟ ⎜⎜ x − ⎟ 3 ⎠⎝ 3 ⎟⎠ ⎝ ⎛ 5 ⎞ Intercepts (0, 0) and ⎜⎜ ± , 0 ⎟⎟ ⎝ 3 ⎠ Symmetric about the origin. y′ = 15 x 4 − 15 x 2 = 15 x 2 ( x + 1)( x − 1) CV: x = 0 and x = ±1 Increasing on (–∞, –1) and (1, ∞); decreasing on (–1, 0) and (0, 1); relative maximum at (–1, 2); relative minimum at (1, –2). ⎡ 2⎤⎡ 2⎤ y ′′ = 60 x3 − 30 x = 60 x ⎢ x + ⎥ ⎢x − ⎥ 2 ⎦⎣ 2 ⎦ ⎣

y

x 5

2 . 2 ⎛ ⎛ 2⎞ 2⎞ Concave down on ⎜⎜ −∞, − ⎟; ⎟⎟ and ⎜⎜ 0, 2 ⎠ 2 ⎟⎠ ⎝ ⎝ ⎛ ⎛ 2 ⎞ 2 ⎞ concave up on ⎜⎜ − , 0 ⎟⎟ and ⎜⎜ , ∞ ⎟⎟ ; ⎝ 2 ⎠ ⎝ 2 ⎠ ⎛ 2 ⎞ 7 2 inflection points at ⎜⎜ ,− ⎟ , (0, 0), and 2 8 ⎟⎠ ⎝ ⎛ 2 7 2⎞ , ⎜⎜ − ⎟. 2 8 ⎟⎠ ⎝

Possible inflection points at x = 0 and x = ±

53. y = 3x 4 − 4 x3 + 1 Intercepts (0, 1) and (1, 0) [the latter is found by inspection of the equation]. No symmetry. y′ = 12 x3 − 12 x 2 = 12 x 2 ( x − 1) CV: x = 0 and x = 1 Decreasing on (–∞, 0) and (0, 1); increasing on (1, ∞); relative minimum at (1, 0). y ′′ = 36 x 2 − 24 x = 12 x(3 x − 2)

Possible inflection points at x = 0 and x =

2 . 3

3

⎛2 ⎞ Concave up on (–∞, 0) and ⎜ , ∞ ⎟ ; concave 3 ⎝ ⎠ ⎛ 2⎞ down on ⎜ 0, ⎟ ; inflection points at (0, 1) and ⎝ 3⎠ ⎛ 2 11 ⎞ ⎜ , ⎟. ⎝ 3 27 ⎠

y

x 3

492


Section 13.3

55. y = 4 x 2 − x 4 = x 2 (2 + x)(2 − x) Intercepts (0, 0) and (±2, 0) Symmetric about the y-axis.

(

3

y′ = 8 x − 4 x = 4 x 2 − x = 4x

(

2+x

)(

2−x

)

2

1⎞ ⎛ 1 ,− ⎟ ⎜± 4⎠ 2 ⎝ 1⎞ ⎛ y ′′ = 12 x 2 − 2 = 12 ⎜ x 2 − ⎟ 6⎠ ⎝ ⎡ 1 ⎤⎡ 1 ⎤ = 12 ⎢ x + ⎥ ⎢x − ⎥ 6⎦⎣ 6⎦ ⎣

)

CV: x = 0, ± 2

( ) ( ) decreasing on ( − 2, 0 ) and ( 2, ∞ ) ; relative maxima at ( ± 2, 4 ) ; relative minimum at (0, 0).


Increasing on −∞, − 2 and 0, 2 ;

y

2 . x=± 3 ⎛ 2 ⎞ , ∞ ⎟⎟ ; and ⎜⎜ ⎝ 3 ⎠

⎛ 2⎞ Concave down on ⎜⎜ −∞, − ⎟ 3 ⎟⎠ ⎝ ⎛ 2 2⎞ concave up on ⎜⎜ − , ⎟⎟ ; inflection points at ⎝ 3 3⎠ ⎛ 2 20 ⎞ ⎜⎜ ± , ⎟⎟ . ⎝ 3 9 ⎠ 5

57. y = x1/ 3 ( x − 8) = x 4 / 3 − 8 x1/ 3 Intercepts (0, 0) and (8, 0) 4 8 y′ = x1/ 3 − x −2 / 3 3 3 4⎡ 2 ⎤ 4( x − 2) = ⎢ x1/ 3 − ⎥= 2 3⎣ x /3 ⎦ 3x2 / 3 CV: x = 0, 2 Decreasing on (–∞, 0) and (0, 2); increasing on (2, ∞); relative minimum at

x

( 2, − 63 2 ) ≈ ( 2, − 7.56) .

2

56. y = x − x = x ( x + 1)( x − 1) Intercepts (0, 0), (−1, 0), and (1, 0) Symmetric about the y-axis. y′ = 4 x 3 − 2 x = 2 x

CV: x = 0, ±

(

)(

2x + 1

3

x

y

2

.

2

5

4

6

⎛ ⎛ 1 ⎞ 1 ⎞ Concave up on ⎜ −∞, − , ∞⎟ ; ⎟ and ⎜ 6⎠ ⎝ 6 ⎠ ⎝ ⎛ 1 1 ⎞ concave down on ⎜ − , ⎟ ; inflection 6 6⎠ ⎝ ⎛ 1 5 ⎞ points at ⎜ ± , − ⎟. 36 ⎠ 6 ⎝

⎡2 ⎤ y ′′ = 8 − 12 x 2 = 12 ⎢ − x 2 ⎥ ⎣3 ⎦ ⎛ 2 ⎞⎛ 2 ⎞ = 12 ⎜⎜ − x ⎟⎟ ⎜⎜ + x ⎟⎟ ⎝ 3 ⎠⎝ 3 ⎠

Possible inflection points when

1

4 −2 / 3 16 −5 / 3 + x x 9 9 4⎡ 1 4 ⎤ 4( x + 4) = ⎢ + = 9 ⎣ x 2 / 3 x5 / 3 ⎥⎦ 9 x5 / 3 Possible inflection points when x = –4, 0. Concave up on (–∞, –4) and (0, ∞); concave y ′′ =

)

2x −1

1 2

(

1 ⎞ 1 ⎞ ⎛ ⎛ Decreasing on ⎜ −∞, − ⎟ and ⎜ 0, ⎟; 2⎠ 2⎠ ⎝ ⎝ ⎛ 1 ⎞ ⎛ 1 ⎞ increasing on ⎜ − , 0 ⎟ and ⎜ , ∞ ⎟; 2 ⎠ ⎝ ⎝ 2 ⎠ relative maximum at (0, 0); relative minima at

down on (–4, 0); inflection points at −4, 12 3 4

)

and (0, 0). Observe that at the origin the tangent line exists but it is vertical.

493


21


y

2

59. y = 4 x1/ 3 + x 4 / 3 = x1/ 3 (4 + x) Intercepts (0, 0) and (–4, 0) ⎤ 4 4 4⎡ 1 y′ = x −2 / 3 + x1/ 3 = ⎢ + x1/ 3 ⎥ 2 / 3 3 3 3 ⎣x ⎦ 4(1 + x) = 3x 2 / 3 CV: x = 0, –1 Decreasing on (–∞, –1); increasing on (–1, 0) and (0, ∞); rel. min at (–1, –3) 8 4 4⎡ 1 2 ⎤ − y ′′ = − x −5 / 3 + x −2 / 3 = ⎢ ⎥ 2 / 3 5 9 9 9 ⎣x x /3 ⎦

x 16

8 3

(2, –6 2 )

58. y = ( x − 1) 2 ( x + 2)2 Intercepts (0, 4), (1, 0), (–2, 0) y′ = ( x − 1)2 [2( x + 2)] + ( x + 2)2 [2( x − 1)] = 2(x – 1)(x + 2)(2x + 1) 1 CV: x = −2, − , 1 2 ⎛ 1 ⎞ Decreasing on (–∞, –2) and ⎜ − , 1⎟ ; increasing ⎝ 2 ⎠

=

4( x − 2)

9 x5 / 3 Possible inflection points when x = 0, 2. Concave up on (–∞, 0) and (2, ∞); concave down on

(

)

(0, 2); inflection point at (0, 0) and 2, 6 3 2 .

1⎞ ⎛ on ⎜ −2, − ⎟ and (1, ∞); relative maximum at 2⎠ ⎝

Observe that at the origin the tangent line exists but it is vertical.

⎛ 1 81 ⎞ ⎜ − 2 , 16 ⎟ ; relative minima at ⎝ ⎠

y

7

3

(2, 6 2 )

(–2, 0) and (1, 0); y′ = 2(2 x3 + 3 x 2 − 3x − 2), so

(

)

y ′′ = 6 2 x 2 + 2 x − 1 . Setting y ′′ = 0 and using x

the quadratic formula gives possible inflection −1 ± 3 . Concave up on points at x = 2 ⎛ ⎛ −1 + 3 ⎞ −1 − 3 ⎞ , ∞ ⎟ ; concave ⎜⎜ −∞, ⎟⎟ and ⎜⎜ ⎟ 2 ⎠ 2 ⎝ ⎝ ⎠

–4

Intercepts (0, 2), (−1, 0) and (−4, 0) 1 y ′ = ( x + 1) ⋅ + x + 4(1) 2 x+4 1 [( x + 1) + 2( x + 4)] = 2 x+4 3( x + 3) = 2 x+4 CV: x = −3, −4 Decreasing on (−4, −3); increasing on (−3, ∞); relative minimum at (−3, −2) 1 3 x + 4(1) − ( x + 3) ⋅ 2 x + 4 y ′′ = ⋅ 2 2 x+4

−1 ± 3 2

8

5

60. y = ( x + 1) x + 4 [Note: x ≥ −4]

⎛ −1 − 3 −1 + 3 ⎞ down on ⎜ , ⎟ ; inflection points ⎜ 2 2 ⎟⎠ ⎝

when x =

–1

y

x

(

5

)

3 2( x + 4) − ( x + 3) 3( x + 5) ⋅ = 3/ 2 4 ( x + 4) 4( x + 4)3 / 2 No possible inflection point. Concave up on (−4, ∞). =

494


5

Section 13.3 Observe that at the origin the tangent line exists but it is vertical.

y

8

y

x 5

x

61. y = 6 x 2 / 3 −

8

⎛ x1/ 3 ⎞ x = 6 x 2 / 3 ⎜1 − ⎟ ⎜ 2 12 ⎟⎠ ⎝

63.

Intercepts (0, 0) and (1728, 0) ⎞ 1⎛8− 3 x ⎞ 1 1⎛ 8 = ⎜ − 1⎟ = ⎜ ⎟ 2 2 ⎝ 3 x ⎠ 2 ⎜⎝ 3 x ⎟⎠ CV: x = 0, 512 Increasing on (0, 512); decreasing on (−∞, 0) and (512, ∞); relative maximum at (512, 128); relative minimum at (0, 0). 4 4 y ′′ = − x −4 / 3 = − 3 3x 4 / 3 Possible inflection point at x = 0. Concave down on (−∞, 0) and (0, ∞). Observe that at the origin the tangent line exists but it is vertical. y ′ = 4 x −1/ 3 −

y

8

y

4 x 2

64.

8

8

y

4

x

300 4

x 2500

65.

5

8

y

1

x 1

62. y = 5 x 2 / 3 − x5 / 3 = x 2 / 3 (5 − x) Intercepts (0, 0) and (5, 0) ⎤ 10 −1/ 3 5 2 / 3 5 ⎡ 2 − x = ⎢ − x2 / 3 ⎥ y′ = x 1/ 3 3 3 3⎣x ⎦ =

66.

8

5

y

5(2 − x)

3 x1/ 3 CV: x = 0, 2 Increasing on (0, 2); decreasing on (–∞, 0) and (2, ∞); relative minimum at (0, 0); relative

(

3

4 x

)

3

maximum at 2, 3 4 ≈ (2, 4.76) 10 −4 / 3 10 −1/ 3 10(1 + x) − x =− x 9 9 9 x4 / 3 Possible inflection point when x = 0, –1. Concave up on (–∞, –1); concave down on (–1, 0), and (0, ∞); inflection point at (–1, 6). y ′′ = −

495



100 q+2

67. p =

g ′′( x) = −

dp 100 =− < 0 for q > 0, so p is dq (q + 2)2 d2 p

=

200

= > 0 for q > 0, dq 2 (q + 2)3 the demand curve is concave up.

decreasing. Since

=

c ′′ =

1

e

U0 A

2 2 ⎡ −x ⎛ x ⎞ −x ⎤ ⎢x ⋅ e 2A ⎜ − ⎟ + e 2A ⎥ A ⎣⎢ ⎝ A⎠ ⎦⎥

2

⋅e

− 2x A

⋅e

− 2x A

2

(x

2

−A

)

( x + A )( x − A )

A , then g ′′( x) < 0 , so the graph is

71. y = 12.5 + 5.8(0.42) x

q2

y′ = 5.8(0.42) x ln(0.42) Since ln(0.42) < 0, we have y′ < 0 , so the function is decreasing. y ′′ = 5.8(0.42) x ln 2 (0.42) > 0 , so the function is concave up.

2

q3 Since c′′ > 0 for q > 0, the graph of the average cost function is concave up for q > 0.

69. S = f ( A) = 12 4 A , 0 ≤ A ≤ 625 . For the given

72. H = 1.00 ⎡1 − e −(0.0464t + 0.0670 ⎤ ⎣ ⎦ dH − (0.0464t + 0.0670) = 0.0464e > 0 , so H is dt increasing. d 2H = −(0.0464)2 e −(0.0464t + 0.0670) < 0 , so H is dt 2 concave down.

− 34

values of A we have S ′ = 3 A > 0 and ⎛ 9 ⎞ −7 S ′′ = − ⎜ ⎟ A 4 < 0 . Thus y is increasing and ⎝4⎠ concave down. 60

A2

U0 A

concave down. If x > A , then g ′′( x) > 0 , so the graph is concave up.

c 1 = q+2+ q q

c′ = 1−

U0 A

A2 If 0 ≤ x <

68. c = q 2 + 2q + 1 c=

e

e

S

73. n = f (r ) = 0.1ln(r ) +

a.

A 625 U0

−x

2

70. g ( x) = e A e 2 A , A > 0, x ≥ 0 (since x represents quantity). g ′( x) = −

e

U0 A

⎡ − x2 ⎤ ⎢ xe 2 A ⎥ A ⎢⎣ ⎥⎦

496

7 − 0.8 , 1 ≤ r ≤ 10 r

dn 0.1 7 0.1r − 7 0.1(r − 70) = − = = <0 dr r r2 r2 r2 for 1 ≤ r ≤ 10. Thus the graph of f is always falling. Also, d 2n 0.1 14 14 − 0.1r =− + = 2 dr r 2 r3 r3 0.1(140 − r ) = >0 r3 for 1 ≤ r ≤ 10. Thus the graph is concave up.


b.

10

Section 13.3

f(r)

⎛ 2a ⎞ concave downon ⎜ , 0 ⎟ . In either case, y has ⎝ 3 ⎠ 2a two points of inflection, when x = 0 and x = . 3

6

2

76. r 1

10 –3

c.

dn dr

3

= −0.26 , so the rate of decrease is r =5

0.26.

–2

74.

Two inflection points

150

77. y = x3 − 2 x 2 + x + 3 y′ = 3 x 2 − 4 x + 1 When x = 2, then y = 5 and y′ = 5 . Thus an equation of the tangent line at x = 2 is y – 5 = 5(x – 2), or y = 5x – 5. Graphing the curve and the tangent line indicates that the curve lies above the tangent line around x = 2. Thus the curve is concave up at x = 2.

20

–20 –50

a.

One relative maximum point

10

b. One relative minimum point c.

One inflection point 10

75.

–2 –2

3 0

4

78.

f ( x ) = 2 x3 + 3 x 2 − 6 x + 1 f ′( x) = 6 x 2 + 6 x − 6 f ′′( x) = 12 x + 6

–12

Two inflection points

The relative minimum of f ′ occurs at a value of x for which ( f ′( x))′ = f ′′( x) = 0. Around this value of x, ( f ′( x))′ goes from − to +. Since ( f ′( x))′ = f ′′( x), the concavity of f must change from concave down to concave up.

y = x5 ( x − a) = x 6 − ax5 y ′ = 6 x5 − 5ax 4 y ′′ = 30 x 4 − 20ax3 = 10 x3 (3 x − 2a) Possible inflection points when x = 0 and 2a x= . If a > 0, y is concave up on (−∞, 0) and 3 ⎛ 2a ⎞ ⎛ 2a ⎞ ⎜ , ∞ ⎟ ; concave down on ⎜ 0, ⎟ . If a < 0, 3 ⎠ ⎝ 3 ⎠ ⎝ 2a ⎞ ⎛ y is concave up on ⎜ −∞, ⎟ and (0, ∞); 3 ⎠ ⎝

79.

f ( x ) = x 6 + 3 x5 − 4 x 4 + 2 x 2 + 1 f ′( x) = 6 x5 + 15 x 4 − 16 x3 + 4 x f ′′( x) = 30 x 4 + 60 x3 − 48 x 2 + 4

Inflection points of f when x ≈ –2.61, –0.26.

497


80.

f ( x) =


x +1

1 . 4 Because there is only one relative extremum and f is continuous, the relative maximum is an absolute maximum.

Thus there is a relative maximum when x =

2

x +1

f ′( x ) = −

x2 + 2 x − 1

( x + 1) 2 ( x + 3 x − 3x − 1) f ′′( x) = ( x + 1) 2

2

3

4. y = 3x 2 − 5 x + 6 y′ = 6 x − 5

2

2

3

CV: x =

Inflection points of f when x ≈ −3.73, −0.27, 1.00.

y ′′ = 6 ⎛5⎞ y ′′ ⎜ ⎟ = 6 > 0 ⎝6⎠

Problems 13.4 1. y = x 2 − 5 x + 6 y′ = 2 x − 5 CV: x = y ′′ = 2

5 6

5 . 6 Because there is only one relative extremum and f is continuous, the relative minimum is an absolute minimum.

Thus there is a relative minimum when x =

5 2

⎛5⎞ y ′′ ⎜ ⎟ = 2 > 0 ⎝2⎠

5. y =

1 3 x + 2 x2 − 5x + 1 3

y ′ = x 2 + 4 x − 5 = ( x + 5)( x − 1)

5 . 2 Because there is only one relative extremum and f is continuous, the relative minimum is an absolute minimum.

Thus there is a relative minimum when x =

CV: x = −5, 1 y ′′ = 2 x + 4 y ′′(−5) = −6 < 0 ⇒ relative maximum when x = −5 y ′′(1) = 6 > 0 ⇒ relative minimum when x = 1

2

2. y = 5 x + 20 x + 2 y′ = 10 x + 20 CV: x = −2 y ′′ = 10 y ′′(−2) = 10 > 0

6. y = x3 − 12 x + 1 y′ = 3x 2 − 12 = 3( x + 2)( x − 2) CV: x = ±2 y ′′ = 6 x y ′′(−2) = −12 < 0 ⇒ relative maximum when x = –2 y ′′(2) = 12 > 0 ⇒ relative minimum when x=2

Thus there is a relative minimum when x = −2. Because there is only one relative extremum and f is continuous, the relative minimum is an absolute minimum. 3. y = −4 x 2 + 2 x − 8 y′ = −8 x + 2

7. y = − x3 + 3 x 2 + 1

1 CV: x = 4 y ′′ = −8

y′ = −3 x 2 + 6 x = −3x( x − 2) CV: x = 0, 2 y ′′ = −6 x + 6 y ′′(0) = 6 > 0 ⇒ relative minimum when x = 0

⎛1⎞ y ′′ ⎜ ⎟ = −8 < 0 ⎝4⎠

y ′′(2) = −6 < 0 ⇒ relative maximum when x = 2

498


Section 13.4

8. y = x 4 − 2 x 2 + 4

⎛1⎞ y ′′ ⎜ ⎟ = 60 > 0 ⇒ relative minimum when ⎝3⎠ 1 x= 3

y′ = 4 x3 − 4 x = 4 x( x + 1)( x − 1) CV: = 0, ±1 y ′′ = 12 x 2 − 4 y ′′(0) = −4 < 0 ⇒ relative maximum when x = 0 y ′′(1) = 8 > 0 ⇒ relative minimum when x = 1

12. y =

y ′′(−1) = 8 > 0 ⇒ relative minimum when x = –1

y ′ = 55 x 2 − 2 x − 21 = (5 x + 3)(11x − 7) 3 7 CV: x = − , 5 11 y ′′ = 110 x − 2

9. y = 7 − 2 x 4 y′ = −8 x3 CV: x = 0 y ′′ = −24 x 2

⎛ 3⎞ y ′′ ⎜ − ⎟ = −68 < 0 ⇒ relative maximum when ⎝ 5⎠ 3 x=− 5 ⎛7⎞ y ′′ ⎜ ⎟ = 68 > 0 ⇒ relative minimum when ⎝ 11 ⎠

Since y ′′(0) = 0 , the second-derivative test fails. Using the first-derivative test, we see that f increases for x < 0 and f decreases for x > 0, so there is a relative maximum when x = 0.

x=

10. y = −2 x 7 y′ = −14 x 6 CV: x = 0 y ′′ = −84 x5

(

11. y = 81x5 − 5 x

(

)(

2

)

CV: x = –2, –5, −

)

(

7 2

)

y ′′ = 2 ⎡ x 2 + 7 x + 10 (2) + (2 x + 7)(2 x + 7) ⎤ ⎣⎢ ⎦⎥ ′′ y (−5) = 18 > 0 ⇒ relative minimum when x = –5 ⎛ 7⎞ y ′′ ⎜ − ⎟ = −9 < 0 ⇒ relative maximum when ⎝ 2⎠

)

)

= 5(3 x + 1)(3x − 1) 9 x 2 + 1

CV: x = ±

2

= 2( x + 2)( x + 5)(2 x + 7)

= 5 9x −1 9x + 1

(

(

)

y′ = 2 x 2 + 7 x + 10 (2 x + 7)

y′ = 81 ⋅ 5 x 4 − 5 = 5 81x 4 − 1

(

11 7

13. y = x 2 + 7 x + 10

Since y ′′(0) = 0 , the second-derivative test fails. However, using the first-derivative test, we see that f decreases for x < 0 and for x > 0, so there is neither a relative maximum nor a relative minimum when x = 0.

2

55 3 x − x 2 − 21x − 3 3

7 2 y ′′(−2) = 18 > 0 ⇒ relative minimum when x = –2 x=−

1 3

y ′′ = 81 ⋅ 5 ⋅ 4 x3

14. y = − x3 + 3 x 2 + 9 x − 2

⎛ 1⎞ y ′′ ⎜ − ⎟ = −60 < 0 ⇒ relative maximum when ⎝ 3⎠ 1 x=− 3

y ′ = −3x 2 + 6 x + 9 = −3( x 2 − 2 x − 3) = −3( x + 1)( x − 3)

CV: x = −1, 3 y ′′ = −6 x + 6 y ′′(−1) = 12 > 0 ⇒ relative minimum when x = −1 499



y ′′(3) = −12 < 0 ⇒ relative maximum when x=3

4 x When x = 0 the denominator is zero but the numerator is not zero, so x = 0 is a vertical asymptote. ⎛4⎞ ⎛4⎞ lim ⎜ ⎟ = 0. Similarly, lim ⎜ ⎟ = 0, so x →∞ ⎝ x ⎠ x →−∞ ⎝ x ⎠ y = 0 is a horizontal asymptote.

5. y = f ( x) =

Problems 13.5 x x −1 When x = 1 the denominator is zero but the numerator is not zero. Thus x = 1 is a vertical asymptote. x x lim = lim = lim 1 = 1 . x →∞ x − 1 x →∞ x x →∞ Similarly lim f ( x) = 1 . Thus the line y = 1 is a

1. y = f ( x) =

=

x2 − 2

x2 x2 When x = 0 the denominator is zero but the numerator is not. Thus x = 0 is a vertical ⎛ 2 ⎞ asymptote. lim ⎜ 1 − ⎟ = 1 − 0 = 1. Similarly x →∞ ⎝ x2 ⎠ lim f ( x) = 1, so y = 1 is a horizontal

x →−∞

horizontal asymptote. x +1 x When x = 0 the denominator is zero but the numerator is not. Thus x = 0 is a vertical x +1 x = lim = lim 1 = 1 . asymptote. lim x →∞ x x →∞ x x →∞ Similarly lim f ( x) = 1 . Thus y = 1 is a

2. y = f ( x) =

x →−∞

asymptote. 1 ( x − 1)( x + 1) x −1 Vertical asymptotes are x = 1 and x = –1. 1 1 lim = lim = 0 . Similarly, 2 x →∞ x − 1 x →∞ x 2 lim f ( x ) = 0 . Thus y = 0 is a horizontal

7. y = f ( x) =

x →−∞

horizontal asymptote. 3.

2

6. y = f ( x) = 1 −

x+2 3x − 5 5 When x = the denominator is zero but the 3 5 numerator is not. Thus x = is a vertical 3 x 1 1 asymptote. lim f ( x) = lim = lim = . x →∞ x →∞ 3 x x →∞ 3 3 1 1 Similarly lim f ( x) = . Thus y = is a 3 3 x →−∞ horizontal asymptote. f ( x) =

1

=

2

x →−∞

asymptote. x ( x − 2)( x + 2) x −4 Vertical asymptotes: x = 2, x = –2. x x 1 lim = lim = lim = 0 . Similarly, 2 2 x →∞ x − 4 x →∞ x x →∞ x lim f ( x ) = 0 . Thus y = 0 is a horizontal

8. y = f ( x) =

x

=

2

x →−∞

asymptote. 9. y = f ( x) = x 2 − 5 x + 5 is a polynomial function, so there are no horizontal or vertical asymptotes.

2x + 1 4. y = f ( x) = 2x + 1 Observe that both the numerator and 1 1 denominator are zero for x = − . For x ≠ − , 2 2 we have f(x) = 1. Thus f is a constant function 1 for x ≠ − . Hence there are no vertical or 2 horizontal asymptotes.

10. y = f ( x) =

=

500

x4 x3 − 4

=

x4 x3 −

(3 4)

3

x4 ( x − 22 / 3 )( x 2 + 22 / 3 x + 24 / 3 )

=

x4 x3 − (22 / 3 )3


Section 13.5

Vertical asymptote: x = 22 / 3. x4

4x

= x+

so the line y = x is an

x3 − 4 x3 − 4 oblique asymptote.

11.

2 x2

16.

=

17.

asymptote. x3 is a polynomial function, so there are 5 no horizontal or vertical asymptotes. f ( x) =

13. y =

2 x 2 + 3x + 1 2

x −5

=

2 x2 + 3x + 1

( x − 5 )( x + 5 )

2 x2

x →∞

x →−∞

asymptote. 14. y = f ( x) =

x2 − 1 (2 x − 1)( x − 4)

3 − x4

=

3 − x4

x3 + x 2 x 2 ( x + 1) Vertical asymptotes are x = 0 and x = –1. 3 − x4 3 − x2 so the line y = −x + 1 = −x +1+ x3 + x 2 x3 + x 2 is an oblique asyptote. x 2 + 4 x3 + 6 x 4

3x2 Observe that both the numerator and the denominator are zero when x = 0. For x ≠ 0, we have x2 1 f ( x) = (1 + 4 x + 6 x 2 ) = (1 + 4 x + 6 x 2 ). 2 3 3x Thus f is a polynomial function for x ≠ 0. Hence there are neither horizontal nor vertical asymptotes.

= lim 2 = 2 x →∞ x2 Similarly, lim = 2. Thus y = 2 is a horizontal x →∞

f ( x) =

18. y = f ( x) =

Vertical asymptotes are x = − 5 and x = 5. lim f ( x) = lim

2x2 − 9 x + 4

=

1 and x = 4. 2 x2 1 1 = lim = ,and lim f ( x) = lim 2 x →∞ x →∞ 2 x 2 x →∞ 2 1 1 lim f ( x) = . Thus y = is a horizontal 2 2 x →−∞ asymptote.

x →−∞

12.

x2 − 1

Vertical asymptotes are x =

2 x2 x 2 + x − 6 ( x + 3)( x − 2) Vertical asymptotes are x = –3 and x = 2. 2 x2 lim f ( x) = lim = lim 2 = 2 , and x →∞ x →∞ x 2 x →∞ lim f ( x) = 2 . Thus y = 2 is a horizontal f ( x) =

f ( x) =

2 x3 + 1 3x(2 x − 1)(4 x − 3)

19. y = f ( x) =

1 , and 2 3 2 x3 1 1 = lim = . x = . lim f ( x) = lim 4 x→∞ x →∞ 24 x3 x →∞ 12 12 1 1 . Thus y = is a Similarly, lim f ( x) = 12 12 x →−∞ horizontal asymptote.

x 2 − 3x − 4

x2 − 3x − 4

(1 + 2 x)2 1 From the denominator, x = − is a vertical 2 asymptote. x2 1 1 = lim = , and lim f ( x) = lim 2 4 x →∞ x →∞ 4 x x →∞ 4 1 1 lim f ( x) = , so y = is a horizontal 4 4 x →−∞ asymptote.

Vertical asymptotes are x = 0, x =

2 5 x − 13 +5 = x−3 x−3 From the denominator, x = 3 is a vertical asymptote. 5x = lim 5 = 5, and lim f ( x) = lim x →∞ x →∞ x x →∞ lim f ( x) = 5. Thus, y = 5 is a horizontal

15. y = f ( x) =

20. y = f ( x) =

1 + 4 x + 4 x2

=

x4 + 1 1 − x4

=

x4 + 1

(1 + x ) (1 − x)(1 + x) 2

From the denominator, vertical asymptotes are x = 1 and x = –1.

x →−∞

asymptote.

501


lim f ( x) = lim

x4


24.

= lim − 1 = −1 , and

− x 4 x →∞ lim f ( x) = −1 . Thus y = –1 is a horizontal

x →∞

x →∞

x →∞

asymptote. 9 x 2 − 16

=

(3 x + 4)(3 x − 4)

2(3 x + 4) 2(3 x + 4)2 4 When x = − , both the numerator and 3 denominator are zero. Since 3x − 4 lim f ( x) = lim = −∞ , the + + 2(3 x + 4) x →−4 / 3 x →−4 / 3

line x = − lim

x →∞

2

3 x Symmetric about the origin. Vertical asymptote 3 3 , so y = 0 is a is x = 0. lim = 0 = lim x →∞ x x →−∞ x horizontal asymptote. 3 y′ = − x2 CV: None, however x = 0 must be included in the inc.-dec. analysis. Decreasing on (–∞, 0) and (0, ∞). 6 y ′′ = x3 No possible inflection point, but we include x = 0 in the concavity analysis. Concave down on (–∞, 0); concave up on (0, ∞).

25. y =

4 is a vertical asymptote. 3

9 x 2 − 16 2(3x + 4)

= lim

2

9 x2

x →∞ 18 x

Similarly, lim f ( x) = x →−∞

2

1 1 = . 2 x →∞ 2

= lim

1 1 . Thus y = is a 2 2

horizontal asymptote. 2 2x 24 x 2 + 20 x − 4 + = 5 12 x 2 + 5 x − 2 5(12 x 2 + 5 x − 2) 4( x + 1)(6 x − 1) = 5(3 x + 2)(4 x − 1)

22. y = f ( x) =

5

5

1 24 x 2 2 2 = lim = . x = . lim f ( x) = lim 2 4 x→∞ x →∞ 60 x x →∞ 5 5 2 2 Similarly, lim f ( x) = . Thus, y = is a 5 5 x →−∞ horizontal asymptote.

26. y =

2 2x − 3

2⎞ ⎛ Intercept: ⎜ 0, − ⎟ 3⎠ ⎝ 3 . 2 lim y = 0 = lim y, so y = 0 is a horizontal

Vertical asymptote is x =

23. y = f ( x) = 2e x + 2 + 4 We have lim f ( x) = +∞ and

x →∞

x →−∞

asymptote. 4 y′ = − (2 x − 3) 2

x →∞

lim f ( x) = 2 ⋅ lim e x + lim 4 x →−∞

y

x

2 1 When x = − or x = , the denominator is 0, 3 4 but the numerator is not. Thus, vertical 2 asymptotes are x = − and 3

x →−∞

x →−∞

is a horizontal asymptote. There is no vertical asymptote because f(x) neither increases nor decreases without bound around any fixed value of x.

x →−∞

21. y = f ( x) =

f ( x) = 12e− x lim f ( x) = 0 and lim f ( x) = +∞ . Thus y = 0

x →−∞

= 2(0) + 4 = 4 Thus y = 4 is a horizontal asymptote. There is no vertical asymptote because f(x) neither increases nor decreases without bound around any fixed value of x. 502


Section 13.5

3 must be considered in the 2 3⎞ ⎛ inc. dec. analysis. Decreasing on ⎜ −∞, ⎟ and 2⎠ ⎝

(Note: x > 0) x lim y = 0 , so y = 0 is a horizontal asymptote.

x →∞

lim y = +∞ , so the line x = 0 is a vertical

x →0+

⎛3 ⎞ ⎜ 2 , ∞ ⎟. ⎝ ⎠ 16 y ′′ = (2 x − 3)3

asymptote. 5 y′ = − < 0 for x > 0. Decreasing on (0, ∞). x3 15 y ′′ = > 0 for x > 0. Concave up on (0, ∞). 2 x5

3 must be 2 considered in the concavity analysis. Concave 3⎞ ⎛ ⎛3 ⎞ down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ . 2⎠ ⎝ ⎝2 ⎠

No possible inflection point, but x =

5

10

28. y =

CV: None, but x =

16

y

y x 16

x 5

29. y = x 2 + x x −1 Intercept (0, 0) Vertical asymptote is x = 1 lim y = 1 = lim y , so y = 1 is a horizontal

x4 + 1

x 2 = 0 gives x = 0 as the only vertical asymptote. Because the degree of the numerator is greater than the degree of the denominator, no horizontal asymptote exists.

x →−∞

asymptote. ( x − 1)(1) − x(1) 1 y′ = =− 2 ( x − 1) ( x − 1)2 CV: None, but x = 1 must be included in the inc.-dec. analysis. Decreasing on (–∞, 1) and (1, ∞). 2 y ′′ = ( x − 1)3 No possible inflection point, but x = 1 must be included in concavity analysis. Concave up on (1, ∞), concave down on (−∞, 1). y

=

x2 x2 x ≠ 0, so there is no y-intercept. Setting y = 0 ⇒ no x-intercept. Replacing x by –x yields symmetry about the y-axis. Setting

27. y =

x →∞

1

y = x 2 + x −2 y′ = 2 x − 2 x =

(

2

)

−3

= 2x −

2 x

3

2 x + 1 ( x + 1)( x − 1)

=

2 x4 − 2 x

3

=

(

)

2 x4 − 1 x

3

. x3 CV: x = ±1, but x = 0 must be included in the inc.-dec. analysis. Decreasing on (–∞, –1) and (0, 1); increasing on (–1, 0) and (1, ∞); relative minima at (–1, 2) and (1, 2), 6 y ′′ = 2 + >0 x4 for all x ≠ 0. Concave up on (–∞, 0) and (0, ∞).

5

x 5

503


8


y

y

50

x 5 x 5

3x2 − 5 x − 1 x−2 ⎛ 1⎞ Intercept: ⎜ 0, ⎟ ⎝ 2⎠ Vertical asymptote is x = 2. 3x 2 − 5 x − 1 1 = 3x + 1 + so y = 3x + 1 is an x−2 x−2 oblique asymptote.

1 ( x 1)( x − 1) + x −1 Intercept (0, –1) Symmetric about the y-axis. Vertical asymptotes are x = –1 and x = 1. 1 1 = 0 = lim , so y = 0 is a lim x →∞ x 2 − 1 x →−∞ x 2 − 1 horizontal asymptote. 2x y′ = − 2 2 x −1

31. y =

30. y =

y′ = =

( x − 2)(6 x − 5) − (3 x 2 − 5 x − 1)(1)

( x − 2)2

6± 3 , 3 but x = 2 must be included in the inc.-dec. ⎛ 6− 3 ⎞ analysis. Increasing on ⎜⎜ −∞, ⎟ and 3 ⎟⎠ ⎝

( x − 1) y ′′ = −2 ⋅ 2

=

2

(

)

(1) − x ⎡ 4 x x 2 − 1 ⎤ ⎢⎣ ⎥⎦

( x − 1) ( x − 1) ⎡⎣⎢( x − 1) − 4x ⎤⎦⎥ = −2 ⋅ ( x − 1) 2 ( 3 x + 1) 2 ( 3 x + 1) = = ( x − 1) [( x + 1)( x − 1)] 2

⎛6+ 3 ⎞ ⎛6− 3 ⎞ , ∞ ⎟⎟ ; decreasing on ⎜⎜ , 2 ⎟⎟ and ⎜⎜ ⎝ 3 ⎠ ⎝ 3 ⎠ ⎛ 6+ 3 ⎞ ⎜⎜ 2, ⎟ ; relative maximum at 3 ⎟⎠ ⎝ ⎛6− 3 ⎞ , 7 − 2 3 ⎟⎟ ; relative minimum at ⎜⎜ ⎝ 3 ⎠ ⎛6+ 3 ⎞ , 7 + 2 3 ⎟⎟ . ⎜⎜ ⎝ 3 ⎠

=

)

CV: x = 0, but x = ±1 must be included in the inc.-dec. analysis. Increasing on (–∞, –1) and (–1, 0); decreasing on (0, 1) and (1, ∞); relative maximum at (0, –1).

3 x 2 − 12 x + 11

From the quadratic formula, CV: x =

y ′′ =

=

(

( x − 2)2

2

1

2

2

2

2

2

4

2

2

4

2

3

3

No possible inflection point, but x = ±1 must be considered in the concavity analysis. Concave up on (–∞, –1) and (1, ∞); concave down on (–1, 1). 5

2

( x − 2) (6 x − 12) − (3 x − 12 x + 11)2( x − 2)

y

( x − 2)4 ( x − 2)(6 x − 12) − 2(3 x 2 − 12 x + 11)

x 5

( x − 2)3 2

( x − 2)3 No possible inflection point, but x = 2 must be included in the concavity analysis. Concave down on (−∞, 2); concave up on (2, ∞) 504


Section 13.5

1

32. y =

y ′′ =

2

(1 − x )3 No possible inflection point, but x = 1 must be included in the concavity analysis. Concave up on (–∞, 1); concave down on (1, ∞).

x +1 Intercept (0, 1) Symmetric about the y-axis. 1 1 = 0 = lim , so y = 0 is a lim 2 2 x →∞ x + 1 x →−∞ x + 1 horizontal asymptote. −2 x y′ = 2 2 x +1

(

4

y

3 5x

)

CV: x = 0 Increasing on (–∞, 0); decreasing on (0, ∞); relative maximum at (0, 1) y ′′ =

(

)

2 3x 2 − 1

(

)

x2 + 1

34. y =

3

Possible inflection points at x = ±

1 3

x2 Intercept is (−1, 0) Vertical asymptote is x = 0. 1+ x x 1 = lim = lim = 0 lim x →∞ x 2 x →∞ x 2 x →∞ x 1+ x = lim , so y = 0 is the only horizontal x →−∞ x 2 asymptote. x+2 y′ = − x3 CV: x = −2, but x = 0 must be included in the inc-dec. analysis. Increasing on (−2, 0); decreasing on (−∞, −2) and (0, ∞); relative 1⎞ ⎛ minimum at ⎜ −2, − ⎟ . 4⎠ ⎝

. Concave

⎛ ⎛ 1 ⎞ 1 ⎞ up on ⎜ −∞, − , ∞ ⎟ ; concave ⎟ and ⎜ 3 3 ⎝ ⎠ ⎝ ⎠ ⎛ 1 1 ⎞ down on ⎜ − , ⎟ ; inflection points at 3 3⎠ ⎝ ⎛ 1 3⎞ , ⎟ ⎜± 3 4⎠ ⎝ 5

1+ x

y

x 5

y ′′ =

2(3 + x)

x4 Possible inflection point when x = 3, but x = 0 must be included in the concavity analysis. Concave up on (−3, 0) and (0, ∞); concave down 2⎞ ⎛ on (−∞, −3); inflection point at ⎜ −3, − ⎟ . 9⎠ ⎝

1+ x 1− x Intercepts: (0, 1) and (–1, 0). x = 1 is the only vertical asymptote. Since 1+ x x lim = lim = lim − 1 = −1 x →∞ 1 − x x →∞ − x x →∞ 1+ x = lim x →−∞ 1 − x the only horizontal asymptote is y = –1. (1 − x)(1) − (1 + x )(−1) 2 = y′ = 2 (1 − x) (1 − x) 2 No critical values, but x = 1 must be considered in the ind.-dec. analysis. Increasing on (–∞, 1) and (1, ∞).

33. y =

y

5

x 5

505



x2 7x + 4 Intercept: (0, 0)

x3 + 1 x Intercept: (–1, 0) Vertical asymptote is x = 0. Because the degree of the numerator is greater than the degree of the denominator, no horizontal asymptote exists.

35. y =

36. y =

Vertical asymptote is x = −

4 . 7

1 4 x2 1 4 16 so y = x − = x− + 7 49 7x + 4 7 49 49(7 x + 4) is an oblique asymptote. (7 x + 4)(2 x) − x 2 (7) y′ = (7 x + 4)2 =

7 x2 + 8x (7 x + 4)

2

=

Since y = x 2 + x −1 , y′ = 2 x − x −2 = 2 x −

x(7 x + 8) (7 x + 4) 2

⎛ 1 1⎞ minimum at ⎜ 3 , 33 ⎟ . ⎜ 2 4 ⎟⎠ ⎝

4⎞ ⎛ 8 and (0, ∞); decreasing on ⎜ − , − ⎟ and 7⎠ ⎝ 7

y ′′ = 2 + 2 x

)

2

(

)

(14 x + 8) − 7 x 2 + 8 x [14(7 x + 4)] (7 x + 4)4

(

−3

2

= 2+

3

10

=

(

)

2 x3 + 1 3

y

)

(7 x + 4) ⎡ (7 x + 4)(14 x + 8) − 14 7 x 2 + 8 x ⎤ ⎢⎣ ⎥⎦ = 4 (7 x + 4) =

.

x2

x x Possible inflection point when x = –1, but x = 0 must be included in concavity analysis. Concave up on (–∞, –1) and (0, ∞); concave down on (–1, 0); inflection point at (–1, 0).

16 ⎞ ⎛ 4 ⎞ ⎛ 8 ⎜ − 7 , 0 ⎟ ; relative maximum at ⎜ − 7 , − 49 ⎟ ; ⎝ ⎠ ⎝ ⎠ relative minimum at (0, 0). +4

x2

2 x3 − 1

1 , but x = 0 must be included in inc.2 dec. analysis. Decreasing on (–∞, 0) and ⎛ ⎛ 1 ⎞ 1⎞ ⎜⎜ 0, 3 ⎟⎟ ; increasing on ⎜⎜ 3 , ∞ ⎟⎟ ; relative 2⎠ ⎝ ⎝ 2 ⎠

8 4 , but x = − must be included in 7 7 8⎞ ⎛ the inc.-dec. analysis. Increasing on ⎜ −∞, − ⎟ 7⎠ ⎝

2

=

CV: x = 3

CV: x = 0, −

(7x y ′′ =

1

x 10

32 (7 x + 4)3

4 must be 7 included in concavity analysis. Concave down 4⎞ ⎛ ⎛ 4 ⎞ on ⎜ −∞, − ⎟ ; concave up on ⎜ − , ∞ ⎟ . 7 7 ⎝ ⎠ ⎝ ⎠

No possible inflection point but x = −

37. y =

9 2

9x − 6x − 8

9 (3x + 2)(3 x − 4)

=

9⎞ ⎛ Intercept: ⎜ 0, − ⎟ 8⎠ ⎝ 2 4 Vertical asymptotes: x = − , x = 3 3 9 1 lim y = lim = lim = 0 = lim y x →∞ x →∞ 9 x 2 x →∞ x 2 x →−∞ Thus y = 0 is a horizontal asymptote. Since

y 3

x 3

(

y = 9 9 x2 − 6 x − 8

506

)

−1

,


(

y′ = 9(−1) 9 x 2 − 6 x − 8 =−

)

−2

Section 13.5

(18 x − 6)

54(3x − 1) [(3x + 2)(3x − 4)]2

1 2 4 , but x = − and x = must be included in inc.-dec. analysis. 3 3 3 2⎞ ⎛ ⎛ 2 1⎞ ⎛1 4⎞ ⎛4 ⎞ Increasing on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ; decreasing on ⎜ , ⎟ and ⎜ , ∞ ⎟ ; 3⎠ ⎝3 ⎠ ⎝ ⎝ 3 3⎠ ⎝3 3⎠

CV: x =

⎛1 ⎞ relative maximum at ⎜ , – 1⎟ . Finding y ′′ gives: 3 ⎝ ⎠

(9x y ′′ = −54 ⋅

2

− 6x − 8

)

2

(

)

(3) − (3 x − 1) ⎡ 2 9 x 2 − 6 x − 8 (18 x − 6) ⎤ ⎢⎣ ⎥⎦

(9x

2

− 6x − 8

)

4

)( ) ( 9 x − 6 x − 8) −162 ( −27 x + 18 x − 12 ) 486 ( 9 x − 6 x + 4 ) = = [(3x + 2)(3 x − 4)] 9 x − 6 x − 8 ( ) = −54 ⋅

(

3 9 x 2 − 6 x − 8 ⎡ 9 x 2 − 6 x − 8 − 4(3x − 1)(3x − 1) ⎤ ⎣⎢ ⎦⎥ 4

2

2

2

3

2

3

Since 9 x 2 − 6 x + 4 = 0 has no real roots, y ′′ is never zero. No possible inflection points, but x = −

2 4 and x = must be included in concavity analysis. Concave up on 3 3

⎛4 ⎞ and ⎜ , ∞ ⎟ ; concave down on ⎝3 ⎠ 3

2⎞ ⎛ ⎜ −∞, − 3 ⎟ ⎝ ⎠

⎛ 2 4⎞ ⎜− 3 , 3 ⎟ . ⎝ ⎠

y

x 3

38. y =

8 x2 + 3x + 1 2x2

8 x 2 + 3 x + 1 is never 0 and x cannot be zero. Thus no intercepts. Vertical asymptote is x = 0. lim y = lim

x →∞

x →∞

8x2 2x2

= lim 4 = 4 = lim y x →∞

x →−∞

Thus y = 4 is a horizontal asymptote. Since y = 4 + y′ = −

3 −1 1 −2 x + x , we have 2 2

3 −2 1 3x + 2 x − x −3 = − x −3 (3 x + 2) = − 2 2 2 x3 507



1⎞ ⎛ 4 ⎜ − , − ⎟. 3 12 ⎝ ⎠

2 CV: x = − , but x = 0 must be included in the 3 2⎞ ⎛ inc. dec. analysis. Decreasing on ⎜ −∞, − ⎟ and 3⎠ ⎝ ⎛ 2 (0, ∞); increasing on ⎜ − , ⎝ 3

y ′′ = −3 ⋅

⎞ 0 ⎟ ; relative ⎠

= −3 ⋅

y

y 2

3x + 1 (3 x − 2)2

x 3

⎛ 1 ⎞ ⎛ 1⎞ Intercepts: ⎜ − , 0 ⎟ , ⎜ 0, ⎟ ⎝ 3 ⎠ ⎝ 4⎠ 2 Vertical asymptote is x = . 3 3x 1 lim y = lim = lim = 0 = lim y x →∞ x →∞ 9 x 2 x →∞ 3 x x →−∞ Thus y = 0 is a horizontal asymptote.

y′ = =

3(3 x − 2)2 [(3 x − 2) − 3(3x + 4)]

7 Possible inflection point when x = − , but 3 2 x = must be included in concavity analysis. 3 7⎞ ⎛ Concave down on ⎜ −∞, − ⎟ ; concave up on 3⎠ ⎝ ⎛ 7 2⎞ ⎛2 ⎞ ⎜ − , ⎟ and ⎜ , ∞ ⎟ ; inflection point at 3 3 3 ⎝ ⎠ ⎝ ⎠ 2 ⎞ ⎛ 7 ⎜ − , − ⎟. 27 ⎠ ⎝ 3

x 10

39. y =

(3x − 2)6

(3x − 2)6 3(−6 x − 14) 18(3 x + 7) = −3 ⋅ = (3x − 2) 4 (3x − 2) 4

⎛ 2 23 ⎞ minimum at ⎜ − , ⎟. ⎝ 3 8 ⎠ 3 y ′′ = 3 x −3 + 3x −4 = ( x + 1). x4 Possible inflection point when x = −1, but x = 0 must be included in the concavity analysis. Concave down on (−∞, −1); concave up on (−1, 0) and (0, ∞); inflection point at (−1, 3). 16

(3x − 2)3 (3) − (3x + 4)(3)(3x − 2)2 (3)

40. y =

(6 x + 5)2

⎛ 1 ⎞ ⎛ 1 ⎞ Intercepts: ⎜ − , 0 ⎟ , ⎜ 0, ⎟ ⎝ 3 ⎠ ⎝ 25 ⎠ 5 Vertical asymptote is x = − . 6 3x 1 = lim = 0 = lim y lim y = lim 2 x →∞ x →∞ 36 x x →∞ 12 x x →−∞ Thus y = 0 is horizontal asymptote. (6 x + 5)2 (3) − (3 x + 1)[12(6 x + 5)] y′ = (6 x + 5)4 3(6 x + 5)[(6 x + 5) − 4(3 x + 1)] = (6 x + 5)4 3(−6 x + 1) −3(6 x − 1) = = (6 x + 5)3 (6 x + 5)3

(3x − 2)2 (3) − (3 x + 1)(2)(3x − 2)(3) (3 x − 2) 4 3(3 x − 2)[(3 x − 2) − 2(3 x + 1)]

=−

3x + 1

(3 x − 2) 4 3(3 x + 4) (3 x − 2)3

4 2 CV: x = − , but x = must be included in 3 3 inc.-dec. analysis. 4⎞ ⎛ ⎛2 ⎞ Decreasing on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 3⎠ ⎝ ⎝3 ⎠ 4 2 ⎛ ⎞ increasing on ⎜ − , ⎟ ; relative minimum at ⎝ 3 3⎠ 508


Section 13.5

1 5 , but x = − must be included in 6 6 5⎞ ⎛ inc.-dec. analysis. Decreasing on ⎜ −∞, − ⎟ and 6⎠ ⎝

CV: x =

⎛1 ⎞ ⎜ 6 , ∞ ⎟ ; increasing on ⎝ ⎠

41. y =

⎛ 5 1⎞ ⎜ − 6 , 6 ⎟ ; relative ⎝ ⎠

= −3 ⋅

(

(

)

inc.-dec. analysis. Increasing on − 3, 0 and

( 0, 3 ) ; decreasing on ( −∞, − 3 ) and

4

(

(6 x + 5)4

2 Possible inflection point when x = , but 3 5 x = − must be included in concavity analysis. 6 5⎞ ⎛ ⎛ 5 2⎞ Concave down on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ; 6 ⎝ ⎠ ⎝ 6 3⎠

)

(

)

y ′′ = 2 x −3 − 12 x −5 = 2 x −5 x 2 − 6 =

(

2 x2 − 6 5

)

x Possible inflection points when x = ± 6 , but x = 0 must be included in the concavity analysis.

(

) concave up on ( − 6, 0 ) and (

(

)

Concave down on −∞, − 6 and 0, 6 ;

⎛2 1 ⎞ ⎜ 3 , 27 ⎟ . ⎝ ⎠ y

⎛ −5 6 ⎞ ⎜⎜ − 6, ⎟. 36 ⎟⎠ ⎝

x 3

3

y

3 – 3

509

x 3

)

6, ∞ ;

⎛ 5 6⎞ inflection points at ⎜ 6, ⎟ and ⎜ 36 ⎟⎠ ⎝

( 16 , 241 ) ( 23 , 271 )

6

⎛ 2 3⎞ 3, ∞ ; relative maximum at ⎜ 3, ⎟; ⎜ 9 ⎟⎠ ⎝

⎛ 2 3⎞ relative minimum at ⎜ − 3, − ⎟. ⎜ 9 ⎟⎠ ⎝

⎛2 ⎞ concave up on ⎜ , ∞ ⎟ ; inflection point at ⎝3 ⎠

–5

3 − x2

x4 CV: x = ± 3 , but x = 0 must be included in the

−12 x + 8

3

)

y′ = − x −2 + 3x −4 = x −4 − x 2 + 3 =

(6 x + 5)6 (6 x + 5) 3x − 2

( x + 1)( x − 1)

asymptote. Since y = x −1 − x −3 , then

6(6 x + 5)2 [(6 x + 5) − 3(6 x − 1)]

= −18 ⋅ = 72 ⋅

(6 x + 5)3 (6) − (6 x − 1) ⎡18(6 x + 5)2 ⎤ ⎣ ⎦ 6 (6 x + 5)

3

=

x x3 Intercepts are (–1, 0) and (1, 0). Symmetric about the origin. Vertical asymptote x = 0. x2 − 1 x2 1 lim = lim = lim 3 3 x →∞ x x →∞ x x →∞ x 1− x = 0 = lim , so y = 0 is the only horizontal x →−∞ x 2

⎛1 1 ⎞ maximum at ⎜ , ⎟ . Finding y ′′ gives: ⎝ 6 24 ⎠ y ′′ = −3 ⋅

x2 − 1


42. y =

ISM: Introductory Mathematical Analysis at (0, 1).

3x

( x − 2)2 Intercept (0, 0) Vertical asymptote at x = 2 3x 3x 3 = lim = lim = 0 and lim 2 2 x →∞ x − 4 x + 4 x →∞ x x →∞ x 3x = 0, so y = 0 is the only lim x →−∞ x 2 − 4 x + 4 horizontal asymptote. −3( x + 2) y′ = ( x − 2)3 CV: x = −2, but x = 2 must be included in the inc.-dec. analysis. Decreasing on (−∞, −2) and (2, ∞); increasing on (−2, 2); relative maximum 3⎞ ⎛ at ⎜ −2, − ⎟ 8⎠ ⎝ y ′′ =

y ′′ =

=

(

(

)

( x + 1)(2 x + 2) − x 2 + 2 x [2] 3

=

2

y 5

x 5

6( x + 4)

16

44. y =

3x 4 + 1

x3 No intercepts Symmetric about the origin.

Vertical asymptote is x = 0.

y

Since y = 3 x + x −3 ,

1 x2 + x + 1 = x +1 x +1 Intercept: (0, 1). x = –1 is the only vertical asymptote. y = x is an oblique asymptote.

43. y = x +

)

( x + 1)(2 x + 1) − x 2 + x + 1 ( x + 1)

x2 + 2 x 2

=

−4

= 3−

3

2

10

4

=

(

3

= 3x +

1 x3

so

)

3 x 2 + 1 ( x + 1)( x − 1)

x x4 CV: ±1, but x = 0 must be considered in the inc.dec. analysis. Increasing on (–∞, –1) and (1, ∞); decreasing on (–1, 0) and (0, 1); relative maximum at (–1, –4); relative minimum at (1, 4). 12 y ′′ = x5 No possible inflection point, but x = 0 must be included in the concavity analysis. Concave down on (–∞, 0); concave up on (0, ∞).

x 10

(

3x 4 + 1

x y = 3x is an oblique asymptote.

y′ = 3 − 3 x

=

( x + 1)

4

( x + 1) ( x + 1)3 No possible inflection point, but x = –1 must be included in the concavity analysis. Concave down on (–∞, –1); concave up on (–1, ∞).

( x − 2)4 Possible inflection point when x = −4, but x = 2 must be included in the concavity analysis. Concave down on (−∞, −4); concave up on 1⎞ ⎛ (−4, 2) and (2, ∞); inflection point at ⎜ −4, − ⎟ . 3⎠ ⎝

y′ =

)

( x + 1)2 (2 x + 2) − x 2 + 2 x [2( x + 1)]

y

x( x + 2)

( x + 1) ( x + 1)2 CV: 0 and –2, but x = –1 must be included in the inc.-dec. analysis. Increasing on (–∞, –2) and (0, ∞); decreasing on (–2, –1) and (–1, 0); relative maximum at (–2, –3); relative minimum

x 10

510


45. y =

−3x 2 + 2 x − 5

−3 x 2 + 2 x − 5 (3 x + 1)( x − 1)

=

3x2 − 2 x − 1

Section 13.5

Note that −3x 2 + 2 x − 5 is never zero. Intercept: (0, 5) 1 Vertical asymptotes are x = − and x = 1. 3 lim y = lim

−3 x 2

= lim − 1 = −1 = lim y x →∞ x →−∞ 3x 2 Thus y = –1 is horizontal asymptote. x →∞

x →∞

( 3x y′ =

2

)

(

(3x − 2 x − 1) 2(3x − 1) ⎡( 3x − 2 x − 1) (−1) − ( −3 x ⎢⎣ = (3x − 2 x − 1) 2

2

2

12(3x − 1)

( 3x

2

)

− 2x −1

2

2

)

+ 2x − 5 ⎤ ⎥⎦

2

2

=

)

− 2 x − 1 (−6 x + 2) − −3x 2 + 2 x − 5 (6 x − 2)

12(3 x − 1)

=

[(3x + 1)( x − 1)]2

1 1 , but x = − and x = 1 must be included in inc.-dec. analysis. 3 3 1⎞ ⎛ ⎛ 1 1⎞ ⎛1 ⎞ Decreasing on ⎜ −∞, − ⎟ and ⎜ − , ⎟ ; increasing on ⎜ , 1⎟ and (1, ∞); relative minimum at 3 3 3 ⎝ ⎠ ⎝ ⎠ ⎝3 ⎠

CV: x =

( 3x y ′′ = 12 ⋅ ( 3x = 12 ⋅ = 12 ⋅

2

2

)

( 3x

) (

2

)

− 2x −1

4

)

− 2 x − 1 ⎡3 3 x 2 − 2 x − 1 − 2(3 x − 1)(6 x − 2) ⎤ ⎣⎢ ⎦⎥

( 3x

2

)

− 2 x − 1 (3) − (3 x − 1) ⎡ 2 3 x 2 − 2 x − 1 (6 x − 2) ⎤ ⎢⎣ ⎥⎦

−27 x 2 + 18 x − 7

( 3x

(

2

⎛1 7⎞ ⎜3, 2⎟. ⎝ ⎠

)

− 2x −1

3

=

2

)

− 2x −1

(

4

−12 27 x 2 − 18 x + 7

)

[(3x + 1)( x − 1)]3

1 and x = 1 must be included 3 1⎞ ⎛ ⎛ 1 ⎞ in concavity analysis. Concave down on ⎜ −∞, − ⎟ and (1, ∞); concave up on ⎜ − , 1⎟ . 3⎠ ⎝ ⎝ 3 ⎠

Since 27 x 2 − 18 x + 7 is never zero, there is no possible inflection point, but x = −

511


10


y

–1 3

5x

1 (3 x + 2)2 + 1 = 3x + 2 3x + 2 9 x 2 + 12 x + 5 = 3x + 2

46. y = 3x + 2 +

Note that 9 x 2 + 12 x + 5 is never zero. ⎛ 5⎞ Intercept: ⎜ 0, ⎟ ⎝ 2⎠ 2 Vertical asymptote is x = − ; oblique asymptote is y = 3x + 2. 3 y′ = 3 − = 3⋅

3

= 3⋅

(3x + 2) 2 2

9 x + 12 x + 3 (3 x + 2)

2

(3 x + 2)2 − 1 (3 x + 2)2

= 9⋅

(3 x + 1)( x + 1) (3x + 2)2

1 2 and x = −1, but x = − must be included in inc.-dec. analysis. Increasing on (−∞, −1) and 3 3 1 2 ⎛ ⎞ ⎛ ⎞ ⎛ 2 1⎞ ⎜ − , ∞ ⎟ ; decreasing on ⎜ −1, − ⎟ and ⎜ − , − ⎟ ; relative maximum at (−1, −2); relative minimum at 3⎠ 3⎠ ⎝ 3 ⎠ ⎝ ⎝ 3 ⎛ 1 ⎞ ⎜ − , 2 ⎟. ⎝ 3 ⎠ 18 y ′′ = −3(−2)(3 x + 2)−3 (3) = (3 x + 2)3 2 2⎞ ⎛ No possible inflection point, but x = − must be included in concavity analysis. Concave down on ⎜ −∞, − ⎟ ; 3⎠ 3 ⎝ ⎛ 2 ⎞ concave up on ⎜ − , ∞ ⎟ . 3 ⎝ ⎠

CV: x = −

y

10

x 5

512


5

Section 13.5

y

52. For y = 6 − 3e− x we have

(

)

⎛ 3 ⎞ lim 6 − 3e− x = lim ⎜ 6 − ⎟ = 6 − 3(0) = 6 x →∞ x →∞ ⎝ ex ⎠ Thus the line y = 6 is a horizontal asymptote for

x 5

the graph of y = 6 − 3e− x . For y = 6 + 3e− x , we

(

)

obtain lim 6 + 3e− x = 6 + 3(0) = 6 , so the line x →∞

48.

5

y = 6 is also a horizontal asymptote for the graph

y

of y = 6 + 3e− x . y

16 x 5

y = 6 + 3e–x

8 4

49.

5

y = 6 – 3e–x

y

(

)

⎛ 76 ⎞ 53. lim 150 − 76e−t = lim ⎜150 − ⎟ t →∞ t →∞ ⎝ et ⎠ = 150 – 0 = 150 Thus y = 150 is a horizontal asymptote.

x 5

54. 50.

5

x 16

1

y –15

15

x 5

–1

x ≈ –0.08, y = 0 55. a a 51. When x = − , then a + bx = 0 so x = − is a b b vertical asymptote. x x 1 1 = lim = lim = lim b x →∞ a + bx x →∞ bx x →∞ b 1 Thus y = is a horizontal asymptote. b

8

–5

5

–8

x ≈ ±2.45, x ≈ 0.67, y = 2 56.

10

–10

10

–10

513



0.15625

relative and absolute maximum when x = The other number is 20 − x =

–1.54255

–6.54255

In the standard window, two vertical asymptotes of the form x = k, where k > 0, are apparent (x ≈ 0.68 and x ≈ 7.32). By zooming around x = –4, another vertical asymptote is apparent (x = –4). Thus three vertical asymptotes exist.

(

A′ =

0

25 0

From the graph, it appears that lim y ≈ 0.48 . x →∞

Stream

Thus a horizontal asymptote is y ≈ 0.48. Algebraically, we have lim

)

1 (9000 − 30 x) 18 Setting A′ = 0 ⇒ x = 300 . Since 1 A′′(300) = (−30) < 0 , we have a maximum at 18 9000 − 15(300) x = 300. Thus y = = 250 . The 18 dimensions are 300 ft by 250 ft.

1

x →∞

40 . 3

3. We are given that 15x + 9(2y) = 9000, or 9000 − 15 x y= . We want to maximize area A, 18 where A = xy. ⎛ 9000 − 15 x ⎞ 1 2 A = xy = x ⎜ ⎟ = 18 9000 x − 15 x 18 ⎝ ⎠

–0.15625

57.

20 . 3

0.34e

0.7 x

4.2 + 0.71e0.7 x 0.34

= lim

x →∞ 4.2 e0.7 x

+ 0.71

=

=

y

0.34e0.7 x e0.7 x lim 0.7 x x →∞ 4.2 + 0.71e 0.7 x e

y x

1200 , and x want to minimize N = 2x + 6y. We have ⎛ 1200 ⎞ N = 2x + 6 y = 2x + 6 ⎜ ⎟, x> 0 ⎝ x ⎠

4. We are given that xy = 1200, or y =

0.34 ≈ 0.48 0 + 0.71

Problems 13.6 1. Let the numbers be x and 82 − x. Then if P = x(82 − x) = 82 x − x 2 , we have P ′ = 82 − 2 x. Setting P ′ = 0 ⇒ x = 41. Since P ′′ = −2 < 0, there is a maximum when x = 41. Because 82 − x = 41, the required numbers are 41 and 41.

N′ = 2 −

7200 x2

Setting N′ = 0 yields x 2 = 3600 , so x = 60. We 14, 400 have N ′′ = , so N ′′(60) > 0 and we have x3 a minimum. If x = 60, then y = 20. Thus N = 2(60) + 6(20) = 240 ft.

2. Let the numbers be x and 20 – x, where 0 ≤ x ≤ 20. Let P = (2 x)(20 − x) 2 = 2 x3 − 80 x 2 + 800 x .

y

dP = 0 gives Setting dx

y

y

y

x

2

P′ = 6 x − 160 x + 800 = 2(3 x − 20)( x − 20) = 0 , 20 or x = 20. P′ > 0 on 3 ⎛ 20 ⎞ ⎛ 20 ⎞ ⎜ 0, 3 ⎟ and P′ < 0 on ⎜ 3 , 20 ⎟ . Thus P has a ⎝ ⎠ ⎝ ⎠

from which x =

514

y

y


Section 13.6

5. c = 0.05q 2 + 5q + 500 Avg. cost per unit = c = c ′ = 0.05 −

500 q2

500

1 . The answer B does not depend on A because A represents the initial value of q, so it doesn’t change q over time.

revenue is maximum when p =

c 500 = 0.05q + 5 + q q

. Setting c ′ = 0 yields 9.

2

, q = 10, 000, q = ±100 . We q2 exclude q = –100 because q represents the 1000 number of units. Since c′′ = > 0 for q > 0, q3 c is an absolute minimum when q = 100 units. 0.05 =

a.

Setting f ′( p) = 0 gives −1 +

900 ( p + 10)2

=0,

= 1 , ( p + 10)2 = 900, ( p + 10)2 p + 10 = ±30 , from which p = 20.

Since f ′′( p ) =

−1800

< 0 for p = 20, we ( p + 10)3 have an absolute maximum of f(20) = 110 grams.

d 2C

= −0.0024 < 0 , a maximum ds 2 occurs when s = 50. Thus a minimum can occur only at an endpoint of the domain. If s = 0, then C = 0.08; if s = 60, then C = 2.96. Thus the minimum cost of $0.08 per hour occurs for s = 0 mi/h and might be due to depreciation, insurance, and so on.

9 , so we have an 11 9 absolute minimum of f (100) = 51 grams. 11

b. f(0) = 70 and f (100) = 51

2 D3 ⎛ C D ⎞ CD 10. R = D 2 ⎜ − ⎟ = − 2 3 ⎝2 3⎠ dR = CD − D 2 . This The rate of change of R is dD is the function to be maximized. Setting C d ⎛ dR ⎞ = C − 2 D = 0 gives D = . Since ⎜ ⎟ dD ⎝ dD ⎠ 2

7. p = –5q + 30 Since total revenue = (price)(quantity), r = pq = (−5q + 30)q = −5q 2 + 30q Setting r ′ = −10q + 30 = 0 ⇒ q = 3 . Since r ′′ = −10 < 0 , r is maximum at q = 3 units, for which the corresponding price is p = –5(3) + 30 = $15.

8. q = Ae

900 , where 0 ≤ p ≤ 100. p + 10

900

6. C = 0.12s − 0.0012 s 2 + 0.08 , where 0 ≤ s ≤ dC 60. Setting = 0 gives 0.12 – 0.0024s = 0 ⇒ ds s = 50. Since

f ( p) = 160 − p −

d 2 ⎛ dR ⎞ ⎜ ⎟ = −2 < 0 , the maximum rate of dD 2 ⎝ dD ⎠ C change occurs when D = . 2

− Bp

Revenue = r = pq = pAe− Bp r ′ = A[e− Bp (1) + pe− Bp (− B)] = A(1 − Bp )e− Bp ⎛1 ⎞ = AB ⎜ − p ⎟ e− Bp ⎝B ⎠ 1 Critical value: p = B 1 If p < , then r ′ > 0 and r is increasing. If B 1 p > , then r ′ < 0 and r is decreasing. Thus B

11. p = 85 − 0.05q c = 600 + 35q Profit = Total Revenue – Total Cost P = pq − c = (85 − 0.05q )q − (600 + 35q ) = −(0.05q 2 − 50q + 600) Setting P ′ = −(0.1q − 50) = 0 yields q = 500. Since P ′′(500) = −0.1 < 0, P is a maximum when q = 500 units. This corresponds to a price of p = 85 − 0.05(500) = $60 and a profit of P = $11,900. 515



12. Cost per unit = $3 10 p= q

MR =

20 1 = = MC. 60 3

2 2 10, 000 q − 40q + 3 q Profit = Total Revenue – Total Cost Since total revenue r = pq and total cost = c = cq , P = pq − cq c=

⎛2 ⎞ = q3 − 100q 2 + 3200q − ⎜ q3 − 40q 2 + 10, 000 ⎟ ⎝3 ⎠ 1 = q3 − 60q 2 + 3200q − 10, 000 3

13. p = 42 – 4q 80 c = 2+ q Total Cost = c = cq = 2q + 80 Profit = Total Revenue – Total Cost P = pq – c = (42 – 4q)q – (2q + 80)

P′ = q 2 − 120q + 3200 = (q − 40)(q − 80) Setting P′ = 0 gives q = 40 or 80. Evaluating profit at q = 0, 40, 80, and 120 gives P(0) = –10,000 130, 000 1 P (40) = = 43,333 3 3 98, 000 2 P (80) = = 32, 666 3 3 P(120) = 86,000 Thus the profit maximizing output is q = 120 units, and the corresponding maximum profit is $86,000.

)

P′ = −(8q − 40)

Setting P′ = −(8q − 40) = 0 gives q = 5. We find that P ′′ = −8 < 0 , so P has a maximum value when q = 5. The corresponding price p is 42 – 4(5) = $22. 14. p =

1 and MC = , then for q = 3600 we 3

15. p = q 2 − 100q + 3200 on [0, 120]

5 −3 Moreover, we have P ′′ = − q 2 < 0 for q > 0, 2 25 . The so P is maximum when q = 9 corresponding price is p = $6.

(

q

have MR =

Profit = Total Revenue – Total Cost P = pq – c ⎛ 10 ⎞ P=⎜ ⎟ q − (3q) = 10 q − 3q ⎜ q⎟ ⎝ ⎠ 25 5 − 3 = 0 yields q = Setting P′ = . 9 q

= − 4q 2 − 40q + 80

20

40

16. a.

q

c = cq = 2q3 − 42q 2 + 228q + 210 dc = 6q 2 − 84q + 228 = 6(q 2 − 14q + 38) dq Using the quadratic formula to solve dc = 0 gives q = 7 − 11 ≈ 3.68 or dq

1 2000 c= + 3 q q + 2000 3 Profit = Total Revenue − Total Cost q P = pq − c = 40 q − − 2000 3 20 1 − = 0 yields q = 3600. Setting P ′ = q 3 Total cost = c = cq =

q = 7 + 11 ≈ 10.32. Evaluating c at q = 3, 7 − 11, 7 + 11, and 12 gives 570, 434 + 44 11 ≈ 579.93, 434 − 44 11 ≈ 288.07, and 354, respectively. Thus the minimum cost is when q = 7 + 11 ≈ 10.32. c(10) = 290 and c(11) = 298, so production should be fixed at q = 10 for a minimum cost of $290.

Since P ′′ = −10q −3 / 2 < 0 for q > 0, it follows that P is a maximum when q = 3600. The 40 ≈ $0.67. Since corresponding price is p = 60

516


Section 13.6

b. c(7) = 434, so the minimum cost still occurs when q = 7 + 11 ≈ 10.32 and production should again be fixed at 10 units.

20. Note that as the number of units produced and sold increases from 0 to 600, the profit increases from 0 to (600)(400) = $24,000. Let q = number of units produced and sold beyond 600. Then the total profit P is given by P = (600)(40) + (40 − 0.05q )q

17. Total fixed costs = $1200, material-labor costs/unit = $2, and the demand 100 . equation is p = q

= 24, 000 + 40q − 0.05q 2 P ′ = 40 − 0.10q Setting P ′ = 0 yields q = 400. Since P ′′ = −0.10 < 0, P is a maximum when q = 400, that is, the total number of units = 600 + 400 = 1000.

Profit = Total Revenue – Total Cost P = pq – c 100 P= ⋅ q − (2q + 1200) q = 100 q − 2q − 1200

(

= 2 50 q − q − 600

21. See the figure in the text. Given that x 2 y = 32 ,

)

we want to minimize S = 4( xy ) + x 2 . Since

⎛ 25 ⎞ − 1⎟ = 0 yields q = 625. We Setting P′ = 2 ⎜ ⎜ q ⎟ ⎝ ⎠

y=

32 x2

, where x > 0, we have

⎛ 32 ⎞ 128 S = 4 x ⎜ ⎟ + x2 = + x 2 , from which 2 x ⎝x ⎠ 128 S′ = − + 2 x . Setting S′ = 0 gives x2 256 2 x3 = 128 , x3 = 64 , x = 4. Since S ′′ = +2, x3 we get S ′′(4) > 0 , so x = 4 gives a minimum. If

− 32

see that P ′′ = −25q < 0 for q > 0, so P is maximum when q = 625. When q = 625, 50 MR = = 2 = MC. When q = 625, then 625 p = $4. 18. Let x = number of $10 per month increases so the monthly rate is 400 + 10x and the number of rented apartments is 100 – 2x. Monthly revenue r is given by r = (rent/apt.) (no. of apt. rented) r = (400 + 10x)(100 – 2x) r ′ = (400 + 10 x)(−2) + (100 − 2 x)(10) = 200 – 40x = 40(5 – x) Setting r′ = 0 yields x = 5. Since r ′′ = −40 < 0 , then r is maximum when x = 5. This results in a monthly rate for an apartment of 400 + 10(5) = $450.

32 = 2 . The dimensions are 16 4 ft × 4 ft × 2 ft.

x = 4, then y =

22. See the figure in the text. We want to maximize V = x 2 y given that 4 xy + x 2 = 192 , or 192 − x 2 4x ⎛ 192 − x 2 ⎞ 1 V = x2 ⎜ ⎟ = 192 x − x3 , x > 0 ⎜ 4x ⎟ 4 ⎝ ⎠ 1 3 2 V ′ = 192 − 3 x = 64 − x 2 4 4 Setting V ′ = 0 gives x = 8. Since ⎛3⎞ V ′′ = ⎜ ⎟ (−2 x) , then V ′′(8) < 0 , so x = 8 gives ⎝4⎠ a maximum. If x = 8, then y = 4. The dimensions are 8 ft × 8 ft × 4 ft. y=

(

19. If x = number of $0.50 decreases, where 0 ≤ x ≤ 36, then the monthly fee for each subscriber is 18 – 0.50x, and the total number of subscribers is 4800 + 150x. Let r be the total (monthly) revenue. revenue = (monthly rate)(number of subscribers) r = (18 – 0.50x)(4800 +150x) r ′ = (18 − 0.50 x)(150) + (4800 + 150 x)(−0.50) = 300 –150x = 150(2 – x) Setting r ′ = 0 yields x = 2. Evaluating r when x = 0, 2, and 36, we find that r is a maximum when x = 2. This corresponds to a monthly fee of 18 – 0.50(2) = $17 and a monthly revenue r of $86,700.

(

) (

The volume is 82 (4) = 256 ft 3 .

517

)

)


23. V = x ( L − 2 x )


2

If S′ = 0 , then πr 3 − K = 0 , πr 3 = K ,

= L2 x − 4 Lx 2 + 4 x3 L where 0 < x < . 2

r=3

π

= 12 x 2 − 8 Lx + L2 = (2x − L)(6x − L) L L For 0 < x < , setting V ′ = 0 gives x = . 6 2 L ⎛ ⎞ Since V ′ > 0 on ⎜ 0, ⎟ and V ′ < 0 on ⎝ 6⎠ L ⎛L L⎞ ⎜ , ⎟ , V is maximum when x = . Thus the 6 2 6 ⎝ ⎠ L length of the side of the square must be in., 6 which results in a volume of 2

V = πr 2 h

240 , x > 0. We want x

(

K − πr 2 . Thus Equation 2πr

Setting V ′ = 0 gives r = h=

=

K − π 3Kπ

2π 3Kπ 2 3

2π

K K 3π

⋅

= K 3π K 3π

2 3

K . Thus 3π

K

2π 3Kπ =

K 3π

Note that since V ′′ = −3πr < 0 for r > 0, we have a maximum. 27. p = 600 − 2q c = 0.2q 2 + 28q + 200 Profit = Total Revenue – Total Cost P = pq – c

(

P = (600 − 2q )q − 0.2q 2 + 28q + 200

(2) K πr

> 0 for r > 0, we

(2) becomes Kr − πr 3 V= 2 dV K − 3πr 2 . = 2 dr

25. See the figure in the text. V = K = πr 2 h (1)

becomes 2K S= + πr 2 r

r3

(2)

From Equation (1), h =

⎛ 240 ⎞ A = ( x + 10)( y + 6) = ( x + 10) ⎜ + 6⎟ ⎝ x ⎠ 2400 = 300 + 6 x + x 2400 A′ = 6 − x2 Setting A′ = 0 gives x = 20. Since 4800 A′′ = > 0 for x = 20, we have a minimum. x3 Thus y = 12, so the dimensions are 20 + 10 by 12 + 6, that is, 30 in. × 18 in.

From Equation (1) h =

4K

26. See the figure in the text. S = K = 2πrh + πr 2 (1)

to minimize A where

S = 2πrh + πr

2 3

have a minimum.

L⎛ L⎞ 2L in 3 . ⎜L− ⎟ = 6⎝ 3⎠ 27

2

( Kπ )

K ⎛ K ⎞3 . =⎜ ⎟ =3 π π ⎝ ⎠

Note that since S ′′ = 2π +

3

24. Since xy = 240, then y =

1

K

h=

V ′ = L2 − 8Lx + 12 x 2

K . Thus π

2

(

= − 2.2q 2 − 572q + 200

. Thus Equation (2)

)

)

P′ = −(4.4q − 572) Setting P′ = 0 yields q = 130. Since P ′′ = −4.4 < 0 , P is maximum when q = 130 units. The corresponding price is p = 600 – 2(130) = $340, and the profit is P = $36,980. If a tax of $22/unit is imposed on the manufacturer, then the cost equation is

)

2 πr 3 − K dS 2K =− + 2πr = . dr r2 r2 518


Section 13.6 minimize the sum C of carrying costs and set-up costs. ⎡ ⎛ q ⎞⎤ ⎛ 1000 ⎞ C = 0.128 ⎢10 ⎜ ⎟ ⎥ + 40 ⎜ ⎟ 2 ⎣ ⎝ ⎠⎦ ⎝ q ⎠

c1 = 0.2q 2 + 28q + 200 + 22q = 0.2q 2 + 50q + 200 . The demand equation remains the same. Thus P1 = pq − c1

(

= (600 − 2q)q − 0.2q 2 + 50q + 200

(

= − 2.2q 2 − 550q + 200

)

)

= 0.64q +

C ′ = 0.64 −

P′1 = −(4.4q − 550)

c = 0.2q 2 + 28q + 200 . Revenue, both before and after the license fee, is given by

30. c = 0.004q3 + 20q + 5000 p = 450 – 4q Profit = Total Revenue – Total Cost P = pq − c

r = pq = 600q − 2q 2 . After the license fee, the cost equation is

(

c1 = c + 1000 = 0.2q 2 + 28q + 1200 and the profit is P1 = r − c1

(

= 600q − 2q

) − ( 0.2q

+ 28q + 1200

q2

40, 000 = 62,500 , 0.64 80, 000 >0, q = 250 (since q > 0). Since C ′′ = q3 C is minimum when q = 250. Thus the economic lot size is 250/lot (4 lots).

28. Original data: p = 600 –2q,

2

40, 000

Setting C′ = 0 yields q 2 =

Setting P′1 = 0 yields q = 125. Since P1′′ = −4.4 < 0 , P1 is maximum when q = 125 units. The corresponding price is p = $350 and the profit is P1 = $34,175 .

2

40, 000 q

= (450 − 4q )q − 0.004q3 + 20q + 5000

( P′ = − ( 0.012q + 8q − 430 ) = −2 ( 0.006q + 4q − 215 )

P = − 0.004q3 + 4q 2 − 430q + 5000

)

)

)

2

As in Problem 27, we find that P1 has a maximum when q = 130 units, which gives p = $340. Thus the profit-maximizing price and output remain the same. Since Profit = r − c1 = r − (c + 1000) = (r − c) − 1000, when q = 130 we have Profit = 36,980 – 1000 (from Problem 27) = $35,980

2

Setting P′ = 0 yields 0.006q 2 + 4q − 215 = 0 −4 ± 21.16 −4 ± 4.6 = 0.012 0.012 −4 + 4.6 = 50 . Since P Since q ≥ 0, choose q = 0.012 is increasing on [0, 50) and decreasing on (50, ∞), P is maximum when q = 50 units. q=

29. Let q = number of units in a production run. Since inventory is depleted at a uniform rate, q assume that the average inventory is . The 2 q ⎛ ⎞ value of average inventory is 10 ⎜ ⎟ , and ⎝2⎠

31. Let x = number of people over the 30. Note: 0 ≤ x ≤ 10. Revenue = r = (number attending)(charge/person) = (30 + x)(50 – 1.25x)

⎡ ⎛ q ⎞⎤ carrying costs are 0.128 ⎢10 ⎜ ⎟ ⎥ . The number ⎣ ⎝ 2 ⎠⎦ 1000 of production runs per year is , and total q

= 1500 + 12.5 x − 1.25 x 2 r ′ = 12.5 − 2.5 x Setting r′ = 0 yields x = 5. Since r ′′ = −2.5 < 0 , r is maximum when x = 5, that is, when 35 attend.

⎛ 1000 ⎞ set-up costs are 40 ⎜ ⎟ . We want to ⎝ q ⎠ 519


ISM: Introductory Mathematical Analysis 34. Let q = level of production. Total Cost Average Cost = c = q For 0 ≤ q ≤ 5000, we have 30q + 10q + 20, 000 20, 000 . c= = 40 + q q Note that total cost for 5000 units is 220,000. For q > 5000, ⎛ cost for those ⎞ (cost for first 5000) + ⎜ ⎟ units beyond 5000 ⎝ ⎠ c= q

32. Let N = horsepower of motor. (Total annual cost) = C = (Annual cost to lease) + (Annual operating cost) ⎛ 0.008 ⎞ C = (200 + 0.40 N ) + 80, 000 ⎜ ⎟ ⎝ N ⎠ 640 = 200 + 0.04 N + N 640 C ′ = 0.4 − N2 Setting C ′ = 0 yields N 2 = 1600, so N = 40 (since N > 0). Since C ′′ =

1280

> 0 for N > 0, C N3 is a minimum when N = 40 horsepower.

=

33. The cost per mile of operating the truck is s 0.165 + . Driver’s salary is $18/hr. The 200 700 number of hours for 700 mi trip is . Driver’s s 12, 600 ⎛ 700 ⎞ salary for trip is 18 ⎜ , or . The cost ⎟ s ⎝ s ⎠ of operating the truck for the trip is s ⎤ ⎡ 700 ⎢0.165 + . 200 ⎥⎦ ⎣ Total cost of trip is 12, 600 s ⎞ ⎛ C= + 700 ⎜ 0.165 + 200 ⎟⎠ s ⎝

220, 000 + [45(q − 5000) + 10(q − 5000)] q

c = 55 −

55, 000 q

If 0 < q ≤ 5000, then c ′ = −

20, 000

< 0 and q2 thus c is decreasing. If q > 5000, then 55, 000 > 0 and thus c is increasing. c′ = q2 Hence c is minimum when q = 5000 units.

35. Profit P is given by P = Total revenue – Total cost = Total revenue – (salaries + fixed cost) = 50q – (1000m + 3000)

( ) = 50 ( m − 15m + 72m − 60 ) , where 0 ≤ m ≤ 8 P′ = 50 ( 3m − 30m + 72 ) = 150 ( m − 10m + 24 ) = 150(m − 4)(m − 6) = 50 m3 − 15m2 + 92m − 1000m − 3000

12, 600

7 Setting C ′ = − + = 0 yields s 2 = 3600 , 2 2 s 25, 200 >0 or s = 60 (since s > 0). Since C ′′ = s3 for s > 0, C is a minimum when s = 60 mi/h.

3

2

2

2

Setting P′ = 0 gives the critical values 4 and 6. We now evaluate P at these critical values and also at the endpoints 0 and 8. P(0) = –3000 P(4) = 2600 P(6) = 2400 P(8) = 3400 Thus Ms. Jones should hire 8 salespeople to obtain a maximum weekly profit of $3400.

520


Section 13.6

36. Profit P is given by P = Total revenue – Total cost = pq – Total cost

P /ton, then 2 P ⎛ 24 − 6 x ⎞ ⎡ 12 − 3 x ⎤ = P ⎢x + PT = Px + ⎜ 2 ⎝ 5 − x ⎟⎠ 5 − x ⎦⎥ ⎣

= 400q − 50q 2 − Total cost. (q in hundreds) dP d = 400 − 100q − (Total cost) dq dq = 400 – 100q – Marginal cost 800 = 400 − 100q − q+5 = =

⎡ x 2 − 10 x + 22 ⎤ P′T = P ⎢ ⎥ 2 ⎢⎣ (5 − x) ⎥⎦ Setting P′T = 0 and using an argument similar

to that above, we find that PT is a maximum

400(q + 5) − 100q(q + 5) − 800 q+5

when x = 5 − 3 tons.

2

−100q − 100q + 1200 q+5

38. x = number of floors. Let R = rate of return. Total Revenue R= Total Cost 60, 000 x = (10 x)[120, 000 + 3000( x − 1)] + 1, 440, 000 2x = 2 x + 39 x + 48

−100(q + 4)(q − 3) q+5 Setting P′ = 0 gives the critical value 3 (since q > 0). We find that P′ > 0 for 0 < q < 3, and P′ < 0 for q > 3. Thus there is a maximum profit when q = 3000 jackets. =

R′ = 2 ⋅

37. x = tons of chemical A (x ≤ 4), 24 − 6 x y= = tons of chemical B, profit on 5− x A = $2000/ton, and profit on B = $1000/ton. ⎛ 24 − 6 x ⎞ Total Profit = PT = 2000 x + 1000 ⎜ ⎟ ⎝ 5− x ⎠

( x 2 + 39 x + 48) 2

R ′ = 0 when x = 48 = 4 3 (x ≥ 0). Since R is

(

)

increasing on 0, 4 3 and decreasing on

(4

)

3, ∞ , R is a maximum when

x = 4 3 ≈ 6.93. The number of floors in the building must be an integer, so we evaluate R when x = 6 and x = 7: R(6) ≈ 0.0377; R(7) ≈ 0.0378. Thus 7 floors should be built to maximize the rate of return.

⎡ 12 − 3 x ⎤ = 2000 ⎢ x + 5 − x ⎥⎦ ⎣ ⎡ (5 − x)(−3) − (12 − 3x)(−1) ⎤ P′T = 2000 ⎢1 + ⎥ (5 − x) 2 ⎣⎢ ⎦⎥ ⎡ 3 ⎤ = 2000 ⎢1 − ⎥ 2 ⎣⎢ (5 − x ) ⎦⎥

39. P ( j ) = Aj

L4 V 3 L2 +B V 1+ j

dP AL4 BV 3 L2 = − =0 dj V (1 + j )2

⎡ x 2 − 10 x + 22 ⎤ = 2000 ⎢ ⎥ 2 ⎢⎣ (5 − x) ⎥⎦ Setting P′T = 0 yields (by the quadratic formula) 10 ± 2 3 x= = 5± 3 2 Because x ≤ 4, choose x = 5 − 3 . Since PT is

Solving for (1 + j ) 2 gives (1 + j ) 2 =

40. a.

BV 4 AL2

d ⎛ 2al ⎞ 2al −2atr + v − = 1+ = 0 when ⎜ ⎟ dv ⎝ v ⎠ v2 v = −2al . Note that

)

increasing on ⎡ 0, 5 − 3 and decreasing on ⎣ 5 − 3, 4 ⎤ , PT is a maximum for x = 5 − 3 ⎦ tons. If profit on A is P/ton and profit on B is

(

48 − x 2

d2 ⎛ 2al ⎞ −4al −2atr + v − = > 0 for 2⎜ v ⎟⎠ dv ⎝ v3

521



a < 0, l > 0, and v > 0. Thus −2atr + v −

42. The profit function is given by

2al v

P = TR − TC = q3 − 20q 2 + 160q − (30q + 50) = q3 − 20q 2 + 130q − 50 where P is in thousands of dollars, q is in tons, and 0 ≤ q ≤ 12. From the graph, the maximum profit occurs when q = 12 tons. The corresponding maximum profit is $358,000 and the selling price per ton is $64,000.

is a minimum for v = −2al . b.

v = −2(−19.6)(20) = 784 = 28 ft/s.

c.

N=

−2(−19.6) (−2)(−19.6)(0.5) + 28 −

2( −19.6)(20) 28

500

≈ 0.5 cars/s = 0.5(3600) cars/h = 1800 cars/h d. When v = −2al , then −2a N = N (l ) = −2atr + −2al + −2a

=

–100

a

= −2atr + 2 −2al atr − −2al The relative change in N when l is reduced N (15) − N (20) . from 20 ft to 15 ft is N (20)

Chapter 13 Review Problems 3x 2 x 2 − 16 ( x + 4)( x − 4) When x = ±4 the denominator is zero and the numerator is not zero. Thus x = 4 and x = –4 are vertical asymptotes. 3x 2 3x 2 lim = lim = lim 3 = 3 x →∞ x 2 − 16 x →∞ x 2 x →∞ Similarly, lim y = 3 . Thus y = 3 is the only

1. y =

With a = −19.6 ft/s 2 and tr = 0.5 s, then −19.6

N (20) =

(−19.6)(0.5) − −2(−19.6)(20) ≈ 0.5185 −19.6 N (15) = (−19.6)(0.5) − −2(−19.6)(15) ≈ 0.5756 The relative change is N (15) − N (20) 0.5756 − 0.5158 ≈ ≈ 0.1101 N (20) 0.5158

41. c =

=

(

=

x →−∞

x+3 x 3 (3 − x) 9 x − 3x When x = 0 or x = 3, the denominator is zero and the numerator is not zero. Thus x = 0 and x = 3 are vertical asymptotes. x 1 1 lim y = lim = − lim = 0 3 x →∞ x x →∞ x →∞ −3 x 2 Similarly, lim y = 0. Thus y = 0 is the only

2. y =

dc 18 120 3q 2 − 18q − 120 = 3− − = dq q q2 q2 =

3x 2

horizontal asymptote.

c 120 = 3q + 50 − 18ln(q ) + ,q>0 q q

3 q 2 − 6q − 40

12

0

−2 al −2 al

x+3

2

=

x →−∞

)

horizontal asymptote.

2

q 3(q − 10)(q + 4)

3. y =

5x2 − 3

=

5x2 − 3

(3x + 2) 2 9 x 2 + 12 x + 4 2 When x = − , the denominator is zero and the 3 2 numerator is not zero. Thus x = − is a vertical 3 asymptote.

2

q Critical value is q = 10 since q ≥ 0. dc dc Since < 0 for 0 < q < 10, and > 0 for dq dq q > 10, we have a minimum when q = 10 cases. This minimum average cost is 3(10) + 50 – 18 ln 10 + 12 ≈ $50.55. 522


5x2

Chapter 13 Review

5 5 = 9 x →∞ x →∞ 9 x x →∞ 9 5 5 Similarly, lim y = . Thus y = is the only horizontal asymptote. 9 9 x →−∞ lim y = lim

4. y =

2

= lim

− x 2 − 30 x − 6 4 x + 1 3x + 1 − = 3x − 5 2 x − 11 (3x − 5)(2 x − 11)

5 5 11 11 or x = , the denominator is zero and the numerator is not zero. Thus x = and x = are 3 3 2 2 vertical asymptotes. − x2 1 ⎛ 1⎞ lim y = lim = lim ⎜ − ⎟ = − 2 6 x →∞ x →∞ 6 x x →∞ ⎝ 6 ⎠

When x =

Similarly, lim y = − x →−∞

5.

f ( x) = f ′( x) =

1 1 . Thus y = − is the only horizontal asymptote. 6 6

5x2 3 − x2 (3 − x 2 )(10 x) − 5 x 2 (−2 x) 2 2

=

(3 − x ) Thus x = 0 is the only critical value.

(

10 x(3 − x 2 + x 2 ) 2 2

(3 − x )

=

30 x (3 − x 2 ) 2

)

Note: Although f ′ ± 3 is not defined, ± 3 are not critical values because ± 3 are not in the domain of f. 6.

f ( x) = 8( x − 1)2 ( x + 6)4 f ′( x) = 8(2)( x − 1)( x + 6) 4 + 8( x − 1)2 (4)( x + 6)3 = 16( x − 1)( x + 6)3 [ x + 6 + 2( x − 1)] = 16( x − 1)( x + 6)3 (3x + 4)

Thus x = 1, x = –6, and x = − 7.

f ( x) =

4 are the critical values. 3

3

x +1 3 − 4x

1 −2 ⎤ ⎡ (3 − 4 x) ⎢ 13 ( x + 1) 3 ⎥ − ( x + 1) 3 (−4) ⎣ ⎦ f ′( x) = = (3 − 4 x) 2

2

1 ( x + 1) − 3 [(3 − 4 x) + 12( x + 1)] 3 2

(3 − 4 x)

=

8 x + 15 2

3( x + 1) 3 (3 − 4 x)2 15 3 3 f ′( x) is zero when x = − ; f ′( x) is not defined when x = –1 or x = . However is not in the domain of f. 8 4 4 15 Thus x = − and x = –1 are critical values. 8

523



− 5x

8.

13xe 6 f ( x) = 6x + 5

⎡ ⎛ ⎤ − 5x ⎞ − 5x − 5x (6 x + 5) ⎢ x ⎜ − 56 e 6 ⎟ + e 6 (1) ⎥ − xe 6 (6) ⎠ ⎣ ⎝ ⎦ f ′( x ) = 13 ⋅ (6 x + 5) 2 13 −e = ⋅ 6

− 56x

{(6 x + 5)[5 x − 6] + 36 x} (6 x + 5)2

(

=

)

{

}

2 13 − 30 x + 25 x − 30 ⋅ 5x 6 e 6 (6 x + 5) 2

2 13 −5 6 x + 5 x − 6 −65(2 x + 3)(3x − 2) = ⋅ = 5x 5x 6 2 e 6 (6 x + 5) 6e 6 (6 x + 5) 2

2 5 5 3 or x = . Although f ′( x) is not defined when x = − , − is not in the 3 6 6 2 2 3 domain of f. Thus x = − and x = are the only critical values. 3 2 f ′( x) is zero when x = −

9.

5 f ( x) = − x3 + 15 x 2 + 35 x + 10 3 f ′( x) = −5 x 2 + 30 x + 35 = −5( x 2 − 6 x − 7) = −5( x − 7)( x + 1) CV: x = −1 and x = 7. Decreasing on (−∞, −1) and (7, ∞); increasing on (−1, 7)

10.

2 x2

f ( x) =

( x + 1) 2 4 x( x + 1)2 − 2 x 2 (2)( x + 1)

f ′( x) =

4x

=

( x + 1) ( x + 1)3 CV: x = 0, but x = –1 is also considered in the inc.-dec. analysis. Increasing on (–∞, –1) and (0, ∞); decreasing on (–1, 0).

11.

f ( x) =

4

6x4 x2 − 3

( x − 3)( 4 x ) − x (2 x) f ′( x) = 6 ⋅ ( x − 3) 12 x ⎡ 2 ( x − 3) − x ⎤ 12 x ( x − 6 ) ⎣⎢ ⎦⎥ = = ( x − 3) ( x − 3) 2

3

2

2

3

2

2

=

(

4

2

2

3

2

2

2

)( x − 6 ) 2 3 )( x − 3 ) ⎤ ⎦

12 x3 x + 6

(

⎡ x+ ⎣ CV: x = 0, ± 6 , but x = ± 3 must also be considered in the inc.-dec. analysis. Decreasing on

( −∞, − 6 ) , ( 0, 3 ) , and (

)

(

)(

)

3, 6 ; increasing on − 6, − 3 , − 3, 0 and 524

(

)

6, ∞ .


12.

Chapter 13 Review

1 . Concave 2 1⎞ ⎛ ⎛1 ⎞ down on ⎜ −∞, ⎟ ; concave up on ⎜ , ∞ ⎟ . 2 2 ⎝ ⎠ ⎝ ⎠

3

f ( x ) = 4 5 x3 − 7 x

f ′′( x) is not defined when x =

1 f ′( x) = 4 ⋅ (5 x3 − 7 x) −2 / 3 (15 x 2 − 7) 3 4(15 x 2 − 7) = 3(5 x3 − 7 x)2 / 3 = =

4

(

15 x + 7

)(

15 x − 7

2

( 3⎡x ( ⎣ 4

16.

)

f ′( x) = 3x 2 + 4 x − 5 f ′′( x) = 6 x + 4 = 2(3 x + 2)

2/3

3[ x (5 x − 7)]

)( 15x − 7 ) 2/3 7 )( 5 x − 7 ) ⎤ ⎦

15 x + 7 5x +

2 . Concave down on 3 2⎞ ⎛ ⎛ 2 ⎞ ⎜ −∞, − 3 ⎟ ; concave up on ⎜ − 3 , ∞ ⎟ . ⎝ ⎠ ⎝ ⎠ f ′′( x) = 0 when x = −

7 7 , 0, ± 15 5 ⎛ 7⎞ ⎛ 7 7 ⎞ Increasing on ⎜⎜ −∞, − ⎟⎟ , ⎜⎜ − , − ⎟⎟ , 5 5 15 ⎝ ⎠ ⎝ ⎠ ⎛ 7 ⎛ 7 ⎞ 7⎞ , , ∞ ⎟⎟ ; decreasing on ⎜⎜ ⎟⎟ , and ⎜⎜ ⎝ 15 5 ⎠ ⎝ 5 ⎠ ⎛ ⎛ 7 ⎞ 7 ⎞ , 0 ⎟⎟ and ⎜⎜ 0, ⎜⎜ − ⎟. 15 ⎟⎠ ⎝ 15 ⎠ ⎝

CV: x = ±

13.

17.

= 3(2 x + 1)2 (2 x + 1 + 6 x + 4) = 3(2 x + 1)2 (8 x + 5) f ′′( x) = 3{(2 x + 1) 2 (8) + (8 x + 5)[2(2 x + 1)(2)]} = 12(2 x + 1)[2(2 x + 1) + 8 x + 5] = 12(2 x + 1)(12 x + 7)

f ( x) = x 4 − x3 − 14

1 7 or x = − . Concave 2 12 7⎞ ⎛ ⎛ 1 ⎞ up on ⎜ −∞, − ⎟ and ⎜ − , ∞ ⎟ ; concave 12 ⎠ ⎝ ⎝ 2 ⎠ 1⎞ ⎛ 7 down on ⎜ − , − ⎟ . 2⎠ ⎝ 12 f ′′( x) = 0 when x = −

f ′′( x) = 12 x 2 − 6 x = 6 x(2 x − 1) 1 . Concave up on 2 ⎛1 ⎞ ⎛ 1⎞ (–∞, 0) and ⎜ , ∞ ⎟ ; concave down on ⎜ 0, ⎟ . ⎝2 ⎠ ⎝ 2⎠ f ′′( x) = 0 when x = 0 or x =

f ( x) = f ′( x) =

x−2 x+2 ( x + 2)(1) − ( x − 2)(1)

f ′′( x) = −

( x + 2)2

18.

=

)

(

2

)

(

)

= 2 2 x3 − 3 x 2 − x + 1

4 ( x + 2) 2

(

)

f ′′( x) = 2 6 x 2 − 6 x − 1

8

f ′′( x) = 0 when 6 x 2 − 6 x − 1 = 0 ; by the

quadratic formula x =

on (−∞, −2); concave down on (−2, ∞)

1 15 . Concave up on ± 2 6

⎛ ⎛1 ⎞ 1 15 ⎞ 15 , ∞ ⎟ ; concave ⎜⎜ −∞, − ⎟⎟ and ⎜⎜ + ⎟ 2 6 ⎠ 6 ⎝ ⎝2 ⎠

1 = (2 x − 1) −1 f ( x) = 2x −1

⎛1 15 1 15 ⎞ down on ⎜ − , + ⎟. ⎜2 6 2 6 ⎟⎠ ⎝

f ′( x) = −2(2 x − 1) −2 f ′′ = 8(2 x − 1)3 =

(

f ( x) = x 2 − x − 1

f ′( x) = 2 x 2 − x − 1 (2 x − 1)

( x + 2)3 f ′′( x) is not defined when x = −2. Concave up

15.

f ( x) = (2 x + 1)3 (3 x + 2) f ′( x) = (2 x + 1)3 (3) + (3 x + 2)[3(2 x + 1) 2 (2)]

f ′( x) = 4 x3 − 3x 2

14.

f ( x ) = x3 + 2 x 2 − 5 x + 2

8 (2 x − 1)3 525


19.


f ( x) = 2 x3 − 9 x 2 + 12 x + 7

(

f ′( x) = 6 x 2 − 18 x + 12 = 6 x 2 − 3 x + 2

23.

)

f ( x) = f ′( x) =

5 23 2 − 13 1 − 13 5x + 2 x + x = x (5 x + 2) = 1 3 3 3 3x 3 2 CV: x = 0 and x = − 5 2⎞ ⎛ Increasing on ⎜ −∞, − ⎟ and (0, ∞); decreasing 5⎠ ⎝

2x +1

2 ⎛ 2 ⎞ on ⎜ − , 0 ⎟ . Relative maximum when x = − ; 5 5 ⎝ ⎠ relative minimum when x = 0.

x2 x 2 (2) − (2 x + 1)(2 x)

x4 2 x[ x − (2 x + 1)] 2(− x − 1) −2( x + 1) = = = x4 x3 x3 CV: x = –1, but x = 0 must be considered in inc.dec. analysis. Decreasing on (–∞, –1) and (0, ∞); increasing on (–1, 0). Relative minimum when x = –1.

21.

f ( x) =

24.

= x 2 ( x − 2)3 [4 x + 3( x − 2)] = x 2 ( x − 2)3 (7 x − 6)

CV: x = 0, 2,

x10 x5 + 10 5

f ′( x) =

=

x2

25. y = x5 − 5 x 4 + 3x

x2 − 4

(

)

( x − 4) 2

( ) ( x − 4)

2 x ⎡ x2 − 4 − x2 ⎤ ⎣⎢ ⎦⎥

=−

y′ = 5 x 4 − 20 x3 + 3

x 2 − 4 (2 x) − x 2 (2 x)

2

6 7

⎛ 6⎞ Increasing on (−∞, 0), ⎜ 0, ⎟ , and (2, ∞); ⎝ 7⎠ ⎛6 ⎞ decreasing on ⎜ , 2 ⎟ . Relative maximum when ⎝7 ⎠ 6 x = ; relative minimum when x = 2. 7

CV: x = 0 and x = −1 Decreasing on (−∞, −1); increasing on (−1, 0) and (0, ∞); relative minimum when x = −1 f ( x) =

f ( x) = x3 ( x − 2) 4 f ′( x) = x3 [4( x − 2)3 (1)] + ( x − 2)4 (3x 2 )

f ′( x) = x9 + x 4 = x 4 ( x5 + 1)

22.

2

f ′( x) =

= 6(x – 1)(x – 2) CV: x = 1 and x = 2 Increasing on (–∞, 1) and (2, ∞); decreasing on (1, 2). Relative maximum when x = 1; relative minimum when x = 2. 20.

5

2

f ( x) = x 3 ( x + 1) = x 3 + x 3

2

y ′′ = 20 x3 − 60 x 2 = 20 x 2 ( x − 3) Possible inflection points occur when x = 0 or x = 3. Concave down on (–∞, 0) and (0, 3); concave up on (3, ∞). Concavity changes at x = 3, so there is an inflection point when x = 3.

2

=

−8 x

( x − 4) 2

2

x2 + 2 1 2 = x + x −1 5x 5 5 1 y ′ = (1 − 2 x −2 ) 5 4 4 y ′′ = x −3 = 5 5 x3 y ′′ is never zero. Although y ′′ is not defined when x = 0, y is not continuous there. Thus there is no inflection point.

26. y =

8x

[( x + 2)( x − 2)]2 CV: x = 0, but x ±2 must be considered in inc.dec. analysis. Increasing on (–∞, –2) and (–2, 0); decreasing on (0, 2) and (2, ∞). Relative maximum when x = 0.

526


(

Chapter 13 Review

)

Possible inflections points occur when x = ±2 or 2 2 5 x=± =± . Concave up on (–∞, –2), 5 5

27. y = 4(3 x − 5) x 4 + 2 = 12 x5 − 20 x 4 + 24 x − 40 y′ = 60 x 4 − 80 x3 + 24 y ′′ = 240 x3 − 240 x 2 = 240 x 2 ( x − 1) Possible inflection points occur when x = 0 or x = 1. Concave down on (–∞, 0) and (0, 1); concave up on (1, ∞). Inflection point when x = 1.

28. y = x 2 + 2 ln(− x) y′ = 2 x +

⎛ 2 5⎞ ⎜⎜ −2, − ⎟ , and 5 ⎟⎠ ⎝

(Note: x < 0)

when x = ±2, ±

2 x 2

y ′′ = 2 −

⎛ 2 5 2 5⎞ , ⎜⎜ − ⎟ , and (2, ∞); concave down on 5 5 ⎟⎠ ⎝

2x − 2

=

x3 e

x

2

=

31.

2( x + 1)( x − 1)

= x3e − x

32.

y ′ = x3 (−e− x ) + e− x (3x 2 ) = −e− x ( x3 − 3 x 2 ) = e − x ( x3 − 6 x 2 + 6 x ) = xe− x ( x 2 − 6 x + 6) y ′′ is defined for all x and y ′′ is zero only when

x = 0 or x 2 − 6 x + 6 = 0. Using the quadratic formula on the second equation, the possible points of inflection occur when x = 0, 3 ± 3.

)

(

)

33.

Concave up on 0, 3 − 3 and 3 + 3, ∞ ;

(

)

(

)

(

)

2

(

)

= 36( x + 2)( x − 2)

x

and f is continuous on [–2, 0].

(5 x − 6) 2

(5 x − 6) 2 (1) − x[10(5 x − 6)] (5 x − 6) 4

(5 x − 6)[(5 x − 6) − 10 x] (5 x − 6)

4

=

−5 x − 6 (5 x − 6)3

5x + 6 (5 x − 6)3

6 . 5 Evaluating f at this value and at the endpoints 1 1 ⎛ 6⎞ and , f ⎜− ⎟ = − gives f (−2) = − 128 120 ⎝ 5⎠ f(0) = 0. Absolute maximum: f(0) = 0; absolute 1 ⎛ 6⎞ . minimum: f ⎜ − ⎟ = − 5 120 ⎝ ⎠

The only critical value on (–2, 0) is x = −

) (

)

2 ⎧ ⎫ y ′′ = 36 ⎨ x ⎡ 4 x x 2 − 4 ⎤ + x 2 − 4 (1) ⎬ ⎢ ⎥ ⎣ ⎦ ⎩ ⎭ 2 2 2 2 = 36 x − 4 ⎡ 4 x + x − 4 ⎤ = 36 x − 4 5 x 2 − 4 ⎣⎢ ⎦⎥

(

f ( x) =

=−

3

y′ = 36 x x 2 − 4

f ( x) = 2 x3 − 15 x 2 + 36 x and f is continuous on [0, 3].

=

Inflection points when x = 0, 3 ± 3. 30. y = 6 x − 4

f ( x) = 3x 4 − 4 x3 and f is continuous on [0, 2].

f ′( x) =

concave down on (−∞, 0) and 3 − 3, 3 + 3 .

2

2 5 . 5

f ′( x) = 6 x 2 − 30 x + 36 = 6( x − 2)( x − 3) The only critical value on (0, 3) is x = 2. Evaluating f at this value and at the endpoints gives f(0) = 0, f(2) = 28, f(3) = 27. Absolute maximum: f(2) = 28; absolute minimum: f(0) = 0.

y ′′ = −e− x (3 x 2 − 6 x) − ( x3 − 3 x 2 )(−e− x )

(

⎞ 2 ⎟ . Inflection points ⎟ ⎠

f ′( x) = 12 x3 − 12 x 2 = 12 x 2 ( x − 1) The only critical value on (0, 2) is x = 1. Evaluating f at this value and at the endpoints gives f(0) = 0, f(1) = –1, and f(2) = 16. Absolute maximum: f(2) = 16; absolute minimum: f(1) = –1.

x x x2 Possible inflection point occurs when x = –1. Concave up on (–∞, –1); concave down on (–1, 0). Inflection point when x = – 1.

29. y =

2

2

⎛2 5 , ⎜⎜ ⎝ 5

(

(

5x + 2

)

)(

(

5x − 2

)

)(

)

527


34.


f ( x) = ( x + 1)2 ( x − 1) 2 / 3 and f is continuous on [2, 3]. ⎡2 ⎤ f ′( x) = ( x + 1) 2 ⎢ ( x − 1) −1/ 3 ⎥ + ( x − 1)2 / 3 [2( x + 1)] ⎣3 ⎦ 2 −1/ 3 [( x + 1) + 3( x − 1)] = ( x + 1)( x − 1) 3 4 4( x + 1)(2 x − 1) = ( x + 1)( x − 1) −1/ 3 (2 x − 1) = 3 3( x − 1)1/ 3

There are no critical values on [2, 3]. Evaluating f at the endpoints gives f(2) = 9 and f (3) = 16(22 / 3 ) ≈ 25.4. Absolute maximum f (3) = 16(22 / 3 ) ≈ 25.4; absolute minimum: f(2) = 9 35.

(

)

f ( x) = x 2 + 1 e− x

a.

( )( ) ⎡ x + 1 − 2 x ⎤ = −e ) ⎦⎥ (x ⎣⎢(

f ′( x) = x 2 + 1 −e− x + e− x (2 x) = −e − x

−x

2

2

)

− 2x + 1

= −e− x ( x − 1)2 CV: x = 1 Decreasing on (–∞, 1) and (1, ∞). No relative extrema.

b.

{

(

f ′′( x) = − e− x [2( x − 1)] + ( x − 1)2 −e− x

)}

= e − x ( x − 1)[−2 + ( x − 1)] = e − x ( x − 1)( x − 3) Possible inflection points when x = 1, 3. Concave up on (–∞, 1) and (3, ∞); concave down on (1, 3).

(

)

(

)

Inflection points at (1, f (1)) = 1, 2e−1 and (3, f (3)) = 3, 10e−3 . 36. Let y = f ( x) =

a.

x 2

x −1

.

Replacing x by –x and y by –y yields − y =

x2 + 1

=−

x2 + 1

( x 2 − 1) 2 [( x + 1)( x − 1)]2 decreasing on (−∞, −1), (−1, 1), and (1, ∞).

d.

2

, or y =

x 2

x −1 (− x) − 1 graph is symmetric about the origin. No other symmetry exists.

b. Since f ′( x) = −

c.

−x

, which is the original equation. Thus the

, there are no critical values. f ′( x) < 0 for all x, so f(x) is

From (b), There are no relative extrema. 1 = 0 . Similarly, lim f ( x) = 0 . Thus the line x →−∞ x →−∞ x x →∞ y = 0 is a horizontal asymptote to the graph of f. lim f ( x) = lim

x

x →−∞ x 2

= lim

528

ISM: Introductory Mathematical Analysis y

e.

Chapter 13 Review

5

y

25

x

x

5

f.

5

From the graph it is clear that no absolute extrema exist.

39. y = x3 − 12 x + 20 Intercept: (0, 20) No symmetry; no asymptotes

37. y = x 2 − 2 x − 24 = ( x + 4)( x − 6) Intercepts: (–4, 0), (6, 0), (0, –24) No symmetry. No asymptotes. y′ = 2 x − 2 = 2( x − 1) CV: x = 1 Increasing on (1, ∞); decreasing on (–∞, 1); relative minimum at (1, –25). y ′′ = 2 No possible inflection point. Concave up on (–∞, ∞). 25

y′ = 3x 2 − 12

(

)

= 3 x 2 − 4 = 3( x + 2)( x − 2)

CV: x = ±2 Increasing on (–∞, –2) and (2, ∞); decreasing on (–2, 2); relative maximum at (–2, 36); relative minimum at (2, 4). y ′′ = 6 x Possible inflection point when x = 0. Concave up on (0, ∞); concave down on (–∞, 0); inflection point at (0, 20).

y

40

x 25

20

–2

(1, –25)

38. y = 2 x3 + 15 x 2 + 36 x + 9 Intercept: (0, 9) No symmetry; no asymptotes

2

x 10

40. y = x 4 − 4 x3 − 20 x 2 + 150 Intercept: (0, 150) No symmetry. No asymptotes.

y ′ = 6 x 2 + 30 x + 36 = 6( x 2 + 5 x + 6) = 6( x + 3)( x + 2)

y′ = 4 x3 − 12 x 2 − 40 x = 4 x( x 2 − 3 x − 10) = 4 x( x + 2)( x − 5) CV: x = 0, –2, 5. Increasing on (–2, 0) and (5, ∞); decreasing on (–∞, –2) and (0, 5); relative maximum at (0, 150); relative minima at (–2, 118) and (5, –225).

CV: x = −3, −2 Increasing on (−∞, −3) and (−2, ∞); decreasing on (−3, −2); relative maximum at (−3, −18); relative minimum at (−2, −19) y ′′ = 12 x + 30 = 6(2 x + 5)

(

y ′′ = 12 x 2 − 24 x − 40 = 4 3x 2 − 6 x − 10

5 Possible inflection point when x = − . 2 5⎞ ⎛ Concave down on ⎜ −∞, − ⎟ ; concave up on 2⎠ ⎝ ⎛ 5 ⎞ ⎜ − 2 , ∞ ⎟ ; inflection point at ⎝ ⎠

y

Possible inflection points when x = 1 ± ⎛ 39 ⎞ Concave up on ⎜⎜ −∞, 1 − ⎟ and 3 ⎟⎠ ⎝ ⎛ ⎞ 39 , ∞ ⎟⎟ ; concave down on ⎜⎜1 + 3 ⎝ ⎠

37 ⎞ ⎛ 5 ⎜− 2 , − 2 ⎟ ⎝ ⎠

529

)

39 . 3


⎛ ⎜⎜1 − ⎝ ⎛ ⎜⎜1 − ⎝ ⎛ ⎜⎜1 + ⎝

ISM: Introductory Mathematical Analysis horizontal asymptote. 5 y′ = − ( x − 3)2 CV: None, but x = 3 must be considered in the inc.-dec. analysis. Decreasing on (–∞, 3) and (3, ∞). 10 y ′′ = ( x − 3)3 No possible inflection point, but x = 3 must be considered in concavity analysis. Concave up on (3, ∞); concave down on (–∞, 3).

39 39 ⎞ , 1+ ⎟ ; inflection points at 3 3 ⎟⎠ ⎞ 39 298 , + 16 39 ⎟⎟ ≈ (−1.08, 133.03) and 3 9 ⎠ ⎞ 39 298 , − 16 39 ⎟⎟ ≈ (3.08, − 66.81) . 3 9 ⎠ 300

y (0, 150) x 10

5

y

(5, –225)

(

x 8

)

41. y = x3 − x = x x 2 − 1 = x( x + 1)( x − 1) Intercepts (0, 0), (−1, 0), and (1, 0) Symmetric about the origin. No asymptotes. y′ = 3 x 2 − 1 =

CV: ±

(

)(

3x + 1

)

3x − 1

43. y = f ( x) =

x2 Intercept: (–5, 0) No symmetry. x = 0 is the only vertical asymptote. x 1 lim y = 100 lim = 100 lim = 0 , and 2 x →∞ x →∞ x x →∞ x lim y = 0 , so y = 0 is the only horizontal

3 3

⎛ ⎛ 3 ⎞ 3⎞ Increasing on ⎜⎜ −∞, − , ∞ ⎟⎟ ; ⎟⎟ and ⎜⎜ 3 ⎠ ⎝ ⎝ 3 ⎠ ⎛ 3 3⎞ decreasing on ⎜⎜ − , ⎟. 3 ⎟⎠ ⎝ 3 y ′′ = 6 x Possible inflection point when x = 0. Concave down on (–∞, 0); concave up on (0, ∞); inflection point at (0, 0). y

x →−∞

asymptote. y = 100 ⎡ x −1 + 5 x −2 ⎤ ⎣ ⎦ ⎡ 1 10 ⎤ y′ = 100 ⎡ − x −2 − 10 x −3 ⎤ = −100 ⎢ + ⎥ ⎣ ⎦ ⎣ x 2 x3 ⎦ −100( x + 10) = x3 CV: x = –10 but x = 0 must be included in inc.dec. analysis. Increasing on (–10, 0); decreasing on (–∞, –10) and (0, ∞); relative minimum at (–10, –5). ⎡ 1 15 ⎤ y ′′ = 100 ⎡ 2 x −3 + 30 x −4 ⎤ = 200 ⎢ + ⎥ ⎣ ⎦ ⎣ x3 x 4 ⎦

5

x 5

42. y =

100( x + 5)

x+2 x −3

=

200( x + 15)

x4 Possible inflection point when x = –15, but x = 0 must also be considered in concavity analysis. Concave up on (–15, 0) and (0, ∞); concave down on (–∞, –15); inflection point at

2⎞ ⎛ Intercepts: ⎜ 0, − ⎟ , (–2, 0) 3⎠ ⎝ Vertical asymptote is x = 3. x+2 x+2 lim = 1 = lim , so y = 1 is a x →∞ x − 3 x →−∞ x − 3 530


Chapter 13 Review

40 ⎞ ⎛ ⎜ −15, − ⎟ 9 ⎠ ⎝ 20

lim y = lim

x →∞

f(x)

(

=

)

(

y ′′ = −2 ⋅ = −2 ⋅

)

3

(3x − 1)8 6(3 x − 1)3 [(3 x − 1) − 2(6 x + 1)]

1 Possible inflection point when x = − , but 3 1 x = must be considered in concavity analysis. 3 1⎞ ⎛ ⎛1 ⎞ Concave up on ⎜ −∞, − ⎟ and ⎜ , ∞ ⎟ ; 3 3 ⎝ ⎠ ⎝ ⎠ ⎛ 1 1⎞ concave down on ⎜ − , ⎟ ; inflection point at ⎝ 3 3⎠ ⎛ 1 1⎞ ⎜ − , ⎟. ⎝ 3 12 ⎠

No possible inflection point, but x = ±1 must be considered in concavity analysis. Concave up on (–1, 1); concave down on (–∞, –1) and (1, ∞). 10

(3x − 1)4 (6) − (6 x + 1)[4(3x − 1)3 (3)]

(3 x − 1)8 −12(−9 x − 3) 36(3x + 1) = = (3 x − 1)5 (3 x − 1)5

−6 3 x 2 + 1

( x2 − 1)

2(3x − 1) 2 [(3x − 1) − 9 x]

1 1 CV: x = − , but x = must be considered in 3 6 1⎞ ⎛ inc.-dec. analysis. Increasing on ⎜ −∞, − ⎟ ; 6⎠ ⎝ ⎛ 1 1⎞ ⎛1 ⎞ decreasing on ⎜ − , ⎟ and ⎜ , ∞ ⎟ ; relative 6 3 3 ⎝ ⎠ ⎝ ⎠ ⎛ 1 8⎞ maximum at ⎜ − , ⎟ . ⎝ 6 81 ⎠

CV: x = 0 but x = ±1 must also be considered in inc.-dec. analysis. Increasing on (0, 1) and (1, ∞); decreasing on (–∞, –1) and (–1, 0); relative minimum at (0, 4). y ′′ =

2 1 =0 lim 27 x→∞ x 2

=

(3 x − 1)6 2(−6 x − 1) −2(6 x + 1) = = (3x − 1) 4 (3 x − 1)4

( x + 2)( x − 2) x − 1 ( x + 1)( x − 1) Intercepts: (0, 4), (2, 0), (–2, 0) Symmetric about the y-axis. Vertical asymptotes are x = 1 and x = –1. x2 lim y = lim = 1 = lim y , so y = 1 is the x →−∞ x →−∞ x 2 x →∞ only horizontal asymptote. 6x y′ = 2 x2 − 1 2

27 x = lim y,

3

x →−∞

=

x2 − 4

2x

so y = 0 is a horizontal asymptote. (3x − 1)3 (2) − 2 x[3(3x − 1)2 (3)] y′ = (3 x − 1)6

x 20

44. y =

x →∞

y

x 5

y

45. y =

1

2x

(3x − 1)3 Intercept: (0, 0) No symmetry

x 2

1 Vertical asymptote is x = . 3

531



1

46. y = 6 x 3 (2 x − 1)

47.

⎛1 ⎞ Intercepts: (0, 0), ⎜ , 0 ⎟ ⎝2 ⎠ No symmetry. No vertical asymptote.

As x → ∞ , both 6x1/ 3 and 2 x − 1 → ∞ . As 1

Setting f ′( x) = 0 ⇒ e x = e− x ⇒ x = − x ⇒ x = 0 CV: x = 0 Increasing on (0, ∞); decreasing on (–∞, 0); relative minimum at (0, 1). Finding f ′′( x) gives:

x → −∞ , both 6x 3 and 2 x − 1 → −∞ . Thus lim y = ∞ = lim y . So no horizontal

x →∞

x →−∞

(

)

asymptote exists. Since y = 6 2 x 4 / 3 − x1/ 3 ,

e x + e− x . f ′′( x) > 0 for all x. No 2 possible inflection point. Concave up on (–∞, ∞).

1 ⎛8 ⎞ y′ = 6 ⎜ x1/ 3 − x −2 / 3 ⎟ = 2 x −2 / 3 (8 x − 1) 3 3 ⎝ ⎠ 2(8 x − 1) = x2 / 3 1 CV: x = 0, 8 ⎛ 1⎞ Decreasing on (–∞, 0) and ⎜ 0, ⎟ ; increasing on ⎝ 8⎠ 1 9⎞ ⎛ ⎞ ⎛1 ⎜ , ∞ ⎟ ; relative minimum at ⎜ , − ⎟ . 4⎠ ⎝8 ⎠ ⎝8 2 ⎛8 ⎞ 4 y ′′ = 2 ⎜ x −2 / 3 + x −5 / 3 ⎟ = x −5 / 3 (4 x + 1) 3 3 ⎝ ⎠ 3 4(4 x + 1) = 3 x5 / 3 1 Possible inflection points when x = − , 0 . 4 1⎞ ⎛ Concave up on ⎜ −∞, − ⎟ and (0, ∞); concave 4⎠ ⎝ ⎛ 1 ⎞ down on ⎜ − , 0 ⎟ ; inflection points at ⎝ 4 ⎠ 3 ⎛ 1 9 2⎞ ⎜⎜ − , ⎟⎟ and (0, 0). ⎝ 4 2 ⎠ 10

e x + e− x 2 Intercept: (0, 1) Symmetric about the y-axis. No asymptotes. e x − e− x f ′( x) = 2 f ( x) =

f ′′( x) =

3

f(x)

x 3

48. y = f ( x) = 1 − ln( x3 ) = 1 − 3ln x

( ) x-intercept is ( e1/ 3 , 0 ) . Since x ≠ 0, there is no y = 0 ⇒ ln x3 = 1 ⇒ x3 = e ⇒ x = e1/ 3 , so the

y-intercept. No symmetry. Since lim y = ∞, x →0+

x = 0 is a vertical asymptote. No horizontal asymptote. 3 f ′( x) = − x CV: None. Decreasing on (0, ∞). 3 f ′′( x) = x2 No possible inflection points. Concave up on (0, ∞).

y

y

x

5

1 x 5

532


49. a.

Chapter 13 Review

False. f ′ ( x0 ) = 0 only indicates the

2

b.

possibility of a relative extremum at x0 , For example, if f ( x) = x3 , then f ′( x) = 3x 2 and f ′(0) = 0 . However there is no relative extremum at x = 0. Then x1 < x2 and f ( x1 ) = −1 < f ( x2 ) = 1 .

c.

True. The absolute minimum is f(0) = 0 and the absolute maximum is f(1) = 1. 3

f(x)

1

d.

x –1

1

f ′( x) =

2π f ′ is defined for all x; f ′( x) = 0 only when x = 0. Thus x = 0 is a critical value. If x < 0, then f ′( x) > 0 ; if x > 0 then f ′( x) < 0 .

3

From (b), f has a relative maximum when x = 0. The coordinates of this relative ⎛ 1 ⎞ maximum are ⎜ 0, ⎟. 2π ⎠ ⎝

x →−∞

e.

=

x0 = 0 , then f ′′ ( x0 ) = 0 , but ( x0 , f ( x0 ) )

− x2

− x2

− x2

1

= =

2π 1

(0) = 0

(0) = 0

2π

2 2 ⎤ 1 ⎡ − x2 −x ⎢ xe (− x) + e 2 (1) ⎥ 2π ⎣⎢ ⎦⎥

( x2 − 1) = e−

x2 2

( x + 1)( x − 1)

2π 2π ′′ f is defined for all x; f ′′( x) = 0 when

is not an inflection point. See graph in part (c).

x = ±1. f is concave up on (–∞, –1) and (1, ∞); f is concave down on (–1, 1).

False. Consider the function f whose graph is shown. On (–2, 2) it has exactly one relative maximum [at the point (0, 1)] but no absolute maximum. 3

e

f ′′( x) = −

e

e

2

2π

2

point. For example, consider f ( x) = x 4 . If

2π 1

x →∞

d. False. If concavity does not change around x0 , then ( x0 , f ( x0 ) ) is not an inflection

2

1

lim

lim

e.

− x2

Thus f is increasing on (–∞, 0) and is decreasing on (0, ∞).

b. False. For example, let x1 = −1 and x2 = 1 .

c.

− xe

f.

f(x)

x 3

From (e), f changes concavity at x = ±1. Also f is continuous there. Thus f has inflection points at x = ±1; the coordinates −1 ⎞ −1 ⎞ ⎛ ⎛ e 2 ⎟ e 2 ⎟ ⎜ ⎜ and 1, are −1, ⎜ ⎜ 2π ⎟ 2π ⎟ ⎝ ⎠ ⎝ ⎠

g.

f(x) 1 x

50. Let y = f ( x) = a.

1 2π

2

e

− x2

3

.

h. Absolute maximum: f (0) =

Replacing x by –x yields the original equation. Thus the graph is symmetric about the y-axis. No other symmetry exists.

No absolute minimum.

533

1 2π



51. c = q3 − 6q 2 + 12q + 18 dc = 3q 2 − 12q + 12 . Marginal dq cost is increasing when its derivative, which is d 2c , is positive. dq 2

b. R(0.5) ≈ 18.5%

Marginal cost =

d 2c dq

2

c.

d 2R

= 6q − 12 = 6(q − 2)

R 1 x

52. r = 320q3 / 2 − 2q 2

1

dr = 480q1/ 2 − 4q. dq Marginal revenue is increasing when its d 2r is positive. derivative, which is dq 2 Marginal revenue =

dq 2 d 2r 2

= 240q =0⇒

240

−4 =

240 q

55.

f (t ) = At 3 + Bt 2 + Ct + D f ′(t ) = 3 At 2 + 2 Bt + C f ′′(t ) = 6 At + 2 B , which gives an inflection B . 3A This value of a must be such that f ′(a) = 0 .

point when 6At + 2B = 0, that is for a = −

−4

2

⎛ B ⎞ ⎛ B ⎞ 3A ⎜ − ⎟ + 2B ⎜ − ⎟+C = 0 ⎝ 3A ⎠ ⎝ 3A ⎠ 1 ⎛ B2 ⎞ 2 ⎛ B2 ⎞ ⎜ ⎟− ⎜ ⎟+C = 0 3 ⎜⎝ A ⎟⎠ 3 ⎜⎝ A ⎟⎠ 1 ⎛ B2 ⎞ C= ⎜ ⎟ 3 ⎜⎝ A ⎟⎠

− 4 = 0 ⇒ 240 = 4 q

q dq ⇒ q = 60 ⇒ q = 3600 d 2r

> 0 for 0 < q < 3600. Thus marginal dq 2 revenue is increasing on (0, 3600).

53. p = 200 −

=

29.92

> 0 for 0 ≤ x ≤ 1. dx (4.4 − 3.4 x)3 We obtain the following graph: 2

> 0 for q > 2. Thus marginal cost is dq 2 increasing for q > 2.

−1/ 2

R(1) = 100%

dR 4.4 = > 0 for 0 ≤ x ≤ 1 dx (4.4 − 3.4 x)2

d 2c

d 2r

R(0) = 0%

3AC = B 2 , which was to be shown.

q , q > 0. The revenue function r is 5

given by

56. a.

3

⎛ q⎞ q2 r = pq = ⎜ 200 − q = 200q − . ⎟ ⎜ 5 ⎟⎠ 5 ⎝ 3 1 r ′ = 200 − q 2 10 3 −1 3 r ′′ = − q 2 = − 20 20 q

Let a = p + q. Then S becomes ⎡ ⎤ ⎥ ma 2 ⎢ e − at S= ⎢ ⎥ 2 p ⎢ q − at ⎥ e + 1 ⎣⎢ p ⎦⎥

(

Since r ′′ < 0 for q > 0, the graph of the revenue function is concave down for q > 0.

534

)


⎡ dS ma ⎢ = ⎢ dt p ⎢ ⎢⎣ 2

(

(

b.

+1

2

− at

( e + 1) + 1) + 2 e + 1)

− at

⎡2 ⎣⎢

q − at p

ma 2 = ae − at p

=

) ( −ae ) − e ( e ⎛q ⎞ −( e e + 1 ) ⎜⎝ p ⎟ ⎠ (

q − at e p

ma3 − at = e p

Chapter 13 Review

(

q − at p

)(

+1

− aq − at e p

4

q − at e p 4

)

⎡ q e−( p + q )t − 1⎤ ⎢⎣ p ⎥⎦

q − ( p + q )t e p

)

+1

3

dS = 0 when dt ⎡q ⎤ m ( p + q )3 e−( p + q )t ⎢ e− ( q + p )t − 1⎥ = 0 p ⎣p ⎦

Since m, p + q, and e −( p + q )t are nonzero, we must have q − ( p + q )t −1 = 0 e p e − ( p + q )t =

p q

⎛ p⎞ −( p + q )t = ln ⎜ ⎟ ⎝q⎠

() () p

q

ln q ln p t=− = p+q p+q 15

57.

–5

)⎤⎦⎥ ⎤⎥⎥ ⎥ ⎥⎦

q − at e a −1 p 3 q − at e +1 p

m ( p + q )3 e − ( p + q ) t p

(

q − at p

− at

q − at p

5 –5

Relative maximum (–1.32, 12.28); relative minimum (0.44, 1.29)

535

Chapter 13: Curve Sketching 58.


5

63. p = 500 − q , where 100 ≤ q ≤ 200. Total revenue = r = pq = q 500 − q

–1

⎛1⎞ r ′ = q ⎜ ⎟ (500 − q )−1/ 2 (−1) + 500 − q (1) ⎝2⎠ 1 = (500 − q )−1/ 2 [− q + 2(500 − q)] 2 1000 − 3q = 2 500 − q

1

–5

Maximum: (1, 1); minimum: (−0.60, −2.24) 5

59.

=

–5

(

3 1000 −q 3

)

2 500 − q No critical values on (100, 200). r(100) = 2000; r(200) ≈ 3464, so 200 units should be produced for maximum revenue.

5

–5

The x-value of the inflection point of f corresponds to the x-intercept of f ′′ . Thus the

64. c = 0.01q 2 + 5q + 100 Avg. cost c =

x-value of the inflection point is x ≈ –0.60. 60.

dc 100 1 100 q 2 − 1002 = 0.01 − = − = dq q 2 100 q 2 100q 2

5

=

10

–10

c 100 = 0.01q + 5 + q q

(q − 100)(q + 100)

100q 2 We find that c is decreasing on (0, 100) and increasing on (100, ∞), so average cost is minimum when q = 100.

–5

Horizontal asymptote y = 0; vertical asymptote x ≈ –0.25

65. p = 500 − 3q

61. q = 80m2 − 0.1m 4

c = q + 200 +

(

)

dq = 160m − 0.4m3 = 0.4m 400 − m2 dm dq = 0.4m(20 + m)(20 − m) . Setting = 0 yields dm m = 0 or m = 20 (for m ≥ 0). We find that q is increasing on (0, 20) and decreasing on (20, ∞), so q is maximum at m = 20.

1000 q

Total Cost = c = cq = q 2 + 200q + 1000 Profit = Total Revenue – Total Cost P = pq − c = (500 − 3q )q − (q 2 + 200q + 1000) = −4(q 2 − 75q + 250) P ′ = −4(2q − 75) Setting P ′ = 0 yields q = 37.5. Since P ′′ = −8 < 0, P is maximum when q = 37.5. In reality, whole units are likely. Since P(37) = P(38) = 4624, the maximum profit is $4624.

62. p = 100e −0.1q Total revenue = r = pq = 100qe−0.1q r ′ = 100[e −0.1q (1) + q(−0.1)e−0.1q ] = 10e −0.1q (10 − q ) r ′ = 0 when q = 10. Since r is increasing when q < 10 and decreasing when q > 10, revenue is maximized when q = 10.

66. V = (10 – 2x)(16 – 2x)x

(

= 4 x3 − 13x 2 + 40 x

Note: 0 < x < 5.

536

)


(

V ′ = 4 3x 2 − 26 x + 40

Chapter 13 Review

)

= 4(x – 2)(3x – 20) 20 . On (0, 5), x = 2 is the only critical value. At x = 2 in., 3 V ′′ = 4(6 x − 26) = 4(12 − 26) = −56 < 0 , so V is maximum at x = 2 in.

Setting V ′ = 0 gives x = 2 or x =

x

x

x

x 10 – 2x

x

x x

x 16 – 2x

67. 2x + 4y = 800; thus x = 400 – 2y Area = A = xy = (400 − 2 y ) y = 400 y − 2 y 2 dA = 400 − 4 y = 4(100 − y ) dy d2A dA = −4 < 0 , A is maximum when y = 100. When y = 100, then x = 200. = 0 gives y = 100. Since dy dy 2 The dimensions are 200 ft by 100 ft.

Setting

y

y

y

y

x

500 x Printed area = A = (x – 8)(y – 10) ⎛ 500 ⎞ = ( x − 8) ⎜ − 10 ⎟ x ⎝ ⎠

68. xy = 500, so y =

4000 ,x>0 x 4000

= 580 − 10x − A′ = −10 +

x2

Setting A′ = 0 gives x = 20. When x = 20, A′′ = −

8000 x

3

Thus the dimensions are 20 in. by 25 in. 6 y 4

4 4 x

537

< 0 , so A is maximum. When x = 20, then y =

500 = 25 . 20



c = 2q3 − 9q 2 + 12q + 20 , where

69. a.

(

3 ≤ q ≤ 6. 4

)

dc = 6q 2 − 18q + 12 = 6 q 2 − 3q + 2 dq = 6(q – 1)(q – 2) dc Setting = 0 gives q = 1 or 2. Evaluating c at these critical values and the endpoints: dq ⎛ 3 ⎞ 793 c⎜ ⎟ = ≈ 24.78 , c(1) = 25, c(2) = 24, c(6) = 200. Thus a minimum occurs at q = 2, which corresponds ⎝ 4 ⎠ 32 24, 000 = $120 . to 200 stands and a total cost of $24,000. This gives an average cost per stand of 200

b. There are no critical values of c in 3 ≤ q ≤ 6, so we only evaluate c at the endpoints: c(3) = 29, c(6) = 200. Thus a minimum occurs at q = 3, which gives 300 stands. 70. N = N′ = =

12,100 + 110t + 100t 2 121 + t 2

, where t ≥ 0.

(121 + t 2 )(110 + 200t ) − (12,100 + 110t + 100t 2 )(2t ) (121 + t 2 ) 2 110(121 − t 2 )

121 + t 2 Setting N ′ = 0 gives t = 11, from which N = 105. Since N ′ > 0 for 0 ≤ t < 11 and N ′ < 0 for t > 11, there is an absolute maximum when t = 11.

Mathematical Snapshot Chapter 13 1. Figure 13.74 does not readily show how long it takes for the population to reach its final size. Figure 13.75 shows that this takes about 45 days. 2. The population declines until it stabilizes between 311 and 312 (as can be verified by inspecting the final state of dP < 0 for all P ≥ 312 . the variable P). This is consistent with the fact that dt 3. Even if the graph starts out exactly coinciding with the ideal curve, a line segment tangent to the curve at one end must (in general) lie slightly off the curve at the other end. This introduces errors that accumulate over successive iterations. The amount of cumulative error could be reduced by taking smaller time steps, such as 1 month instead of 1 year, and correspondingly drawing shorter line segments.

538


11. ∆y = [4 – 7(3.02)] – [4 – 7(3)] = –0.14 dy = –7 dx = –7(0.02) = –0.14

1. y = 5x – 7 d dy = (5 x − 7)dx = 5 dx dx

12. ∆y = ⎡5(−1.02) 2 − 5(−1)2 ⎤ = 0.202 ⎣ ⎦ dy = 10x dx = 10(–1)(–0.02) = 0.2

2. dy = y′dx = 0 dx = 0

13. ∆y = [2(−1.9)2 + 5(−1.9) − 7] − [2(−2) 2 + 5(−2) − 7] = −0.28 dy = (4x + 5)dx = [4(−2) + 5](0.1) = −0.3

1 −1 3. d [ f ( x)] = f ′( x)dx = ( x 4 − 9) 2 (4 x3 )dx 2 2 x3 dx = x4 − 9

4. d [ f ( x)] = f ′( x)dx

(

= 3(8 x − 5) 4 x 2 − 5 x + 2

5. u = x du =

)

2

14. ∆y = [3(−1.03) + 2] − [3(−1) + 2]2 = 0.1881 dy = 6(3x + 2) dx = 6[3(–1) + 2](–0.03) = 0.18 2

15. ∆y = 32 − (3.95) 2 − 32 − (42 ) ≈ 0.049

dx

( )

d −2 2 x dx = −2 x −3 dx = − dx dx x3

du = u′dx = −

1 −3 / 2 x dx 2

(

17. a.

)

d ⎡ 1 ln x 2 + 7 ⎤ dx = (2 x)dx ⎢ ⎥ 2 ⎣ ⎦ dx x +7 2x = dx 2 x +7

7. dp =

f ′(1) =

+3

= 3e 2 x

2

+3

=

18. a.

[(3x + 1)(4 x) + 3]dx

(12x

2

dx =

−4

(−0.05) = 0.050

16

x+5 x +1 ( x + 1)(1) − ( x + 5)(1)

( x + 1)

2

=

−4 ( x + 1) 2

−4 = −1 4

b. We use f(x + dx) ≈ f(x) + dy with x = 1, dx = 0.1. f (1.1) = f (1 + 0.1) ≈ f (1) + f ′(1)dx

9. dy = y′dx 2 2 = ⎡(9 x + 3)e2 x +3 (4 x) + e2 x +3 (9) ⎤ dx ⎢⎣ ⎥⎦ 2

f ( x) =

f ′( x) =

d ⎛ x 3 + 2 x −5 ⎞ 2 x3 + 2 x −5 dx ⎜e ⎟ dx = (3x + 2)e dx ⎝ ⎠

= 3e 2 x

32 − x

2

16. ∆y = ln 4.9 – ln 5 ≈ –0.0202 1 1 1 dy = (−1)dx = dx = (0.1) = −0.02 −x x −5

6. u = x −1/ 2

8. dp =

−x

dy =

−2

)

+ 4 x + 3 dx

1 ln( x 2 + 12) 2 1 1 x dy = ⋅ (2 x )dx = dx 2 2 x 2 + 12 x + 12

10. y = ln x 2 + 12 =

6 + (−1)(0.1) = 2.9 2

y = f ( x ) = x3 x Using logarithmic differentiation, ln y = 3x ln x, 1 dy ⎛1⎞ ⋅ = 3 x ⎜ ⎟ + (ln x)(3) = 3(1 + ln x) y dx ⎝x⎠ dy = y[3(1 + ln x)] = 3 x3 x (1 + ln x) dx f ′(1) = 3(1)(1 + 0) = 3

539

Chapter 14: Integration

ISM: Introductory Mathematical Analysis 23. Let y = f(x) = ln x

b. We use f(x + dx) ≈ f(x) + dy with x = 1, dx = –0.02 f (0.98) = f (1 − 0.02) ≈ f (1) + f ′(1)dx

f ( x + dx) ≈ f ( x) + dy = ln( x) +

= 13 + (3)(−0.02) = 0.94

If x = 1 and dx = –0.03, then ln(0.97) = f (1 + (−0.03)) 1 ≈ ln(1) + (−0.03) = −0.03 1

19. Let y = f ( x) = x f ( x + dx) ≈ f ( x) + dy = x +

1

dx

2 x

24. Let y = f(x) = ln x

If x = 289 and dx = ⫺1, then 288 = f (289 − 1) 1 ≈ 289 + (−1) 2 289 577 = 34 ≈ 16.97

f ( x + dx) ≈ f ( x) + dy = ln( x) +

1 ln1.01 = f (1 + 0.01) ≈ ln(1) + (0.01) = 0.01 1

25. Let y = f ( x) = e x

f ( x + dx) ≈ f ( x) + dy = x +

f ( x + dx) ≈ f ( x) + dy = e x + e x dx If x = 0 and dx = 0.001, then

1

dx

2 x

e0.001 = f (0 + 0.001) ≈ e0 + e0 (0.001) = 1.001

If x = 121 and dx = 1, then

= 11

1 2 121

26. Let y = f ( x) = e x

(1)

f ( x + dx) ≈ f ( x) + dy = e x + e x dx If x = 0 and dx = –0.01, then

1 . 22

e −0.01 = f (0 + (–0.01)) ≈ e0 + e0 (−0.01) = 0.99

21. Let y = f ( x) = 3 x

1

65.5 = f (64 + 1.5) ≈ 3 64 + 1.5 3 ⋅ 42

=4

dx

2

( )

2 3 3 64

22. Let y = f ( x) = 4 x . 1 4x

If x = 16 and dx = 0.3, then 4 16.3

4 0.3 3

2

29.

dq = 6 p p2 + 5 dp

30.

dq 1 dp = , so = 2 p+5 dp 2 p + 5 dq

(

31. q = p −1 ,

( ) 4 16

=2

dy dx 1 = 10 x + 3, so = dx dy 10 x + 3

dx

1

= f (16 + 0.3) ≈ 4 16 + = 2+

3 4

28. (1.5)

1 32

f ( x + dx) = f ( x) + dy = 4 x +

dx 1 1 dy = 2 , so = = dy dy 2 dx dx

3x 3

If x = 64 and dx = 1.5, then

= 4+

27.

1

f ( x + dx ) ≈ f ( x) + dy = 3 x +

3

1 dx x

If x = 1 and dx = 0.01, then

20. Let y = f ( x) = x

122 = f (121 + 1) ≈ 121 +

1 dx x

3

(0.3)

32.

3 320

540

)

2

, so

dp = dq

1

(

6 p p2 + 5

)

2

dq −1 dp = −1 p −2 = = − p2 , so 2 dp dq p

dq dp 1 1 = −2e4− 2 p , so =− = − e2 p −4 − p 4 2 dp dq 2 2e


33.

Section 14.1

dx 1 1 = = dy dy 14 x − 6

39. p =

10 q

dx

dx 1 = dy 36

If x = 3,

34.

p(q + dq ) ≈ p + dp =

x

dx 3 = . If x = 3, dy 2

35. p =

= 2+

500 q+2

q =18

(q + 2) 2 500

=− q =18

4 5

36. p = 50 − q dp 1 =− dq 2 q

5 q3

dq

1 51 = = 2.04 25 25

q +8

q4 + 3q + 400 2 If q = 10 and dq = 2,

41. c =

dq = −2 q dp dq dp

−

We approximate p when q = 40. 200 100 p(q + dq ) ≈ p + dp = − dq 3 q + 8 (q + 8) 2 If q = 41 and dq = 1, then 200 100 p(40) = p(41 − 1) ≈ − (1) 3 49 (49) 2 200 100 9700 = − = ≈ 28.28 7 343 343

dq (q + 2) 2 =− dp 500 =−

q

200

40. p =

dp −500 = dq (q + 2) 2

dq dp

10

If q = 25 and dq = –1, then 10 5 − p(24) = p(25 + (−1)) ≈ (−1) 25 (25)3

dx 1 1 x = = = dy dy 2 2 dx

. We approximate p when q = 24.

= −2 q q =100

q =100

= −20

3

4

q 2

=

+ 3q + 400

(2003)(2) ≈ 0.7 5430

42. S = 20 I , I decreases from 45 to 44

37. P = 397 q − 2.3q 2 − 400, q changes from 90 to 91. ∆P ≈ dP = P ′dq = (397 − 4.6q )dq Choosing q = 90 and dq = 1, ∆P ≈ [397 − 4.6(90)](1) = −17. True change is P(91) − P(90) = 16,680.7 − 16,700 = −19.3.

∆S ≈ dS = S ′dI =

10 I

dI

Choosing I = 45 and dI = − ∆S ≈

38. r = 250q + 45q 2 − q3 , q increases from 40 to 41.

(

( 2q + 3) dq

dc = c

)

43. V =

∆r ≈ dr = r ′dq = 250 + 90q − 3q 2 dq

Choosing q = 40 and dq = 1, ∆r ≈ (–950)(1) = –950 True change is r(41) – r(40) = 16,974 – 18,000 = –1026

1 , then 2

10 ⎛ 1 ⎞ ⎜ − ⎟ ≈ −0.745. 45 ⎝ 2 ⎠

4 3 πr 3

∆V ≈ dV = V ′dr = 4πr 2 dr

(

) (

dr = 6.6 × 10−4 − 6.5 × 10−4 = 0.1× 10−4 = 10−5

541

)

1 . 2


(

∆V ≈ 4π 6.5 × 10−4


) (10 ) = (1.69 ×10 ) π cm 2

−5

−11

3

.

44. (P + a)(v + b) = k k P= −a v+b dP = −k (v + b)−2 dv

45. a.

We substitute q = 40 and p = 20 402 4000 = 2+ 200 202 2 + 8 = 10 10 = 10

b. We differentiate implicitly with respect to p. 1 ⎛ dq ⎞ 8000 0+ ⎜ 2q ⎟ = − 3 200 ⎝ dp ⎠ p From part (a) q = 40 when p = 20. Substituting gives dq ⎞ 1 ⎛ 8000 ⎜ 2 ⋅ 40 ⎟ = − 3 dp ⎠ 200 ⎝ 20 dq = −2.5 dp

c.

46. a.

q( p + dp) ≈ q ( p) + dq = q( p ) + q′( p)dp q(19.20) = q (20 + (−0.8)) ≈ q (20) + q′(20)dp = 40 + (−2.5)(−0.8) = 42 units

Profit = TR − TC = pq − cq P=

1 3 80, 000 ⎞ 1 3 ⎛ q − 66q 2 + 7000q − ⎜ 500q − q 2 + = q − 65q 2 + 6500q − 40, 000 2 2 ⎟⎠ 2 ⎝

If q = 100, then P =

1 (100)3 − 65(100) 2 + 6500(100) − 40, 000 = 460, 000 2

b. We use P(q + dq) ≈ P(q) + dP with q = 100 and dq = −2. P (98) = P (100 + (−2)) ⎛3 ⎞ ≈ P (100) + ⎜ q 2 − 130q + 6500 ⎟ dq ⎝2 ⎠ ⎡3 ⎤ 2 = 460, 000 + ⎢ (100) − 130(100) + 6500 ⎥ (−2) 2 ⎣ ⎦ = $443, 000

542


Section 14.2

Principles in Practice 14.2 1.

2.

4.

24 24 ∫ 5 x dx = 5∫ x dx = 5 ⋅

5.

∫ 5x

t3 + C = 0.04t 3 + C 3 The form of the revenue function is

x 24+1 +C 24 + 1 x 25 x 25 = 5⋅ +C = +C 25 5

2 ∫ 0.12t dt = 0.12

R (t ) = 0.04t 3 + C .

3. Let S(t) = the number of subscribers t months after the competition entered the market, then 480 S ′(t ) = − . t3 480 S (t ) = ∫ − dt = −480 ∫ t −3 dt t3 ⎛ t −2 ⎞ 240 = −480 ⎜ ⎟ + C = 240t −2 + C = 2 + C ⎜ −2 ⎟ t ⎝ ⎠ 240 +C . The number of subscribers is S (t ) = t2 4.

8 ∫ x dx =

∫ 28.3 dq = 28.3q + C The form of the cost function is 28.3q + C.

x8+1 x9 +C = +C 8 +1 9

3.

∫(

= 5⋅

6.

7.

)

3

= 500t +

3 2

8.

2 3 The population is N (t ) = 500t + t 2 + C 3

∫ ( 2.1t

2

)

dS

∫ dt dt .

9.

dx = 2 ∫ x −10 dx = 2 ⋅

7

∫ x4

dx = 7 ∫ x −4 dx = 7 ⋅ 7

3 x3

1

∫ t 7 / 4 dt = ∫ t 4 3t

3/ 4

S (t ) = 0.7t 3 − 32.7t 2 + 491.6t + C

1 1

t −7 / 4+1 − 74 + 1

−9

−5 4

=−

∫ 2 x dx = 2 ∫ x dx = 2 ln x + C

dt =

+C =

+C

7 x 4 = ⋅ +C 2 −5

Problems 14.2

2.

−7 / 4

+1

7 −9 7 x4 10. ∫ 9 dx = ∫ x 4 dx = ⋅ +C 2 2 −9 + 1 2x 4 4

= 0.7t − 32.7t + 491.6t + C The amount of money saved is

∫ 7 dx = 7 x + C

x −4+1 7 x −3 +C = +C −4 + 1 −3

+C

7

2

1.

x −10+1 +C −10 + 1

2 x −9 2 +C = − +C −9 9 x9

=−

⎛ t3 ⎞ ⎛ t2 ⎞ = 2.1⎜ ⎟ − 65.4 ⎜ ⎟ + 491.6t + C ⎜3⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠

1

x −6 5 +C = − +C −6 6 x6

∫ x10

=−

− 65.4t + 491.6 dt

3

x −7 +1 +C −7 + 1

z −3 1 1 z −3+1 dz = ∫ z −3 dz = ⋅ +C 3 3 3 −3 + 1

2

=

2 3 + C = 500t + t 2 + C 3

5. The amount of money saved is

∫

dx = 5∫ x −7 dx = 5 ⋅

1 z −2 1 = ⋅ +C = − +C 3 −2 6z2

1 ⎛ ⎞ 500 + 300 t dt = ∫ ⎜ 500 + 300t 2 ⎟ dt ⎝ ⎠

t2

−7

14 5

5x 4 1

543

+C

t −3 / 4 − 43

+C


11.

∫ (4 + t )dt = ∫ 4 dt + ∫ t dt = 4t + = 4t +

12.

13.


∫(

t2 +C 2

19.

)

r 3 + 2r dr = ∫ r 3 + 2 ∫ r dr

=

r 3+1 r1+1 + 2⋅ +C 3 +1 1+1

=

r4 + r2 + C 4

∫( y =

t t +1 +C 1+1

20.

)

5

− 5 y dy = ∫ y 5 dy − ∫ 5 y dy

y5+1 y1+1 − 5⋅ +C 5 +1 1+1

y6 y2 y6 5 y2 = − 5⋅ +C = − +C 6 2 6 2

14.

∫ ( 5 − 2w − 6w ) dw = ∫ 5 dw − 2∫ w dw − 6 ∫ w2 dw 2

= 5w − 2 ⋅

15.

∫ ( 3t = 3⋅

16.

23.

2

6

24.

∫ (5 − 2

−1

=

2 x3 8 x5 ⋅ − ⋅ +C 7 3 3 5

=

2 x3 8 x5 − +C 21 15 x

dx = π ∫ e x dx = πe x + C

⎛ ex

⎞ 1 + 2 x ⎟ dx = ∫ e x dx + 2∫ x dx ⎟ 3 3 ⎝ ⎠

ex + x2 + C 3

∫(x

8.3

)

− 9 x6 + 3 x −4 + x −3 dx

=

x9.3 x7 x −3 x −2 − 9⋅ + 3⋅ + +C 9.3 7 −3 −2

=

x9.3 9 x 7 1 1 − − − +C 3 9.3 7 x 2 x2

∫ (0.7 y

3

+ 10 + 2 y −3 )dy

y4 y −2 + 10 y + 2 ⋅ +C 4 −2 1 = 0.175 y 4 + 10 y − +C y2

t3 t5 t7 + + +C 3 5 7

= 0.7 ⋅

17. Since 7 + e is a constant, ∫ (7 + e)dx = (7 + e) x + C . 18.

⎛ 2 x2 8 4 ⎞ 2 2 8 4 ∫ ⎜⎜ 7 − 3 x ⎟⎟ dx = 7 ∫ x dx − 3 ∫ x dx ⎝ ⎠

=

∫ (1 + t + t + t )dt = ∫ 1 dt + ∫ t 2 dt + ∫ t 4 dt + ∫ t 6 dt =t+

x 2 3 x5 − +C 14 20

∫ ⎜⎜

t t − 4 ⋅ + 5t + C = t 3 − 2t 2 + 5t + C 3 2 4

=

22.

2

2

1 3 4 ⎟ dx = ∫ x dx − ∫ x dx 7 4 ⎠

1 x2 = ex + 2 ⋅ +C 3 2

− 4t + 5 dt = 3∫ t dt − 4 ∫ t dt + ∫ 5 dt

3

4⎞

1 x 2 3 x5 ⋅ − ⋅ +C 7 2 4 5

∫ πe

w2 w3 − 6⋅ +C 2 3

)

3

=

21.

= 5w − w2 − 2w3 + C 2

⎛x

∫ ⎜⎝ 7 − 4 x

1 +1

−2 x 2 1 2 x2 25. ∫ +C dx = − ∫ x 2 dx = − ⋅ 3 3 3 1 +1

)

1⎞ 9 9 ⎛ dx = ∫ ⎜ 5 − ⎟ dx = ∫ dx = x + C 2⎠ 2 2 ⎝

2

3 2

3 2

2 x 4x =− ⋅ +C = − +C 3 3 9 2

26. 544

∫ dz = ∫ 1 dz = 1⋅ z + C = z + C


Section 14.2

− 1 +1

1

1 −1 1 x 4 dx = ∫ x 4 dx = ⋅ 27. ∫ +C 8 4 4 − 1 +1 4 x2 4 3

=

33.

3

1 x4 x4 ⋅ +C = +C 4 3 3

(

4

28.

−4

−4

4

∫ (3x)3 dx = ∫ 27 x3 dx = − 27 ∫ x

−3

4 x ⋅ +C 27 −3 + 1 4 x −2 2 =− ⋅ +C = +C 27 −2 27 x 2

29.

⎛ x3

∫ ⎜⎜

⎝ 3

−

)

(

dx

−3+1

=−

(

3u − 4 1 1 du = ∫ (3u − 4)du = 3∫ u du − 4∫ du 5 5 5 2 ⎛ ⎞ 1 u 3 4 = ⎜ 3 − 4u ⎟ + C = u 2 − u + C ⎟ 5 ⎜⎝ 2 10 5 ⎠ 1 = 2∫ z dz − ∫ 5 dz 7 ⎞ 1 ⎛ z2 1 = ⎜ 2 ⋅ − 5z ⎟ + C = z 2 − 5z + C ⎟ 7 ⎜⎝ 2 7 ⎠

∫

34.

3 ⎞ 1 ⎟ dx = ∫ x3 dx − 3∫ x −3 dx 3⎟ 3 x ⎠

35.

1 ⎛1

1 x ⎞ ⎟ dx = ∫ e dx 36 ⎠ 1 1 = e x dx = e x + C 36 ∫ 36

∫ 12 ⎜⎝ 3 e

∫ (u

e

x

+ eu )du = ∫ u e du + ∫ eu du

1 x3+1 x −3+1 = ⋅ − 3⋅ +C −3 + 1 3 3 +1

=

1 x4 3 x −2 x4 = ⋅ − 3⋅ +C = + +C −2 3 4 12 2 x 2

30.

⎛ 1

1 ⎞

1

∫ ⎜⎝ 2 x3 − x 4 ⎟⎠ dx = 2 ∫ x

−3

36.

dx − ∫ x −4 dx

1 x −2 x −3 ⋅ − +C 2 −2 −3 1 1 =− + +C 2 4x 3 x3

31.

⎛ 3w2 2 ∫ ⎜⎜ 2 − 3w2 ⎝

32.

∫ e− s

y ⎛ 3 2 e y − y + 3 2 ⎜ ∫⎜ 6 ⎝

= 3⋅ =

⎞ 3 2 ⎟ dw = ∫ w2 dw − ∫ w−2 dw ⎟ 2 3 ⎠

37.

1 y e dy 6∫

y4 y3 1 y − 2⋅ + ⋅e + C 4 3 6

3 y 4 2 y3 e y − + +C 4 3 6

∫ (2

)

1 ⎛ 1 ⎞ x − 3 4 x dx = ∫ ⎜ 2 x 2 − 3x 4 ⎟ dx ⎝ ⎠ 1

1

3

= 2⋅ s

⎞ ⎟ dy ⎟ ⎠

= 2 ∫ x 2 dx − 3∫ x 4 dx

3 w3 2 w−1 w3 2 = ⋅ − ⋅ +C = + +C 2 3 3 −1 2 3w 4

u e+1 u +e +C e +1

= 3∫ y 3 dy − 2∫ y 2 dy +

=

)

s

ds = 4∫ e ds = 4e + C

38.

545

x2 3 2

5

− 3⋅

x4 5 4

3

5

4 x 2 12 x 4 +C = − +C 3 5

∫ 0 dt = 0 ⋅ t + C = C

)



⎛ 3 x2 ⎞ 7 39. ∫ ⎜ − − + 6 x ⎟dx ⎜ ⎟ 5 2 x ⎝ ⎠ 2 1 − ⎛ ⎞ x 3 7x 2 = ∫⎜− − + 6 x ⎟ dx ⎜ 5 ⎟ 2 ⎝ ⎠ 2 1 1 7 − = − ∫ x 3 dx − ∫ x 2 dx + 6 ∫ x dx 5 2 5

44.

45.

1

1 x3 7 x2 x2 =− ⋅ − ⋅ + 6⋅ +C 5 5 2 1 2 3x − 7 x + 3x2 + C 25

=−

40.

⎛3

∫ ⎜⎝

u+

1

= ∫ u 3 du + ∫ u =

u

+

4 3

=

z3 + 2z2 + 4z + C 3

∫ (3u + 2)

3

du = ∫ (27u 3 + 54u 2 + 36u + 8)du

u4 u3 u2 + 54 ⋅ + 36 ⋅ + 8u + C 4 3 2

27 4 u + 18u 3 + 18u 2 + 8u + C 4

u

1 2

1 2

46.

⎞ ⎟ du ⎠ − 12

du

2 ⎛ 2 ⎞ ⎛ − 15 ⎞ 1 dx 2 x 1 − = − ⎟ dx ∫ ⎜⎝ 5 x ⎟⎠ ∫ ⎜⎝ ⎠ −1 ⎛ − 25 ⎞ = ∫ ⎜ 4 x − 4 x 5 + 1⎟ dx ⎝ ⎠ 3

= 4⋅

+C

∫(x = =

42.

2

)

(

)

47.

x4 x3 x2 − 3⋅ + 5⋅ − 15 x + C 4 3 2

4

=

( x3 + 8x2 + 7 ) dx = ∫ ( x7 + 8x6 + 7 x4 ) dx

x8 x7 x5 = + 8⋅ + 7⋅ +C 8 7 5 x8 8 x 7 7 x 5 = + + +C 8 7 5

43.

∫

48.

x2 5 2

x2 3 2

+ x+C

v3 v −4 + 3v − 2 ⋅ +C 3 −4

2v 3 1 + 3v + +C 3 2v 4

∫ ⎡⎣6e

u

(

− u3

)

7 ⎡ ⎤ u + 1 ⎤ du = ∫ ⎢ 6eu − u 2 − u 3 ⎥du ⎦ ⎣ ⎦

9

= 6⋅e −

u2 9 2 9

−

u4 +C 4

2u 2 u 4 = 6e − − +C 9 4 u

3

+ 3⋅

4 5

−2

u

1 ⎛ 3 ⎞ x ( x + 3)dx = ∫ ⎜ x 2 + 3x 2 ⎟ dx ⎝ ⎠

5

=

− 4⋅

( 2v4 + 3v2 − 2v−3 ) dv = ∫ ( 2v 2 + 3 − 2v −5 ) dv ∫v

= 2⋅

2

x 5x − x3 + − 15 x + C 4 2

∫x

3 5

4

x5

4 20 x 5 = − 5x 5 + x + C 3

+ 5 ( x − 3)dx = ∫ x3 − 3x 2 + 5 x − 15 dx

4

x5

3

4

1 3u 3 = + 2u 2 + C 4

41.

)

2

1 ⎞ −1 ⎛ 1 ⎟ du = ∫ ⎜ u 3 + u 2 u⎠ ⎝ 4 3

(

dz = ∫ z 2 + 4 z + 4 dz

z3 z2 + 4 ⋅ + 4z + C 3 2

=

1 2

2

=

= 27 ⋅

2

3

5 3

∫ ( z + 2)

+C

5

3 2x 2 = + 2x 2 + C 5

546


49.

∫

z 4 + 10 z 3 2z2

dz =

Section 14.3 Principles in Practice 14.3

1 ⎛ z 4 10 z 3 ⎞ ⎜ + 2 ⎟ dz 2 ∫ ⎜⎝ z 2 z ⎟⎠

(

1. N (t ) = ∫

)

1 = ∫ z 2 + 10 z dz 2 1 ⎛ z3 z2 ⎞ = ⎜ + 10 ⋅ ⎟ + C 2 ⎜⎝ 3 2 ⎟⎠ z3 5z 2 = + +C 6 2

50.

51.

x4 − 5x2 + 2 x

∫

e x + e2 x ex

40, 000 = 800(5) + 200e5 + C , so

(

C = 40, 000 − 4000 + 200e5

N (t ) = 800t + 200et + 6317.37 d ( y′ ) = 84t + 24 dt ⎛ t2 ⎞ y′ = ∫ (84t + 24)dt = 84 ⎜ ⎟ + 24t + C1 ⎜2⎟ ⎝ ⎠

2. Since y ′′ =

⎛ e x e2 x ⎞ dx = ∫ ⎜ + ⎟ dx ⎜ ex ex ⎟ ⎝ ⎠ x = ∫ 1 + e dx

= 42t 2 + 24t + C1 Since y′(8) = 2891 , we have

)

2891 = 42(8) 2 + 24(8) + C1 = 2880 + C1 , so

= x + ex + C

52.

∫

3

( x + 1) x

2

2

dx = ∫

6

C1 = 2891 − 2880 = 11 , and y′ = 42t 2 + 24t + 11 .

3

x + 2x +1

(

)

y (t ) = ∫ y′dt = ∫ 42t 2 + 24t + 11 dt

dx x2 = ∫ ( x 4 + 2 x + x −2 )dx

⎛ t3 ⎞ ⎛ t2 ⎞ = 42 ⎜ ⎟ + 24 ⎜ ⎟ + 11t + C2 ⎜3⎟ ⎜2⎟ ⎝ ⎠ ⎝ ⎠

x5 x 2 x −1 + 2⋅ + +C 5 2 −1 x5 1 = + x2 − + C x 5 =

= 14t 3 + 12t 2 + 11t + C2 Since y(2) = 185, we have 185 = 14(2)3 + 12(2)2 + 11(2) + C2 = 182 + C2 , so C2 = 185 − 182 = 3 .

53. No, F(x) – G(x) might be a nonzero constant. 54. a.

)

= 36, 000 − 200e5 ≈ 6317.37

dx =

(

)

= 800t + 200et + C Since N(5) = 40,000, we have

1 ⎛ 2 2⎞ ⎜ x − 5 + ⎟ dx ∫ 2 5 x⎠ ⎝ 5x 3 ⎞ 1⎛ x = ⎜ − 5 x + 2 ln x ⎟ + C ⎜ ⎟ 5⎝ 3 ⎠

∫

(

dN dt = ∫ 800 + 200et dt dt

y (t ) = 14t 3 + 12t 2 + 11t + 3

( )

d F ( x) = xe x = xe x + e x (1) = e x ( x + 1) dx

b. There is only one. 55. Because an antiderivative of the derivative of a function is the function itself, we have d ⎡ 1 ⎤ 1 ∫ dx ⎢⎢ 2 ⎥⎥ dx = 2 + C . x +1 ⎣ x +1 ⎦

547



Problems 14.3 1.

5. y ′′ = −3x 2 + 4 x y ′ = ∫ (−3x 2 + 4 x)dx = − x3 + 2 x 2 + C1

dy = 3x − 4 dx y = ∫ (3 x − 4)dx =

Using y (−1) =

y ′(1) = 2 implies 2 = −1 + 2 + C1 , so C1 = 1.

2

3x − 4x + C 2

y = ∫ (− x3 + 2 x 2 + 1)dx = −

13 gives 2

1 2 y(1) = 3 implies 3 = − + + 1 + C2 , so 4 3

13 3(−1) 2 = − 4(−1) + C 2 2 13 11 = +C 2 2 3x 2 Thus C = 1, so y = − 4x + 1 . 2

2.

C2 =

(

)

Using y(3) =

x2 + x + C1 2 y′(0) = 0 implies 0 = 0 + 0 + C1 , so C1 = 0 .

y′ = ∫ ( x + 1)dx =

⎡ x2 ⎤ x3 x 2 y = ∫ ⎢ + x ⎥ dx = + + C2 . 6 2 ⎣⎢ 2 ⎦⎥ y(0) = 5 implies 5 = 0 + 0 + C2 , so C2 = 5 . Thus

x3 x 2 − +C 3 2

19 33 32 19 gives = − +C 2 3 2 2

y=

19 9 = +C 2 2

y ′′ = ∫ 2 x dx = x 2 + C1

x3 x 2 − +5. 3 2

y ′′(−1) = 3 implies that 3 = 1 + C1 , so C1 = 2 . x3 + 2 x + C2 3 y′(3) = 10 implies 10 = 9 + 6 + C2 , so C2 = −5 .

(

5

y=∫

x

)

y′ = ∫ x 2 + 2 dx =

x 5

x3 x 2 + +5. 6 2

7. y ′′′ = 2 x

Thus, C = 5, so y =

3. y′ =

x 4 2 x3 19 19 . Thus y = − + + x+ . 4 3 12 12

6. y ′′ = x + 1

dy = x2 − x dx y = ∫ x 2 − x dx =

x 4 2 x3 + + x + C2 4 3

dx = ∫ 5 x

− 12

1

dx = 5 ⋅

x2 1 2

⎛ x3 ⎞ x4 y = ∫ ⎜ + 2 x − 5 ⎟ dx = + x 2 − 5 x + C3 . ⎜ 3 ⎟ 12 ⎝ ⎠ y(0) = 13 implies that 13 = 0 + 0 − 0 + C3 , so

+ C = 10 x + C

y(9) = 50 implies 50 = 10 9 + C , 50 = 30 + C, C = 20. Thus y = 10 x + 20.

C3 = 13 . Therefore y =

y(16) = 10 ⋅ 4 + 20 = 60

8. y ′′′ = e x + 1

4. y′ = − x 2 + 2 x

(

x4 + x 2 − 5 x + 13 . 12

)

y ′′ = ∫ e x + 1 dx = e x + x + C1

x3 y = ∫ − x + 2 x dx = − + x 2 + C 3 8 1 y(2) = 1 implies 1 = − + 4 + C , so C = − . 3 3

y′ = ∫ e x + x dx = e x +

x3 1 + x2 − . 3 3 1 1 1 y (1) = − + 1 − = 3 3 3

⎡ x2 ⎤ x3 y = ∫ ⎢e x + + 1⎥ dx = e x + + x + C3 2 6 ⎥⎦ ⎣⎢

(

2

)

y ′′(0) = 1 implies 1 = 1 + 0 + C1 , so C1 = 0 . x2 + C2 2 y′(0) = 2 implies 2 = 1 + 0 + C2 , so C2 = 1 .

(

Thus y = −

548

)


Section 14.3

y(0) = 3 implies that 3 = 1 + 0 + 0 + C3 , so C3 = 2 . Thus y = e x +

9.

p=

x3 + x+2. 6

function is p = 5000 − 3q −

dr = 0.7 dq

13.

r = ∫ 0.7 dq = 0.7 q + C

14.

dr 1 = 10 − q dq 16

(

When q = 0, then c = 2000, so C = 2000. Thus the cost function is c = q 2 + 75q + 2000 . 15.

(

)

0.08 3 q − 0.8q 2 + 6.5q + C. If q = 0, then 3 c = 8000, from which C = 8000. Hence 0.08 3 c= q − 0.8q 2 + 6.5q + 8000. If q = 25, 3 1 substituting gives c(25) = 8079 or $8079.17. 6

)

= 275q − 0.5q 2 − 0.1q3 + C . When q = 0, r must 2

dc = 0.08q 2 − 1.6q + 6.5 dq

c = ∫ 0.08q 2 − 1.6q + 6.5 dq

Thus r = ∫ 275 − q − 0.3q 2 dq 16.

3

be 0, so C = 0 and r = 275q − 0.5q − 0.1q .

dc = 0.000204q 2 − 0.046q + 6 dq c = ∫ (0.000204q 2 − 0.046q + 6)dq

r = 275 − 0.5q − 0.1q 2 . q Thus the demand function is

Since r = pq, then p =

= 0.000068q3 − 0.023q 2 + 6q + C When q = 0, then c = 15,000, from which C = 15,000. The cost function is

p = 275 − 0.5q − 0.1q 2 .

12.

dc = 2q + 75 dq c = ∫ (2q + 75)dq = q 2 + 75q + C

1 p = 10 − q . 32 dr = 275 − q − 0.3q 2 dq

dc = 1.35 dq

When q = 0, then c = 200, so 200 = 0 + C, or C = 200. Thus c = 1.35q + 200.

1 ⎤ 1 ⎡ r = ∫ ⎢10 − q ⎥ dq = 10q − q 2 + C 16 ⎦ 32 ⎣ When q = 0, then r = 0, so C = 0 and 1 r = 10q − q 2 . Since r = pq, then 32 r 1 p = = 10 − q . The demand function is q 32

11.

3q3 . 2

c = ∫ 1.35dq = 1.35q + C

If q = 0, r must be 0, so 0 = 0 + C, C = 0. Thus r = 0.7q. Since r = pq, we have r 0.7q p= = = 0.7 . The demand function is q q p = 0.7. 10.

r 3q3 = 5000 − 3q − . Therefore the demand q 2

c = 0.000068q3 − 0.023q 2 + 6q + 15, 000. When q = 200, substitution gives c(200) = 15,824.

dr = 5000 − 3(2q + 2q3 ), so dq

P2 ⎡ P ⎤ + 2P + C 17. G = ∫ ⎢ − + 2 ⎥ dP = − 50 ⎣ 25 ⎦ When P = 10, then G = 38, so 38 = –2 + 20 + C, from which C = 20. Thus 1 G = − P 2 + 2 P + 20 . 50

r = ∫ (5000 − 6q − 6q3 )dq

3q 4 +C 2 When q = 0, then r = 0, so C = 0 and 3q 4 r = 5000q − 3q 2 − . Since r = pq, then 2 = 5000q − 3q 2 −

549


18.


dy = −1.5 − x dx x2 +C 2 When x = 1, then y = 57.3, so y = ∫ (−1.5 − x )dx = −1.5 x −

57.3 = –1.5 – 0.5 + C, or C = 59.3. Thus y = −1.5 x − 0.5 x 2 + 59.3 . 19. v = ∫ −

( P1 − P2 ) r

dr = −

2lη

( P1 − P2 ) r 2 4lη

Since v = 0 when r = R, then 0 = − v=−

20.

( P1 − P2 ) r 2 ( P1 − P2 ) R 2 +

4lη

dr = 100 − 3q 2 dq

(

4lη

=

+C

( P1 − P2 ) R 2 4lη

+ C , so C =

( P1 − P2 ) ( R 2 − r 2 ) 4lη

( P1 − P2 ) R 2 4lη

. Thus

.

)

r = ∫ 100 − 3q 2 dq = 100q − q3 + C

When q = 0, then r = 0, so C = 0 and r = 100q − q3 . Since r = pq, then p =

η=

p q dp dq

=

p q

−2q

=−

p 2q 2

When q = 5, then p = 75, so η =

21.

r = 100 − q 2 . q

dc = 0.003q 2 − 0.4q + 40 dq

(

−75 3 =− . 2(25) 2

)

c = ∫ 0.003q 2 − 0.4q + 40 dq = 0.001q3 − 0.2q 2 + 40q + C

When q = 0, then c = 5000, so 5000 = 0 – 0 + 0 + C, or C = 5000. Thus c = 0.001q3 − 0.2q 2 + 40q + 5000 . When q = 100, then c = 8000. Since 8000 Total Cost c dc = $80 . (Observe that knowing = , when q = 100, we have c = = 27.50 100 Quantity q dq when q = 50 is not relevant to the problem.)

Avg. Cost = c =

22.

f ′′( x) = 30 x 4 + 12 x f ′( x) = ∫ (30 x 4 + 12 x)dx = 6 x5 + 6 x 2 + C1

f ′(1) = 10, so 10 = 6 + 6 + C1 and C1 = −2. f ′( x) = 6 x5 + 6 x 2 − 2 f ( x) = ∫ (6 x5 + 6 x 2 − 2)dx = x6 + 2 x3 − 2 x + C2

550


Section 14.4

Thus f (965.335245) − f (−965.335245) = [(965.335245)6 + 2(965.335245)3 − 2(965.335245) + C2 ]

− [(−965.335245)6 + 2(−965.335245)3 − 2(−965.335245) + C2 ]

= 3,598, 280, 000

Principles in Practice 14.4 1. Using the values given,

dT = −0.5(70 − 60)e−0.5t = −5e−0.5t dt

dT dt = ∫ −5e −0.5t dt = 10e−0.5t + C dt

T (t ) = ∫

2. The number of words memorized is v(t). 35 v(t ) = ∫ dt = 35ln t + 1 + C . t +1 Problems 14.4 1. Let u = x + 5 ⇒ du = 1dx = dx

∫ ( x + 5) 2.

7

[dx] = ∫ u 7 du =

∫ 15( x + 2)

4

u8 ( x + 5)8 +C = +C 8 8

dx = 15∫ ( x + 2)4 [dx] = 15 ⋅

( x + 2)5 + C = 3( x + 2)5 + C 5

3. Let u = x 2 + 3 ⇒ du = 2 x dx

(

(

)

5

)

5

2 2 5 ∫ 2 x x + 3 dx = ∫ x + 3 [2 x dx] = ∫ u du =

( x 2 + 3) =

u6 +C 6

6

6

+C

4. Let u = x3 + 5 x 2 + 6 ⇒ du = (3x 2 + 10 x)dx.

∫ ( 3x

2

)(

)

+ 10 x x3 + 5 x 2 + 6 dx

(

) ( 1

)

= ∫ x3 + 5 x 2 + 6 ⎡⎢ 3x 2 + 10 x dx ⎤⎥ ⎣ ⎦ = ∫ u du =

u2 +C 2

( x3 + 5 x 2 + 6 ) = 2

2

+C

551



(

)

10. Let u = x − 5 ⇒ du = dx. 1 −1 ∫ x − 5 dx = ∫ ( x − 5) 2 [dx]

5. Let u = y 3 + 3 y 2 + 1 ⇒ du = 3 y 2 + 6 y dy

∫(

3 y2 + 6 y

)(

)

y3 + 3 y 2 + 1

(

= ∫ u du =

5 3

(

=

+C

)

3 3 y + 3 y2 + 1 5 2

5 3

3

2

(5t 3 − 3t 2 + t )18 +C 18

12.

5

5 1 ∫ (3x − 1)3 dx = 3 ∫ (3x − 1)3 [3 dx ] 5 1 5 du = ∫ u −3 du ∫ 3 3 u 3

∫

4x

(

dx = ∫ 2 x 2 − 7

( 2x2 − 7 ) −9 2 x2 − 7 ) ( =− +C 10

)

−10

dx =

1 (7 x − 6)4 [7 dx] 7∫

=

1 4 1 u5 u du = ⋅ +C 7∫ 7 5

=

(7 x − 6)5 +C 35

(3x3 + 7 ) dx = 19 ∫ (3x3 + 7 ) 4 3 1 ( 3x + 7 ) = ⋅ +C ∫x

3

2

3

⎡9 x 2 dx ⎤ ⎣ ⎦

4

4

+C

36

13. Let v = 5u 2 − 9 ⇒ dv = 10u du

[4 x dx]

∫ u (5u

2

− 9)14 du =

1 (5u 2 − 9)14 [10u du ] 10 ∫

1 14 1 v15 (5u 2 − 9)15 v dv = ⋅ +C = +C ∫ 10 10 15 150

9. Let u = 2 x − 1 ⇒ du = 2 dx .

14.

1 2

2 x − 1dx = ∫ (2 x − 1) dx

2 ∫ 9 x 1 + 2 x dx =

(

1 1 = ∫ (2 x − 1) 2 [2 dx] 2

=

4

3 x3 + 7 ) ( =

9

∫

+C = 2 x−5 +C

1 2

9

5 u −2 5(3 x − 1)−2 = ⋅ +C = − +C 3 −2 6

8.

( x − 5) 2

∫ (7 x − 6)

17

7. Let u = 3 x − 1 ⇒ du = 3 dx

=

+C

11. Let u = 7 x − 6 ⇒ du = 7 dx

+C

∫ (15t − 6t + 1)(5t − 3t + t ) dt = ∫ (5t 3 − 3t 2 + t )17 [(15t 2 − 6t + 1)dt ] =

1 2

1

5

u3

u1/ 2

−1/ 2 ∫ u du =

) ⎡⎢⎣(3 y 2 + 6 y ) dy ⎤⎥⎦

2 3

6.

dy

2 3

= ∫ y3 + 3 y 2 + 1

=

2 3

2 9 1 + 2x = ⋅ 3 4

3 2

)

3 2

+C

2

3 1 12 1 u 1 u du = ⋅ + C = (2 x − 1) 2 + C ∫ 2 2 3 3

2

=

552

(

3 1 + 2 x2 2

)

(

9 1 + 2 x2 4∫

3 2

+C

)

1 2

[4 x dx]


15. Let u = 27 + x5 ⇒ du = 5 x 4 dx 1 4 4 5 3 5 ∫ 4 x 27 + x dx = 5 ∫ 27 + x

(

)

(

)

1 3

Section 14.4

24. ⎡5 x 4 dx ⎤ ⎣ ⎦

4

(

)

4 3

+C

16. Let u = 4 − 5x ⇒ du = −5dx. 1 9 9 ∫ (4 − 5 x) dx = − 5 ∫ (4 − 5 x) [−5 dx] 1 1 u10 1 = − ∫ u 9 du = − ⋅ + C = − (4 − 5 x)10 + C 5 5 10 50

26.

dt =

∫

t 2 +t

20.

2 − w3

∫ −3w e

2

[(2t + 1) dt ]

+t

+C

3

3 4 x4

3

4

dx = ∫

1

(

)

⎡ 3 x 2 + 4 x3 dx ⎤ ⎥⎦ x + x ⎣⎢ 3

4

6 x2 − 6 x

∫ 1 − 3x2 + 2 x3 dx

3

dw = ∫ e− w ⎡ −3w2 dw⎤ = e− w + C ⎣ ⎦

dx =

3 x 2 + 4 x3

28. Let u = 1 − 3 x 2 + 2 x3 ⇒ du = (−6 x + 6 x 2 )dx.

1

=∫

2

3

[(−6 x + 6 x 2 )dx

1 − 3x + 2 x 1 = ∫ du = ln u + C = ln 1 − 3x 2 + 2 x3 + C u

21. Let u = 7 x ⇒ du = 14 x dx 1 7 x2 1 u 7 x2 ∫ xe dx = 14 ∫ e [14 x dx] = 14 ∫ e du 2 1 1 = eu + C = e7 x + C 14 14

∫x e

[(1 + 2 x + 6 x 2 )dx]

3

= ln x3 + x 4 + C

2

22.

2

dx

x +x 1 = ∫ du = ln u + C u

19. Let u = t 2 + t ⇒ du = (2t + 1)dt t 2 +t

x + x 2 + 2 x3 2

27. Let u = x3 + x 4 ⇒ du = (3x 2 + 4 x3 )dx

5 3t + 7 5 e [3 dt ] = e3t + 7 + C ∫ 3 3

∫ (2t + 1)e dt = ∫ e = ∫ eu du = eu + C = et

∫

12 x 2 + 4 x + 2

= ln[( x + x 2 + 2 x3 )2 ] + C

3x

3t + 7

∫ 5e

1 −6 x 5 ⎡ −30 x 4 dx ⎤ e ⎣ ⎦ 30 ∫

x + x + 2x = 2 ln x + x 2 + 2 x3 + C

∫ 3e dx = ∫ e [3 dx] = ∫ eu du = eu + C = e3 x + C 18.

dx = −

1 −6 x5 e +C 30

=∫

17. Let u = 3 x ⇒ du = 3 dx 3x

e

25. Let u = x + 5 ⇒ du = dx 1 1 ∫ x + 5 [dx] = ∫ u du = ln u + C = ln x + 5 + C

3

3 27 + x5 5

4 −6 x5

=−

4 1 4 u3 +C = ∫ u 3 du = ⋅ 5 5 4 =

∫x

29. Let u = z 2 − 6 ⇒ du = 2 z dz 6z

∫ ( z 2 − 6)5 = 3∫ ( z

1 4 x4 ⎡ 3 ⎤ 16 x dx e ⎣ ⎦ 16 ∫

= 3∫ u −5 du = 3

2

− 6)−5 [2 z dz ]

u −4 3 + C = − ( z 2 − 6)−4 + C −4 4

4

4 1 e4 x = ⋅ e4 x + C = +C 16 16

30.

3

3

∫ (5v − 1)4 dv = 5 ∫ (5v − 1)

−4

[5 dv]

3 (5v − 1)−3 ⋅ +C −3 5 1 = − (5v − 1)−3 + C 5

23. Let u = −3x ⇒ du = −3dx. 4 −3 x −3 x ∫ 4e dx = − 3 ∫ e [−3 dx] 4 4 4 = − ∫ eu du = − eu + C = − e−3 x + C 3 3 3

=

553


31.

4


39. Let u = x 2 − 4 ⇒ du = 2 x dx

1

∫ x dx = 4∫ x dx = 4 ln x + C

x

∫

3 1 1 32. ∫ [2 dy ] dy = 3 ⋅ ∫ 1+ 2y 2 1+ 2 y 3 = ln 1 + 2 y + C 2

x2 − 4

dx =

(

)

− 12 1 2 − 4 [2 x dx] x 2∫ 1

=

1 − 12 1 u2 +C u du = ⋅ ∫ 2 2 1 2

2

= x −4 +C

33. Let u = s 3 + 5 ⇒ du = 3s 2 ds s2

1

1

∫ s3 + 5 ds = 3 ∫ s3 + 5 ⎡⎣3s =

34.

2

40. Let u = 1 − 3x ⇒ du = −3 dx. 9 1 ∫ 1 − 3x dx = −3∫ 1 − 3x [−3 dx] 1 = −3∫ du = −3ln u + C = −3ln 1 − 3x + C u

ds ⎤ ⎦

1 1 1 1 du = ln u + C = ln s3 + 5 + C ∫ 3 u 3 3 2 x2

⎛

1⎞

1

∫ 3 − 4 x3 dx = 2 ⎜⎝ − 12 ⎟⎠ ∫ 3 − 4 x3 ⎡⎣−12 x

2

dx ⎤ ⎦

41. Let u = y 4 + 1 ⇒ du = 4 y3 dy

∫ 2y

1 = − ln 3 − 4 x3 + C 6

1 1 36. ∫ dt = 7 ⋅ ∫ [10t dt ] 2 2 10 5t − 6 5t − 6 7 = ln 5t 2 − 6 + C 10

=

5 x dx = 5 ∫ x1/ 2 dx = 5

4 dy = 2 ∫ y 3e y +1dy

1 y 4 +1 ⎡ 3 ⎤ e 4 y dy ⎣ ⎦ 4∫ 1 1 = ∫ eu du = eu + C 2 2 1 y 4 +1 = e +C 2

7t

∫

e

= 2⋅

35. Let u = 4 − 2x ⇒ du = −2 dx 5 5 1 ∫ 4 − 2 x dx = − 2 ∫ 4 − 2 x [−2 dx] 5 1 5 5 = − ∫ du = − ln u + C = − ln 4 − 2 x + C 2 u 2 2

37.

3 y 4 +1

x

3 2

3 2

42.

∫2

1

2 x − 1dx = ∫ (2 x − 1) 2 [2 dx] 3

= =

+C

(2 x − 1) 2 3 2

+C

3 2 (2 x − 1) 2 + C 3

43. Let u = −2v3 + 1 ⇒ du = −6v 2 dv 3 1 2 −2v3 +1 dv = − ∫ e−2v +1 ⎡ −6v 2 dv ⎤ ∫v e ⎣ ⎦ 6

2 5 32 x +C 3

1 u 1 e du = − eu + C 6∫ 6 3 1 = − e−2v +1 + C 6 =−

1

1 38. ∫ dx = ∫ (3 x)−6 [3 dx] 6 3 (3x ) 1 (3 x)−5 ⋅ +C 3 −5 1 = − (3 x)−5 + C 15 =

554


44.

∫3

x2 2 x3 + 9

(

1 = ⋅ 6 =

45.

dx =

2 x3 + 9

)

(

1 2 x3 + 9 4

)

)

− 13 1 3 ⎡6 x 2 dx ⎤ 2 9 x + ⎣ ⎦ 6∫

52.

2 3

= +C

53.

−5 x x −5 x x ∫ ( e + 2e ) dx = ∫ e dx + 2∫ e dx

∫ 4 3 y + 1dy = 4∫ ( y + 1) = 4⋅

47.

( y + 1)

4 3

4 3

1 3

54. [dy ]

3

55.

4

(7 − 2 x − 5 x) +C 4

3 y2

∫ 2 ye 2

49.

x +2

dy = 2 ⋅

∫ x3 + 6 x

dx =

2 1 3 y2 1 e [6 y dy ] = e3 y + C ∫ 6 3

(

56.

)

1 1 ⎡ 2 3 x + 6 dx ⎤⎥ ∫ 3 ⎦ 3 x + 6 x ⎣⎢

x

−3 x

5x

∫ (e + 2e − e )dx 2 1 = ∫ e x dx − ∫ e−3 x [(−3)dx] − ∫ e5 x [5 dx] 3 5

57.

16 s − 4

5

∫ (8w

∫ −(x

)

+1

−1

dx = ∫

x 2

2x + 1

dx

)

+ w2 − 2)(6w − w3 − 4 w6 ) −4 dw

)( x3 − x6 ) dx −10 1 ⎡ 3 x 2 − 6 x5 dx ⎤ = − ∫ ( x3 − x 6 ) ) ⎦⎥ ⎣⎢( 3 3 6 −9 −9 1 (x − x ) 1 3 x − x6 ) + C =− ⋅ +C = ( −9 3 27 2

3

−10

− 2 x5

∫ 5 (v − 2)e

∫ ( 2x

2 − 4v + v 2

dv

3

)(

)

+ x x 4 + x 2 dx

( ) ⎡⎣⎢( 4 x3 + 2 x ) dx ⎤⎦⎥ 4 2 2 2 1 (x + x ) 1 = ⋅ + C = ( x4 + x2 ) + C 2 2 4 =

2 1 = e x − e−3 x − e5 x + C 3 5

51.

2

2 3 1 = ⋅ ∫ e2− 4v + v [(2v − 4) dv] 5 2 2 3 = e2− 4v + v + C 10

1 = ln x3 + 6 x + C 3

50.

∫ x (2x

1 (6 w − w3 − 4 w6 ) −4 [(6 − 3w2 − 24w5 )dw] 3∫ 1 (6 w − w3 − 4w6 )−3 =− ⋅ +C 3 −3 1 = (6 w − w3 − 4w6 )−3 + C 9

1 = − (7 − 2 x 2 − 5 x) 4 + C 2

48.

2 3 2 (t + t + 1)7 + C 7

=−

∫ (8 x + 10)(7 − 2 x − 5 x) dx = −2 ∫ (7 − 2 x 2 − 5 x)3 [(−4 x − 5)dx] = −2 ⋅

6

(t 3 + t 2 + 1)7 +C 7

(

4

2

2

1 1 [4 x dx] ∫ 2 4 2x +1 1 = ln 2 x 2 + 1 + C 4

+ C = 3( y + 1) 3 + C

2

3

=

1 = − ∫ e −5 x [−5 dx] + 2∫ e x dx 5 1 = − e−5 x + 2e x + C 5

46.

2

∫ (6t + 4t )(t + t + 1) dt = 2 ∫ (t 3 + t 2 + 1)6 [(3t 2 + 2t )dt ] = 2⋅

+C

2 3 2 3

(

Section 14.4

1

∫ 3 − 2s + 4s 2 dx = 2∫ 3 − 2s + 4s 2 [(8s − 2)ds] = 2 ln 3 − 2 s + 4s 2 + C

555

1 x4 + x2 2∫

1


58.

∫ (e

)

3.1 2


dx = ∫ e6.2 dx = e6.2 x + C , because e6.2

65.

is a constant. 7 + 14 x

59.

dx

2 5

x −x ∫ (e − e )

2

1 ⎞ 1 dx ⎟ dx = ∫ 2 xdx − ∫ 2x ⎠ 2x 1 1 1 −1 = ∫ (2 x) 2 [2 dx] − ∫ (2 x) 2 [2 dx] 2 2 2x −

3

∫ (4 − x − x ) 2 −5 = −7 ∫ (4 − x − x ) [(−1 − 2 x)dx]

2

2

1 2 2 32 = x − 2x 2 + C 3

(

66.

)

dx = ∫ e2 x − 2 + e−2 x dx

1 2x ⎛ 1⎞ e [2 dx] − ∫ 2 dx + ⎜ − ⎟ ∫ e−2 x [−2 dx] 2∫ ⎝ 2⎠ 1 1 = e2 x − 2 x − e−2 x + C 2 2 1 2 x −2 x = e −e − 2x + C 2

∫3

x4 ex

5

dx = 3∫ x 4 e− x dx = −

5

67.

)

=

61. u = 4 x3 + 3x 2 − 4

(

68.

)

du = 12 x 2 + 6 x dx = 6 x(2 x + 1)dx 4x ∫ x(2 x + 1)e

3

2

+3 x − 4

63.

3

(

2 1 8 − 5x =− ⋅ 5 10

)

5 2

2

64.

∫e

− 7x

dx = −7 ∫ e

− 7x

+C = −

(

3 2

69.

⎡

)

5 2

2

− 16

)

2

−

1 ⎤ dx 2 x + 5 ⎥⎦

2 1 1 1 x 2 − 16 [2 x dx] − ∫ [2 dx] ∫ 2 2 2x + 5

⎡

2

)

3

1 − ln 2 x + 5 + C 2 ⎤

x5

x

∫ ⎢⎢ x2 + 1 + ( x6 + 1)2 ⎥⎥dx ⎣

=∫

[−10 x dx]

1 8 − 5x2 25

)

x5 2 x3 + + x+C 5 3

(

u 4 1 6−3u 2 + e [−6u du ] 4 6∫ 2 1 1 = u 4 + e6−3u + C 4 6

(8 − 5x2 ) dx = − 101 ∫ (8 − 5x2 )

(

2

2 3 1 = x 2 − 16 6

3 6 −3u ∫ (u − ue )du =

∫x

)

+ 1 dx = ∫ x 4 + 2 x 2 + 1 dx

( ) 3 2 1 ( x − 16 ) 1 = ⋅ − ln 2 x + 5 + C

dx

1 4 x3 + 3 x 2 − 4 e [6 x(2 x + 1)dx] 6∫ 3 2 1 1 1 = ∫ eu du = eu + C = e4 x +3 x − 4 + C 6 6 6 2

2

∫ ⎢⎣ x ( x =

=

62.

∫(x

3 − x5 e [−5 x 4 dx] 5∫

5 3 = − e− x + C 5

=

(

3

1

1 (2 x) 2 1 (2 x) 2 (2 x) 2 = ⋅ − ⋅ +C = − 2x + C 3 1 2 2 3

(4 − x − x 2 ) −4 = −7 +C −4 7 = (4 − x − x 2 )−4 + C 4

60.

⎛

∫ ⎜⎝

=

⎦

x x2 + 1

dx + ∫

x5

(

)

x6 + 1

2

dx

(

)

1 1 1 [2 x dx] + ∫ x6 + 1 ∫ 2 2 x +1 6

(

)

−1

−2

6 1 1 x +1 2 = ln x + 1 + ⋅ +C −1 2 6 1 1 = ln x 2 + 1 − +C 6 2 6 x +1

+C

( (

− 7x ⎡ 1 ⎤ − = − +C dx e 7 ⎢ 7 ⎥ ⎣ ⎦

556

) )

(

)

⎡6 x5 dx ⎤ ⎣ ⎦


70.

⎡ 3

3

∫ ⎢⎢ x − 1 + ( x − 1)2 ⎥⎥ dx = ∫ x − 1 [dx] + ∫ ( x − 1) ⎣

⎦

= 3ln x − 1 +

71.

⎤

1

Section 14.4

⎡ 2

−2

[dx]

( x − 1) −1 1 + C = 3ln x − 1 − +C x −1 −1

⎤ − 8 x5 )( x3 − x6 )−8 ⎥ dx ⎦ 1 1 4 3 6 −8 = ∫ [4 dx] − ∫ ( x − x ) [(3 x 2 − 6 x5 )dx] 2 4x + 1 3 1 4 ( x3 − x 6 )−7 = ln 4 x + 1 − ⋅ +C −7 2 3 1 4 = ln 4 x + 1 + ( x3 − x 6 ) −7 + C 2 21

∫ ⎢⎣ 4 x + 1 − (4 x

72.

∫ (r

73.

∫ ⎢⎣

⎡

3

+5

)

2

2

(

)

dr = ∫ r 6 + 10r 3 + 25 dr =

3x + 1 −

1 7 5 4 r + r + 25r + C 7 2

1 1 ⎤ x 1 1 1 2 dx = ∫ (3 x + 1) 2 [3 dx] − ∫ [2 x dx] ⎥ dx = ∫ (3x + 1) dx − ∫ 2 2 3 2 x +3 x +3 x + 3⎦

x

2

3

(

)

3 2 1 (3 x + 1) 2 1 = ⋅ − ln x 2 + 3 + C = (3 x + 1) 2 − ln x 2 + 3 + C 3 9 3 2

2

74.

⎡

x2

x

⎤

1

1

1

∫ ⎢⎢ 3x2 + 5 − ( x3 + 1)3 ⎥⎥ dx = 6 ∫ 3x2 + 5 [6 x dx] − 3 ∫ ( x

3

+ 1) −3 [3x 2 dx]

⎣ ⎦ 1 1 ( x3 + 1)−2 1 1 = ln 3x 2 + 5 − ⋅ + C = ln 3x 2 + 5 + ( x3 + 1)−2 + C 6 3 6 6 −2

75. Let u = x ⇒ du =

∫

1 − 12 1 x dx = dx . 2 2 x

e x

⎡ 1 ⎤ dx = 2∫ e x ⎢ dx ⎥ x ⎣2 x ⎦

= 2 ∫ eu du = 2eu + C = 2e x + C

76.

∫ (e

77.

∫

5

)

(

)

− 3e dx = e5 − 3e x + C , because e5 − 3e is a constant.

1 + e2 x 4e x

(

dx =

1 ⎛ 1 e2 x ⎞ ⎜ + ⎟ dx 4 ∫ ⎜⎝ e x e x ⎟⎠

)

1 e− x + e x dx ∫ 4 1 1 = − ∫ e− x [−1 dx] + ∫ e x dx 4 4 1 −x 1 x = − e + e +C 4 4 =

557


78.


1

1 ⎛1 ⎞2 ⎡ 1 ⎤ + 9 dt = −2∫ ⎜ + 9 ⎟ ⎢ − dt ⎥ t ⎝t ⎠ ⎣ t2 ⎦

2

∫ t2

= −2

(

1 +9 t 3 2

)

83. y ′′ =

+C

y′(−2) = 3 implies 3 =

3

)

1

∫ x2 + 2 x ln ( x x +1

2

(

2

x + 2x

)

(2 x + 2)dx

+ 2 x dx

)

=

⎡ 2x + 2 ⎤ 1 dx ⎥ ln x 2 + 2 x ⎢ 2∫ ⎣ x2 + 2 x ⎦

=

1 1 u2 1 u du = ⋅ + C = ln 2 x 2 + 2 x + C ∫ 2 2 2 4

(

4

3

80. Let u = 8 x 4 = 2 x 3 ⇒ du =

)

8 13 x dx 3

84. y ′′ = ( x + 1)3 / 2 3

y ′ = ∫ ( x + 1) 2 dx =

4

3 2x3 ⎡8 1 ⎤ 3 u ∫ xe dx = 8 ∫ e ⎢⎣ 3 x 3 dx ⎥⎦ = 8 ∫ e du 3 3 3 4 = eu + C = e 8 x + C 8 8 3

3

8 x4

2 64 ⋅ 32 + C1 ⇒ C1 = − , so 5 5 5 2 64 y ′ = ( x + 1) 2 − 5 5 5 64 ⎤ ⎡2 y = ∫ ⎢ ( x + 1) 2 − ⎥ dx 5 5⎦ ⎣

1 (3 − 2 x)2 [−2 dx] 2∫ 1 (3 − 2 x)3 1 =− ⋅ + C = − (3 − 2 x)3 + C 2 3 6 1 11 y(0) = 1 implies 1 = − (27) + C , so C = . 6 2 1 3 11 Thus y = − (3 − 2 x) + . 6 2

(

7

=

y = ln

2 ( x + 1) 2 64 ⋅ − x + C2 7 5 5 2

7 4 64 = x + C2 ( x + 1) 2 − 35 5 4 64 y(3) = 0 implies 0 = ⋅128 − (3) + C2 , so 35 5 7 832 4 64 832 C2 = . Thus y = ( x + 1) 2 − x+ . 35 35 5 35

)

1 1 1 [2 x dx] = ln x 2 + 6 + C ∫ 2 2 x +6 2 1 1 y(1) = 0 implies 0 = ln(7) + C , so C = − ln 7 . 2 2 1⎡ Thus y = ⎢ln x 2 + 6 − ln 7 ⎤⎥ , or ⎦ 2⎣

(

5 2 ( x + 1) 2 + C1 5

y ′(3) = 0 ⇒ 0 =

81. y = ∫ (3 − 2 x) 2 dx = −

82. y =

1 5 + C1 , so C1 = . Thus 2 2

5 y′ = − x −1 + . 2 5⎞ 1 5 ⎛ y = ∫ ⎜ − x −1 + ⎟ dx = − ∫ dx + ∫ dx 2⎠ 2 x ⎝ 5 = − ln x + x + C2 2 5 y(1) = 2 implies that 2 = 0 + + C2 , so 2 1 C2 = − . Thus 2 5 1 1 5 1 y = − ln x + x − = ln + x − . 2 2 x 2 2

4 ⎛1 ⎞2 = − ⎜ + 9⎟ + C 3⎝t ⎠

(

x2

y′ = ∫ x −2 dx = − x −1 + C1

3 2

79. Let u = ln x 2 + 2 x ⇒ du =

1

)

x2 + 6 7

558


85. V (t ) = ∫ =

Section 14.5 Problems 14.5

dV dt = ∫ 8e0.05t dt dt

8 e0.05t [0.05 dt ] 0.05 ∫

1.

dx 2x2 ⎛ 2 x6 8 x 4 4x = ∫⎜ + − ⎜ 2 x2 2 x2 2 x2 ⎝

⎞ ⎟ dx ⎟ ⎠ 1 4 2 = ∫ x dx + 4 ∫ x dx − 2∫ dx x x5 4 3 = + x − 2 ln x + C 5 3

= 160e0.05t + C The house cost $350,000 to build, so V(0) = 350. 350 = 160e0 + C = 160 + C 190 = C V (t ) = 160e0.05t + 190 dl 12 dt = ∫ dt dt 2t + 50 = 6 ln 2t + 50 + C Since the expected life span was 63 years in 1940, l(0) = 63. 63 = 6 ln 50 + C C = 63 – 6 ln 50 ≈ 39.53 l (t ) = 6 ln 2t + 50 + 39.53

86. l (t ) = ∫

2.

3.

9 x2 + 5 5 ⎞ ⎛ ∫ 3x dx = ∫ ⎜⎝ 3x + 3x ⎟⎠ dx 3 5 = x 2 + ln x + C 2 3

∫ ( 3x

2

+2

)

2 x3 + 4 x + 1dx

( ) ⎡⎢⎣( 6 x2 + 4) dx ⎤⎥⎦ 3 1 ( 2 x + 4 x + 1) = ⋅ +C

=

l (58) = 6 ln 166 + 39.53 ≈ 70.20 The expected life span for people born in 1998 (58 years after 1940) is about 70 years.

1 2

1 2 x3 + 4 x + 1 2∫

3 2

3 2

2

87. Note that r > 0. B Rr ⎡ Rr B1 ⎤ C = ∫⎢ + ⎥ dr = ∫ dr + ∫ 1 dr 2K r ⎣ 2K r ⎦ 1 R r dr + B1 ∫ dr = 2K ∫ r 2 R r = ⋅ + B1 ln r + B2 2K 2 Thus we obtain C =

∫

2 x6 + 8 x 4 − 4 x

=

4.

(

x

∫4

x2 + 1

dx =

2 1 ( x + 1) = ⋅

2

Rr + B1 ln r + B2 . 4K

3 4

2

1 3 − 3x)dx = e3 x + 2 − x 2 + C 88. f ( x) = ∫ (e 3 2 1 1 1 ⎛ ⎞ f ⎜ ⎟ = 2 implies 2 = e3 − + C , so 3 6 ⎝3⎠ 13 1 C = − e3 . Thus, 6 3 1 3 13 1 f ( x ) = e3 x + 2 − x 2 + − e3 , 3 2 6 3 1 13 1 f (2) = e8 − 6 + − e3 3 6 3 1 8 3 = (2e − 2e − 23) ≈ 983.12 6

=

3x+2

5.

)

1 2 x3 + 4 x + 1 3

(

)

2 2 x +1 3

3 4

3 2

+C

(

)

−1 1 x 2 + 1 4 [2 x dx] ∫ 2

3 4

+C +C

9

dx = 9∫ (2 − 3 x)−1/ 2 dx 2 − 3x ⎛ 1⎞ = 9 ⎜ − ⎟ ∫ (2 − 3x) −1/ 2 [−3 dx] ⎝ 3⎠ (2 − 3 x)1/ 2 = −3 + C = −6 2 − 3 x + C

∫

1 2

6.

∫

2 xe x e

x2

−2

= ln e 559

2

x2

dx = ∫

1 e

−2 +C

x2

⎡ 2 xe x 2 dx ⎤ ⎢ ⎥⎦ −2⎣


7.

∫4

7x

( )

dx = ∫ eln 4


7x

dx = ∫ e(ln 4)(7 x ) dx

13.

1 e(ln 4)(7 x ) [7 ln 4 dx] 7 ln 4 ∫ 1 = ⋅ e(ln 4)(7 x ) + C 7 ln 4

5e 2 x

5

=

=

8.

9.

( )

1 e 7 ln 4 t

∫5

ln 4 7 x

+C =

=

14.

7x

4 +C 7 ln 4

7

1 1 (ln 5)t = e(ln 5)t [ln 5 dt ] = ⋅e +C ln 5 ∫ ln 5 5t = +C ln 5 x2 ⎞ ⎛ 4 dx = 14 x − 2 xe ⎟⎟ ∫ ⎜⎜ ⎠ ⎝

15.

x x x2 − 4 ⋅ e 4 + C = 7 x 2 − 4e 4 + C 2 2

e⎞ ⎛ x + x e + ex + ⎟ dx x⎠

1 = ∫ e x dx + ∫ x e dx + e ∫ x dx + e ∫ dx x x e+1 ex 2 + + e ln x + C e +1 2

x

2

dx = −

+C ⎡ 7 ⎤ ⎢ − 2 dx ⎥ ⎣ x ⎦

1 7x e 7∫

Note that since x 2 + 9 > 0 for all values of x, the absolute value bars are not needed.

6 x 2 − 11x + 5 2 = 2x − 3 + . 3x − 1 3x − 1

18. By using long division on the integrand, 5 − 4x2 4 ⎞ ⎛ ∫ 3 + 2 x dx = ∫ ⎜⎝ −2 x + 3 − 3 + 2 x ⎟⎠ dx 1 [2 dx] = ∫ (−2 x + 3)dx − 2 ∫ 3 + 2x

6 x 2 − 11x + 5 2 ⎞ ⎛ dx = ∫ ⎜ 2 x − 3 + ⎟ dx 3x − 1 3 x −1 ⎠ ⎝ 1 1 [3 dx] = 2 ∫ x dx − ∫ 3 dx + 2 ⋅ ∫ 3 3x − 1 2 = x 2 − 3x + ln 3x − 1 + C 3

Thus

∫ x2

1

2

17. By using long division on the integrand, 5 x3 45 x ⎞ ⎛ ∫ x2 + 9 dx = ∫ ⎜⎝ 5 x − x2 + 9 ⎟⎠ dx 45 1 [2 x dx] = ∫ 5 x dx − ∫ 2 2 x +9 5 45 = x 2 − ln( x 2 + 9) + C 2 2

∫ ⎜⎝ e

11. By long division,

7

dx = ∫ e x ⋅

)

16. By using long division on the integrand, 2 x 4 − 6 x3 + x − 2 dx ∫ x−2 16 ⎞ ⎛ = ∫ ⎜ 2 x3 − 2 x 2 − 4 x − 7 − ⎟ dx x−2⎠ ⎝ 1 2 = x 4 − x3 − 2 x 2 − 7 x − 16 ln x − 2 + C . 2 3

⎞ ⎟⎟ dx ⎠

2

= ex +

dx = − ∫ e8−6 x [−6 dx]

1 = − e +C 7

x ⎡1 ⎤ = 14 ∫ x dx − 2 ⋅ 2∫ e 4 ⎢ x dx ⎥ 2 ⎣ ⎦

10.

ex

(2)dx]

7 x

x2

2

)

4 −3 x 2

(

= 14 ∫ x dx − 2∫ xe 4 dx

= 14 ⋅

∫ 6 (e

2x

5 ln(7e2 x + 4) + C 14

= −e8−6 x + C = − e4−3 x

dt = ∫ (eln 5 )t dt = ∫ e(ln 5)t dt

x2 ⎛ 2 x 7 − e ∫ ⎜⎜ 4 ⎝

1

∫ 7e2 x + 4 dx = 14 ∫ 7e2 x + 4[7e

∫

= − x 2 + 3x − 2 ln 3 + 2 x + C

19.

(3 x + 2)( x − 4) 3 x 2 − 10 x − 8 dx = ∫ dx 12. ∫ x −3 x−3 11 ⎞ 3 2 ⎛ = ∫ ⎜ 3x − 1 − ⎟ dx = x − x − 11ln x − 3 + C 2 x −3⎠ ⎝

∫

(

x +2

2

3 x

2 = ⋅ 3 560

)

(

dx =

x +2 3

)

2 3∫

(

x +2

3

+C =

2 9

(

)

2

⎡ 1 ⎤ dx ⎥ ⎢ ⎣2 x ⎦

x +2

)

3

+C


20.

5e s

∫ 1 + 3es

Section 14.5

( )

5 1 [3e s ds ] ∫ 3 1 + 3e s 5 = ln(1 + 3e s ) + C 3 4

⎛ 1 ⎞ 5⎜ x3 + 2 ⎟ 4 ⎠ dx = 3 5 ⎛ x 13 + 2 ⎞ ⎡ 1 x − 23 dx ⎤ 21. ∫ ⎝ ⎟ ⎢ ∫ ⎜⎝ ⎥ 3 2 ⎠ ⎣3 ⎦ x

=

5

⎛ 1 ⎞ = 3⎜ x 3 + 2 ⎟ + C ⎝ ⎠

22.

∫

1

1+ x

1 2 ⎛ ⎞ dx = 2∫ ⎜ 1 + x 2 ⎟ ⎝ ⎠

x

)

3 2

+C

2 ln x ⎡ 1 ⎤ (ln x) dx = ∫ (ln x) ⎢ dx ⎥ = +C x 2 ⎣x ⎦ 1 = ln 2 x + C 2

24.

0.6

dt = −

2 ⎡ 3 ⎤ (3 − t 3 / 2 )0.6 ⎢ − t1/ 2 dt ⎥ ∫ 3 ⎣ 2 ⎦

31.

1.6 2 (3 − t 3 / 2 )1.6 5 3−t t =− ⋅ +C = − +C 3 1.6 12

(

25.

26.

)

∫

32.

2

9 x5 − 6 x 4 − ex3 7x

2

dx = 2 ∫

1

2

⎡2 ⎤ ⎢ x dx ⎥ ⎣ ⎦

3 3 e x +1 dx = ∫ x 2 (e x +1 )1/ 2 dx 3

3

2 x 2+1 ⎡ 3 2 ⎤ 2 x 2+1 e ⎢ 2 x dx ⎥ = 3 e dx 3∫ ⎣ ⎦

8

⎡ 1

1

⎤

∫ ( x + 3) ln( x + 3) dx = 8∫ ln( x + 3) ⎢⎣ x + 3 dx ⎥⎦ = 8ln ln( x + 3) + C

ln (r + 1) ⎛ 1 ⎞ dr = ∫ [ln(r + 1)]2 ⎜ dr ⎟ r +1 r + 1 ⎝ ⎠ 1 3 = ln (r + 1) + C 3

∫

2

2

30. By using long division on the integrand, x+3 3 ⎞ ⎛ ∫ x + 6 dx = ∫ ⎜⎝1 − x + 6 ⎟⎠ dx = x − 3ln x + 6 + C

)

∫ t (3 − t t )

4

∫x =

∫

(

∫ x ln

29.

2

23.

( )

ln ( 2 x ) ( 2x ) = 2 ln ln ( 2x 2 ) + C

3

(

1 ln 3 ln x 3ln x e +C = +C ln 3 ln 3

28. ⎡ 1 − 12 ⎤ ⎢ 2 x dx ⎥ ⎣ ⎦

1 2 ⎛ 2 ⎞ ⎜1 + x ⎟ ⎝ ⎠ + C = 4 1+ x = 2⋅ 3 3

ln x

eln 3 3ln x 27. ∫ dx = ∫ dx x x 1 ⎡ ln 3 ⎤ = e(ln 3) ln x ⎢ dx ⎥ ∫ ln 3 ⎣ x ⎦ 1 (ln 3) ln x = ⋅e +C ln 3

ds =

⎛ e2 + x e − 2 x ⎞ dx = ee2 x + 1 xe +1 − x 2 + C ⎟ ⎠ e +1

∫ ⎜⎝ e

33. By using long division on the integrand, 2x ⎞ x3 + x 2 − x − 3 ⎛ ∫ x2 − 3 dx = ∫ ⎜⎝ x + 1 + x 2 − 3 ⎟⎠ dx 1 [2 x dx] = ∫ ( x + 1)dx + ∫ 2 x −3 x2 = + x + ln x 2 − 3 + C 2

e ⎞ 6 ⎛9 dx = ∫ ⎜ x3 − x 2 − x ⎟ dx 7 7 ⎠ ⎝7 9 4 2 3 e 2 = x − x − x +C 28 7 14

34.

∫

4 x ln 1 + x 2 1 + x2

(

= ∫ ln 1 + x

561

2

)

dx = ∫

(

4 x ⋅ 12 ln 1 + x 2 1 + x2

(

) dx

)

2 2 ⎡ 2x ⎤ ln 1 + x dx ⎥ = +C ⎢ 2 ⎣1 + x 2 ⎦


35.

∫

6 x 2 ln( x3 + 1) 2 3

x +1


42.

dx

⎡ 6 x2 ⎤ = ∫ [2 ln( x3 + 1)]1/ 2 ⎢ dx ⎥ 3 ⎣⎢ x + 1 ⎦⎥ =

36.

[2 ln( x3 + 1)]3 / 2 3 2

(

2 ∫3 x + 2

= 3∫ e

37.

38.

⎛

)

− 12

( x +2) 2

1 2

2 xe x + 2 dx

43.

− 12 2 ⎡ ⎤ 2 x x 2 dx ⎥ = 3e x + 2 + C + ⎢ ⎣ ⎦

∫ x2 + 2 x −1 dx = 2 ∫ x2 + 2 x−1 ⎡⎢⎣( 2 x − 2 x =

∫

)

=

⎞ − ln 7 ⎟ dx ⎟ 4 ⎝ x − 4x ⎠ 1 1 4 2 = ∫ ( x − 4 x) [(4 x3 − 4)dx] − ln 7 ∫ dx 4 3 1 4 = ( x − 4 x) 2 − (ln 7) x + C 6 x − x −2

1

( x + 1) ln ( x2 + 1)

1

dx

⎡ 2x ⎤ dx ⎥ ⎢ 2 ln x + 1 ⎣ x + 1 ⎦

(

1

)

2

(

)

= ln ln x 2 + 1 + C

x3 − 1

∫ ⎜⎜

2x

2

=∫

2 + C = ln 3 / 2 ( x3 + 1) 2 + C 3

(

∫

44.

xe x

2

2

ex + 2 1 ⋅ 2

(

(

)

− 12 1 x2 ⎡ 2 xe x 2 dx ⎤ dx = ∫ e + 2 ⎣⎢ ⎦⎥ 2

2

ex + 2

)

1 2 2

+ C = ex + 2 + C

1 2

5

∫ (3x + 1)[1 + ln(3x + 1)]2 dx 5 ⎡ 1 ⎤ ⋅ 3 dx ⎥ [1 + ln(3 x + 1)]−2 ⎢ ∫ 3 ⎣ 3x + 1 ⎦ 5 =− +C 3[1 + ln(3 x + 1)] =

−2

)⎤⎥⎦ dx

1 ln x 2 + 2 x −1 + C 2

45.

∫

( e− x + 6 ) e

x

2

(

dx = − ∫ e − x + 6

e− x + 6 ) ( =−

)

2

⎡ −e− x dx ⎤ ⎣ ⎦

3

39.

∫

2 x 4 − 8 x3 − 6 x 2 + 4 x3

dx

6 4 ⎞ ⎛ = ∫ ⎜ 2 x − 8 − + ⎟ dx x x3 ⎠ ⎝ 1 = 2 ∫ x dx − ∫ 8 dx − 6 ∫ dx + 4 ∫ x −3 dx x

⎡ ⎤ ⎢ 1 ⎥ 1 46. ∫ ⎢ − ⎥ dx 2 ⎢ 8 x + 1 e x 8 + e− x ⎥ ⎣ ⎦ −2 1 1 ⎡ −e− x dx ⎤ = ∫ [8 dx ] − (−1) ∫ 8 + e − x ⎣ ⎦ 8 8x + 1

(

−2

2

40.

e x + e− x

∫ e x − e− x dx = ∫ e x − e− x ⎡⎢⎣( e 1

x

)

(

x x − 8 x − 6 ln x + 4 ⋅ +C 2 −2 2 = x 2 − 8 x − 6 ln x − +C x2 = 2⋅

+C

3

(

)

−1

8 + e− x 1 = ln 8 x + 1 + +C 8 −1 1 1 = ln 8 x + 1 − +C 8 8 + e− x

)

+ e− x dx ⎤⎥ ⎦

= ln e x − e − x + C

41. By using long division on the integrand, x 1 ⎞ ⎛ ∫ x + 1 dx = ∫ ⎜⎝1 − x + 1 ⎟⎠ dx = x − ln x + 1 + C

562

)


47.

∫(x

3

+ ex

)

(

x 2 + e dx = ∫ x x 2 + e x 2 + e

(

1 = ∫ x2 + e 2 =

48.

(

1 2 x +e 5

∫3

x ln x

)( 2 1 ( x + e) [2 x dx] = ⋅

Section 14.5

)

)

5 2

3 2

1 2

53. eln( x

+C

54.

+C

55.

( )

x ln x (1 + ln x )dx = ∫ eln 3 (1 + ln x )dx

1 e(ln 3) x ln x [(ln 3)(1 + ln x)dx] ln 3 ∫ 1 (ln 3) x ln x 1 ln 3 x ln x = ⋅e +C = e +C ln 3 ln 3

=

49.

∫ =

56.

3x ln x +C ln 3 1 2

⎛

2 3⋅8

3 2

3 2

3

3 2

1 2

57.

1 2

⎞ ⎡ 32 3 12 ⎤ ⎢8 ⋅ 2 ⋅ x dx ⎥ ⎣ ⎦

3

∫ ⎜⎝ 8 2 x 2 + 3 ⎟⎠

=

3 ⋅16 2

⎛ 32 32 ⎞2 ⎜ 8 x + 3⎟ ⎠ +C ⋅⎝

50.

3

2

− 23

∫

∫e

f ( x ) + ln( f ′( x ))

dx = ∫ e f ( x ) ⋅ eln( f ′( x )) dx = ∫ e f ( x ) [ f ′( x) dx]

dr 200 = dq (q + 2)2

r = pq, then p =

⎡1 ⎤ ⎢ x dx ⎥ ⎣ ⎦

r 100 = . q q+2

The demand function is p =

1 3

= 6(ln x) + C s

3

2 2 ds = − ∫ e− s 51. ∫ 3 3 e s 3 2 = − e− s + C 3

(q + 2)−1 +C −1

200 +C q+2 When q = 0, then r = 0, so 0 = –100 + C, or 200 100q + 100 = . Since C = 100. Hence r = − q+2 q+2

3 2

∫ x(ln x)2 / 3 dx = 2∫ (ln x)

ln x + ln e x dx x x ln x + x ⎛ ln x ⎞ =∫ + 1⎟ dx dx = ∫ ⎜ x ⎝ x ⎠

∫

=−

3 2 ⎡ 2 + 3⎤ + C (8 ) = x ⎢ ⎥ ⎦ 36 2 ⎣

1

( ) dx =

ln xe x

r = ∫ 200(q + 2)−2 dq = 200 ⋅

3

2

∫ dx = ∫ 1 dx = x + C

= e f ( x) + C

⎛ ⎞ x (8 x ) + 3dx = ∫ ⎜ 8 x + 3 ⎟ ⋅ x dx ⎝ ⎠ 3 2

+1)

ln 2 x ⎡1 ⎤ = ∫ (ln x) ⎢ dx ⎥ + ∫ 1 dx = + x+C 2 ⎣x ⎦

( )

=

2

is simply x 2 + 1. Thus 1 ln( x 2 +1) dx = ∫ ( x 2 + 1)dx = x3 + x + C ∫e 3

dx

5 2

5 2

2

)

58.

⎡ 3 12 ⎤ ⎢ − 2 s ds ⎥ ⎣ ⎦

100 . q+2

dr 900 = dq (2q + 3)3 r = ∫ 900(2q + 3)−3 dq

ln 3 x 1 ⎡1 ⎤ dx = ∫ (ln x)3 ⎢ dx ⎥ 52. ∫ 3x 3 ⎣x ⎦

= 900 ⋅

1 (2q + 3) −3 [2 dq] 2∫

= 450 ⋅

(2q + 3)−2 225 +C = − +C −2 (2q + 3)2

When q = 0, then r = 0, so 0 = –25 + C or 225 C = 25. Hence r = − + 25 . Since (2q + 3) 2

1 (ln x)4 1 = ⋅ + C = ln 4 x + C 3 4 12

563



r 25 225 = − q q q(2q + 3)2

r = pq, then p =

62.

−1 ⎞ ⎛ 1 (2 I ) 2 ⎟ ⎜ − C=∫ dI ⎜2 2 ⎟ ⎝ ⎠ 1 1 −1 = ∫ dI − ∫ (2 I ) 2 [2 dI ] 2 4

⎤ 25 ⎡ 9 The demand function is p = ⎢1 − ⎥. 2 q ⎢⎣ (2q + 3) ⎥⎦

59.

dc 20 = dq q + 5

1

1 1 (2 I ) 2 = ⋅I − ⋅ + C1 1 2 4

20 1 dq = 20 ∫ dq = 20 ln q + 5 + C q+5 q+5 When q = 0, then c = 2000, so 2000 = 20 ln(5) + C, or C = 2000 – 20 ln 5. Hence c = 20 ln q + 5 + 2000 − 20 ln 5 c=∫

2

2I I = − + C1 2 2 3 3 4 + C1 , so implies = 1 − 4 4 2 I The consumption function is C = − 2 C (2) =

q+5 = 20 ( ln q + 5 − ln 5 ) + 2000 = 20 ln + 2000 5 q+5 + 2000. The cost function is c = 20 ln 5

60.

63.

dc = 3e0.002q dq c = ∫ 3e0.002 q dq = 3 ⋅

1 e0.002q [0.002 dq] ∫ 0.002

The cost function is c = 1500e

0.002 q

C = ∫I

⎤ 1 ⎥ dI = 3 dI − 1 I − 2 dI ∫ ∫ ⎥ 4 6 ⎦

1

3 1 I2 3 I = I− ⋅ + C1 = I − + C1 1 4 6 4 3 2

+ 500.

Thus C =

dC 1 = dI I − 12

3 I + C1 . I− 4 3

C(25) = 23 implies that 23 = 1

dI =

I2 1 2

71 . 12 The consumption function is 3 1 71 C= I− I+ . 4 3 12

C(9) = 8 implies that 8 = 2 ⋅ 3 + C1 , or C1 = 2 .

(

)

I +1 .

The consumption function is C = 2

3 5 ⋅ 25 − + C1 , so 4 3

C1 =

+ C1 = 2 I + C1

Thus C = 2 I + 2 = 2

3 . 4 2I 3 + . 2 4

C1 =

dC 3 1 = − dI 4 6 I −1 ⎡ 3 I 2 C = ∫⎢ − ⎢4 6 ⎣

= 1500e0.002q + C When q = 0, then c = 2000, so 2000 = 1500 + C, or C = 500.

61.

dC 1 1 = − dI 2 2 2 I

(

)

I +1 .

64.

dc 100 = 10 − dq q + 10 ⎛ 100 ⎞ c = ∫ ⎜ 10 − ⎟ dq = 10q − 100 ln q + 10 + C q + 10 ⎠ ⎝ ln q + 10 C c Avg. cost = = 10 − 100 + q q q When q = 100, then avg. cost = 50, so ln(110) C + 50 = 10 − 100 , or 100 100 C = 100(40 + ln (110)). Thus c = 10q − 100 ln q + 10 + 100(40 + ln(110))

564


Section 14.5

Evaluating c when q = 0 gives fixed cost: c(0) = –100 ln(10) + 100(40 + ln (110)) ≈ 4240. The fixed cost is $4240.

b.

1

dc 100q − 3998q + 60 = dq q 2 − 40q + 1

a.

3 ⎛ ⎞2 0.04q 2 + 4 ⎟ ⎜ 0.9 ⎝ ⎠ +C = ⋅ 3 0.06

dc 100(40)2 − 3998(40) + 60 = dq q = 40 (40) 2 − 40(40) + 1 = $140 per unit

2

3

3 ⎛ ⎞2 Thus c = 10 ⎜ 0.04q 2 + 4 ⎟ + C . When ⎝ ⎠

dc by using long dq

3

q = 0, then c = 360, so 360 = 10(4) 2 + C , or 3

division: 100q 2 − 3998q + 60 c=∫ dq q 2 − 40q + 1

3 ⎛ ⎞2 C = 280. Hence c = 10 ⎜ 0.04q 2 + 4 ⎟ + 280 . ⎝ ⎠ 3

When q = 25, then c = 10(9) 2 + 280 = $550 .

⎛ 2q − 40 ⎞ = ∫ ⎜ 100 + ⎟ dq 2 ⎜ q − 40q + 1 ⎟⎠ ⎝ 1 = ∫ 100 dq + ∫ [(2q − 40)dq] 2 q − 40q + 1

c.

Thus c = 100q + ln q 2 − 40q + 1 + C. When q = 0, then c = 10,000, so 10,000 = 0 + ln(1) + C, so C = 10,000.

If c = f(q), then f(q + dq) ≈ f(q) + dc dc = f (q) + dq . Letting q = 25 and dq dq = –2, we have dc f (23) = f (25 − 2) ≈ f (25) + ⋅ (−2) dq q = 25 = 550 + 13.50(–2) = $523

2

Hence c = 100q + ln q − 40q + 1 + 10, 000 . When q = 40, then c = 4000 + ln(1) + 10,000 = $14,000.

67.

dV 8t 3 = dt 0.2t 4 + 8000 8t 3

V =∫

If c = f(q), then dc f (q + dq ) ≈ f (q ) + dc = f (q) + dq dq Letting q = 40 and dq = 2, we have dc f (42) = f (40 + 2) ≈ f (40) + ⋅ (2) dq q = 40

0.2t 4 + 8000

(

= 10 ∫ 0.2t 4 + 8000

( 0.2t 4 + 8000) = 10 1 2

= 14,000 + 140(2) = $14,280

)

dt − 12

⎡ 0.8t 3 ⎤ dt ⎣ ⎦

1 2

+C

Thus V = 20 0.2t 4 + 8000 + C . If t = 0, then V = 500, so 500 = 20 8000 + C , 500 = 20 1600 ⋅ 5 + C , 500 = 800 5 + C , or C = 500 − 800 5 . Hence

3 dc 9 = q 0.04q 2 + 4 66. dq 10

a.

1 ⎡ 2 dq ⎤ 0.06 q ⎢⎣ ⎥⎦

3

b. To find c, we integrate

c.

3 9 q 0.04q 2 + 4dq 10

3 2 0.9 ⎛ 2 + 4⎞ = 0.04 q ⎜ ⎟ ∫ 0.06 ⎝ ⎠

2

65.

c=∫

dc 9 9 27 25 9 = ⋅ 5 ⋅ 3 = = 10 2 dq q = 25 10 = $13.50 per unit

V = 20 0.2t 4 + 8000 + 500 − 800 5 . When t = 10, then V = 20 10, 000 + 500 − 800 5 = 20(100) + 500 − 800 5 ≈ $711 per acre.

565


68.


dr a ae− q ae− q = = = dq e q + b (eq + b)e− q 1 + be− q r=∫

ae− q −q

b.

1 ⎛ 1⎞ dq = ⎜ − ⎟ a ∫ [−be − q dq ] ⎝ b ⎠ 1 + be− q

1

Thus S =

I I − 5.4 3 + C1 . When I = 24, 2 3

then S = 3, so 3 = 12 − 5.4 3 8 + C1 , or I I − 5.4 3 + 1.8 . If C is 2 3 the total national consumption (in billions of dollars), then C + S = I, or ⎛I ⎞ I C = I − S = I − ⎜⎜ − 5.4 3 + 1.8 ⎟⎟ . 3 ⎝2 ⎠ C1 = 1.8 . Thus S =

dS 5 dI = ∫ dI dI ( I + 2) 2

Therefore, C =

( I + 2) −1 + C1 −1

c.

5 + C1 . If C is the total national I +2 consumption (in billions of dollars), then 5 − C1 . C + S = I, or C = I – S. Hence C = I + I +2 1 When I = 8, then C = 7.5, so 7.5 = 8 + − C1 , or 2 5 C1 = 1 . Thus S = 1 − . If S = 0, then I +2 5 5 0 = 1− ⇒ =1⇒ 5 = I + 2 ⇒ I = 3 I +2 I +2

Thus S = −

70. a.

⎞ ⎟ dI ⎟ ⎠

I 1.8 I 3 I 5.4 3 = − ⋅ + C1 = − ⋅ I + C1 2 33 1 2 33 3

a Now r = 0 when q = 0, so 0 = − ln(1 + b) + C , b a or C = ln(1 + b). Hence b a a r = − ln(1 + be − q ) + ln(1 + b) b b a 1+ b = ln b 1 + be − q r a 1+ b p= = ln q bq 1 + be − q

= 5∫ ( I + 2) −2 dI = 5 ⋅

⎛1 1.8 dS dI = ∫ ⎜ − ⎜2 3 2 dI 3I ⎝

⎛ 1 1.8 − 23 ⎞ I ⎟ dI = ∫⎜ − 3 3 ⎝2 ⎠

1 + be a = − ln(1 + be − q ) + C b

69. S = ∫

S=∫

I I + 5.4 3 − 1.8 . 2 3

From (b), when I = 81, then 81 81 C = + 5.4 3 − 1.8 = 40.5 + 16.2 − 1.8 2 3 = 54.9 Thus consumption is $54.9 billion when income is $81 billion.

d. If C = f(I), then f(I + dI) ≈ f(I) + dC = f ( I ) +

dC dI . Let dI

I = 81 and dI = –3. Then f (78) = f (81 − 3) ≈ f (81) +

dC (−3) dI I =81

17 (−3) = 53.2 . Thus when income 30 is $78 billion, then consumption is approximately $53.2 billion. = 54.9 +

If C is total national consumption (in billions of dollars), then ⎛1 dC dS 1.8 ⎞ ⎟ . Thus = 1− = 1− ⎜ − ⎜2 3 2 ⎟ dI dI 3I ⎠ ⎝ ⎛1 dC 1.8 ⎞⎟ = 1− ⎜ − ⎜ 2 3 3(81) 2 ⎟ dI I =81 ⎝ ⎠ ⎛ 1 1.8 ⎞ 17 = 1− ⎜ − . ⎟= ⎝ 2 27 ⎠ 30

566


Section 14.6

Principles in Practice 14.6 1. Divide the interval [0, 10] into n subintervals of equal length ∆x, so ∆x =

10 . The endpoints of the subintervals n

10 ⎛ 10 ⎞ ⎛ 10 ⎞ ⎛ 10 ⎞ ⎛ 10 ⎞ , 2 ⎜ ⎟ , 3 ⎜ ⎟ , ... , (n − 1) ⎜ ⎟ , and n ⎜ ⎟ = 10 . Letting Sn denote the sum of the areas of the n ⎝ n⎠ ⎝ n⎠ ⎝ n⎠ ⎝ n⎠ rectangles corresponding to right-hand endpoints, we have 10 ⎛ 10 ⎞ 10 ⎡ ⎛ 10 ⎞ ⎤ 10 ⎡ ⎛ 10 ⎞ ⎤ Sn = R ⎜ ⎟ + R ⎢ 2 ⎜ ⎟ ⎥ +…+ R ⎢ n ⎜ ⎟ ⎥ n ⎝ n ⎠ n ⎣ ⎝ n ⎠⎦ n ⎣ ⎝ n ⎠⎦

are 0,

=

10 ⎡ ⎧ ⎧ ⎛ 10 ⎞ ⎫ ⎧ ⎛ 10 ⎞ ⎫ ⎛ 10 ⎞ ⎫⎤ ⎢ ⎨600 − 0.5 ⎜ ⎟ ⎬ + ⎨600 − 0.5(2) ⎜ ⎟ ⎬ +… + ⎨600 − 0.5(n) ⎜ ⎟ ⎬⎥ n ⎣⎩ ⎝ n ⎠⎭ ⎩ ⎝ n ⎠⎭ ⎝ n ⎠ ⎭⎦ ⎩

=

⎤ 10 ⎡ ⎛ 10 ⎞ 600n − 0.5 ⎜ ⎟ {1 + 2 +…+ n}⎥ ⎢ n ⎣ ⎝ n⎠ ⎦

=

10 ⎡ ⎛ 10 ⎞ n(n + 1) ⎤ 600n − 0.5 ⎜ ⎟ ⎢ ⎥ n ⎣ ⎝ n⎠ 2 ⎦

10 [600n − 2.5(n + 1)] n ⎛ n +1⎞ = 6000 − 25 ⎜ ⎟ ⎝ n ⎠ Now take the limit of Sn as n → ∞ =

⎡ ⎡ ⎛ n + 1 ⎞⎤ ⎛ 1 ⎞⎤ lim Sn = lim ⎢6000 − 25 ⎜ ⎟ ⎥ = lim ⎢6000 − 25 ⎜ 1 + ⎟ ⎥ = 6000 − 25 = 5975 n n →∞ ⎣ n →∞ ⎝ ⎠⎦ ⎝ n ⎠⎦ ⎣ The total revenue for selling 10 units is $5975. n →∞

Problems 14.6 1. f(x) = x, y = 0, x = 1 1 S3 , ∆x = 3 1 ⎛1⎞ 1 ⎛ 2⎞ 1 S3 = f ⎜ ⎟ + f ⎜ ⎟ + 3 ⎝3⎠ 3 ⎝ 3⎠ 3

⎛ 3 ⎞ 1 ⎡1 2 3⎤ 1 6 2 f⎜ ⎟ = ⎢ + + ⎥= ⋅ = ⎝ 3 ⎠ 3 ⎣3 3 3⎦ 3 3 3 2 The area is approximately sq unit. 3

2. f(x) = 3x, y = 0, x = 1 1 S5 , ∆x = 5 1 ⎛1⎞ 1 ⎛2⎞ 1 S5 = f ⎜ ⎟ + f ⎜ ⎟ + f 5 ⎝5⎠ 5 ⎝ 5⎠ 5 9 The area is approximately 5

⎛ 3 ⎞ 1 ⎛ 4 ⎞ 1 ⎛ 5 ⎞ 1 ⎡ 3 6 9 12 15 ⎤ 1 45 9 = ⎜ ⎟ + f ⎜ ⎟+ f ⎜ ⎟ = ⎢ + + + + ⎥ = ⋅ ⎝ 5 ⎠ 5 ⎝ 5 ⎠ 5 ⎝ 5 ⎠ 5 ⎣5 5 5 5 5 ⎦ 5 5 5

sq units.

567


3.

ISM: Introductory Mathematical Analysis 6. f(x) = 3x + 2; [0, 3] 3 ∆x = n 3 ⎛3⎞ 3 ⎛ 3⎞ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠ 3⎡ ⎛3⎞ ⎛ 3 ⎞⎤ = ⎢ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ ⎥ n⎣ ⎝n⎠ ⎝ n ⎠⎦

f ( x) = x 2 , y = 0, x = 1 1 4 1 ⎛1⎞ 1 ⎛2⎞ 1 S4 = f ⎜ ⎟ + f ⎜ ⎟ + f 4 ⎝4⎠ 4 ⎝4⎠ 4 1⎡1 4 9 16 ⎤ = ⎢ + + + ⎥ 4 ⎣ 16 16 16 16 ⎦ 1 30 15 = ⋅ = 4 16 32 15 The area is approximately 32 S 4 , ∆x =

4.

⎛3⎞ 1 ⎛4⎞ ⎜ 4⎟+ 4 f ⎜ 4⎟ ⎝ ⎠ ⎝ ⎠

=

3 ⎧9 ⎫ ⎨ (1 +…+ n) + 2n ⎬ n ⎩n ⎭ 3 ⎧ 9 n(n + 1) ⎫ 27(n + 1) = ⎨ ⋅ + 2n ⎬ = +6 n ⎩n 2 2n ⎭

sq units.

=

f ( x) = x 2 + 1 , y = 0, x = 0, x = 1 1 2 1 ⎛1⎞ 1 S2 = f ⎜ ⎟ + 2 ⎝2⎠ 2

3 ⎧⎡ ⎛ 3 ⎞ ⎤ ⎡ ⎛ 3 ⎞ ⎤⎫ ⎨ ⎢3 ⎜ ⎟ + 2 ⎥ +…+ ⎢3 ⎜ n ⋅ ⎟ + 2 ⎥ ⎬ n ⎩⎣ ⎝ n ⎠ ⎦ ⎣ ⎝ n ⎠ ⎦⎭

S 2 , ∆x =

7. a.

⎛ 2 ⎞ 1 ⎡ 5 8 ⎤ 1 13 13 f ⎜ ⎟= ⎢ + ⎥= ⋅ = 8 ⎝ 2 ⎠ 2 ⎣4 4⎦ 2 4

Sn =

1 ⎡⎛ 1 ⎞ ⎛ 2 ⎞ ⎛ n ⎞⎤ ⎜ + 1⎟ + ⎜ + 1⎟ +…+ ⎜ + 1⎟ ⎥ ⎢ n ⎣⎝ n ⎠ ⎝ n ⎠ ⎝ n ⎠⎦

1 ⎡1 ⎤ (1 + 2 +…+ n) + n ⎥ n ⎢⎣ n ⎦ 1 ⎡ 1 n(n + 1) ⎤ = ⎢ ⋅ + n⎥ n ⎣n 2 ⎦ n +1 = +1 2n =

The area is approximately

13 sq units. 8

5. f(x) = 4x; [0, 1] 1 ∆x = n 1 ⎛1⎞ 1 ⎛ 1⎞ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠ 1⎡ ⎛1⎞ ⎛ 1 ⎞⎤ = ⎢ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ ⎥ n⎣ ⎝n⎠ ⎝ n ⎠⎦

b.

1⎡ 1 n⎤ 4 ⋅ +…+ 4 ⋅ ⎥ n ⎢⎣ n n⎦ 4 4 n(n + 1) = ⋅ [1 +…+ n] = 2 2 n n2 2(n + 1) = n =

⎡ n +1 ⎤ lim Sn = lim ⎢ + 1⎥ n →∞ ⎣ 2n ⎦ 1 1 ⎡ ⎤ = lim ⎢ + + 1⎥ n→∞ ⎣ 2 2n ⎦ 1 3 = + 0 +1 = 2 2 n →∞

2 ⎡⎛ 2 ⎞ ⎛ 2 ⎞ ⎛ 2⎞ ⎢⎜ ⎟ + ⎜ 2 ⋅ ⎟ +…+ ⎜ n ⋅ ⎟ n ⎢⎝ n ⎠ ⎝ n ⎠ ⎝ n⎠ ⎣ 2

8. a.

Sn =

2

2 22 ⎡ 2 1 + 22 +…+ n 2 ⎤ ⋅ ⎦ n n2 ⎣ 8 n(n + 1)(2n + 1) = ⋅ 6 n3 =

=

568

4(n + 1)(2n + 1) 3n 2

2⎤

⎥ ⎥⎦


lim Sn = lim

b.

4(n + 1)(2n + 1)

Section 14.6

4 ⎡ 2n 2 + 3n + 1⎤ ⎦ = lim ⎣ n→∞ 3n 2

3n 2 ⎡4 ⎛ 3 1 ⎞⎤ 4 8 = lim ⎢ ⎜ 2 + + ⎟ ⎥ = (2 + 0 + 0) = 2 n n ⎠⎦ 3 3 n→∞ ⎣ 3 ⎝

n →∞

n →∞

9. f(x) = x, y = 0, x = 1 1 ∆x = n 1 ⎛1⎞ 1 Sn = f ⎜ ⎟ +…+ n ⎝n⎠ n 1 n +1 1 ⎡ 1 ⎤ = ⋅ = ⎢1 + ⎥ 2 n 2 ⎣ n⎦ 1 lim Sn = 2 n →∞ 1 The area is sq unit. 2

n⎤ 1 1 n(n + 1) ⎛ 1 ⎞ 1 ⎡1 [1 +…+ n] = f ⎜ n ⋅ ⎟ = ⎢ +…+ ⎥ = ⋅ 2 n⎦ n 2 ⎝ n ⎠ n ⎣n n2

10. f(x) = 3x, y = 0, x = 1 1 ∆x = n 1 ⎛1⎞ 1 ⎛ 1⎞ 1⎡ 1 n⎤ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢3 ⋅ +…+ 3 ⋅ ⎥ n ⎝n⎠ n ⎝ n⎠ n⎣ n n⎦ 3 3 n(n + 1) 3 n + 1 3 ⎡ 1 ⎤ = ⋅ [1 +…+ n] = = ⋅ = ⎢1 + ⎥ 2 2 2 n 2 ⎣ n⎦ n n2 3 lim Sn = 2 n →∞ 3 The area is sq units. 2 11.

f ( x) = x 2 , y = 0, x = 1 ∆x =

1 n

2 2 1 ⎛1⎞ 1 ⎛ 1 ⎞ 1 ⎡⎛ 1 ⎞ ⎛ 1⎞ ⎤ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢⎜ ⎟ +…+ ⎜ n ⋅ ⎟ ⎥ n ⎝n⎠ n ⎝ n ⎠ n ⎢⎝ n ⎠ ⎝ n ⎠ ⎥⎦ ⎣ 1 ⎡2 1 n(n + 1)(2n + 1) = 1 +…+ n 2 ⎤ = ⋅ 3⎣ ⎦ 6 n n3

Sn =

=

1 2n 2 + 3n + 1 1 ⎡ 3 1 ⎤ ⋅ = ⎢2 + + ⎥ 2 6 6⎣ n n2 ⎦ n

lim Sn =

n →∞

1 3

The area is

1 sq unit. 3

569



12. y = x 2 , y = 0, x = 1, x = 2 1 n 1 ⎛ 1⎞ 1 ⎛ 1⎞ 1 ⎛ 1⎞ Sn = f ⎜1 + ⎟ + f ⎜1 + 2 ⋅ ⎟ + " f ⎜ 1 + n ⋅ ⎟ n ⎝ n⎠ n ⎝ n⎠ n ⎝ n⎠ 2 2⎫ ⎧ 1 ⎪⎡ 1 ⎤ 1⎤ ⎪ ⎡ = ⎨ ⎢1 + ⎥ + " + ⎢1 + n ⋅ ⎥ ⎬ n ⎪⎣ n ⎦ n⎦ ⎪ ⎣ ⎩ ⎭ ⎧ ⎡ ⎤ ⎡ 1⎪ 1 1 1 1 ⎤ ⎫⎪ = ⎨ ⎢1 + 2 ⋅ + ⎥ + " + ⎢1 + 2n ⋅ + n 2 ⋅ ⎥ ⎬ 2 n ⎪⎩ ⎣ n n ⎦ n n 2 ⎦ ⎪⎭ ⎣

∆x =

1⎧ 2 1 2 2 ⎫ ⎨n + (1 + " + n) + 2 (1 + " + n ) ⎬ n⎩ n n ⎭ 2 n(n + 1) 1 n(n + 1)(2n + 1) = 1+ ⋅ + ⋅ 2 6 n2 n3 =

n + 1 1 2n 2 + 3n + 1 + ⋅ n 6 n2 3 1 ⎤ ⎡ 1⎤ 1 ⎡ = 1 + ⎢1 + ⎥ + ⎢ 2 + + ⎥ n n2 ⎦ ⎣ n⎦ 6 ⎣

= 1+

lim Sn = 1 + 1 +

n →∞

The area is 13.

1 7 = 3 3

7 sq units. 3

f ( x) = 3x 2 , y = 0, x = 1 ∆x =

1 n

Sn =

2 2 1 ⎛1⎞ 1 ⎛ 1⎞ 1⎡ ⎛1⎞ ⎛ 1⎞ ⎤ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ = ⎢3 ⎜ ⎟ +…+ 3 ⎜ n ⋅ ⎟ ⎥ n ⎝n⎠ n ⎝ n⎠ n⎢ ⎝n⎠ ⎝ n ⎠ ⎥⎦ ⎣

3 ⎡2 3 n(n + 1)(2n + 1) 1 2n 2 + 3n + 1 1 ⎡ 3 1 ⎤ 1 +…+ n 2 ⎤ = ⋅ = ⋅ = ⎢2 + + ⎥ 3 2 3⎣ ⎦ 6 2 2⎣ n n2 ⎦ n n n lim Sn = 1

=

n →∞

The area is 1 sq unit.

570


14.

Section 14.6

f ( x) = 9 − x 2 , y = 0, x = 0

20

3 n 3 ⎛3⎞ 3 ⎛ 3⎞ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠ 2 ⎡ ⎛ 3 ⎞ 2 ⎤ ⎫⎪ 3 ⎧⎪ ⎡ ⎛ 3 ⎞ ⎤ = ⎨ ⎢9 − ⎜ ⎟ ⎥ +…+ ⎢9 − ⎜ n ⋅ ⎟ ⎥ ⎬ n ⎪⎢ ⎝ n ⎠ ⎥ n ⎠ ⎥⎪ ⎦ ⎣⎢ ⎝ ⎦⎭ ⎩⎣ 2 ⎫⎪ 3 ⎧⎪ ⎛3⎞ = ⎨9n − ⎜ ⎟ ⎡12 +…+ n 2 ⎤ ⎬ ⎦ n⎪ ⎝n⎠ ⎣ ⎩ ⎭⎪

y

∆x =

x

16.

5

4

∫0 9 dx Let f(x) = 9 4 ∆x = n 4 ⎛4⎞ 4 ⎛ 4⎞ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠ 4 4 = [9 +…+ 9] = ⋅ 9n = 36 n n

⎡ 27 n(n + 1)(2n + 1) ⎤ = 27 − ⎢ ⋅ ⎥ 6 ⎣ n3 ⎦ 2 ⎡ ⎤ 9 2n + 3n + 1 9⎡ 3 1 ⎤ = 27 − ⎢ ⎥ = 27 − ⎢ 2 + + 2 ⎥ 2 2 ⎢⎣ 2 n n n ⎦ ⎣ ⎥⎦ lim Sn = 27 − 9 = 18

n →∞

4

Sn = 36 ∫0 9 dx = nlim →∞

The area is 18 sq units. 15.

3

1

3

∫1 5x dx

10

Let f(x) = 5x. 2 ∆x = n 2 ⎛ 2⎞ 2 ⎛ 2⎞ Sn = f ⎜1 + ⎟ + " + f ⎜1 + n ⋅ ⎟ n ⎝ n⎠ n ⎝ n⎠ 2⎡ ⎛ 2⎞ 2 ⎛ ⎞⎤ = ⎢5 ⎜ 1 + ⎟ + " + 5 ⎜ 1 + n ⋅ ⎟ ⎥ n⎣ ⎝ n⎠ n ⎠⎦ ⎝ 10 ⎡ 2 ⎤ = ⎢ (1 + " + 1) + (1 + " + n) ⎥ n ⎣ n ⎦ 10 ⎡ 2 n(n + 1) ⎤ = ⎢n + ⋅ n ⎣ n 2 ⎥⎦ 10 = [n + n + 1] n 10 = (2n + 1) n 10 = 20 + n

y

4

17.

x 10

3

∫0 −4x dx Let f(x) = –4x. 3 ∆x = n 3 ⎛3⎞ 3 ⎛ 3⎞ Sn = f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠ 3⎡ ⎛3⎞ 36 ⎛ 3 ⎞⎤ = ⎢ −4 ⎜ ⎟ −…− 4 ⎜ n ⋅ ⎟ ⎥ = − [1 +…+ n] n⎣ ⎝n⎠ ⎝ n ⎠⎦ n2 =−

3

Sn = 20 ∫1 5 x dx = nlim →∞

3

36 n(n + 1) n +1 ⎡ 1⎤ ⋅ = −18 ⋅ = −18 ⎢1 + ⎥ 2 2 n ⎣ n⎦ n

Sn = −18 ∫0 −4 x dx = nlim →∞

571


15


y

Sn = 3

2 ⎡⎛ 1 ⎞ 2 1 ⎧⎪ ⎡⎛ 1 ⎞ 1⎤ 1 ⎤ ⎫⎪ ⎨ ⎢⎜ ⎟ + ⎥ +…+ ⎢⎜ n ⋅ ⎟ + n ⋅ ⎥ ⎬ n ⎪ ⎢⎝ n ⎠ n⎥ n ⎥⎪ ⎢⎣⎝ n ⎠ ⎦ ⎦⎭ ⎩⎣

=

x 15

1 ⎛1⎞ 1 ⎛ 1⎞ f ⎜ ⎟ +…+ f ⎜ n ⋅ ⎟ n ⎝n⎠ n ⎝ n⎠

2 ⎫⎪ 1 ⎧⎪⎛ 1 ⎞ ⎡ 2 1 2 ⎨⎜ ⎟ ⎣1 +…+ n ⎤⎦ + [1 +…+ n]⎬ n ⎪⎝ n ⎠ n ⎩ ⎭⎪ 1 n(n + 1)(2n + 1) 1 n(n + 1) = ⋅ + ⋅ 6 2 n3 n2

=

18.

4

∫1 (2 x + 1)dx Let f(x) = 2x + 1. 4 −1 3 ∆x = = n n 3 ⎛ 3⎞ 3 ⎛ 3⎞ Sn = f ⎜1 + ⎟ + " + f ⎜1 + n ⋅ ⎟ n ⎝ n⎠ n ⎝ n⎠ ⎡ ⎛ 3 ⎧⎡ ⎛ 3 ⎞ ⎤ 3 ⎞ ⎤⎫ = ⎨ ⎢ 2 ⎜1 + ⎟ + 1⎥ + " + ⎢ 2 ⎜1 + n ⋅ ⎟ + 1⎥ ⎬ n ⎩⎣ ⎝ n ⎠ ⎦ n ⎠ ⎦⎭ ⎣ ⎝ 3⎧ 6 ⎫ = ⎨2n + (1 + 2 + " + n) + n ⎬ n⎩ n ⎭ 18 n(n + 1) = 6+ ⋅ +3 2 n2 n +1 = 9 + 9⋅ n ⎛ 1⎞ = 9 + 9 ⎜1 + ⎟ ⎝ n⎠ 4

∫1

)

+ x dx = lim Sn = n→∞

1 1 5 + = 3 2 6

y

x

20.

5

2

∫1 ( x + 2)dx Let f(x) = x + 2. 1 ∆x = n 1 ⎛ 1⎞ 1 ⎛ 1⎞ Sn = f ⎜ 1 + ⎟ +…+ f ⎜ 1 + n ⋅ ⎟ n ⎝ n⎠ n ⎝ n⎠ 1 ⎧ ⎡⎛ 1 ⎞ ⎤ ⎡⎛ 1 ⎞ ⎤⎫ = ⎨ ⎢⎜1 + ⎟ + 2 ⎥ +…+ ⎢⎜ 1 + n ⋅ ⎟ + 2 ⎥ ⎬ n ⎩ ⎣⎝ n ⎠ ⎦ n ⎠ ⎦⎭ ⎣⎝ 1⎧ 1 ⎫ = ⎨n + (1 + " + n) + 2n ⎬ n⎩ n ⎭ 1 n(n + 1) = 1+ ⋅ +2 2 n2 1 n +1 = 3+ ⋅ 2 n 1 ⎡ 1⎤ = 3 + ⎢1 + ⎥ 2 ⎣ n⎦ 2 1 7 Sn = 3 + = ∫1 ( x + 2)dx = nlim 2 2 →∞

x 6

)

+ x dx

Let f ( x) = x 2 + x . ∆x =

2

1

n →∞

2

1⎡ 3 1 ⎤ 1 ⎡ 1⎤ 2 + + ⎥ + ⎢1 + ⎥ ⎢ 6⎣ n n2 ⎦ 2 ⎣ n ⎦

5

10

1

=

1

(2 x + 1)dx = lim Sn = 9 + 9 = 18

∫0 ( x

1 2n 2 + 3n + 1 1 n + 1 ⋅ + ⋅ 6 2 n n2

∫0 ( x

y

19.

=

1 n

572


Section 14.7

y

5

area is composed of three subareas, A1 , A2 , and A3 and A = A1 + A2 + A3 . A1 = area of rectangle = (2)(1) = 2 sq units

x

21.

3

1 1 (1)(1) = sq unit 2 2 1 ⎛1⎞ 1 A3 = area of triangle = (1) ⎜ ⎟ = sq unit 2 ⎝2⎠ 4 Since 1 1 11 sq units, we A = A1 + A2 + A3 = 2 + + = 2 4 4 3 11 have ∫ f ( x)dx = . −1 4 A2 = area of triangle =

5

1 2

x 2 + 1 dx is simply a real number. Thus

∫2

3 Dx ⎡⎢ ∫ x 2 + 1dx ⎤⎥ = Dx (real number) = 0. ⎣ 2 ⎦

22.

0 ≤ x <1

2 if ⎧ ⎪ f ( x) = ⎨ 4 − 2 x if ⎪5 x − 10 if ⎩

y

3

1≤ x < 2

y = f(x) A3

2≤ x≤3

1

A 1 2 A2

f is continuous and f(x) ≥ 0 on [0, 3]. Thus

x 3

3

∫0 f ( x) dx gies the area A bounded by y = f(x), y = 0, x = 0 and x = 3. From the diagram, this area is composed of three subareas, A1 , A2 and A3 , and A = A1 + A2 + A3 . A1 = area of rectangle = (1)(2) = 2 sq unit

24. 44.6 sq units 25. 14.77 sq units

1 (1)(2) = 1 sq unit 2 1 A3 = area of triangle = (1)(10) = 5 sq unit 2 Since A = A1 + A2 + A3 = 2 + 1 + 5 = 8 sq units, A2 = area of triangle =

we have 10

26. 1.7 sq units 27. 2.4 28. 0.7

3

∫0 f ( x) dx = 8.

29. –25.5 30. 0.39

y

Principles in Practice 14.7 5 y = f(x)

1.

A3 A1

A2 1

x 2

3

∫3 10, 000e

(

0.02t

⎛ e0.02t dt = ⎜ 10, 000 ⎜ 0.02 ⎝

= 500, 000e0.02t

3

6

⎞ ⎟ ⎟ ⎠3

)3

6

( ) = 500, 000 ( e0.12 − e0.06 ) ≈ 32,830 = 500, 000 e0.02(6) − e0.02(3)

⎧1 if x ≤ 1 ⎪⎪ 23. f ( x) = ⎨2 − x if 1 ≤ x ≤ 2 ⎪ x ⎪⎩−1 + 2 if x > 2 f is continuous and f(x) ≥ 0 on [–1, 3]. Thus

∫−1 f ( x)dx

6

The total income for the chain between the third and sixth years was about $32,830.

gives the area A bounded by y = f(x),

y = 0, x = –1, and x = 3. From the diagram, this 573



2. The total cost for the first 5 years is M(5) or 5

9.

M (5) − M (0) = ∫ M ′( x )dx 0

5

⎛ x3 ⎞ 90 x + 5000 dx = ⎜ 90 + 5000 x ⎟ ∫0 ⎜ ⎟ 3 ⎝ ⎠0 5

(

)

2

(

= 30 x3 + 5000 x

) 0 = 30(5)3 + 5000(5) − 0 5

Problems 14.7

2.

3

∫0

∫2

(1 − e)dx = (1 − e) x

2

∫1

x2 5 x dx = 5 ⋅ 2

2

11.

∫1 3t

1

∫−3 (2 x − 3)dx = ( x 1

13.

3 3 1 ⎛ 3⎞ 4 dt = − ⋅ t −2 = − − ⎜ − ⎟ = 1 2 6 ⎝ 2⎠ 3

2 x −2

x −1 dx = − 2 2

∫1

2

2

=− 1

∫2 ( 3

1 2x 1

∫4 (2t − 3t

= −160 − (−10) = −150 2 2

)

)dt = (t

−8

8

− 3x

1

) −3 = −2 − 18 = −20

14.

3

3

∫1/ 2 ( 3/ 2

=

1

⎞ 1 ⎛ 17 ⎞ ⎟ = − −⎜− ⎟ ⎟ 2 ⎝ 2⎠ ⎠ −1

1 y 2 − 2 y + 1 dy = ∫ ( y − 1)2 dy = ( y − 1)3 2 3 2

2

8

x dx = ∫ x 4 / 3 dx 8 −8

3 ⋅128 3(−128) = − 7 7 768 = 7

8 1 7 − = 3 3 3

1

8 3 4

∫−8

3x7 / 3 = 7

16 = =8 2

8.

−3

1 ⎛ 1⎞ 1 = − −⎜− ⎟ = 4 ⎝ 2⎠ 4

4

⎛ 1 9 y2 6. ∫ (4 − 9 y ) = ⎜ 4 y − ⎜ −1 2 ⎝

=

3

5 15 = 10 − = 2 2

x2 4. ∫ −5 x dx = −5 ⋅ 2 2

7.

9

9

2

8

5.

9

∫8 dt = ∫8 1 dt = t 8 = 9 − 8 = 1

12.

= 4(1 − e) − 2(1 − e) = 2(1 − e)

3.

)

10.

3

5 dx = 5 x = 5(3) − 5(0) = 15 − 0 = 15 0

4

−1

⎛ ⎞ w2 3w − w − 1 dw = ⎜ w3 − − w⎟ ⎜ ⎟ 2 ⎝ ⎠ −2 2

1 15 = − − (−8) = 2 2

= 3750 + 25,000 = 28,750 The total cost for the first 5 years is $28,750.

1.

∫−2 ( −1

3/ 2

⎛ x3 x 2 ⎞ x + x + 1 dx = ⎜ + + x⎟ ⎜ 3 ⎟ 2 ⎝ ⎠ 1/ 2

)

2

15 2 37 − = 4 3 12

3

1

15.

∫1/ 2 x2

16.

∫9 ( 36

3

dx = −

1 1 5 = − − (−2) = x 1/ 2 3 3

)

( z + 1)6 17. ∫ ( z + 1) dz = −1 6 1

2

1 − t3 ) 4

= 0 − (−48) = 48

574

36

⎛2 3 ⎞ x − 2 dx = ⎜ x 2 − 2 x ⎟ = 72 − 0 = 72 3 ⎝ ⎠9 1

5

= −1

32 32 −0 = 3 3


Section 14.7

8

18.

8⎛

∫1 ⎜⎝ x

1 3

−x

− 13

2 ⎛ 43 ⎞ 3x 3x 3 ⎟ ⎞ ⎜ − ⎟ dx = ⎜ 4 2 ⎟ ⎠ ⎝ ⎠1

27.

⎛ 3 ⎞ 27 = 6−⎜− ⎟ = ⎝ 4⎠ 4

19.

1

∫0 2 x =

20.

21.

2

( x3 − 1)

3

(

)

1 3 x −1 6

3

∫2

(

)

3 2 1 3 x − 1 ⎡3x 2 dx ⎤ ∫ ⎣ ⎦ 3 0

= 0−

1 1 =− 6 6

0

( x + 2)4 4

3

28.

2

dy = 4 ln y

29.

−1

24.

25.

26.

∫

e +1

∫2

30.

1

1

e dx =

2

)(

+ 4 x x3 + 2 x 2

(

= ∫ x3 + 2 x 2 0

) ( 4

)

4

31.

∫1/ 3

1

∫−1

10 − 3 pdp = −

(

q q 2 + 3dq = q2 + 3

)

3 2

1 1 2 2 [ −3 dp ] (10 − 3 p ) 3 ∫1/ 3

(

)

1 1 1 2 2 + q 3 [2q dq] 2 ∫−1

1

=

3

8 8 − =0 3 3

∫

=

0

(

(

)

)

3

1 7x +1 ⋅ 4 21

4 3

1

( 7 x + 1) 3

=

4 3

0

dx

)

32.

243 243 −0 = 5 5

0

16 1 15 − = 28 28 28

⎞ x ⎜⎜ 3x − 2 ⎟ dx ( x + 2)4 / 3 ⎟⎠ ⎝ 7 1 7 = ∫ 3 x dx − ∫ ( x 2 + 2)−4 / 3 [2 x dx] 0 2 0

∫0

7⎛

3 = x2 2

0

7 0

1 ( x 2 + 2)−1/ 3 − ⋅ 2 −1 3

7

0

3 3 = (7 − 0) + [9−1/ 3 − 2−1/ 3 ] 2 2 3⎛ 1 1 ⎞ = ⎜7 + − ⎟ 3 3 2⎝ 9 2⎠ 575

1

28

3

=

1

3 1 1 + 1dx = ∫ 7 x3 + 1 ⎡ 21x 2 dx ⎤ ⎣ ⎦ 0 21

1 23 x 7 x3 0

1

⎡ 3 x 2 + 4 x dx ⎤ ⎢⎣ ⎥⎦

=

5

1 1 20 / 3 2 [3 dx ] x + (3 5) 3 ∫−1/ 3

−1

51

x3 + 2 x 2 ) ( =

4

3 x + 5 dx =

0

2 x3

1

2

= e5 x = e5 − 0 = e5

1 e +1 dx = ln x − 1 = ln e − ln1 = 1 − 0 = 1 2 x −1

∫0 ( 3x

∫−1/ 3

=

5 1 x3 2 5 3 e [3 x dx] = e x ∫ 3 0 3 5 1 0 5 = (e − e ) = (e − 1) 3 3 1

∫0 5 x

20 / 3

1 3 = − − (−1) = 4 4

3 2 2 38 = − (10 − 3 p) 2 = − (8 − 27) = 9 9 9 1/ 3

1

1 5 e dx 0

( x − 3)

2

2

= 4(ln 8 − ln1)

6 22. ∫ e dx = 6 ln x = 6 ln1 − 6 ln ee −e x e −e = 0 − 6e = −6e

23.

1

20

=4(ln 8 – 0) = 4 ln 8 −1

5

5

−3

3 3 2 (3x + 5) 2 9 − 13 2 = (125 − 8) = 26 9

625 369 − 64 = 4 4

=

5

=

8

y

∫4

( x − 3)−2 dx = 2 ∫ ( x − 3) dx = 2 ⋅ 4 −2 ( x − 3)3 2

=−

dx =

( x + 2)3 dx =

84

∫1

41

5

4


33.

2 x3 + x

1

∫0 x 2 + x4 + 1


39.

dx

(

)

=

1 1 1 ⎡ 4 x3 + 2 x dx ⎤ ⎥⎦ 2 ∫0 x 4 + x 2 + 1 ⎣⎢

=

1 ln x 4 + x 2 + 1 2

(

1

(

n = m(b − a ) + b 2 − a 2 2

36.

37.

1 ex

∫0

⎞ n ⎟ = my + y 2 ⎟ 2 ⎠a a

0

a

)

x

95

∫1

ln e

42.

2

1

∫−11 + e x

= −2 ∫

−2

⎞ ⎟ = 8 ⎧[ 0 − (−2) ] + ⎛ 1 − 0 ⎞ ⎫ ⎨ ⎜ ⎟⎬ ⎟⎟ ⎝2 ⎠⎭ ⎩ 0⎠

= 2 ln

43.

2

∫0

+1 1

1

⋅

e− x e− x

dx

dx

(

)

1/ 2

1/ 2

2

+x

= 9 2 −6

4 x 2 dx + ∫

0

1/ 2 3

2

2 x dx

1/ 2

2

1 ⎞ 47 ⎛1 ⎞ ⎛ = ⎜ − 0⎟ + ⎜ 4 − ⎟ = 6 4 ⎠ 12 ⎝ ⎠ ⎝ 3

44. ⎛⎜ ∫ x dx ⎞⎟ − ∫ ⎝ 1 ⎠

)

3 3 x dx 1

3 ⎛ 2 3⎞ x ⎟ x4 ⎜ = − ⎜⎜ 2 ⎟⎟ 4 1 1 ⎝ ⎠ 3 ⎛ 9 1 ⎞ ⎛ 81 1 ⎞ = ⎜ − ⎟ −⎜ − ⎟ ⎝2 2⎠ ⎝ 4 4⎠

= 43 − 20 = 44 576

e +1 1 +1 e

e2 + e e(e + 1) = 2 ln = 2 ln e = 2 1+ e 1+ e

0

3

1

) −1

⎡ −e− x dx ⎤ = −2 ln e− x + 1 ⎦ +1 ⎣

f ( x)dx = ∫

4 x3 = 3

1 ⎞ 38. ∫ ⎜ 6 x − ⎟ dx 1 ⎝ 2x ⎠ 2 1 1 2 −1 = 6 ∫ x 2 dx − ∫ (2 x) 2 [2 dx] 1 2 1

) (

−1 1 + e x

(

2⎛

(

1

2

1

= −2 ln e−1 + 1 + 2 ln(e + 1) = 2 ln

⎛ x −1 x − 2 x − 3 ⎞ −2 −3 −4 + − = + − x x x dx 3( ) 3 ⎜ ⎟ ∫π ⎜ −1 −2 −3 ⎟ ⎝ ⎠π ⎛ 1 1 1 ⎞ ⎛ 1 1 1 ⎞ = 3⎜ − − + + ⎟ − 3⎜ − − ⎟ ⎝ e 2e 2 3e3 ⎠ ⎝ π 2π2 3π3 ⎠ 1 3 3 3 3 1 = − − + + − 3 e 2 π 2 π3 2e 2π e e

dx = ∫

−1 e− x

e

2

95

1

−1 e− x

1

1 ⎡ 3 ⎤ = ⎢ 4 x 2 − (2 x) 2 ⎥ = 8 2 − 2 − 4 − 2 ⎣ ⎦1

)

95 x 95 dx = ∫ 1 dx = x 1 1 x = 95 − 1 = 94

dx = ∫

e− x

⎛ 2 x = 8⎜ − ⎜⎜ 2 ⎝

x2 2

(

= (6 + ln 19) – 0 = 6 + ln 19

= 2∫

+

)

2

∫−2 8 x dx = 8 ⎜⎝ ∫−2 − x dx + ∫0 x dx ⎟⎠ 0

x

(

⎡ x4 x2 ⎤ =⎢ + + ln x3 + 5 x + 1 ⎥ 2 ⎣⎢ 4 ⎦⎥ 0

⎞

1

1 3 x2 + 2 x [(2 x + 2)dx] e 2 ∫1

41. Using long division on the integrand 2 x 6 + 6 x 4 + x3 + 8 x 2 + x + 5 dx ∫0 x3 + 5 x + 1 2⎡ 3x 2 + 5 ⎤ = ∫ ⎢ x3 + x + ⎥dx 0 x3 + 5 x + 1 ⎥⎦ ⎢⎣

1

⎛

dx =

1 x2 + 2 x 1 e3 12 e e −1 = e15 − e3 = 2 2 2 1

b

b

b

40.

− e− x 1 dx = (e x + e− x ) 2 2 0 1 = [(e + e −1 ) + (1 + 1)] 2 1⎛ 1 ⎞ = ⎜e + + 2⎟ 2⎝ e ⎠

1

x2 + 2 x 3

=

) 0 = 12 [ln 3 − ln1] = 12 ln 3

⎛ b ny 2 34. ∫ (m + ny )dy = ⎜ my + ⎜ a 2 ⎝

35.

3

∫1 ( x + 1)e


45.

x

1

1

t2

f ( x) = ∫ 3

x

dt = −

1⎛

1

Section 14.7

3 3 3 = − + 3 = 3− t1 x x

3⎞

∫e f ( x)dx = ∫e ⎜⎝ 3 − x ⎟⎠ dx = ( 3x − 3ln x )

51.

46.

∫

∫

1 3 2

dx = 0 + = =

1

1 3 2 1

3 2 1 = 3

47.

3

∫1

2

∫1

2

3

1 3

2 3

1

2

∫0

3 2

2

e

2

3

1 dx

2

48.

∫1

4

∫1

0

2 −0

)

49.

3 x3 3⎛

d

3 x

∫2 ⎜⎝ dx ∫2 e 50.

f ( x) = ∫

3

x et

54.

2 3

4

1

2

2

et + e − t x 1

e

=∫

e

t

e +e

−t

= ln et + e−t

=

1 e + e− x e x − e− x x

10

∫0

−4

x

− 12

(

3

∫2 f ( x) dx = 7 − 6 = 1 .

55.

d 3 x3 e dx = 0. Thus dx ∫2

31

1⎛

1⎞

⎝

⎠

2

0

1 ⎡ 1 ⎛ 1 ⎞⎤ 1 = ⎢ − ⎜ − ⎟ ⎥ = . Thus 3 ⎣ 8 ⎝ 8 ⎠ ⎦ 12

dx =

)

x

10−4

1 2

1 2 0

5

∫0 2000e

−0.06t

=2 x

dt = 2000 ⋅

2000 −0.06t e 0.06 ≈ $8639

5

=−

dt [(et − e−t ) dt ] x e

= ln(e x + e− x ) − ln(ee − e−e ) f ′( x) =

0

1 1 −0 = 2 2

10−4 0

= 2 10−4 − 0

= 2 10−2 = 0.02

3 ⎞ 3 dx ⎟ dx = ∫ 0 dx = C 2 = C − C = 0 2 ⎠

− e −t

=

a

1 3

dx is a constant, so

0

53. The total number receiving between a and b dollars equals the number N(a) receiving a or more dollars minus the number N(b) receiving b or more dollars. Thus

f ( x)dx = ∫ f ( x)dx − ∫ f ( x)dx + ∫ f ( x)dx

∫2 e

0

1

x2 2

1 1 µ = ; σ2 = 2 12

f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx

3

1

b

4

2

1

N (a) − N (b) = ∫ − Ax − B dx .

2

6 = 2 − ∫ f ( x)dx + 5 , so

5

0

1⎛ 1⎞ = ⎜x− ⎟ 3⎝ 2⎠

= ∫ f ( x)dx + ∫ f ( x)dx = 4 + 3 = 7 4

5

= α 2T − 0 = α 2T

1

f ( x)dx = ∫ f ( x)dx − ∫ f ( x)dx 3

T

σ 2 = ∫ ( x − µ ) 2 f ( x)dx = ∫ ⎜ x − ⎟ dx 0 0 2

f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx , so

1

5

52. µ = ∫ ( x ⋅1)dx = ∫ x dx =

x

(

∫0

5

α 2 dt = α 2 t

1

= (3 – 0) – (3e – 3) = 6 – 3e 7 x2 2 e dx + 7 0

T

[e x + e− x (−1)] − 0

e x + e− x

577

=− 0

5 −0.06t 1 e [−0.06 dt ] −0.06 ∫0

(

)

2000 −0.03 e −1 0.06


56.

∫0 ( e t

− aτ


)

− e−bτ dτ

61.

1 t − aτ 1 t −bτ e [− a dτ ] − e [−b dτ ] −a ∫0 −b ∫0

=

⎛ e− aτ e−bτ ⎞ = ⎜− + ⎟ ⎜ a b ⎟⎠ ⎝ 0 ⎛ e− at e−bt ⎞ ⎛ 1 1 ⎞ = ⎜− + ⎟−⎜− + ⎟ ⎜ a b ⎟⎠ ⎝ a b ⎠ ⎝ =

57.

− at

1− e a

64

−

∫36 10, 000

62.

1− e b

64

36

63. 1 2

(100 − t ) [(−1) dt ]

3 2 = − (10, 000)(100 − t ) 2 3

t

∫0

64 36

3000e0.05τ dτ = 3000 ⋅

= 60, 000e0.05τ

59.

t

0

64.

180

∫90

3

800

=

q

3

∫10 ( 250 + 90q − 3q 20

dq

200 q1/ 2 = ⋅ 1 3 2

400

500

(

2

800

500

)

800 − 500 ≈ $1367.99 20

) dq = ( 250q + 45q2 − q3 ) 10

(

2

+ 8q

(

)6

12

(8t + 10)dt = 4t 2 + 10t 81× 106

∫0

(300 + t )

)

4

12

(

dt = 81× 106

(300 + t ) −3 −3

= 696 − 0 = 696 = 696 − 204 = 492

) ∫0700 (300 + t )−4 dt

700 0 700

(

) (3001+ t )3

(

)

(

)

= − 27 × 106

)

)0

(8t + 10)dt = 4t 2 + 10t

700

12

(

12

∫0

(

1 t 0.05τ e [0.05 dτ ] 0.05 ∫0

0

1 ⎞ ⎛ 1 = − 27 × 106 ⎜ − ⎟ 3 3003 ⎠ ⎝ 1000 1 ⎞ ⎛ 1 = − 27 × 106 ⎜ − ⎟ 9 27 ⋅106 ⎠ ⎝ 10 27 27 973 =− +1 = − +1 = = 0.973 3 1000 1000 10

) 65 75

= 1162.5 – 942.5 = $220 60.

∫

= 81× 106

= 60, 000 e0.05t − 1

∫65 (0.2q + 8)dq = ( 0.1q 75

2000 10 3q

800 −1/ 2 q dq 500

400

∫6

2 = − (10, 000)[216 − 512] 3 ≈ 1,973,333

58.

800

500

dq = ∫

= 15,000 − 6000 = $9000

100 − tdt

= (−1)(10, 000) ∫

3

=

−bt

300q

200

=

t

2000

800

∫500

(0.004q 2 − 0.5q + 50) dq 180

0.004 3 q − 0.25q 2 + 50q 3 90 = 8676 − 3447 = $5229 =

65. G = ∫

R

−R

578

i dx = ix

R −R

= iR − (−iR) = 2 Ri


66. E = ∫

Section 14.7

i ⎡ −k ( R − x) −k ( R+ x) ⎤ e dx +e ⎦

R

−R 2 ⎣

i ⎡ 1 R −k ( R− x) 1 R −k ( R+ x) ⎤ e [k dx] + e [− k dx]⎥ ∫ ⎢ − R 2 ⎣k −k ∫− R ⎦ R i ⎡ R −k ( R− x) = [k dx] − ∫ e− k ( R + x ) [− k dx]⎤⎥ e −R 2k ⎢⎣ ∫− R ⎦ =

R

=

i ⎡ −k ( R − x) −k ( R + x) ⎤ e −e ⎦ 2k ⎣ −R

i 2k i = 2k =

(

) (

) ⎡ 2 − 2e −2 kR ⎤ = i (1 − e−2kR ) ⎣ ⎦ k

⎡ 1 − e − k (2 R ) − e− k (2 R ) − 1 ⎤ ⎣⎢ ⎦⎥

∫ (m + x)[1 − (m + x)]dx = ∫0 ( m + x − m − 2mx − x A= 0 R R ∫0 [1 − (m + x)]dx ∫0 (1 − m − x)dx R

R

67.

=

⎡ mx + ⎢⎣

x2 2

− m2 x − mx 2 −

x3 ⎤ 3 ⎥⎦

2

R

0

R

⎡ x − mx − x 2 ⎤ 2 ⎦⎥ ⎣⎢ 0

⎡ mR + R 2 − m2 R − mR 2 − R3 ⎤ − 0 2 3 ⎥⎦ ⎢ =⎣ 2 ⎡ R − mR − R ⎤ − 0 2 ⎦⎥ ⎣⎢ 2 R ⎡⎢ m + R2 − m2 − mR − R3 ⎤⎥ m + R − m 2 − mR − R 2 ⎣ ⎦= 2 3 = R R 1− m − 2 R ⎡⎣1 − m − 2 ⎤⎦

68.

69.

3.5

∫2.5 (1 + 2 x + 3x

4

∫0

2

)dx = ( x + x 2 + x3 )

3.5 2.5

= 58.625 − 24.375 = 34.25

1 4 1 (4 x + 4) −1 dx = ∫ (4 x + 4)−2 [4 dx] = ⋅ 4 0 4 −1 (4 x + 4) 2 1

4

4 0

4

1 1 1 1 1 ⎛1 ⎞ 1 =− ⋅ =− ⋅ = − ⎜ − 1⎟ = = 0.05 4 4x + 4 0 16 x + 1 0 16 ⎝ 5 ⎠ 20

70.

1 3t e dt 0

∫

=

1 1 3t e3t e [3 dt ] = ∫ 3 0 3

1

= 0

(

)

1 3 e − 1 ≈ 6.36 3

71. 3.52 72. 0.23 579

2

) dx


ISM: Introductory Mathematical Analysis estimate of the amount the culture grew over the first four hours is 4 0.5 0.2t 2 ∫0 0.3e dt ≈ 3 (34.5956) ≈ 5.77 grams.

73. 14.34 74. 3.64 Principles in Practice 14.8

Problems 14.8

60

1. In this case, f (t ) =

, n = 5, a = 0, and

t2 + 9 b−a 5−0 b = 5. Thus h = = = 1 . The terms to n 5 be added are 60 60 f (0) = = = 20 2 3 0 +9 2 f (1) = 2 f (2) = 2 f (3) = 2 f (4) =

2(60)

=

2

1 +9 2(60)

2

3 +9 2(60)

10

=

2

2 +9 2(60)

120

13 120

=

18

=

2

120

1.

≈ 37.9473 ≈ 33.2820 ≈ 28.2843

2.

2

2. In this case, f (t ) = 0.3e0.2t , n = 8, a = 0, and b−a 4 = = 0.5 . The terms to be 8 n

added are f (0) = 0.3e0 = 0.3 4 f (0.5) = 4(0.3)e 2 f (1) = 2(0.3)e

0.2

4 f (1.5) = 4(0.3)e 2 f (2) = 2(0.3)e

1.8

2 f (3) = 2(0.3)e

4 f (3.5) = 4(0.3)e

≈ 1.2615 ≈ 1.8820

≈ 1.3353

4 f (2.5) = 4(0.3)e

≈ 4.1884

≈ 3.6298

2.45

1

170

, n = 6, a = –2, b = 4 1 + x2 Simpson’s b − a 4 − (−2) 6 = = =1 h= n 6 6 f (−2) = 34 = 34 4 f (−1) = 4(85) = 340 2 f (0) = 2(170) = 340 4 f (1) = 4(85) = 340 2 f (2) = 2(34) = 68 4 f (3) = 4(17) = 68 f (4) = 10 = 10 1200 170

1

∫−2 1 + x2 dx ≈ 3 (1200) = 400

≈ 0.7328

1.25

170

f ( x) =

4

0.45

0.8

, n = 6, a = –2, b = 4. Trapezoidal 1 + x2 b − a 4 − (−2) 6 h= = = =1 n 6 6 f (−2) = 34 = 34 2 f (−1) = 2(85) = 170 2 f (0) = 2(170) = 340 2 f (1) = 2(85) = 170 2 f (2) = 2(34) = 68 2 f (3) = 2(17) = 34 f (4) = 10 = 10 826 4

120 = 24 5

0.05

170

∫−2 1 + x2 dx ≈ 2 (826) = 413

4 +9 60 60 f (5) = = ≈ 10.2899 2 34 5 +9 The sum of the above terms is 153.8035. The estimate of the radius after 5 seconds is 5 60 1 ∫0 2 dt ≈ 2 (153.8035) ≈ 76.90 feet. t +9

b = 4. Thus, h =

f ( x) =

≈ 13.9060

3.2

f (4) = 0.3e ≈ 7.3598 The sum of the above terms is 34.5956. The 580


3.

1 2

dx ≈

4

∫1 6.

0.2 (3.4000) = 0.340 2

Actual value:

1 2 x dx 0

∫

x3 = 3

1

= 0

1 ≈ 0.333 3

f ( x) = x 2 , n = 4, a = 0, b = 1 Simpson’s b − a 1− 0 h= = = 0.25 n 4 f (0) = 0.0000 4 f (0.25) = 0.2500 2 f (0.50) = 0.5000 4 f (0.75) = 2.2500 f (1) = 1.0000 4.0000 1 2

∫0 x

dx ≈

f ( x) =

1 2 x dx 0

∫

x3 = 3

1

= 0

4

1 x

2

1 1 = − − (−1) = 0.750 x1 4

1 , n = 6, a = 1, b = 4 x Trapezoidal b − a 4 −1 h= = = 0.5 n 6 f (1) = 1.0000 2 f (1.5) = 1.3333 2 f (2) = 1.0000 2 f (2.5) = 0.8000 2 f (3) = 0.6667 2 f (3.5) = 0.5714 f (4) = 0.2500 5.6214 f ( x) =

0.5 dx ≈ (5.6214) ≈ 1.405 x 2 Actual value: 41

∫1

4

dx = ln x

x ≈ 1.386

7.

1 ≈ 0.333 3

1

dx ≈

x2

4

dx = −

41

, n = 4, a = 1, b = 4 x2 Simpson’s b − a 4 −1 h= = = 0.75 n 4 f (1) = 1.0000 4 f (1.75) = 1.3061 2 f (2.50) = 0.3200 4 f (3.25) = 0.3787 f (4) = 0.0625 3.0673

∫1

1

∫1

0.25 1 (4.0000) = ≈ 0.333 3 3

Actual value:

5.

Actual value:

f ( x) = x 2 , n = 5, a = 0, b = 1 Trapezoidal b − a 1− 0 1 h= = = = 0.2 n 5 5 f (0) = 0.0000 2 f (0.2) = 0.0800 2 f (0.4) = 0.3200 2 f (0.6) = 0.7200 2 f (0.8) = 1.2800 f (1) = 1.0000 3.4000

∫0 x

4.

Section 14.8

= ln 4 − ln1 = ln 4 – 0 = ln 4 1

x , n = 4, a = 0, b = 2 x +1 Trapezoidal b−a 2−0 h= = = 0.5 n 4 f (0) = 0.0000 2 f (0.5) = 0.6667 2 f (1) = 1.0000 2 f (1.5) = 1.2000 f (2) = 0.6667 3.5334 f ( x) =

Thus 2 x 0.5 ∫0 x + 1 dx ≈ 2 (3.5334) ≈ 0.883 8.

0.75 (3.0673) ≈ 0.767 3

581

f ( x) =

1

, n = 4, a = 2, b = 4 x + x2 Simpson’s b−a 4−2 h= = = 0.5 n 4


ISM: Introductory Mathematical Analysis 12. a = 2, b = 5, h = 0.5 0 f (2) = 4 f (2.5) = 24 2 f (3) = 20 4 f (3.5) = 44 2 f (4) = 28 4 f (4.5) = 60 f (5) = 16 192

f (2) = 0.1667 4 f (2.5) = 0.4571 2 f (3) = 0.1667 4 f (3.5) = 0.2540 f (4) = 0.0500 1.0945 4

dx

∫2 x + x2 ≈ 9.

70

∫45 l (t )dt ,

0.5 (1.0945) ≈ 0.182 3

0.5 (192) = 32 3 The area is about 32 square units. 5

∫2 f ( x) dx ≈

males, n = 5, a = 45, b = 70

70 − 45 =5 5 l (45) = 93, 717 2l (50) = 183, 232 2l (55) = 177, 292 2l (60) = 168,376 2l (65) = 155, 094 l (70) = 68,375 846, 086 h=

70

13.

55

∫35

3 −1 = 0.5 4 = f (1) = 1 4 f (1.5) = 4(2) = 2 f (2) = 2(2) = 4 f (2.5) = 4(0.5) = = f (3) = 1

5

3

∫1

l (t )dt , females, n = 4, a = 35, b = 55

55 − 35 =5 4 l (35) = 97,964 2l (40) = 194, 796 2l (45) = 193,164 2l (50) = 190, 784 l (55) = 93,562 770, 270 h=

55

f ( x)dx , n = 4, a = 1, b = 3

h=

∫45 l (t )dt ≈ 2 (846, 086) = 2,115, 215 10.

3

∫1

14.

f ( x)dx ≈

f ( x) =

1 8 4 2 1 16

0.5 8 (16) = 3 3

2 1+ x

, a = 1, b = 3, n = 4

3 −1 = 0.5 4 Simpson’s f (1) ≈ 1.4142 4 f (1.5) ≈ 5.0596 2 f (2) ≈ 2.3094 4 f (2.5) ≈ 4.2762 f (3) = 1.0000 14.0594 h=

5

∫35 l (t )dt ≈ 2 (770, 270) = 1,925, 675 11. a = 1, b = 5, h = 1 f (1) = 0.4 = 0.4 4 f (2) = 4(0.6) = 2.4 2 f (3) = 2(1.2) = 2.4 4 f (4) = 4(0.8) = 3.2 f (5) = 0.5 = 0.5 8.9

2 0.5 (14.0594) ≈ 2.343 dx ≈ 3 1+ x For the actual value, we have 3 2 3 −1/ 2 ∫1 1 + x dx = 2∫1 (1 + x) dx 3

∫1

3

(

)

= 2[2(1 + x)1/ 2 ] = 4 2 − 2 ≈ 2.343

1 ∫1 f ( x)dx ≈ 3 (8.9) ≈ 3.0 The area is about 3.0 square units. 5

1

582


15.

Section 14.8 =0 = f (0) = 0.5 − 0.5 4 f (0.5) = 4(2.3 − 0.3) = 4(2) = 2 f (1) = 2(2.2 − 0.7) = 2(1.5) = = 4(2) = 4 f (1.5) = 4(3 − 1) 2 f (2) = 2(2.5 − 0.5) = 2(2) = 4 f (2.5) = 4(2.2 − 0.2) = 4(2) = 2 f (3) = 2(1.5 − 0.5) = 2(1) = 4 f (3.5) = 4(1.3) − 0.8) = 4(0.5) = =0 = f (4) = 1 − 1

f ( x) = 1 − x 2 , a = 0, b = 1, n = 4 1− 0 = 0.25 4 Simpson’s f (0) = 1.0000 4 f (0.25) = 3.8730 2 f (0.50) = 1.7321 4 f (0.75) = 2.6458 f (1) = 0.0000 9.2509 h=

1

∫0

4

Area ≈ ∫ f ( x)dx ≈

0.25 1 − x dx ≈ (9.2509) ≈ 0.771 3 2

0

18. a.

dr 16. ∫ dq = r (80) − r (0) = r (80) 0 dq [since r(0) = 0] Using Simpson’s rule with h = 10 and dr f (q) = : dq 80

f (0) = 10 = 4 f (10) = 4(9) = 2 f (20) = 2(8.5) = 4 f (30) = 4(8) = 2 f (40) = 2(8.5) = 4 f (50) = 4(7.5) = 2 f (60) = 2(7) = 4 f (70) = 4(6.5) = f (80) = 7 =

MC =

0 8 3 8 4 8 2 2 0 35

0.5 35 km2 (35) = 3 6

dc dq

dc dq dq = c (100) − c(0) = (total cost of 100 units) − (fixed costs) = total variable costs of 100 units Using the trapezoidal rule with h = 20 and dc f (q) = to estimate the integral: dq f (0) = 260 2 f (20) = 500 2 f (40) = 480 2 f (60) = 400 2 f (80) = 480 f (100) = 250 2370 100 dc 20 ∫0 dq dq ≈ 2 (2370) = $23, 700 100

∫0

10 36 17 32 17 30 14 26 7 189

dr 10 dq ≈ (189) = 630 dq 3 The total revenue is about $630. 80

∫0

17. Let f(x) = distance from near to far shore at point

b.

4

x on highway. Then area ≈ ∫ f ( x)dx . Using 0

MR =

dr dq

dr dq = r (100) − r (0) = r (100) dq [since r(0) = 0] = total revenue from sale of 100 units Using the trapezoidal rule with h = 20 and dr g (q) = to estimate the integral: dq 100

∫0

Simpson’s rule with h = 0.5:

583



= = = = = =

g (0) 2 g (20) 2 g (40) 2 g (60) 2 g (80) g (100)

3. y = 5x + 2, x = 1, x = 4

410 700 600 500 540 250 3000 100 dr 20 ∫0 dq dq ≈ 2 (3000) = $30, 000

c.

4

Area = ∫ (5 x + 2)dx 1

4

⎛ 5x2 ⎞ 9 87 =⎜ + 2 x ⎟ = 48 − = ⎜ 2 ⎟ 2 2 ⎝ ⎠1 25

y

x 5

At q = 100: total revenue = 30,000 total cost = (total var. costs) + (fixed costs) = 23, 700 + 2000 = 25, 700 Thus maximum profit = (total revenue) − (total costs) = 30,000 − 25,700 = $4300.

4. y = x + 5, x = 2, x = 4 4

⎛ x2 ⎞ Area = ∫ ( x + 5)dx = ⎜ + 5x ⎟ ⎜ ⎟ 2 ⎝ 2 ⎠2 = 28 – 12 = 16 4

Problems 14.9

10

In Problems 1–34, answers are assumed to be expressed in square units.

y

1. y = 4x, x = 2

x

2

Area = ∫ 4 x dx = 0

10

2 2 x2 0

5

=8–0=8

y

5. y = x – 1, x = 5 5

⎛ x2 ⎞ 5 Area = ∫ ( x − 1)dx = ⎜ − x⎟ ⎜ 2 ⎟ 1 ⎝ ⎠1 x 2

2. y =

=

5

5

3 x + 1, x = 0, x = 16 4

y

16

⎛ 3x 2 ⎞ ⎞ Area = ∫ ⎜ x + 1⎟ dx = ⎜ + x⎟ ⎜ ⎟ 0 ⎝4 ⎠ ⎝ 8 ⎠0 = 112 – 0 = 112 16 ⎛ 3

20

15 ⎛ 1 ⎞ 16 −⎜− ⎟ = =8 2 ⎝ 2⎠ 2

x 1

y

x 20

584

5

8


Section 14.9

6. y = 3 x 2 , x = 1, x = 3

9. y = x 2 + 2 , x = –1, x = 2 3

3

Area = ∫ 3 x 2 dx = x3 = 27 − 1 = 26 1

40

Area = ∫

1

2

−1

y

=

(

2

⎛ x3 ⎞ x + 2 dx = ⎜ + 2 x ⎟ ⎜ 3 ⎟ ⎝ ⎠ −1

)

2

20 ⎛ 7 ⎞ 27 −⎜− ⎟ = =9 3 ⎝ 3⎠ 3 y

8

x 5

x

7. y = x 2 , x = 2, x = 3 3 2 x dx 2

Area = ∫

16

x3 = 3

–1

3

= 9− 2

8 19 = 3 3

2

5

10. y = x + x 2 + x3 , x = 1

y

1

⎛ x 2 x3 x 4 ⎞ 1 Area = ∫ ( x + x 2 + x3 )dx = ⎜ + + ⎟ ⎜ 2 0 3 4 ⎟⎠ ⎝ 0 = x 2 3

13 13 −0 = 12 12 3

5

y

8. y = 2 x 2 − x , x = –2, x = –1 Area = ∫

−1

−2

(

x 3

−1

⎛ 2 x3 x 2 ⎞ 2 x − x dx = ⎜ − ⎟ ⎜ 3 2 ⎟⎠ ⎝ −2 2

)

7 ⎛ 44 ⎞ 37 = − −⎜− ⎟ = 6 ⎝ 6 ⎠ 6 16

11. y = x 2 − 2 x , x = –3, x = –1

y

Area = ∫

−1

−3

1

)

4 50 = − − (−18) = 3 3

x –2 –1

(

−1

⎛ x3 ⎞ x − 2 x dx = ⎜ − x 2 ⎟ ⎜ 3 ⎟ ⎝ ⎠ −3 2

18

5

y

x –3

585

–1

2

5



12. y = 3x 2 − 4 x , x = –2, x = –1 Area = ∫

−1

−2

15. y = 2 − x − x3 , x = –3, x = 0

(3x2 − 4 x ) dx = ( x3 − 2 x2 ) −2 −1

Area = ∫

y

(2 − x − x ) 3

−3

= –3 – (–16) = 13 32

0

0

⎛ x2 x4 ⎞ dx = ⎜ 2 x − − ⎟ ⎜ 2 4 ⎟⎠ ⎝ −3

⎛ 123 ⎞ 123 = 0−⎜− ⎟= 4 ⎝ 4 ⎠ y

40

x –2 –1

5 x

13. y = 2 − x − x

2

5

1

⎛ x 2 x3 ⎞ Area = ∫ (2 − x − x )dx = ⎜ 2 x − − ⎟ ⎜ −2 2 3 ⎟⎠ ⎝ −2 7 ⎛ 10 ⎞ = −⎜− ⎟ 6 ⎝ 3⎠ 9 = 2 1

5

2

16. y = e x , x = 1, x = 3 3

3

1

1

Area = ∫ e x dx = e x 32

= e3 − e

y

y

x 1

x

3

5

17. A = 3 + 2 x − x 2 4 , x = 1, x = 2 x 24 Area = ∫ dx = 4 ln x 1 x = ln 16

Area = ∫

−1

(3 + 2x − x2 ) dx 3

14. y =

5

3

2 1

⎛ x3 ⎞ = ⎜ 3x + x 2 − ⎟ ⎜ 3 ⎟⎠ ⎝ −1

= 4 ln(2) – 0 = 4 ln 2

⎛ 5 ⎞ 32 = 9−⎜− ⎟ = ⎝ 3⎠ 3

y

5

y

x 1 2

x

5

5

586


18. y =

1 ( x − 1)

Area = ∫

2

21. y = x + 9 , x = –9, x = 0

, x = 2, x = 3 1

3

Area = ∫

( x − 1)

( x − 1)−1 = −1

3

dx = ∫ ( x − 1) −2 dx 2

=

3

⎛ 1 ⎞ = ⎜− ⎟ ⎝ x −1 ⎠ 2 2 1 1 = − − (−1) = 2 2 5

0

( x + 9)

−9

0

3 2

3 2

1

0

x + 9dx = ∫ ( x + 9) 2 dx

−9

3

2

2

Section 14.9

3

−9

2( x + 9) 2 = 3

= 18 – 0 = 18 10

0

−9

y

y x –9

6

x 5

22. y = x 2 − 4 x , x = 2, x = 6

5

4

6

2

4

Area = ∫ −( x 2 − 4 x)dx + ∫ ( x 2 − 4 x)dx

1 19. y = , x = 1, x = e x e1 e Area = ∫ dx = ln x = ln e – ln 1 = 1 – 0 = 1 1 1 x

4

y

20

y

x 1

e

5

x 7

1 , x = 1, x = e2 x e2 1 Area = ∫ dx = ln x 1 x

20. y =

23. y = 2 x − 1 , x = 1, x = 5

e2 1

Area = ∫

2

5

1

= ln e − ln1 = 2 − 0 = 2 y

=

1

3

e2

587

2 x − 1 dx

1 1 5 (2 x − 1) 2 [2 dx ] ∫ 1 2

(2 x − 1) 2 = 3

x 1

6

⎛ x3 ⎞ ⎛ x3 ⎞ = ⎜ − + 2x2 ⎟ + ⎜ − 2x2 ⎟ ⎜ 3 ⎟ ⎜ 3 ⎟ ⎝ ⎠2 ⎝ ⎠4 ⎡ 32 16 ⎤ ⎡ ⎛ 32 ⎞ ⎤ = ⎢ − ⎥ + ⎢0 − ⎜ − ⎟ ⎥ = 16 ⎣ 3 3 ⎦ ⎣ ⎝ 3 ⎠⎦

5

= 9− 1

1 26 = 3 3


5


y

26. y = x 2 + 4 x − 5, x = −5, x = 1 1

Area = ∫ −( x 2 + 4 x − 5) dx −5

x 1

5

24. y = x3 + 3x 2 , x = –2, x = 2

10

2

⎛ x4 ⎞ 2 + x3 ⎟ Area = ∫ x3 + 3 x 2 dx = ⎜ ⎜ 4 ⎟ −2 ⎝ ⎠ −2 = 12 – (–4) = 16

(

25

1

⎛ x3 ⎞ = − ⎜ + 2 x2 − 5x ⎟ ⎜ 3 ⎟ ⎝ ⎠ −5 8 100 ⎛ ⎞ = −⎜− − ⎟ ⎝ 3 3 ⎠ = 36

)

y

x 10

y

x

27. y = e x + 1, x = 0, x = 1

5

1

1

Area = ∫ (e x + 1)dx = (e x + x) = (e1 + 1) − 1 = e 0

0

8

y

25. y = 3 x , x = 2 Area = ∫

23

0

4

3x 3 xdx = ∫ x dx = 0 4 2

1 3

( ) = 332

2

0

4

3(2) 3 = −0 4 x 5

3

=

3 2 2 4

2

3

28. y = x , x = –2, x = 2

y

Area = ∫

2

−2

x2 =− 2

x 2

0 −2

0

2

−2

0

x dx = ∫ (− x)dx + ∫ x dx

x2 + 2

2 0

= [0 – (–2)] + [2 – 0] = 4 3

y

x –2

588

2


29. y = x +

Section 14.9

2 , x = 1, x = 2 x

32. y = x − 2 , x = 2, x = 6 3

2( x − 2) 2 Area = ∫ x − 2 dx = 2 3

2

= (2 + 2 ln 2) − 5

y

1 3 3 = + 2 ln 2 = + ln 4 2 2 2

y

x

x 2

2

5

6

8

33. y = 2 x − x 2 , x = 1, x = 3

30. y = 4 + 3 x − x 2

Area = ∫

4

Area = ∫ (4 + 3 x − x 2 )dx

2

1

−1

4 ⎛ 3 x 2 x3 ⎞ = ⎜ 4x + − ⎟ ⎜ 2 3 ⎟⎠ ⎝ −1 56 ⎛ 13 ⎞ = −⎜− ⎟ 3 ⎝ 6⎠ 125 = 6 10

2

16 16 = −0 = 3 3 8

1

6

6

⎛x ⎞ 2⎛ 2⎞ + 2 ln x ⎟ Area = ∫ ⎜ x + ⎟ dx = ⎜ ⎜ ⎟ 1 ⎝ x⎠ ⎝ 2 ⎠1 2

( 2 x − x2 ) dx + ∫23 − ( 2 x − x2 ) dx 2

⎛ ⎛ x3 ⎞ x3 ⎞ = ⎜ x2 − ⎟ − ⎜ x2 − ⎟ ⎜ ⎜ 3 ⎟⎠ 3 ⎟⎠ ⎝ 1 ⎝

3

2

⎡4 2⎤ ⎡ 4⎤ 6 = ⎢ − ⎥ − ⎢0 − ⎥ = = 2 ⎣3 3⎦ ⎣ 3⎦ 3 5

y

y 3

x

1 2

x 10

5

34. y = x 2 + 1 , x = 0, x = 4 31. y = x3 , x = –2, x = 4 0

4 3 x dx 0

3

Area = ∫ − x dx + ∫ −2

x4 =− 4

0 −2

x4 + 4

4

Area = ∫

0

=

0

= [0 – (–4)] + [64 – 0] = 68 64

4

y

(

)

76 76 −0 = 3 3 25

–2

4

⎛ x3 ⎞ x + 1 dx = ⎜ + x ⎟ ⎜ 3 ⎟ ⎝ ⎠0 2

y

x

x

4

4

589


35.


⎧ 2 if f ( x) = ⎨3x ⎩16 − 2 x if

0≤ x<2 x≥2

3

2

38. a.

Area = ∫ f ( x)dx = ∫ 3x dx + ∫ (16 − 2 x)dx 0

=

2 x3 0

0

(

+ 16 x − x 2

2

)2 3

3

(1 − x)2 dx

1 1 = − (−1 − 0) = 9 9

= [8 – 0] + [39 – 28] = 19 sq units 16

1

2 1 1 (1 − x)3 = (−1) ∫ (1 − x)2 [− dx] = − ⋅ 1 3 3 3

3

2

21

P (1 ≤ x ≤ 2) = ∫

y

b.

36. y =

5

c.

1 b−a 1 x dx = ab−a b−a

Area = ∫ =

t

t

d. a

39. a. x a

37. a.

c.

x2 P (0 ≤ x ≤ 1) = ∫ x dx = 08 16 1 16

1

= 0

x2 P (2 ≤ x ≤ 4) = ∫ x dx = 28 16 41

P ( x ≥ 3) = ∫

41

3

8

x dx =

3

1

3

∫0 f ( x)dx = ∫0 f ( x)dx + ∫1

2 4

x 16

3

f ( x)dx

1 + P( x ≥ 1) 9

P (3 ≤ x ≤ 7) = ∫

4 2

1 −0 16

b.

c. 1 3 = 1− = 4 4

= 1−

71

3

= ln 7 − ln 3 = ln

b

11

=

b.

t

11

8 Thus, P ( x ≥ 1) = . 9

y

1 b–a

1

1 (1 − x)2 dx = − (1 − x)3 03 9 0 1 1 = − (0 − 1) = 9 9 P ( x ≤ 1) = ∫

1=

t a t −a − = sq units b−a b−a b−a

51

P ( x ≤ 5) = ∫

x

dx = ln x

7 3

dx = ln x

e x = ln(5) − ln e = ln(5) − 1

P ( x ≥ 4) = ∫

e2

4

7 3

5 e

1 dx = ln x x

e2 4

2

= ln e − ln 4 = 2 − ln 4

d.

9 7 = 16 16

(

)

P e ≤ x ≤ e2 = ∫ = ln x

e2 e

e2

e

1 dx x

= ln e 2 − ln e

=2–1=1 40. a.

590

1

5/ 2 1 5⎞ ⎛ P ⎜1 ≤ x ≤ ⎟ = ∫ (1 − x )2 dx 1 2⎠ 3 ⎝ 5/ 2 1 1 ⎛ 27 ⎞ 3 = − (1 − x)3 = − ⎜ − − 0⎟ = 9 9⎝ 8 ⎠ 8 1

x 2 3

2

r

∫1

1 x2

r

dx = −

1 1 1 = − +1 = 1− x1 r r


Section 14.9

y

b.

y = 12 x

x 1

c.

r

⎛ 1⎞ dx = lim ⎜1 − ⎟ [from part (a)] →∞ r ⎝ r⎠ x =1– 0=1 r

1

r →∞ 1

2

∫

lim

d.

y

y = 12 x

x 1

41. 1.89 sq units 42. 7.18 sq units 43. The x-intercept on [1, 3] is A ≈ 2.190327947

(

A

)

Area = ∫ − x 4 − 2 x3 − 2 dx + ∫ 1

3 A

( x4 − 2 x3 − 2) dx

≈ 11.41 sq units y

50

x 5

44. The x-intercepts are A ≈ –0.3294085282 and B ≈ 1.539613346 Area = ∫

B A

(1 + 3x − x4 ) dx ≈ 3.53 sq units

10

y x 5

–20

591



Problems 14.10 1. Area = ∫

b

a

3

( yUPPER − yLOWER ) dx = ∫−2 ⎡⎣( x + 6) − x 2 ⎤⎦ dx

( yUPPER − yLOWER ) dx = ∫0 ( 2x − x 2 ) dx a

2. Area = ∫

2

b

3. Intersection points: x 2 − x = 2 x, x 2 − 3 x = 0, x( x − 3) = 0 ⇒ x = 0 or x = 3

Area = ∫

3

0

4

( yUPPER − yLOWER ) dx + ∫3 ( yUPPER − yLOWER )dx

(

)

(

)

3 4 = ∫ ⎡⎢ 2 x − x 2 − x ⎤⎥ dx + ∫ ⎡⎢ x 2 − x − 2 x ⎤⎥ dx 0⎣ 3 ⎣ ⎦ ⎦

4. Intersection points: x( x − 3) 2 = 2 x, x( x − 3) 2 − 2 x = 0 , x ⎡ ( x − 3) 2 − 2 ⎤ = 0 , x( x 2 − 6 x + 7) = 0 ⇒ x = 0 , 3 ± 2 ⎣ ⎦ (from the quadratic formula) Area = ∫

3− 2

0

=∫

3− 2 ⎡

0

3+ 2

( yUPPER − yLOWER ) dx + ∫3− 2 ( yUPPER − yLOWER ) dx

3+ 2 x ( x − 3) − 2 x ⎥⎤dx + ∫ 3− ⎣⎢ ⎦

2 2

⎡ 2 x − x( x − 3)2 ⎤ dx ⎣ ⎦

5. The graphs of y = 1 − x 2 and y = x – 1 intersect when 1 − x 2 = x − 1 , 0 = x 2 + x − 2 , 0 = ( x − 1)( x + 2) ⇒ x = 1 or x = –2. When x = 1, then y = 0. We use horizontal elements, where y ranges from 0 to 1. Solving y = x – 1 for x gives x = y + 1, and solving y = 1 − x 2 for x gives x 2 = 1 − y , x = ± 1 − y . We must choose x = 1 − y because x is not negative over the given region. 1

1

Area = ∫ ( xRIGHT − xLEFT )dy = ∫ ⎡⎣ ( y + 1) − 1 − y ⎤⎦ dy 0 0 6. The graphs of y = 2x and y = –2x – 8 intersect when 2x = –2x – 8, 4x = –8, x = –2. When x = –2, then y = –4. We y use horizontal elements, where y ranges from –4 to 4. Solving y = 2x for x gives x = ; solving y = –2x – 8 for x 2 −y −8 gives 2x = –y – 8, x = . 2 4 4 ⎡ y ⎛ − y − 8 ⎞⎤ Area = ∫ ( xRIGHT − xLEFT )dy = ∫ ⎢ − ⎜ ⎟ dy −4 −4 ⎣ 2 ⎝ 2 ⎠ ⎥⎦

592


Section 14.10

7. The graphs of y = x 2 − 5 and y = 7 − 2 x 2

8

y

intersect when x 2 − 5 = 7 − 2 x 2 , 3x 2 = 12, x 2 = 4, so x = ± 4 = ±2. We use vertical elements. 2

Area = ∫ ( yUPPER − yLOWER ) dx

x

1 2

5

= ∫ [(7 − 2 x 2 ) − ( x 2 − 5)] dx 1

10

10. y = x, y = –x + 3, y = 0. Region appears below. 3 Intersection: x = –x + 3, 2x = 3, x = 2

y

Area = ∫

x 10

3/ 2

0

x2 = 2

3/ 2 0

x dx + ∫

3

(− x + 3)dx

3/ 2 3

⎛ x2 ⎞ +⎜− + 3x ⎟ ⎜ 2 ⎟ ⎝ ⎠ 3/ 2

⎡ 9 ⎤ ⎡⎛ 9 ⎞ ⎛ 9 9 ⎞⎤ 9 = ⎢ − 0 ⎥ + ⎢⎜ − + 9 ⎟ − ⎜ − + ⎟ ⎥ = ⎣ 8 ⎦ ⎣⎝ 2 ⎠ ⎝ 8 2 ⎠⎦ 4

8. The curves y 2 = x and 2y = 3 − x (or x = 3 − 2y) intersect when y 2 = 3 − 2 y, y 2 + 2 y − 3 = 0,

5

(y + 3)(y − 1) = 0 ⇒ y = −3 or 1. We use horizontal elements.

y

1

Area = ∫ ( xRIGHT − xLEFT ) dy 0 1

= ∫ [(3 − 2 y ) − y 2 ] dy 0

5

x 3 2

y

5

11. y = x 2 + 1, x ≥ 0, x = 0, y = 3. Region appears below.

x 5

Intersection: x 2 + 1 = 3 , so x = ± 2 2

Area = ∫

0

[3 − ( x 2 + 1)]dx = ∫

0

2

9. y = x 2 , y = 2 x Region appears below.

5

y

Intersection: x 2 = 2 x, x 2 − 2 x = 0 , x(x – 2) = 0, so x = 0 or 2. Area = ∫

0

(2x − x ) 2

(2 − x 2 )dx

⎛ x3 ⎞ 4 2 4 2 = ⎜ 2x − ⎟ = −0 = ⎜ ⎟ 3 3 3 ⎝ ⎠0

In Problems 9–34, the answers are assumed to be expressed in square units.

2

2

2

⎛ x3 ⎞ dx = ⎜ x 2 − ⎟ ⎜ 3 ⎟⎠ ⎝ 0

x 3

8⎞ 4 ⎛ = ⎜4− ⎟−0 = 3 3 ⎝ ⎠

593



14. y 2 = x + 1 , x = 1. Region appears below.

12. y = x 2 + 1 , y = x + 3. Region appears below.

Intersection: y 2 = 2, y = ± 2

Intersection: x 2 + 1 = x + 3, x 2 − x − 2 = 0 , (x + 1)(x – 2) = 0, so x = –1, 2

(

2

)

Area = ∫ ⎡⎢ ( x + 3) − x 2 + 1 ⎤⎥ dx −1 ⎣ ⎦ x + 2 − x 2 ) dx −1 (

=∫

−

2

(

)

⎛ x2 x3 ⎞ =⎜ + 2x − ⎟ ⎜ 2 3 ⎟⎠ ⎝ −1

5

8⎞ ⎛1 1⎞ 9 ⎛ = ⎜2+ 4− ⎟−⎜ − 2+ ⎟ = 3 2 3⎠ 2 ⎝ ⎠ ⎝ y

x 5

– 2

15. x = 8 + 2y, x = 0, y = –1, y = 3. Region appears below.

5

(

3

Area = ∫ (8 + 2 y )dy = 8 y + y 2 −1

2

10

Intersection: 10 − x 2 = 4, x 2 = 6, so x = ± 6 6

3

) −1

= (24 + 9) – (–8 + 1) = 40

13. y = 10 − x , y = 4. Region appears below.

− 6 6

y

[(10 − x 2 ) − 4] dx

=∫

− 6

2

y

2

x

Area = ∫

2

⎛ 2 2⎞ ⎛ 2 2⎞ 8 2 = ⎜⎜ 2 2 − ⎟⎟ − ⎜⎜ −2 2 + ⎟= 3 ⎠ ⎝ 3 ⎟⎠ 3 ⎝

2

8

3⎞ ⎛ ⎡1 − y 2 − 1 ⎤ dy = ⎜ 2 y − y ⎟ ⎥⎦ ⎜ 2 ⎣⎢ 3 ⎟⎠ ⎝ −

2

Area = ∫

x 16

2

(6 − x ) dx

6 ⎛ x3 ⎞ = ⎜ 6x − ⎟ ⎜ 3 ⎟⎠ ⎝ − 6 ⎛ 6 6⎞ ⎛ 6 6⎞ = ⎜⎜ 6 6 − ⎟⎟ − ⎜⎜ −6 6 + ⎟ 3 3 ⎟⎠ ⎝ ⎠ ⎝ =8 6

16. y = x − 6, y 2 = x . Region appears below. Intersection: y 2 = y + 6, y 2 − y − 6 = 0, (y + 2)(y − 3) = 0, so y = −2, 3. 3 Area = ∫ ⎡ ( y + 6) − ( y 2 ) ⎤dy ⎦ −2 ⎣

y

3

10

⎛ y2 y3 ⎞ =⎜ + 6y − ⎟ ⎜ 2 3 ⎟⎠ ⎝ −2

x 10

8 ⎞ 125 ⎛9 ⎞ ⎛ = ⎜ + 18 − 9 ⎟ − ⎜ 2 − 12 + ⎟ = 2 3⎠ 6 ⎝ ⎠ ⎝

594


10

Section 14.10

y

19. y 2 = 4 x, y = 2x − 4. Region appears below. ⎛y ⎞ Intersection: y 2 = 4 ⎜ + 2 ⎟ , y 2 − 2 y − 8 = 0, ⎝2 ⎠ (y + 2)(y − 4) = 0, so y = −2 or 4. 2 4 ⎡⎛ y ⎞ y ⎤ Area = ∫ ⎢⎜ + 2 ⎟ − ⎥ dy −2 ⎝ 2 ⎠ 4 ⎥⎦ ⎢⎣

(9, 3) x 16 (4, –2)

4

⎛ y2 y3 ⎞ =⎜ + 2y − ⎟ ⎜ 4 12 ⎟⎠ ⎝ −2 16 ⎞ ⎛ 2⎞ ⎛ = ⎜ 4 + 8 − ⎟ − ⎜1 − 4 + ⎟ 3⎠ ⎝ 3⎠ ⎝ =9

17. y = 4 − x 2 , y = –3x. Region appears below. Intersection: −3x = 4 − x 2 , x 2 − 3x − 4 = 0 , (x + 1)(x – 4) = 0, so x = –1 or 4. Area

∫−1 ⎡⎣⎢( 4 − x 4

2

) − (−3x)⎤⎦⎥ dx

5

4

y

⎛ x3 3 x 2 ⎞ = ⎜ 4x − + ⎟ ⎜ 3 2 ⎟⎠ ⎝ −1

(4, 4) x

64 1 3 ⎞ 125 ⎛ ⎞ ⎛ = ⎜ 16 − + 24 ⎟ − ⎜ −4 + + ⎟ = 3 3 2⎠ 6 ⎝ ⎠ ⎝ 6

(1, –2)

5

y

20. y = x3 , y = x + 6, x = 0 Region appears below.

x 10

Intersection: x3 = x + 6, x3 − x − 6 = 0, ( x − 2)( x 2 + 2 x + 3) = 0 ⇒ x = 2 x3 = 0 ⇒ x = 0 2

Area = ∫ [( x + 6) − x3 ] dx

2

18. x = y + 2, x = 6 . Region appears below. 2

0

Intersection: y + 2 = 6, y = 4 , y = ±2

(

)

(

)

2 2 Area = ∫ ⎡⎢ 6 − y 2 + 2 ⎤⎥ dy = ∫ 4 − y 2 dy −2 ⎣ −2 ⎦ 2

⎛ y3 ⎞ 8 ⎞ 32 ⎛ 8⎞ ⎛ = ⎜ 4y − ⎟ = ⎜ 8 − ⎟ − ⎜ −8 + ⎟ = ⎜ ⎟ 3 ⎠ 3⎠ ⎝ 3⎠ 3 ⎝ ⎝ −2 5

2

⎛ x2 x4 ⎞ =⎜ + 6x − ⎟ ⎜ 2 4 ⎟⎠ ⎝ 0 = (2 + 12 − 4) − (0) = 10

2

10

y

x 10

y

x 8

595


ISM: Introductory Mathematical Analysis 2 y + 15 ⎞ ⎛ 23. y 2 = 3 x, 3x − 2y = 15 ⎜ or x = ⎟ . Region 3 ⎠ ⎝ appears below. ⎛ 2 y + 15 ⎞ Intersection: y 2 = 3 ⎜ ⎟, ⎝ 3 ⎠

21. 2 y = 4 x − x 2 , 2y = x – 4. Region appears below. Intersection: x − 4 = 4 x − x 2 , x 2 − 3 x − 4 = 0 , (x + 1)(x – 4) = 0, so x = –1 or 4. Note that the 4 x − x2 y-values of the curves are given by y = 2 x−4 and y = . 2 4 ⎡⎛ 4 x − x 2 ⎞ ⎛ x − 4 ⎞ ⎤ Area = ∫ ⎢⎜ ⎟−⎜ ⎟ ⎥ dx ⎟ −1 ⎢⎜ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦ ⎞ 4 ⎛3 x2 = ∫ ⎜ x− + 2 ⎟ dx ⎟ −1 ⎜ 2 2 ⎝ ⎠

y 2 − 2 y − 15 = 0, (y + 3)(y − 5) = 0, so y = −3 or 5. 2 5 ⎡⎛ 2 ⎞ y ⎤ Area = ∫ ⎢⎜ y + 5 ⎟ − ⎥ dy −3 ⎝ 3 ⎠ 3 ⎥⎦ ⎢⎣ 5

⎛1 y3 ⎞ = ⎜ y2 + 5 y − ⎟ ⎜3 9 ⎟⎠ ⎝ −3 125 ⎞ ⎛ 25 = ⎜ + 25 − ⎟ − (3 − 15 + 3) 9 ⎠ ⎝ 3 256 = 9

4

⎛ 3x 2 x3 ⎞ =⎜ − + 2x ⎟ ⎜ 4 ⎟ 6 ⎝ ⎠ −1 64 ⎛ ⎞ ⎛3 1 ⎞ = ⎜ 12 − + 8 ⎟ − ⎜ + − 2 ⎟ 6 ⎝ ⎠ ⎝4 6 ⎠ 125 = 12 5

8

y

( 253 , 5)

y

x 9 (3, –3)

x 8

24. y = 2 − x 2 , y = x. Region appears below. Intersection: x = 2 − x 2 , x 2 + x − 2 = 0 , ( x + 2)( x − 1) = 0 ⇒ x = −2 or 1. 22. y = x , y = x 2 . Region appears below. 2

4

Intersection: x = x , x = x , x − x = 0 ,

(

)

x x3 − 1 = 0 , so x = 0, 1.

Area = ∫

1

0

(

x−x

2

)

(

)

1 1⎞ ⎛ 8 ⎛ ⎞ 9 = ⎜ 2 − − ⎟ − ⎜ −4 + − 2 ⎟ = 3 2⎠ ⎝ 3 ⎝ ⎠ 2

1

⎛ 32 ⎞ 2x x3 dx = ⎜ − ⎟ ⎜ 3 3 ⎟ ⎝ ⎠0

5

1 ⎛ 2 1⎞ = ⎜ − ⎟−0 = 3 3 3 ⎝ ⎠ 5

1

⎛ x3 x 2 ⎞ Area ∫ ⎡⎢ 2 − x 2 − x ⎤⎥ dx = ⎜ 2 x − − ⎟ ⎜ −2 ⎣ ⎦ 3 2 ⎟⎠ ⎝ −2 1

4

y

(1, 1) (–2, –2)

y

(1, 1)

x 5

596

x 5


Section 14.10

25. y = 8 − x 2 , y = x 2 , x = –1, x = 1. Region appears below.

27. y = x 2 , y = 2, y = 5. Region appears below. Area

Intersection: x 2 = 8 − x 2 , 2 x 2 = 8, x 2 = 4 , so x = ±2.

(

1

Area = ∫ ⎡⎢ 8 − x −1 ⎣ ⎛ 2x = ⎜ 8x − ⎜ 3 ⎝

3

8

2

)− x

1

2⎤

(

dx = ∫ 8 − 2 x −1 ⎦⎥

2

(

3

)

) dx

2 2

3

1

4y2 = 3

⎞ 2⎞ ⎛ 2 ⎞ 44 ⎛ ⎟ = ⎜ 8 − ⎟ − ⎜ −8 + ⎟ = ⎟ 3⎠ ⎝ 3⎠ 3 ⎠ −1 ⎝

5

4⋅5 5 4⋅ 2 2 4 − = 5 5−2 2 3 3 3

(

= 2

y

y

8

x=– y

x= y x

x –1

1

5

5

28. y = x3 + x, y = 0 (x-axis), x = −1, x = 2 Region appears below.

26. y 2 = 6 − x, 3y = x + 12. Region appears below. y 2 = 6 − (3 y − 12), y 2 + 3 y − 18 = 0, 3

Area = ∫ [(6 − y 2 ) − (3 y − 12)] dy

2

0

0

2

⎛ x4 x2 ⎞ ⎛ x4 x2 ⎞ = ⎜− − ⎟ +⎜ + ⎟ ⎜ 4 ⎜ 4 2 ⎟⎠ 2 ⎟⎠ ⎝ ⎝ −1 0 ⎡ ⎛ 1 1 ⎞⎤ = ⎢ 0 − ⎜ − − ⎟ ⎥ + [(4 + 2) − 0] ⎣ ⎝ 4 2 ⎠⎦ 27 = 4

−6 3

= ∫ (18 − y 2 − 3 y ) dy 3

⎛ y3 3 y 2 ⎞ = ⎜18 y − − ⎟ ⎜ 3 2 ⎟⎠ ⎝ −6 27 ⎞ ⎛ = ⎜ 54 − 9 − ⎟ − (−108 + 72 − 54) 2 ⎠ ⎝ 243 = 2 10

0

−1

Area = ∫ −( x3 + x)dx + ∫ ( x3 + x)dx

(y + 6)(y − 3) = 0, so y = −6, 3

−6

5

5 y2 = ∫ ⎡ y − − y ⎤ dy = ∫ 2 ydy = 2 ⋅ 3 ⎣ ⎦ 2 2 5

15

y

x

y 3

x 50

597

)



(

)

29. y = x3 − 1 , y = x ⫺ 1. Region appears below. Intersection: x3 − 1 = x − 1, x3 − x = 0 , x x 2 − 1 = 0 , x(x + 1)(x – 1) = 0, so x = 0 or x = ±1. 0

1

−1

0

Area = ∫ [ x3 − 1 − ( x − 1)] dx + ∫ [ x − 1 − ( x3 − 1)] dx =∫

0

−1

( x3 − x ) dx + ∫01( x − x3 ) dx 0

1

⎛ x4 x2 ⎞ ⎛ x2 x4 ⎞ =⎜ − ⎟ +⎜ − ⎟ ⎜ 4 ⎜ 2 2 ⎟⎠ 4 ⎟⎠ ⎝ ⎝ −1 0 ⎡ ⎛ 1 1 ⎞ ⎤ ⎡⎛ 1 1 ⎞ ⎤ 1 = ⎢ 0 − ⎜ − ⎟ ⎥ + ⎢⎜ − ⎟ − 0 ⎥ = ⎣ ⎝ 4 2 ⎠ ⎦ ⎣⎝ 2 4 ⎠ ⎦ 2 2

y x 3

(

)

30. y = x3 , y = x . Region appears below. Intersection: x3 = x , x6 = x, x6 − x = 0 , x x5 − 1 = 0 , x = 0, 1 Area = ∫

1

0

(

x−x

3

)

1

⎛ 32 ⎞ 2x x4 dx = ⎜ − ⎟ ⎜ 3 4 ⎟ ⎝ ⎠0

5 ⎛2 1⎞ = ⎜ − ⎟−0 = 12 ⎝3 4⎠ 3

y

(1, 1)

x 3

1 −17 − 4 x 1 . Region appears below. Intersection: = , −17 x − 4 x 2 = 4 , 4 x 2 + 17 x + 4 = 0 , x x 4 1 (4x + 1)(x + 4) = 0, so x = − or –4. 4

31. 4 x + 4 y + 17 = 0, y =

598


Area = ∫

Section 14.10 −1/ 4

⎛ 17 x2 ⎞ ⎛ −17 − 4 x ⎞ ⎤ − dx ln x x = + + ⎜ ⎟ ⎟⎥ ⎢x ⎜ ⎜ 4 2 ⎟⎠ 4 ⎝ ⎠⎦ ⎣ ⎝ −4

−1/ 4 ⎡ 1

−4

255 ⎛ 1 17 1 ⎞ = ⎜ ln − + ⎟ − ( ln 4 − 17 + 8 ) = − 4 ln 2 4 16 32 32 ⎝ ⎠ 1

y

x

–1

1

4

32. y 2 = − x − 2, x − y = 5, y = −1, y = 1. Region appears below. Intersection: y 2 = − x − 2 intersects y = ±1 when x = −3; x − y = 5 intersects y = 1 when x = 6; x − y = 5 intersects y = −1 when x = 4 1

⎛ y2 1 1 y3 ⎞ Area = ∫ [( y + 5) − (− y 2 − 2)] dy = ∫ ( y + 7 + y 2 ) dy = ⎜ + 7y + ⎟ ⎜ 2 −1 −1 3 ⎟⎠ ⎝ −1 1⎞ ⎛1 1 ⎞ 44 ⎛1 = ⎜ +7+ ⎟−⎜ −7− ⎟ = 3⎠ ⎝ 2 3⎠ 3 ⎝2 y

5

x

5

33. y = x – 1, y = 5 – 2x. Region appears below. Intersection: x – 1 = 5 – 2x, 3x = 6, so x = 2. 2

4

2

4

0

2

0

2

Area = ∫ [(5 − 2 x) − ( x − 1)]dx + ∫ [( x − 1) − (5 − 2 x)]dx = ∫ (6 − 3x)dx + ∫ (3x − 6)dx =−

1 2 1 4 (6 − 3 x)2 (6 − 3x)[−3 dx] + ∫ (3x − 6)[3 dx] = − ∫ 3 0 3 2 6

= –[0 – 6] + [6 – 0] = 6 + 6 = 12 5

y

y = 5 – 2x y=x–1

x

8

599

2

+ 0

(3x − 6) 2 6

4 2



34. y = x 2 − 4 x + 4, y = 10 − x 2 . Region appears below. Intersection: x 2 − 4 x + 4 = 10 − x 2 , 2 x 2 − 4 x − 6 = 0 , x 2 − 2 x − 3 = 0 , (x – 3)(x + 1) = 0, so x = 3, –1.

(

) (

)

(

) (

)

3 4 Area = ∫ ⎡⎢ 10 − x 2 − x 2 − 4 x + 4 ⎤⎥ dx + ∫ ⎡⎢ x 2 − 4 x + 4 − 10 − x 2 ⎤⎥dx 2⎣ 3 ⎣ ⎦ ⎦ 2(

)

3

3 4 2 x 2 − 4 x − 6 ) dx = 2 {∫ ( 3 + 2 x − x 2 ) dx + ∫ ( x 2 − 2 x − 3) dx} ( 3 2 3

= ∫ 6 + 4 x − 2 x 2 dx + ∫

4

3 4 ⎧⎛ 3⎞ ⎛ x3 ⎞ ⎫⎪ ⎧ ⎡ 22 ⎤ ⎡ 20 ⎪ ⎤⎫ 2 x 2 = 2 ⎨⎜ 3x + x − ⎟ + ⎜ − x − 3 x ⎟ ⎬ = 2 ⎨ ⎢9 − ⎥ + ⎢ − − (−9) ⎥ ⎬ = 2{4} = 8 ⎜ ⎟ ⎜ ⎟ 3 ⎠ 3 3⎦ ⎣ 3 ⎦⎭ ⎩⎣ ⎪⎝ ⎠ 3 ⎪⎭ 2 ⎝ ⎩ y

10

x 5

x ∫0 ⎡⎢⎣ x − ( 14 15 1

Area between curve and diag. 35. = Area under diagonal

2

)

1 x ⎤ dx + 15 ⎥⎦

1

∫0 x dx 1

14 ⎤ 14 1 14 ⎛ x 2 x3 ⎞ 14 ⎡⎛ 1 1 ⎞ ⎤ 14 1 7 Numerator = ∫ ⎢ x − x 2 ⎥ dx = ∫ x − x 2 dx = ⎜ − ⎟ = ⎢⎜ − ⎟ − 0 ⎥ = ⋅ = ⎜ ⎟ 0 ⎣ 15 15 ⎦ 15 0 15 ⎝ 2 3 ⎠ 15 ⎣⎝ 2 3 ⎠ ⎦ 15 6 45 0

(

1 ⎡14

x2 Denominator = ∫ x dx = 0 2 1

Coefficient of inequality =

1

0 7 45 1 2

=

1 2

=

14 45

)

11 x ∫0 ⎡⎣⎢ x − ( 12 1

Area between curve and diag. 36. = Area under diagonal

2

)

1 x ⎤ dx + 12 ⎦⎥

1

∫0 x dx 1

11 1 11 ⎛ x 2 x3 ⎞ 11 ⎡⎛ 1 1 ⎞ ⎤ 11 1 11 Numerator = ∫ x − x 2 dx = ⎜ − ⎟ = ⎢⎜ − ⎟ − 0 ⎥ = ⋅ = ⎜ ⎟ 0 12 12 ⎝ 2 3 ⎠ 12 ⎣⎝ 2 3 ⎠ ⎦ 12 6 72 0

(

Denominator =

)

1 (see Problem 35). 2

Coefficient of inequality =

11 72 1 2

=

11 36

600


Section 14.10

37. y 2 = 3 x, y = mx Intersection: (mx)2 = 3 x, m2 x 2 = 3x 3 m2

3

If x = 0, then y = 0; if x =

m2

, then y =

28 3 32 3

3 . m

With horizontal elements,

3 / m2

0 5

(

2

3

⋅100 = 87.5%

x 2 = k , x = ± k . Equating areas gives

∫−

k k

( k − x2 ) dx = 12 ∫−22 ( 4 − x2 ) dx

⎛ x3 ⎞ ⎜ kx − ⎟ ⎜ 3 ⎟⎠ ⎝ −

)

3 x − mx dx.

2

k

k

1⎛ x3 ⎞ = ⎜ 4x − ⎟ 2 ⎜⎝ 3 ⎟⎠ −2

4 16 k = 3 3 3 2

y

3

2

( )

k 2 = 4 ⇒ k = 4 3 = 22 x

8

5

2 3

y

y = x 2 − 1 , y = 2x + 2

x

2

– k

Intersection: x − 1 = 2 x + 2 , x 2 − 2 x − 3 = 0 , (x – 3)(x + 1), so x = 3 and –1. The area is

∫−1 ⎡⎣⎢2 x + 2 − ( x

2

− 1 ⎤⎥dx ⎦

41. 4.76 sq units 42. Two integrals are involved. Answer: 36.65 sq units

3

⎛ x3 ⎞ 32 = ⎜ − + x2 + 3x ⎟ = ⎜ 3 ⎟ ⎝ ⎠ −1 3 8

5

40. 0.23 sq units

) 3 = ∫ ( − x 2 + 2 x + 3) dx −1 3

k

43. Two integrals are involved. Answer: 7.26 sq units

y

44. Three integrals are involved. Answer: 358.18 sq units

x 5

601

4

= 2 3 ≈ 2.52

k

38. a.

) dx = 43 . Thus

39. y = x 2 and y = k intersect when

3 m

⎛y y y ⎞ y ⎞ − ⎜ − ⎟ dy = ⎜ ⎟ ⎜m 3 ⎟ ⎜ 2m 9 ⎟ 0 ⎝ ⎠ ⎝ ⎠0 9 3 3 = − = square units 2 m3 m3 2 m3 Note: With vertical elements, Area = ∫

3/ m ⎛

2

2

32 4 28 . Hence the − = 3 3 3 percentage above the x-axis is

.

Area = ∫

1

the area above is

m2 x 2 − 3x = 0, x(m2 x − 3) = 0, x = 0 or x=

∫−1(1 − x

b. The area below is



Problems 14.11

PS = ∫

1. D : p = 22 − 0.8q ⎫ S : p = 6 + 1.2q ⎬⎭ CS = ∫

5

⎛ q2 ⎞ = ⎜ 0.5q − ⎟ = (2.5 − 1.25) − 0 = 1.25 ⎜ 20 ⎟⎠ ⎝ 0

[ f (q) − p0 ] dq

0

8

8

= ∫ [(22 − 0.8q ) − 15.6] = ∫ (6.4 − 0.8q) dq 0

0

(

= 6.4q − 0.4q 2 q0

PS = ∫

0

4. D : p = 400 − q 2 ⎫ ⎬ S : p = 20q + 100 ⎭

) 0 = (51.2 − 25.6) − 0 = 25.6 8

Equilibrium pt. ( q0 , p0 ) = (10, 300)

[ p0 − g (q)] dq

8

8

0

0

(

(

=∫

8

10

PS = ∫ =∫

(900 − q 2 ) dq 30

0

5

5

3.5

3.5

= 100[(50 − 25) − (35 − 12.25)] = 225 PS = ∫

3.5

0.5

5

Equilibrium pt. = ( q0 , p0 ) = (5, 5) CS = ∫

= (2000 − 1000) − 0 = 1000

CS = ∫ 100(10 − 2 p )dp = 100(10 p − p 2 )

50 ⎫ ⎪ q +5 ⎬ q + 4.5⎪⎭ 10

q0

10

Equilibrium pt. = ( q0 , p0 ) = (300, 3.5) We use horizontal strips and integrate with respect to p.

⎛ q3 ⎞ = ⎜ 900q − ⎟ ⎜ 3 ⎟⎠ ⎝ 0 = (27, 000 − 9000) − 0 = 18, 000

S: p=

)0

5. D : q = 100(10 − 2 p ) ⎫ S : q = 50(2 p − 1) ⎬⎭

30

3. D : p =

(200 − 20q )dq

(

PS = ∫ [1300 − (400 + q 2 )] dq 0

[300 − (20q + 100)]dq

= 200q − 10q 2

30

⎛ q3 ⎞ = ∫ (900 − q 2 ) dq = ⎜ 900q − ⎟ ⎜ 0 3 ⎟⎠ ⎝ 0 = (27, 000 − 9000) − 0 = 18, 000

=∫

10

0

CS = ∫ [(2200 − q ) − 1300]

0 30

10

0

2

30

(100 − q2 ) dq

⎛ q3 ⎞ 1000 ⎞ 2000 ⎛ = ⎜ 100q − ⎟ = ⎜ 1000 − ⎟−0 = ⎜ ⎟ 3 ⎠ 3 ⎠ 3 ⎝ ⎝ 0

2. D: p = 2200 − q 2 ⎪⎫ ⎬ S : p = 400 + q 2 ⎪⎭ Equilibrium point = (q0 , p0 ) = (30, 1300) 0

10

0

) 0 = (76.8 − 38.4) − 0 = 38.4

30

)

10 CS = ∫ ⎡⎢ 400 − q 2 − 300 ⎤⎥ dq 0 ⎣ ⎦

= ∫ [15.6 − (6 + 1.2q )]dq = ∫ (9.6 − 1.2q )dq = 9.6q − 0.6q 2

[ p0 − g (q)] dq

5⎡ 5⎛ q⎞ ⎛q ⎞⎤ = ∫ ⎢5 − ⎜ + 4.5 ⎟ ⎥ dq = ∫ ⎜ 0.5 − ⎟ dq 0⎣ 0 10 ⎠ ⎝ 10 ⎠⎦ ⎝

Equilibrium pt. = ( q0 , p0 ) = ( 8, 15.6 ) q0

q0

0

p

50(2 p − 1)dp = 50( p 2 − p )

= 50[(12.25 − 3.5) − (0.25 − 0.5)] = 450

CS

S (300, 3.5)

PS

[ f (q) − p0 ] dq

D

5⎡

5 ⎤ 50 =∫ ⎢ − 5⎥ dq = ( 50 ln q + 5 − 5q ) 0 q+5 0 ⎣ ⎦ = [50 ln(10) – 25] – [50 ln(5)] = 50[ln(10) – ln(5)] – 25 = 50 ln(2) – 25

q 1000

602

3.5 0.5


Section 14.11

6. D : q = 100 − p ⎫⎪ ⎬ p S : q = 2 − 10 ⎪⎭ Equilibrium pt. = ( q0 , p0 ) = (8, 36) Integrating with respect to p, CS = ∫

100

3 2 = − (100 − p) 2 3

300

⎛ q2 ⎞ = ⎜ 5q − = (1500 − 750) − 0 = 750 ⎟ ⎜ 120 ⎟⎠ ⎝ 0 For CS we integrate with respect to p. From the demand equation, q = 0 ⇒ p = 20 .

100

20 ⎛ p3 ⎞ CS = ∫ 400 − p dp = ⎜ 400 p − ⎟ ⎜ 10 3 ⎟⎠ ⎝ 10 2 8000 ⎞ ⎛ 1000 ⎞ ⎛ = ⎜ 8000 − ⎟ − ⎜ 4000 − ⎟ = 1666 3 3 ⎠ ⎝ 3 ⎠ ⎝ 20

36

⎛ 2 ⎞ 1024 = 0 − ⎜ − ⋅ 512 ⎟ = 3 3 ⎝ ⎠ 36 ⎡ p ⎤ PS = ∫ ⎢ − 10 ⎥ dp 20 2 ⎣ ⎦

p = 211−5 = 64 CS = ∫

20

=−

S (8, 36)

10. a.

10

7. We integrate with respect to p. From the demand equation, when q = 0, then p = 100. 100

84

=∫

100

84

=−

1

100

(10 + 10)(30 + 20) = 1000, (20)(50) = 1000, 1000 = 1000 30 − 4(10) + 10 = 0, 30 − 40 + 10 = 0, 0 = 0 1000 , q + 20

1000 − 10 q + 20

⎤ ⎞ 1000 − 10 ⎟ − 10 ⎥ dq ⎢⎜ ⎠ ⎣⎝ q + 20 ⎦ 30 = [1000 ln(q + 20) − 20q ] 0 = 1000 ln(50) − 600 − [1000 ln(20)] ⎛ 50 ⎞ = 1000 ln ⎜ ⎟ − 600 ⎝ 20 ⎠ ⎛5⎞ = 1000 ln ⎜ ⎟ − 600 ⎝2⎠

CS = ∫

30 ⎡⎛

0

84

3 20 ⎡ 20 ⎤ = − ⎢0 − (16) 2 ⎥ = − (−64) 3 ⎣ 3 ⎦ 2 = 426 ≈ $426.67 3

400 − p 2 +5, 60

60 p = 400 − p 2 + 300 , p 2 + 60 p − 700 = 0 ,

11. CS ≈ 1197; PS ≈ 477

( p + 70)( p − 10) = 0 ⇒ p = 10 and q = 400 − 102 = 300 , so equilibrium pt. is

( q0 ,

)

⎛ 211 ⎞ 64 − 320 − ⎜ − − 0⎟ ⎜ ln 2 ⎟ ln 2 ⎝ ⎠

p=

−10(100 − p ) 2 [− dp ]

8. At equilibrium, p =

5

⎛ 211− q ⎞ − 64 dq = ⎜ − − 64q ⎟ ⎜ ln 2 ⎟ ⎝ ⎠0

b. (p + 10)(q + 20) = 1000, p + 10 =

10 100 − pdp

3 20 (100 − p ) 2 3

(2

11− q

≈ 2542.307 hundred ≈ $254,000

q

CS = ∫

5

0

D

PS

)

211− q = 2q +1 ⇒ 11 − q = q + 1 ⇒ q = 5 , so

p

CS

(

2

9. At equilibrium,

36

⎛ p2 ⎞ =⎜ − 10 p ⎟ = (324 − 360) − (100 − 200) ⎜ 4 ⎟ ⎝ ⎠ 20 = 64 100

⎛ q ⎞⎤ ⎢10 − ⎜ 60 + 5 ⎟ ⎥ dq ⎝ ⎠⎦ ⎣

0

100 − p dp

36

300 ⎡

PS = ∫

p0 ) = (300, 10) . 603



12. Let p = f(q).

5.

40

PS = ∫ [80 − f (q )] dq 0 40

=∫

0

80 dq − ∫

= 3200 − ∫

40

0

40

0

f (q ) dq

6

∫ ( x + 5)3 dx = 6∫ ( x + 5)

−3

dx

6( x + 5)−2 +C −2

=

= −3( x + 5)−2 + C

f (q ) dq

Use the trapezoid rule with h = 10 to estimate 40

∫0

f (q ) dq :

f (0) 2 f (10) 2 f (20) 2 f (30) f (40)

6.

= 25 = = 2(49) = = 2(59) = = 2(71) = = 80 =

25 98 118 142 80 463

7.

8.


)

∫ dx = ∫ 1 dx = 1⋅ x + C = x + C

3.

∫0 (

)

8⎛

9.

⎛ 1⎞

2

∫0 xe

4− x 2

dx = −

=0

2

)

− 6 dx ⎤⎥ ⎦

1 2 4− x 2 [−2 x dx] e 2 ∫0

2 0

13

) (

∫0

3t + 8dt = 4

3

8

4

3 (−3) − 302 302 1

1 (3t + 8) 3 = ⋅ 4 3

3 ⎛ ⎞ 2x 2 ⎜ = 2⋅ + x2 ⎟ ⎜ ⎟ 3 ⎝ ⎠0 3 2 ⎛ ⎞ = ⎜ 2 ⋅ 2 2 + 64 ⎟ − 0 3 ⎝ ⎠ 64 256 = + 64 = 3 3

4.

3 302

∫ x3 − 6 x + 1 dx = 2∫ x3 − 6 x + 1 ⎡⎢⎣( 3x

(

1 2

(

9

)

1 1 = − e0 − e 4 = e 4 − 1 2 2

⎞ 2 x + 2 x dx = ∫ ⎜ 2 x + 2 x ⎟ dx 0⎝ ⎠

8

302

6 x 2 − 12

2 1 = − e4− x 2

x4 x2 + 2⋅ − 7x + C 1. ∫ x3 + 2 x − 7 dx = 4 2 x4 = + x2 − 7 x + C 4

2.

( y − 6)302 dy = 302

= 2 ln x3 − 6 x + 1 + C

f (q ) dq ≈

(

∫3 ( y − 6)

301

=

10 (463) = 2315 2 Thus PS = 3200 − 2315 = $885. 40

∫0

9

(3t + 8) 4

=

)

1

∫ 5 − 3x dx = 4 ⎜⎝ − 3 ⎟⎠ ∫ 5 − 3x [−3 dx]

∫

11.

∫ y( y + 1)

604

0

1

= 0

113 11 −4 4

1

∫0

2

(

)

dy = ∫ y3 + 2 y 2 + y dy

y 4 2 y3 y 2 + + +C 4 3 2

=

12.

1

4 − 2x 4 1 ⎛4 2 ⎞ dx = ∫ ⎜ − x ⎟ dx = x − x 2 + C 7 7 7 ⎝7 7 ⎠

10.

4 = − ln 5 − 3x + C 3

4 3

1 1 1 3 [3 dt ] (3 t + 8) 3 ∫0

1

10−8 dx = 10−8 x = 10−8 − 0 = 10−8 0


13.

⎛ t1/ 5 t1/ 3 ⎞ dt = ∫ t ∫ ⎜⎜ t1/ 2 − t1/ 2 ⎟⎟ dt ⎝ ⎠ 7 /10 t t5 / 6 = ∫ (t −3 /10 − t −1/ 6 )dt = − +C 5

t −3t

10 6 = t 7 /10 − t 5 / 6 + C 7 5

14.

7 10

20.

5 6

21.

3

∫1

2t

22.

2

1 3 1 dt = ∫ [6t 2 dt ] 3 3 1 3 + 2t 3 3 + 2t 3 1 = ln(3 + 2t 3 ) 3 1 1 1 ⎛ 57 ⎞ = [ln(57) − ln(5)] = ln ⎜ ⎟ 3 3 ⎝ 5 ⎠

2 3 ∫ x 3x + 2dx =

(

3 1 3x + 2 = ⋅ 3 9

)

3 2

2

18.

3

)

1 2

(

4

)

3 2

∫ (e

4

2 7/2

(x + x )

2y

7 2

24.

+C

25.

= ln x + 2 ⋅

x −1 +C −1

2

−2

+C

dx

2 +C x

3e3 x

1

2

∫0 1 + e3x dx = ∫0 1 + e3x [3e

(

3x

dx]

2

0

70

∫7

2

dx = x

∫1 5 x

)

4

70

= 70 − 7 = 63

7

5 − x 2 dx = −

5 (5 − x 2 )3 / 2 =− ⋅ 3 2

4 4 ( x + x 2 )7 / 2 + C 7

2

2

1

5 2 (5 − x 2 )1/ 2 [−2 x dx] 2 ∫1

)

1 2y ⎛ 1⎞ e [2 dy ] − ⎜ − ⎟ ∫ e−2 y [−2 dy ] 2∫ ⎝ 2⎠ 1 1 1 = e2 y + e−2 y + C = e2 y + e−2 y + C 2 2 2 =

)

605

2

5 = − (5 − x 2 )3 / 2 3 1

5 5 35 = − (13 / 2 − 43 / 2 ) = − (1 − 8) = 3 3 3

− e−2 y dy

(

1

2 3

1 ⎛ y5 y 2 ⎞ 23. ∫ 10 y − y + 1 dy = 10 ⎜ − + y⎟ ⎜ 5 ⎟ −2 2 ⎝ ⎠ −2 ⎛1 1 ⎞ ⎛ 32 ⎞ = 10 ⎜ − + 1⎟ − 10 ⎜ − − 2 − 2 ⎟ = 111 5 2 5 ⎝ ⎠ ⎝ ⎠

2 5/ 2

+C =

)

+ C = − 7 − 2 x2

2 ⎞

1

⎡9 x 2 dx ⎤ ⎣ ⎦

2 +C = 3 x3 + 2 27

(

2 3

= ln(1 + e6 ) − ln(1 + 1) ⎛ 1 + e6 ⎞ = ln ⎜ ⎟ ⎜ 2 ⎟ ⎝ ⎠

∫ (8 x + 4 x)( x + x ) dx = 2 ∫ ( x 4 + x 2 )5 / 2 [(4 x3 + 2 x)dx] = 2⋅

19.

(

1 3 x3 + 2 9∫

⎛1

)

= ln(1 + e3 x )

4 x2 − x 16. ∫ dx = ∫ (4 x − 1)dx = 2 x 2 − x + C x

17.

)

∫ ⎜⎝ x + x 2 ⎟⎠ dx = ∫ x dx + 2∫ x = ln x −

1 (0.5 x − 0.1)5 ⋅ + C = (0.5 x − 0.1)5 + C 0.2 5

(

−1 8⎛ 1⎞ dx = ⎜ − ⎟ ∫ 7 − 2 x 2 3 [−4 x dx] 3 3⎝ 4⎠ 3 7 − 2 x2

8x

(

(0.5 x − 0.1) dx 0.4 1 1 = ⋅ ( 0.5 x − 0.1)4 [0.5 dx] 0.4 0.5 ∫

∫

∫

2 2 3 7 − 2x =− ⋅ 3 2

4

=

15.

Chapter 14 Review


26.

1

∫0

(

(2 x + 1) x 2 + x

)

4


0(

)

1

4

dx = ∫ x 2 + x [(2 x + 1) dx] =

(

x2 + x

51

)

=

5

25 32 −0 = 5 5

0

⎡ 1 27. ∫ ⎢ 2 x − 2 0⎢ ( x + 1) 3 ⎣ 1

1

1 1 ⎡ ⎤ 2 1 3 ⎤ 1 1 x ( x 1) + − 23 ⎢ ⎥ = ⎡ x 2 − 3( x + 1) 3 ⎤ ⎥ dx = 2 x dx − ( x + 1) [dx] = 2 ⋅ − ∫0 ∫0 ⎢⎣ ⎥⎦ 1 ⎢ 2 ⎥ ⎥ 0 3 ⎦ ⎣ ⎦0

= ⎡1 − 33 2 ⎤ − [ 0 − 3] = 4 − 33 2 ⎣ ⎦

28.

∫3 3 ( 27

)

3 x − 2 x + 1 dx = 3∫

27 ⎛

3

⎜ 3x ⎝

1 2

3 ⎡⎛ 2 ⎞ ⎛ = 3 ⎢⎜ 3 ⋅ 3 3 − 729 + 27 ⎟ − ⎜ 3 ⎠ ⎝ ⎣⎝

( )

29.

30.

t −3

∫

t2

27

2 3 ⎛ ⎞ ⎞ − 2 x + 1⎟ dx = 3 ⎜ 3 ⋅ x 2 − x 2 + x ⎟ 3 ⎠ ⎝ ⎠3 3 2 ⎤ ⎞ 3⋅ 3 − 9 + 3 ⎟ ⎥ = 3(−540) = −1620 3 ⎠⎦

( )

−1 ⎡ 12 ⎤ t 2 t −1 t 3⎥ 3 2 −1 ⎛ − 32 −2 ⎞ ⎢ − 3⋅ + C = −2t 2 + 3t −1 + C = − +C dt = ∫ dt = ∫ ⎜ t − 3t ⎟ dt = − 1 ⎢ t2 t2 ⎥ t − 1 −2 ⎝ ⎠ t ⎣ ⎦

3z3

1 ⎞ ⎛ 2 + z +1+ ⎟ dz z −1 ⎠ ⎛ z3 z 2 ⎞ = 3⎜ + + z + ln z − 1 ⎟ + C ⎜ 3 ⎟ 2 ⎝ ⎠

∫ z − 1 dz = 3∫ ⎜⎝ z

0

⎛ x2 ⎞ 0 ⎛ x2 + 4 x − 1 5 ⎞ 31. ∫ dx = ∫ ⎜ x + 2 − dx = + 2 x − 5ln x + 2 ⎟ ⎜ ⎟ ⎜ ⎟ −1 −1 ⎝ x+2 x+2⎠ ⎝ 2 ⎠ −1 0

⎛1 ⎞ 3 = (−5ln 2) − ⎜ − 2 − 0 ⎟ = − 5ln 2 2 ⎝ ⎠ 2

32.

∫

( x2 + 4) (

2

x2

dx = ∫

x 4 + 8 x 2 + 16

)

x2

dx

= ∫ x 2 + 8 + 16 x −2 dx =

x3 x −1 x3 16 + 8 x + 16 +C = + 8x − + C 3 3 −1 x 1

2 ⎛ 3 ⎞2 ⎡ 3 1 ⎤ 33. ∫ 9 x x + 1 dx = 9 ⋅ ∫ ⎜ x 2 + 1⎟ ⎢ x 2 dx ⎥ 3 ⎝ ⎠ ⎣2 ⎦ 3 2

3

⎛ 32 ⎞2 3 ⎜ x + 1⎟ ⎛ 32 ⎞2 ⎝ ⎠ = 6⋅ + C = 4 ⎜ x + 1⎟ + C 3 ⎝ ⎠ 2

606


Chapter 14 Review

1

34.

e

∫

5x

∫ 3

dx =

3x

5x 2

e

1

x

40.

dx

1 2

1

⎤ 2 ⎡ 5 −1 1∫ e 5 x ⎢ x 2 dx ⎥ 3⋅ 5 ⎣ 2 ⎦ 1 ⎛ ⎞ 2 2 ⎜ e 5x ⎟ + C = ⎟ 15 ⎜⎝ ⎠ 2 e 5x + C = 15 2

=

35.

36.

e eln x

∫1

x = ln e – ln 1 =1–0=1

∫

6x2 + 4

e

3

x +2x

= −2e

e

x

1

2

dx = ∫

2

x

dx = ∫

dx = −2∫ e

(

− x3 + 2 x

(

∫

(1 + e2 x )3

e

−2 x

) +C =

dx = =

38.

∫

3

−2

e

6 + e−3 x

2

∫3

=

1

(3x2 + 2) dx⎤⎦⎥

+C

x+5 ⎛ 5⎞ dx = ∫ ⎜ 1 + ⎟ dx = x + 5ln x + C x ⎝ x⎠ y(1) = 3 implies 3 = 1 + 0 + C, so C = 2. Thus y = x + 5ln x + 2

42. y = ∫

(1 + e2 x )3 +C 8

In Problems 43–58, answers are assumed to be expressed in square units.

dx

( ) −2 = − ∫ ( 6 + e−3 x ) ⎡ −3e−3 x dx ⎤ ⎣ ⎦ − 1 6 + e−3 x ) ( 1 =− +C = 3x

6 + e−3 x

3 x ln10

103 x dx = 3∫ e 2

)

1 2x e [2 dx] + ∫ 3 dx 2∫ 1 = e2 x + 3 x + C 2 1 1 1 y (0) = − implies that − = + 0 + C , so 2 2 2 1 C = –1. Thus y = e2 x + 3 x − 1 2

dx = ln x

⎣⎢

x3 + 2 x

(

e

) ⎡−

dx x 2 + 3x + 7 ⎛ 5 x3 + 15 x 2 + 35 x 2x + 3 ⎞ = ∫⎜ + ⎟ dx ⎜ x 2 + 3x + 7 x 2 + 3x + 7 ⎟⎠ ⎝ 1 [(2 x + 3) dx] = ∫ 5 x dx + ∫ 2 x + 3x + 7 5x2 = + ln x 2 + 3x + 7 + C 2

41. y = ∫ e2 x + 3 dx = ∫ e2 x dx + ∫ 3 dx

1 (1 + e2 x )3 [2e2 x dx] 2∫

−1

39.

x

− x3 + 2 x

e

37.

e1

1

∫

5 x3 + 15 x 2 + 37 x + 3

43. y = x 2 − 1 , x = 2, y ≥ 0. Region appears below. Area = ∫

2

1

( x2 − 1) dx 2

⎛ x3 ⎞ = ⎜ − x⎟ ⎜ 3 ⎟ ⎝ ⎠1 ⎛8 ⎞ ⎛1 ⎞ 4 = ⎜ − 2 ⎟ − ⎜ − 1⎟ = ⎝3 ⎠ ⎝3 ⎠ 3

+C

dx

8

3x 2 ln10 ⎡ 3ln10 ⎤ e2 = 3⋅ ∫ ⎢ 2 dx ⎥ 3ln10 ⎣ ⎦

=

3x 2 32x ln10 2 e +C = 10 2 + C ln10 ln10

=

2 103 x +C ln10

y

x 2

607

5



44. y = 4e x , x = 0, x = 3. Region appears below. 3

3

0

0

Area = ∫ 4e x dx = 4e x 100

= 4(e3 − 1)

y

x 3

5

45. y = x + 4 , x = 0. Region appears below. Area = ∫

0

x + 4dx = ∫ ( x + 4) [dx] =

−4 5

3

1 2

0

−4

y

( x + 4) 2 3 2

0

3

−4

2( x + 4) 2 = 3

0

= −4

16 16 −0 = 3 3

x 5

46. y = x 2 − x − 6, x = −4, x = 3. Region appears below. −2

3

⎛ x3 x 2 ⎞ ⎛ x3 x 2 ⎞ Area = ∫ ( x − x − 6) dx + ∫ −( x − x − 6) dx = ⎜ − − 6x ⎟ − ⎜ − − 6x ⎟ ⎜ 3 ⎟ ⎜ ⎟ −4 −2 2 2 ⎝ ⎠ −4 ⎝ 3 ⎠ −2 9 ⎡⎛ 8 ⎞ ⎛ 64 ⎞ ⎤ ⎡⎛ ⎞ ⎛ 8 ⎞ ⎤ 67 = ⎢⎜ − − 2 + 12 ⎟ − ⎜ − − 8 + 24 ⎟ ⎥ − ⎢⎜ 9 − − 18 ⎟ − ⎜ − − 2 + 12 ⎟ ⎥ = 2 ⎠ ⎝ 3 ⎠ ⎦ ⎣⎝ ⎠ ⎝ 3 ⎠⎦ 2 ⎣⎝ 3 −2

10

3

2

2

y

x 10

608


Chapter 14 Review

47. y = 5 x − x 2 . Region appears below.

50. y = x3 − 1 , x = –1. Region appears below. 5

1

⎛ 5 x 2 x3 ⎞ 5 − ⎟ Area = ∫ 5 x − x 2 dx = ⎜ ⎜ 2 0 3 ⎟⎠ ⎝ 0

⎛ x4 ⎞ 1 Area = ∫ − x3 − 1 dx = − ⎜ − x⎟ ⎜ 4 ⎟ −1 ⎝ ⎠ −1

125 ⎛ 125 125 ⎞ =⎜ − ⎟−0 = 3 ⎠ 6 ⎝ 2

⎛ 3⎞ ⎛5⎞ = −⎜ − ⎟ + ⎜ ⎟ = 2 ⎝ 4⎠ ⎝4⎠

(

)

(

y

10

)

y

3

x 3

x 10

–5

48. y = 4 x , x = 1, x = 16. Region appears below. Area = ∫

16 4

1

=

xdx = ∫

16

1

5

4x 4 x dx = 5 1 4

51. y 2 = 4 x , x = 0, y = 2. Region appears below.

16

Area = ∫

0

1

5

128 4 124 − = 5 5 5 4

y2 y3 dy = 4 12

2

2

= 0

8 2 −0 = 12 3

y

2

y

x 5

2 x 16

8

52. y = 3x 2 − 5, x = 0, y = 4. Region appears below. 3x 2 − 5 = 4, 3x 2 = 9, x 2 = 3, so x = ± 3.

1 + 2, x = 1, x = 4. Region appears below. x 4⎛ 1 4 ⎞ Area = ∫ ⎜ + 2 ⎟ dx = ( ln x + 2 x ) 1 ⎝x 1 ⎠ = [ln(4) + 8] − [0 + 2] = 6 + ln 4

49. y =

Area = ∫

3

[4 − (3x 2 − 5)] dx

=∫

3

[9 − 3 x 2 ] dx = (9 x − x3 )

0

(

0

)

= 9 3 −3 3 −0 = 6 3

y

5

5

y

x 5

x 0

1

4

5

609

3 0



53. y = x 2 + 4 x − 5 , y = 0. Region appears below.

Area = ∫

−2 5/ 2

x 2 + 4 x − 5 = 0 , (x + 5)(x – 1) = 0, so x = –5, 1. 1

(

=∫

)

2

−2

Area = ∫ − x + 4 x − 5 dx −5

[(10 − x 2 ) − ( x 2 − x)]dx (10 + x − 2 x 2 )dx 5/ 2

⎛ x 2 2 x3 ⎞ = ⎜10 x + − ⎟ ⎜ 2 3 ⎟⎠ ⎝ −2 25 125 ⎞ ⎛ 16 ⎞ ⎛ = ⎜ 25 + − ⎟ − ⎜ −20 + 2 + ⎟ 8 12 ⎠ ⎝ 3⎠ ⎝ 243 = 8

1

⎛ x3 ⎞ = − ⎜ + 2 x2 − 5x ⎟ ⎜ 3 ⎟ ⎝ ⎠ −5 ⎛1 ⎞ ⎛ 125 ⎞ = − ⎜ + 2 − 5⎟ + ⎜ − + 50 + 25 ⎟ = 36 ⎝3 ⎠ ⎝ 3 ⎠ y

4

5/ 2

15

x

y

4

x 5

54. y = 2 x 2 , y = x 2 + 9 . Region appears below.

56. y = x , x = 0, y = 3. Region appears below.

2 x 2 = x 2 + 9, x 2 = 9 , so x = ±3 3

(

x = 3 , so x = 9.

) ( )

Area = ∫ ⎡⎢ x 2 + 9 − 2 x 2 ⎤⎥ dx −3 ⎣ ⎦

3 ⎛ 2x 2 ⎜ Area = ∫ 3 − x dx = 3 x − 0 ⎜ 3 ⎝ = (27 − 18) − 0 = 9

9

3

⎛ x3 ⎞ = ∫ 9 − x 2 dx = ⎜ 9 x − ⎟ ⎜ −3 3 ⎟⎠ ⎝ −3 = (27 – 9) – (–27 + 9) = 36 3

(

)

18

5

y

(

)

9

⎞ ⎟ ⎟ ⎠0

y

3 x 16

x –3

2

3

5

57. y = ln x, x = 0, y = 0, y = 1. Region appears below.

2

55. y = x − x, y = 10 − x . Region appears below.

y = ln x ⇒ x = e y

x 2 − x = 10 − x 2 , 2 x 2 − x − 10 = 0,

1

1

0

0

Area = ∫ e y dy = e y

5 (x + 2)(2x − 5) = 0, so x = −2 or . 2

5

= e −1

y

1

x 5

610


Chapter 14 Review

58. y = 2 − x, y = x − 3, y = 0, y = 2. Region appears below.

62.

1000

33

∫10

3q + 70

2

Area = ∫ [( y + 3) − (2 − y )] dy 2

0

0

= (4 + 2) − 0 = 6 3

1000 (3q + 70) 2 = ⋅ 1 3 2

33

x

63.

100

∫0

0.008e−0.008t dt = − ∫

1 3 3 ⎛ ⎞ 2q ⎟ dq = ∫ 100dq − 2 ∫ q 2 dq 59. r = ∫ ⎜ 100 − 2 2 ⎝ ⎠

64.

3

3 3 q2 = 100q − 2⋅ + C = 100q − 2q 2 + C 3 2

5

∫0 4000e =

2

100 0

0.05t

4000 0.05t e 0.05

When q = 0, then r = 0. Thus 0 = 0 – 0 + C, so 1 r = 100 − 2q 2 = 100 − 2q . Thus q

5

=

3

4

3

4

= ∫ (9 − 3x)dx + ∫ (3x − 9)dx 0

q3 7 2 + q + 6q + C 60. c = ∫ q + 7q + 6 dq = 3 2 When q = 0, then c = 2500. Thus 2500 = 0 + 0 + 0 + C, so C = 2500. Hence q3 7 2 c= + q + 6q + 2500 . When q = 6, then 3 2 c = $2734.

)

3

⎛ 3x2 = ⎜ 9x − ⎜ 2 ⎝

3

4

⎞ ⎛ 3x 2 ⎞ − 9x ⎟ ⎟ +⎜ ⎟ ⎜ ⎟ ⎠0 ⎝ 2 ⎠3

⎡⎛ 27 ⎞ ⎤ ⎡ ⎛ 27 ⎞⎤ = ⎢⎜ 27 − ⎟ − 0 ⎥ + ⎢(24 − 36) − ⎜ − 27 ⎟ ⎥ 2 2 ⎝ ⎠ ⎝ ⎠⎦ ⎣ ⎦ ⎣ = 15 square units 16

2

4000 ⎡ 0.25 ⎤ e − 1 ≈ $22, 722 ⎦ 0.05 ⎣

0

3

∫15 (250 − q − 0.2q

1 5 0.05t [0.05 dt ] e 0.05 ∫0

Area = ∫ [(9 − 2 x) − x]dx + ∫ [ x − (9 − 2 x)]dx

p = 100 − 2q . 2

[−0.008 dt ]

= −e−0.8 + 1 ≈ 0.5507

dt = 4000 ⋅

0

e

65. y = 9 – 2x, y = x; from x = 0 to x = 4. Region appears below. Intersection: x = 9 – 2x, 3x = 9, so x = 3.

3

C = 0. Hence r = 100q − 2q 2 . Since r = pq, then p =

100 0.008t

0

= −e−0.008t

25

10

2000 3q + 70 3 10 2000 [13 − 10] = $2000 = 3

5

61.

33

=

y

(

1 33 −1 (3q + 70) 2 [3 dq ] ∫ 3 10 1

0 2

= ∫ (2 y + 1) dy = ( y 2 + y )

dq = 1000 ⋅

y

)dq y = 9 – 2x

25

⎛ q 2 0.2q3 ⎞ = ⎜ 250q − − ⎟ ⎜ 2 3 ⎟⎠ ⎝ 15 3125 ⎞ ⎛ = ⎜ 6250 − 312.5 − ⎟ − (3750 − 112.5 − 225) 3 ⎠ ⎝ ≈ $1483.33

y=x x 4

611

8



1 66. y = 2 x 2 , y = 2 − 5x; from x = −1 to x = . 3 Region appears below.

68. D : p = (q − 5) 2 ⎪⎫ ⎬ S : p = q 2 + q + 3⎪⎭ Equilibrium pt. = ( q0 , p0 ) = (2, 9)

2 x 2 = 2 − 5 x, 2 x 2 + 5 x − 2 = 0, −5 ± 41 x= (from the quadratic formula), 4 x ≈ −2.85 or 0.35. Area = ∫

1/ 3

−1

CS = ∫

2

0

2

3 ⎡ ⎤ ⎡ (q − 5)2 − 9 ⎤ dq = ⎢ (q − 5) − 9q ⎥ ⎣ ⎦ ⎣⎢ 3 ⎦⎥ 0

2 ⎛ −27 ⎞ ⎛ 125 ⎞ =⎜ − 18 ⎟ − ⎜ − − 0 ⎟ = $14 thousand 3 3 3 ⎝ ⎠ ⎝ ⎠ ≈ $14,667

[(2 − 5 x) − 2 x 2 ]dx 1/ 3

⎛ 5 x 2 2 x3 ⎞ = ⎜ 2x − − ⎟ ⎜ 2 3 ⎟⎠ ⎝ −1 5 2⎞ ⎛2 5 2 ⎞ ⎛ = ⎜ − − ⎟ − ⎜ −2 − + ⎟ 2 3⎠ ⎝ 3 18 81 ⎠ ⎝ 340 = square units 81

69.

qn

∫q0

n dq = −(u + v) ∫ dt 0 q − qˆ

ln q − qˆ

qn q0

n

= −(u + v)t 0

ln qn − qˆ − ln q0 − qˆ = −(u + v)n

y

ln q0 − qˆ − ln qn − qˆ = (u + v)n

15

ln

q0 − qˆ = (u + v)n qn − qˆ

n=

x 5

q − qˆ 1 ln 0 u + v qn − qˆ

as was to be shown. 67. D : p = 0.01q 2 − 1.1q + 30 ⎫⎪ ⎬ S : p = 0.01q 2 + 8 ⎪⎭ Equilibrium pt. = ( q0 , p0 ) = (20, 12) CS = ∫

q0

0

=∫

20

=∫

20

0 0

= =

)

⎡ 0.01q 2 − 1.1q + 30 − 12 ⎤ dq ⎣⎢ ⎦⎥

( 0.01q

2

)

20

q0

0

=

=∫

20

0

20

[ p0 − g (q)] dq = ∫0

( 4 − 0.01q ) 2

4η l

π ( P1 − P2 )

R

∫0 2η l π ( P1 − P2 ) R 2η l

(

)

r R 2 − r 2 dr

∫0 ( R

2

)

r − r 3 dr R

⎛ 0.01q3 1.1q 2 ⎞ =⎜ − + 18q ⎟ ⎜ 3 ⎟ 2 ⎝ ⎠0

PS = ∫

0

π ( P1 − P2 ) ⎛ R 2 r 2 r 4 ⎞ = − ⎟ ⎜ ⎜ 2 2η l 4 ⎟⎠ ⎝ 0

− 1.1q + 18 dq

2 ⎛ 80 ⎞ = ⎜ − 220 + 360 ⎟ − 0 = 166 3 ⎝ 3 ⎠

( P1 − P2 ) ( R 2 − r 2 )

70. Q = ∫ 2π rv dr = 2π∫ r ⋅

[ f (q) − p0 ] dq

(

R

π ( P1 − P2 ) ⎛ R 4 ⎞ πR 4 ( P1 − P2 ) ⎜ ⎟= ⎜ 4 ⎟ 2η l 8η l ⎝ ⎠ As was to be shown. =

(

π ( P1 − P2 ) ⎡⎛ R 4 R 4 ⎞ ⎤ − ⎢⎜ ⎟ − 0⎥ 2η l 4 ⎟⎠ ⎦⎥ ⎢⎣⎜⎝ 2

)

⎡12 − 0.01q 2 + 8 ⎤ dq ⎣⎢ ⎦⎥ 20

⎛ 0.01q3 ⎞ dq = ⎜ 4q − ⎟ ⎜ 3 ⎟⎠ ⎝ 0

80 ⎞ 1 ⎛ = ⎜ 80 − ⎟ − 0 = 53 3 ⎠ 3 ⎝ 612

dr



71. Case 1. r ≠ –1

b. Total number of units sold r +1 1/ x

1/ x 1 1/ x u g ( x) = ∫ ku r du = ∫ u r du = 1 1 k r +1

(

)

80

0

(

1

= 40t − 0.25t 2

1 x − r −1 − 1 r +1 1 ⎡ 1 −(r + 1) x − r − 2 ⎤ = − g ′( x) = r ⎣ ⎦ r +1 x +2 Case 2. r = –1 1/ x 1 1 1/ x g ( x) = ∫ ku −1du = ∫ du 1 1 k u 1/ x ⎛1⎞ = ln u = ln ⎜ ⎟ − 0 = − ln x 1 ⎝x⎠ 1 1 g ′( x) = − = − r x x +2 =

R

0

= ∫ f (t )dt = ∫

c.

3. a.

Total revenue R

30

0 30

0

=∫

(

5

0

5

)0

c.

= (500 – 25) – 0 = 475 25

20

(100 − 2t )dt = (100t − t 2 )

= (2500 − 625) − (2000 − 400) = 275

2. a.

25 20

R

0

80

0 80

=∫

0

(50 + 0.2t ) ⋅ (40 − 0.5t )dt

( 2000 − 17t − 0.1t 2 ) dt 80

17 1 ⎞ ⎛ = ⎜ 2000t − t 2 − t 3 ⎟ 2 30 ⎝ ⎠0 = 160, 000 − 54, 400 −

(900 − t 2 ) dt

Average delivered price total revenue = total number of units sold 2, 002,500 = = $111.25 18, 000


Total revenue = ∫ (m + st ) f (t )dt =∫

30

0 30

1 ⎞ ⎛ = ⎜ 900t − t 3 ⎟ = 27, 000 − 9000 = 18, 000 3 ⎠0 ⎝

f (t )dt = ∫ (100 − 2t )dt = 100t − t 2

f (t )dt = ∫

R

0

= ∫ f (t )dt = ∫


25

30

b. Total number of units sold

75. CS ≈ 1148; PS ≈ 251

∫20

(90, 000 + 900t − 100t 2 − t3 ) dt

100 3 1 4 t − t 3 4 0 = 2,700,000 + 405,000 – 900,000 – 202,500 = $2,002,500

74. Two integrals are needed. Answer: 32.75

b.

(100 + t )(900 − t 2 )dt

= 90, 000t + 450t 2 −

73. Two integrals are involved. Answer: 15.08 sq units

∫0

= 3200 − 1600 = 1600

= ∫ (m + st ) f (t )dt = ∫

72. Two integrals are needed. Answer: 101.75 sq units

5

80

Average delivered price total revenue = total number of units sold 88,533.33 ≈ ≈ $55.33 1600

0

1. a.

)0

(40 − 0.5t )dt

51, 200 ≈ $88,533.33 3

613

Chapter 15 Thus,


2

0.1t

(

) (

)

1.

e0.1t +C 0.1

= −40te0.1t + 400e0.1t + C S (t ) = −40te0.1t + 400e0.1t + C and S(0) = 5000 5000 = 0 + 400e0 + C C = 4600

2.

2. P (t ) = ∫ 0.1t (ln t )2 dt Let u = (ln t ) 2 and dv = 0.1t dt, so 2 ln t ⎛1⎞ du = 2(ln t ) ⎜ ⎟ dt = dt and t t ⎝ ⎠

2

t2 = 0.05t 2 2

dt

(

)

⎛ 2 ln t ⎞ = 0.05t 2 (ln t ) 2 − ∫ 0.05t 2 ⎜ ⎟ dt ⎝ t ⎠

3.

(

−x

dx

−x

dx = − xe− x − ∫ −e− x dx

= − xe − x − ∫ e− x [− dx] = − xe− x − e− x + C

)

⎛1⎞ ln t − ∫ 0.05t 2 ⎜ ⎟ dt ⎝t ⎠

= −e − x ( x + 1) + C

= 0.05t 2 ln t − ∫ 0.05t dt = 0.05t 2 ln t − 0.05

∫ xe ∫ xe

1 du = dt and v = 0.05t 2 . t 2

dx

v = −e − x .

∫ 0.1t ln t dt , let u = ln t and dv = 0.1t dt, so

∫ 0.1t ln t dt = 0.05t

3 x +1

Letting u = x, dv = e− x dx , then du = dx,

= 0.05(t ln t )2 − ∫ 0.1t ln t dt

For

∫ xe

If u = x and dv = e3 x +1dx, then du = dx and 1 v = e3 x +1. 3 x 3 x +1 1 3 x +1 3 x +1 ∫ xe dx = 3 e − ∫ 3 e dx x 1 1 = e3 x +1 − ⋅ ∫ e3 x +1[3 dx] 3 3 3 x 3 x +1 1 3 x +1 = e − e +C 3 9 1 = e3 x +1 (3x − 1) + C 9

S (t ) = −40te0.1t + 400e0.1t + 4600

∫ 0.1t (ln t )

∫ f ( x)dx = uv − ∫ v du 3 3 2 2 = x ⋅ ( x + 5) 2 − ∫ ( x + 5) 2 dx 3 3 3 5 2 2 2 = x( x + 5) 2 − ⋅ ( x + 5) 2 + C 3 3 5 3 5 2 4 = x ( x + 5) 2 − ( x + 5) 2 + C 3 15

= −40te0.1t + ∫ 40e0.1t dt

v = ∫ 0.1t dt = 0.1

)

2

Problems 15.1

dt = (−4t ) 10e0.1t − ∫ 10e0.1t (−4)dt

= −40te0.1t + 40

2

= 0.05(t ln t ) − 0.05t ln t + 0.025t + C

Let u = –4t and dv = e0.1t dt , so du = –4 dt, and 1 0.1t v = ∫ e0.1t dt = e = 10e0.1t . 0.1

∫ −4te

(

P (t ) = 0.05(t ln t )2 − 0.05t 2 ln t − 0.025t 2 + C

1. S (t ) = ∫ −4te0.1t dt

t2 +C 2

= 0.05t 2 ln t − 0.025t 2 + C

614


4.

∫ xe

−5x

Section 15.1

dx

8.

Letting u = x, dv = e−5 x dx, then du = dx,

Letting u = t, dv = e−t dt , then du = dt, v = −e −t ⎛ t ⎞ −t −t ∫ ⎜⎝ et ⎟⎠ dt = −te − ∫ −e dt

1 v = − e−5 x . 5

∫ xe

−5 x

xe−5 x 1 − ∫ − e−5 x dx 5 5 −5 x −5 x xe e =− + +C 5 5(−5)

dx = −

=−

5.

∫y

3

e

= −te −t − e−t + C = −e−t (t + 1) + C

9.

−5 x

1⎞ ⎛ x+ ⎟+C 5 ⎜⎝ 5⎠

∫y

6.

3

∫ 3x

y 4 ln y y4 ⎛ 1 ⎞ −∫ ⎜ dy ⎟ 4 4 ⎝y ⎠

y 4 ln y y3 y 4 ln y y 4 −∫ dy = − +C 4 4 4 16

=

y4 4

∫x

2

1⎤ ⎡ ⎢ ln( y ) − 4 ⎥ + C ⎣ ⎦

10.

ln x dx

∫x =

7.

2

1 dx , x

12 x 1+ 4x

dx

then du = 12dx, v =

x3 3 ln x dx =

∫

Letting u = 12x, dv = (1 + 4 x)

Letting u = ln x, dv = x 2 dx , then du = v=

2 x + 3 dx

1 1 = 3x ⋅ (2 x + 3)3 / 2 − ∫ (2 x + 3)3 / 2 ⋅ 3 dx 3 3 3/ 2 1 = x(2 x + 3) − (2 x + 3)5 / 2 + C 5 1 3/ 2 = (2 x + 3) [5 x − (2 x + 3)] + C 5 1 = (2 x + 3)3 / 2 (3x − 3) + C 5 3 = (2 x + 3)3 / 2 ( x − 1) + C 5

y4 4

=

2 x + 3 dx

1 v = (2 x + 3)3 / 2 . 3

ln y dy

ln y dy =

∫ 3x

Letting u = 3x, dv = 2 x + 3 dx, then du = 3dx,

⎛1⎞ Letting u = ln y, dv = y3 dy , then du = ⎜ ⎟ dy , ⎝ y⎠ v=

⎛ t ⎞

∫ ⎜⎝ et ⎟⎠ dt

∫ 3

3

x ln x x ⎛1 ⎞ − ∫ ⎜ dx ⎟ 3 3 ⎝x ⎠

1 (1 + 4 x) 2

1

3

= 4 x − 1[6 x − (1 + 4 x)] + C = (2 x − 1) 4 x + 1 + C

∫ ln(4 x) dx ⎛1⎞ Letting u = ln(4x), dv = dx, then du = ⎜ ⎟ dx , ⎝x⎠ v = x. ⎛1 ⎞ ∫ ln(4 x)dx = x ln(4 x) − ∫ x ⎜⎝ x dx ⎟⎠ = x ln(4 x) − ∫ dx = x ln(4 x) − x + C

= x[ln(4x) – 1] + C 615

dx ,

1 2

1 + 4x (1 + 4 x) 2 dx = 12 x ⋅ −∫ ⋅12dx 2 2 1+ 4x 12 x

= 6 x 1 + 4 x − (1 + 4 x ) 2 + C

x3 ln x x3 x3 ⎡ 1⎤ − +C = ln( x) − ⎥ + C 3 9 3 ⎢⎣ 3⎦

− 12

Chapter 15: Methods and Applications of Integration

11.


x

∫ (5 x + 2)3 dx

14.

1 (5 x + 3)−2 . 10

x

∫ (5 x + 2)3 dx =−

x 10(5 x + 3)

−∫−

2

1 (5 x + 3)−2 dx 10

1 (5 x + 3)−1 ⋅ +C 5(−1) 10(5 x + 3) 2 10 1 x =− − +C 2 50(5 x + 3) 10(5 x + 3) x

=−

12.

+

ln( x + 1)

⎡ 1

1

⎤

∫ 2( x + 1) dx = 2 ∫ ln( x + 1) ⎢⎣ x + 1 dx ⎥⎦

15.

n

(Form: ∫ u du ) ln( x + 1)

∫ 2( x + 1) dx =

3x + 5 e2 x

2

∫1 4 xe

∫

ln x x

2

ln( x + 1) +C 4

∫1

dx

Letting u = ln x, dv = x −2 dx , then du =

2x

dx

Letting u = 4x, dv = e2 x dx , then du = 4dx, 1 v = e2 x 2

2

2

13.

dx

Letting u = 3x + 5, dv = e−2 x dx, then du = 3dx 1 and v = − e−2 x . 2 3x + 5 3x + 5 1 −2 x ∫ e2 x dx = − 2e2 x − ∫ − 2 e ⋅ 3 dx 3 x + 5 3 −2 x =− + ∫ e dx 2e 2 x 2 3 x + 5 3 ⎛ 1 −2 x ⎞ =− + ⎜− e ⎟+C ⎠ 2e 2 x 2 ⎝ 2 1 [2(3x + 5) + 3] + C =− 4e 2 x 1 (6 x + 13) + C =− 4e 2 x

Letting u = x, dv = (5 x + 3)−3 dx, then du = dx and v = −

∫

2

4 xe 2 x dx = ⎡ 2 xe2 x − ∫ 2e2 x dx ⎤ ⎣ ⎦1 2

2

= ⎡ 2 xe2 x − e2 x ⎤ = e2 x (2 x − 1) ⎣ ⎦1 1

1 dx , x

(

)

= e 4 (3) − e2 (1) = e2 3e2 − 1

−1

v = −x . ln x ln x ⎞ −1 ⎛ 1 ∫ x 2 dx = − x − ∫ − x ⎜⎝ x dx ⎟⎠ ln x ln x 1 =− + ∫ x −2 dx = − − +C x x x 1 = − (1 + ln x) + C x

16.

2

∫1 2 xe

−3 x

dx

Letting u = 2x, dv = e−3 x dx, then du = 2 dx and 1 v = − e−3 x . 3

616


2

∫1 2 xe

−3 x

Section 15.1

dx

19. 2

⎡ 2 xe−3 x ⎤ 2 = ⎢− − ∫ − e−3 x dx ⎥ 3 3 ⎢⎣ ⎥⎦ 1

1 1 ⎡ ⎤ = ⎢ −6 x(4 − x) 2 − ∫ −2(4 − x) 2 (3 dx) ⎥ ⎣ ⎦1

2

3 1 ⎡ ⎤ = ⎢ −6 x(4 − x) 2 − 4(4 − x ) 2 ⎥ ⎣ ⎦1

2

1

1 ⎞ ⎤ ⎡ 2e −3 ⎛ ⎜ 2 + 3 ⎟⎥ − ⎢− 3 ⎝ ⎠ ⎦⎥ ⎣⎢

{

17.

∫0 xe

− x2

⎛ 1 ⎞⎤ ⎜1 + 3 ⎟ ⎥ ⎝ ⎠ ⎦⎥

{

18.

∫

(

= 2 9 3 − 10 2

20.

1

3 x3 4 − x2

=− 0

(

) (

1 −1 1 e − 1 = 1 − e −1 2 2

∫e

u

du )

)

∫ (ln x)

)

− 12

4 − x2

)

1 2

2

1 2

1 2

= − 4 − x 2 ⎡3 x 2 ⎣⎢

(

)

⎡ 2 ln x ⎤ dx = x(ln x) 2 − ∫ x ⎢ dx ⎥ ⎣ x ⎦

∫ ln( x)dx , let u = ln x, dv = dx. Then ⎛1

Thus

( ) (4 − x )

= − x2 + 8

2

⎞

∫ ln( x)dx = x ln x − ∫ x ⎜⎝ x dx ⎟⎠ = x[ln( x) − 1] + C1 .

.

dx

= −3 x 2 4 − x 2 = −3 x 2

dx

⎛1⎞ du ⎜ ⎟ dx , v = x, so ⎝x⎠

dx , then

du = 6x dx,

∫

)

= x(ln x)2 − 2 ∫ ln( x)dx

(

(

2 −9 3

⎡ 2 ln x ⎤ du = ⎢ ⎥ dx , v = x. ⎣ x ⎦

For

v = − 4 − x2

2

2

Letting u = (ln x)2 , dv = dx, then

dx

Letting u = 3 x 2 , dv = x 4 − x 2

3 x3

∫ (ln x)

)

2

} 1 = −2 (10

= −2 4 − x ( x + 8)

1 1 2 dx = − ∫ e− x (−2 x dx) (Form: 2 0

2 1 = − e− x 2

}1

= −2 4 − x [3 x + 2(4 − x)]

2e ⎡ 7 3 ⎛ 4 ⎞ ⎤ =− ⎢ − e ⎜ ⎟⎥ 3 ⎣3 ⎝ 3 ⎠⎦ 2 =− [7 − 4e3 ] 6 9e 1

dx , then du = 3dx,

2

2

−6

− 12

v = −2(4 − x) . 2 3x ∫1 4 − x dx

⎡ 2 xe−3 x 2e−3 x ⎤ = ⎢− − ⎥ 3 9 ⎥⎦ ⎢⎣ 1

⎡ 2e−6 = ⎢− 3 ⎣⎢

dx

1 2

2

1 ⎞⎤ ⎛ ⎜ x + 3 ⎟⎥ ⎝ ⎠ ⎦⎥

4− x

Letting u = 3x, dv = (4 − x)

⎡ 2 xe−3 x 2 e−3 x ⎤ = ⎢− + ⋅ ⎥ 3 3 −3 ⎥⎦ ⎢⎣ 1

⎡ 2e−3 x = ⎢− 3 ⎣⎢

3x

2

∫1

(

− ∫ − 4 − x2

)

1 2

(6 x dx)

( ) +C + 2 ( 4 − x )⎤ + C ⎦⎥ − 2 4 − x2

3 2

2

4 − x2 + C

617

∫ (ln x)

2

dx = x ⎡ (ln x)2 − 2 ln( x) + 2 ⎤ + C . ⎣ ⎦


21.


∫ 3(2 x − 2) ln( x − 2)dx

24.

1

e

Letting u = ln x and dv = x 3 dx, then du =

∫ 3(2 x − 2) ln( x − 2)dx

and v =

= 3x ( x − 2) ln( x − 2) − ∫ 3 x dx

1

3 dx x−2

3

Letting u = xe x , dv = ( x + 1)−2 dx , then du = ( x + 1)e x dx , v = −( x + 1) −1 .

∫ ( x + 1)2 =−

xe x + e x dx x +1 ∫

xe x + ex + C x +1

x ⎞ ex ⎛ ⎛ x +1− x ⎞ = e x ⎜1 − = ex ⎜ +C = +C ⎟ ⎟ x +1 ⎝ x +1 ⎠ ⎝ x +1 ⎠

∫x

25.

2 x

Letting u = x 2 , dv = e x dx , then du = 2x dx and v=e .

∫(

2 x 2 x x ∫ x e dx = x e − ∫ e (2 x dx)

26.

x x ∫ xe dx , let u = x, dv = e dx . Then du = dx,

dx = xe x − ∫ e x dx = xe x − e x + C1

= e x ( x − 1) + C1 .

Thus

(

∫x

2

(

)

dx = ∫ x 2 − 2 xe− x + e−2 x dx

x3 e −2 x − − 2∫ xe− x dx 3 2

x − e− x

)

∫x

2

dx =

∫ xe

−x

dx ,

x3 e −2 x − + 2e− x ( x + 1) + C 3 2

2 3x

e dx

Letting u = x 2 , dv = e3 x dx, then du = 2x dx and 1 v = e3 x . 3 1 2 3x 1 3x 2 3x ∫ x e dx = 3 x e − ∫ 3 e ⋅ 2 x dx 1 2 = x 2 e3 x − ∫ xe3 x dx 3 3

v = e x and x

)

x

= x e − 2∫ xe dx

∫ xe

−x

Using Problem 3 for

x

For

∫(x −e =

e dx

2 x

1

3 ⎛ 3 43 3 43 1 ⎤ ⎞⎟ ⎡ ⎜ = 5 ⎢ ln x ⋅ x − ∫ x ⋅ dx ⎥ ⎜⎣ x ⎦e⎟ 4 4 ⎝ ⎠ 3⎞ ⎛ 3ln x 4 3 1 ⎡ ⎤ = 5⎜ ⎢ x 3 − ∫ x 3 dx ⎥ ⎟ ⎜⎣ 4 4 ⎦e⎟ ⎝ ⎠ ⎛ 3ln x 4 9 4 3 ⎞ ⎡ ⎤ = 5⎜ ⎢ x3 − x3 ⎥ ⎟ ⎜⎣ 4 16 ⎦ e ⎟ ⎝ ⎠ ⎛ ⎡ 3ln 3 43 9 43 ⎤ ⎡ 3ln e 43 9 43 ⎤ ⎞ = 5⎜ ⎢ e − e ⎥⎟ 3 − 3 ⎥−⎢ 16 ⎦ ⎣ 4 16 ⎦ ⎠ ⎝⎣ 4 4 4 ⎛3 ⎡ 3⎤ 3 ⎡1⎤⎞ = 5 ⎜ 3 3 ⎢ln 3 − ⎥ − e 3 ⎢ ⎥ ⎟ 4 4⎦ 4 ⎣4⎦⎠ ⎣ ⎝

∫ ( x + 1)2 dx

dx = −

3 43 x . 4

e

xe x

xe x

1 dx x

5∫ x 3 ln x dx

3 = 3x ( x − 2) ln( x − 2) − x 2 + C 2

23.

3

x ln( x5 ) dx = 5∫ x 3 ln x dx

Letting u = 3 ln(x − 2), dv = (2x − 2)dx, then 3 du = dx and v = x 2 − 2 x = x( x − 2). x−2 = 3x ( x − 2) ln( x − 2) − ∫ x( x − 2) ⋅

22.

33

∫e

e dx = x 2 e x − 2 ⎡ e x ( x − 1) ⎤ + C ⎣ ⎦

2 x

)

= e x x2 − 2x + 2 + C .

For

∫ xe

3x

dx, let u = x, dv = e3 x dx, then

1 du = dx, v = e3 x , and 3 1 3x 1 3x 3x ∫ xe dx = 3 xe − ∫ 3 e dx 1 1 = xe3 x − e3 x + C1. 3 9 618


Section 15.1

∫ (2

Thus, 1 2 3x 2 ⎛ 1 3x 1 3x ⎞ ∫ x e dx = 3 x e − 3 ⎜⎝ 3 xe − 9 e ⎟⎠ + C 1 3x e (9 x 2 − 6 x − 2) + C = 27 2 3x

27.

3 x2

∫x e

=

Letting u = x 2 , dv = xe x dx , then du = 2x dx, ⎛1⎞ 2 v = ⎜ ⎟ ex . ⎝2⎠ 2

2

=

28.

30.

2

x2e x ex −∫ (2 x dx) 2 2 2

2

x2e x ex ex − +C = 2 2 2

( x − 1) + C

=

2

5 x ∫ x e dx

2

x4 2 x4 = 2

2

e x − 2 ∫ x3e x dx

Using Problem 27 for

∫x

5 x2

e dx =

3 x2

∫x e

4

x x2 ⎡1 2 ⎤ e − 2 ⋅ ⎢ e x ( x 2 − 1) ⎥ + C 2 ⎣2 ⎦

29.

x

⎛

− 12

dx

1

−2 2 = x ln ⎛⎜ x + x 2 [2 x dx] ⎝ = x ln ⎛⎜ x + x 2 + 1 ⎞⎟ − x 2 + 1 + C ⎝ ⎠

e3

31. Area = ∫ (ln x)dx . Letting u = ln x, dv = dx, 1

∫ ( 2 + x ) dx = ∫ ( 2 + 2 x2 + x ) dx = ∫ 22 x dx + ∫ x 2 x +1 dx + ∫ x 2 dx For ∫ x 2 x +1 dx , let u = x, dv = 2 x +1 dx . Then 2x

x +1

( ) 1 + 1 ⎞⎟ − ∫ ( x + 1) ⎠ 2

2 x4 x2 e − e x ( x 2 − 1) + C 2 1 x2 4 = e ( x − 2 x 2 + 2) + C 2

2

1 2

∫ ln ⎜⎝ x +

= x ln ⎛⎜ x + x 2 + 1 ⎞⎟ − ∫ x x 2 + 1 ⎝ ⎠

dx,

=

x

⎛ x2 + 1 + x ⎞ ⎜ ⎟= x + x 2 + 1 ⎜⎝ x 2 + 1 ⎟⎠ 1

x 2 + 1 ⎞⎟ dx , let ⎠ ⎛ ⎞ 2 u = ln ⎜ x + x + 1 ⎟ , dv = dx. Then ⎝ ⎠ 1 du = dx , v = x, and x2 + 1 ⎛ ⎞ 2 ∫ ln ⎜⎝ x + x + 1 ⎟⎠ dx

2 1 2 e x − ∫ e x ⋅ 4 x3 dx 2 2

dx = ∫ 22 x dx + ∫ x 2 x +1 dx + ∫ x 2 dx

1 x 1 x3 ⋅ 22 x −1 + ⋅ 2 x +1 − ⋅ 2 x +1 + +C ln 2 ln 2 3 ln 2 2

For

x2

Letting u = x and dv = xe dx, then 1 2 du = 4 x3 dx and v = e x . 2 5 x ∫ x e dx =

2

d ⎡ ⎛ ⎤ ln ⎜ x + x 2 + 1 ⎞⎟ ⎥ dx ⎢⎣ ⎝ ⎠⎦ ⎛ 1 x ⎞ ⎜1 + ⎟ = x + x 2 + 1 ⎝⎜ x 2 + 1 ⎠⎟

2

4

)

1 2x 2 [2 dx] + ∫ x 2 x +1 dx + ∫ x 2 dx 2∫ 1 x 1 x3 = ⋅ 22 x + ⋅ 2 x +1 − ⋅ 2 x +1 + +C 2 ln 2 ln 2 3 ln 2 2

2

2

+x

=

dx

3 x ∫ x e dx =

x

2

⎛1⎞ then du = ⎜ ⎟ dx , v = x. ⎝x⎠ e3

e3

1 ⎤ ⎡ ∫1 (ln x)dx = ⎢⎣( x ln x) − ∫ x ⋅ x dx ⎥⎦ 1

1 ⋅ 2 x +1 and ln 2 x 1 x +1 x +1 x +1 ∫ x2 dx = ln 2 ⋅ 2 − ln 2 ∫ 2 dx x 1 = ⋅ 2 x +1 − ⋅ 2 x +1 + C1 . Thus 2 ln 2 ln 2

du = dx, v =

e3

e3

= ⎡( x ln x ) − ∫ dx ⎤ = [ x ln( x) − x] ⎣ ⎦1 = ⎡e ⋅ 3 − e − [1 ⋅ 0 − 1] = 2e + 1 ⎣ ⎦ 3

3⎤

3

The area is (2e3 + 1) sq units. 619

1


ISM: Introductory Mathematical Analysis du = dq, v = −10e− (0.1q +1) , and

1

32. Area = ∫ x 2 e x dx. 0

20

I1 = ⎡ −10(q + 10)e −(0.1q +1) + 10 ∫ e− (0.1q +1) dq ⎤ ⎣ ⎦0

Letting u = x 2 , dv = e x dx, then du = 2x dx and v = ex .

∫x

20

= ⎡ −10(q + 10)e− (0.1q +1) − 100e− (0.1q +1) ⎤ ⎣ ⎦0

2 x

For

e = x 2 e x − 2 ∫ xe x dx

∫ xe

x

= −10e −(0.1q +1) [(q + 10) + 10]

dx, let u = x and dv = e x dx, then

du = dx and v = e x .

∫ xe

x

Thus

x

x

x

x

= −10e

x

= xe − ∫ e dx = xe − e = e ( x − 1). 1 2 x x e 0

∫

2 x

x

dx = ( x e − 2[e ( x − 1)]) = (e x [ x 2 − 2 x + 2])

= e−2 The area is (e − 2) sq units.

1

0

Thus CS = 10 I1 − 300e −3 I 2

(

0

−3

35. a.

1 dx, x

−1

≈ 237.89

Consider

∫ p dq . Letting u = p, dv = dq,

then du =

dp dq , v = q. Thus dq dp

2 2 x ln x dx 1

2

(since r = pq). dp dq . dq Combining the integrals gives ⎛ dp ⎞ r = ∫ ⎜ p + q ⎟ dq . dq ⎠ ⎝

b. From (a), r = ∫ p dq + ∫ q

2

⎛x ⎞ 1 = ⎜ ln x − ∫ x 2 dx ⎟ ⎜ 3 ⎟ 3 ⎝ ⎠1 2

⎛ x3 1 ⎞ = ⎜ ln x − x3 ⎟ ⎜ 3 9 ⎟⎠ ⎝ 1 8⎞ ⎛ 1⎞ ⎛8 = ⎜ ln 2 − ⎟ − ⎜ 0 − ⎟ 9⎠ ⎝ 9⎠ ⎝3 8 7 = ln(2) − 3 9 7⎞ ⎛8 The area is ⎜ ln(2) − ⎟ sq units. 3 9⎠ ⎝

c.

When q = 20, then p = 300e

dp ⎞ ⎜ p + q ⎟ dq dq ⎠ ⎝ q dr =∫ 0 dq = r ( q0 ) − r (0) = r ( q0 ) 0 dq [since r(0) = 0].

.

dx

x

x

x

∫ f ( x)e dx = f ( x)e − ∫ f ′( x)e dx . Thus x x x ∫ f ( x)e dx + ∫ f ′( x)e dx = f ( x)e + C

20

For I1 , let u = q + 10, dv = e

x

v = e x . Using integration by parts,

= 10 ∫ (q + 10)e −(01.q +1) dq − 300e−3 ∫ dq 0 0

I2

− (0.1q +1)

∫ f ( x )e

Letting u = f(x), dv = e x dx , then du = f ′( x)dx ,

⎡10(q + 10)e −(0.1q +1) − 300e−3 ⎤ dq ⎣ ⎦

I1

dr dp . Thus = p+q dq dq

q0 ⎛

36.

20

From (b),

∫0

34. p = 10(q + 10)e−(0.1q +1) −3

dp

∫ p dq = pq − ∫ q dq dq = r − ∫ q dq dq

⎛ x3 x3 1 ⎞ = ⎜ ln x − ∫ ⋅ dx ⎟ ⎜ 3 3 x ⎟⎠ ⎝ 1 3

0

= −400e−3 + 200e−1

)

= −10, 000e + 2000e CS ≈ $237.89

x3 . 3

CS = ∫

0

0

= 10 −400e −3 + 200e−1 − 300e−3 (20)

Letting u = ln x, dv = x 2 dx, then du =

20

20

20

I 2 = q 0 = 20 − 0 = 20

33. Area = ∫

∫

(q + 20)

1

2 2 x ln x dx. 1

v=

− (0.1q +1)

20

dq . Then 620


Section 15.2

37. f and its inverse f −1 satisfy the equation

r (q) = ∫

−1

f ( f ( x)) = x. Differentiating this equation using the Chain Rule we get: f ′( f −1 ( x)) ⋅ ( f −1 )′( x) = 1. Thus ( f −1 )′( x) =

∫

1 f ′( f −1 ( x))

=∫

. Now to evaluate

=

letting u = f −1 ( x) and dv = dx. Then du =

So

f ′( f −1 ( x))

∫f

−1

( x) dx = xf

To evaluate

( x) − ∫

x

∫ f ′( f −1 ( x)) dx

x f ′( f −1 ( x))

1 f ′( f −1 ( x))

dx.

we will use the fact

∫ f ′( f −1 ( x))

.

300t 3 t2 + 6

dx = ∫ f ( f −1 ( x)) ⋅ ( f −1 )′( x) dx

=

5(q + 4) 2

=

300t 3 + 1800t − 1800t

(

)

t2 + 6

300t t 2 + 6 − 1800t 2

t +6

= 300t −

1800t

u = t 2 + 6 , so du = 2t dt


Express

300t 3

t2 + 6 1800t t 2 + 6 is irreducible. To integrate , let t2 + 6

( x) dx = xf −1 ( x) − F ( f −1 ( x)) + C.

1. r (q ) = ∫ r ′(q)dq = ∫

5 3(q + 1)3 ln . 2 q+3

300t 3 by t 2 + 6 to reduce the fraction.

since F ′ = f . Finally, −1

5 1 5 ln + C so C = ln 3 and 2 3 2

dt t2 + 6 Since the degree of the numerator is greater than the degree of the denominator, we first divide

= F ( f −1 ( x))

∫f

5 2

2. V (t ) = ∫ V ′(t )dt = ∫

Hence x

dx

5 (q + 1)3 +C ln 2 q+3

r (q) =

that x = f ( f −1 ( x)) and ( f −1 )′( x) =

dq − ∫

Since r(0) = 0, 0 =

dx and v = x. −1

q + 4q + 3

dq q +1 q+3 15 5 = ln q + 1 − ln q + 3 + C 2 2

f −1 ( x) dx we will use integration by parts, 1

15 2

5(q + 4) 2

300t 3

5(q + 4) 2

q + 4q + 3

1800t

∫ t 2 + 6 dt = ∫ 300t dt − ∫ t 2 + 6 dt

dq

= 150t 2 − 900 ln t 2 + 6 + C

(

)

V (t ) = 150t 2 − 900 ln t 2 + 6 + C

as a sum of partial

q + 4q + 3 fractions. 5(q + 4) 5(q + 4) A B = = + 2 q + 4q + 3 (q + 1)(q + 3) q + 1 q + 3

Problems 15.2 1.

5(q + 4) = A(q + 3) + B(q + 1) 5 . 2 15 When q = –1, we get 5(3) = A(2), so A = . 2

When q = –3, we get 5(1) = –2B, so B = −

621

10 x 2

=

10 x A B = + ( x + 6)( x + 1) x + 6 x + 1

x + 7x + 6 10 x = A( x + 1) + B ( x + 6) If x = –1, then –10 = 5B, or B = –2. If x = –6, then –60 = –5A, or A = 12. 12 2 − Answer x + 6 x +1


2.

3.

4.

5.

6.

x+5


x+5 A B = + 2 x − x + x − x +1 ( 1)( 1) 1 x −1 x + 5 = A(x + 1) + B(x – 1) If x = –1, then 4 = –2B, or B = –2. If x = 1, then 6 = 2A, or A = 3. 3 2 − Answer x −1 x +1 =

2 x2 2

= 2+

7.

2 x 2 − 15

= 2+

−10 x − 12

8.

=

x+4 2

=

2x + 3 x ( x − 1)

=

x +x

x x +1 2

3x2 + 5

(x

2

+4

)

2

=

Ax + B 2

x +4

(

(

9.

A B + x + 2 ( x + 2) 2

x + 4 x + 4 ( x + 2) x + 4 = A(x + 2) + B If x = −2, then 2 = B. If x = 0, then 4 = 2A + B, 2A = 4 − B = 4 − 2 = 2, or A = 1. 1 2 + Answer: x + 2 ( x + 2)2 2

A Bx + C + x x2 + 1

+

Cx + D

(x

2

+4

)

)

2

3x 2 + 5 = Ax3 + Bx 2 + (4 A + C ) x + (4 B + D) Thus A = 0, B = 3, 4A + C = 0, 4B + D = 5. This gives A = 0, B = 3, C = 0, D = –7. 3 7 − Answer: 2 2 x +4 x2 + 4

−10 x − 15

x+4 2

=

2

3x 2 + 5 = ( Ax + B ) x 2 + 4 + (Cx + D)

x + 5x −10 x − 15

f ( x) =

x2 + 3

x 2 + 3 = ( A + B ) x 2 + Cx + A Thus A + B = 1, C = 0, A = 3. This gives A = 3, B = –2, C = 0. 3 2x Answer: − 2 x x +1

(by long division). x2 + 5x −10 x − 15 A B = = + 2 x ( x + 5) x x +5 x + 5x –10x – 15 = A(x + 5) + Bx. If x = 0, then –15 = 5A, or A = –3. If x = –5, then 35 = –5B, or B = –7. 3 7 Answer: 2 − − x x+5 2

=

( ) + 3 = A ( x + 1) + ( Bx + C ) x

3

x2

x2 + 5x + 6 −10 x − 12 A B = = + 2 ( x + 2)( x + 3) x + 2 x +3 x + 5x + 6 −10x − 12 = A(x + 3) + B(x + 2) If x = −3, then 18 = −B, or B = −18. If x = −2, then 8 = A. 8 18 − Answer: 2 + x+2 x+3 x + 5x + 6 −10 x − 12

x2 + 3

5x − 2

)

5x − 2 A B = + x x − x x −1 ( 1) x −x 5x – 2 = A(x – 1) + Bx If x = 1, then 3 = B. If x = 0, then –2 = –A, or A = 2. 5x − 2 3 ⎞ ⎛2 ∫ x2 − x dx = ∫ ⎜⎝ x + x − 1 ⎟⎠ dx 2

=

= 2 ln x + 3ln x − 1 + C = ln x 2 ( x − 1)3 + C

10.

A B C + + 2 x x x −1

2 x + 3 = Ax( x − 1) + B( x − 1) + Cx 2 If x = 0, then 3 = –B, or B = –3. If x = 1, then 5 = C. If x = –1, then 1 = 2A – 2B + C, 1 = 2A + 6 + 5, or A = –5. 5 3 5 + Answer: − − x x2 x − 1

7x + 6

7x + 6 A B = + + +3 x ( x 3) x x x + 3x 7x + 6 = A(x + 3) + Bx If x = −3, then −15 = −3B, or B = 5. If x = 0, then 6 = 3A, or A = 2. 7x + 6 5 ⎞ ⎛2 ∫ x2 + 3x dx = ∫ ⎜⎝ x + x + 3 ⎟⎠ dx = 2 ln x + 5ln x + 3 + C 2

=

= ln x 2 ( x + 3)5 + C

622


11.

x + 10

x + 10 A B = + ( x + 1)( x − 2) x + 1 x −2 x −x−2 x + 10 = A(x – 2) + B(x + 1) If x = 2, then 12 = 3B, or B = 4. If x = –1, then 9 = –3A, or A = –3. x + 10 4 ⎞ ⎛ −3 ∫ x2 − x − 2 dx = ∫ ⎜⎝ x + 1 + x − 2 ⎟⎠ dx 2

=

= −3ln x + 1 + 4 ln x − 2 + C = ln

12.

Section 15.2

2x −1 2

=

( x − 2) 4 ( x + 1)3

+C

2x −1 A B = + ( x − 4)( x + 3) x − 4 x + 3

x − x − 12 2x − 1 = A(x + 3) + B(x − 4) If x = −3, then −7 = −7B, or B = 1. If x = 4, then 7 = 7A, or A = 1. 2x −1 1 ⎞ ⎛ 1 ∫ x2 − x − 12 dx = ∫ ⎜⎝ x − 4 + x + 3 ⎟⎠ dx = ln x − 4 + ln x + 3 + C = ln ( x − 4)( x + 3) + C

13.

3 x3 − 3 x + 4

1 3 x3 − 3 x + 4 ⋅ 4 x2 − 1 4 x2 − 4 1⎛ 4 ⎞ = ⎜ 3x + ⎟ 2 4⎝ x −1 ⎠ 4 4 A B = = + 2 x − x + x − x +1 ( 1)( 1) 1 x −1 4 = A(x + 1) + B(x – 1) If x = –1, then 4 = –2B, or B = –2. If x = 1, then 4 = 2A, or A = 2. 3 x3 − 3 x + 4 1 ⎛ 2 −2 ⎞ ∫ 4 x 2 − 4 dx = 4 ∫ ⎜⎝ 3x + x − 1 + x + 1 ⎟⎠ dx =

2 ⎤ ⎛ 1 ⎞ ⎡ 3x = ⎜ ⎟⎢ + 2 ln x − 1 − 2 ln x + 1 ⎥ + C ⎝ 4 ⎠ ⎣⎢ 2 ⎦⎥ 2 2 x −1 ⎤ ⎛ 1 ⎞ ⎡ 3x ⎢ ⎥+C =⎜ ⎟ + ln x +1 ⎥ ⎝ 4 ⎠ ⎢⎣ 2 ⎦

14.

7(4 − x 2 ) 7(2 + x)(2 − x) = ( x − 4)( x − 2)( x + 3) ( x − 4)( x − 2)( x + 3) −7( x + 2) A B = = + ( x − 4)( x + 3) x − 4 x + 3 −7(x + 2) = A(x + 3) + B(x − 4) If x = −3, then 7 = −7B, or B = −1. If x = 4, then −42 = 7A, or A = −6. 7(4 − x 2 ) −1 ⎞ ⎛ −6 ∫ ( x − 4)( x − 2)( x + 3) dx = ∫ ⎜⎝ x − 4 + x + 3 ⎟⎠ dx = −6 ln x − 4 − ln x + 3 + C = − ln ( x − 4)6 ( x + 3) + C

623


15.

3x − 4 3

=

2

3x − 4 A B C = + + x( x + 1)( x − 2) x x + 1 x − 2

x − x − 2x 3x − 4 = A(x + 1)(x − 2) + Bx(x − 2) + Cx(x + 1) If x = 0, then −4 = −2A, or A = 2. 7 If x = −1, then −7 = 3B, or B = − . 3 1 If x = 2, then 2 = 6C, or C = . 3 7 1 ⎞ ⎛ 3x − 4 2 −3 3 ⎜ dx = + + ∫ x3 − x 2 − 2 x ∫ ⎜ x x + 1 x − 2 ⎟⎟ dx ⎝ ⎠ 7 1 = 2 ln x − ln x + 1 + ln x − 2 + C 3 3 = ln

16.


4− x 4

x −x

2

=

x2 3 x − 2 3

( x + 1)7

4− x 2

x ( x + 1)( x − 1)

=

+C

A B C D + + + x x2 x + 1 x − 1

4 − x = Ax( x + 1)( x − 1) + B( x + 1)( x − 1) + Cx 2 ( x − 1) + Dx 2 ( x + 1) 5 3 If x = 0, then 4 = −B, or B = −4. If x = −1, then 5 = −2C, or C = − . If x = 1, then 3 = 2D, or D = . If x = 2, 2 2 then 2 = 6A + 3B + 4C + 12D, 2 = 6A − 12 − 10 + 18, or 2 = 6A − 4, so A = 1. 3 ⎞ ⎛1 4 − 52 4− x 2 ⎜ dx = − + + ∫ x4 − x2 ∫ ⎜ x x2 x + 1 x − 1 ⎟⎟ dx ⎝ ⎠ 4 5 3 = ln x + − ln x + 1 + ln x − 1 + C x 2 2 =

17.

4 1 x 2 ( x − 1)3 + ln +C x 2 ( x + 1)5

2(3x5 + 4 x3 − x)

∫ x6 + 2 x4 − x2 − 2dx = ∫ x6 + 2 x 4 − x 2 − 2 ⎡⎣⎢( 6 x 1

5

)

+ 8 x3 − 2 x dx ⎤ ⎦⎥

⎛ ⎛1⎞ ⎞ ⎜ Form: ∫ ⎜ ⎟ du ⎟ (Partial fractions not required.) ⎝u⎠ ⎠ ⎝

Answer: ln x6 + 2 x 4 − x 2 − 2 + C

18.

x 4 − 2 x3 + 6 x 2 − 11x + 2

= x +1+

7 x 2 − 13x + 2

x3 − 3 x 2 + 2 x x3 − 3 x 2 + 2 x 7 x 2 − 13 x + 2 7 x 2 − 13 x + 2 A B C = = + + 3 2 x − 3 x + 2 x x( x − 1)( x − 2) x x − 1 x − 2 7 x 2 − 13x + 2 = A( x − 1)( x − 2) + Bx( x − 2) + Cx( x − 1)

If x = 0, then 2 = 2A, or A = 1. If x = 1, then −4 = −B, or B = 4. If x = 2, then 4 = 2C, or C = 2.

624


∫

x 4 − 2 x3 + 6 x 2 − 11x + 2 3

2

x − 3x + 2 x

Section 15.2

1 4 2 ⎞ ⎛ = ∫ ⎜ x +1+ + + ⎟ dx x x −1 x − 2 ⎠ ⎝ x2 + x + ln x + 4 ln x − 1 + 2 ln x − 2 + C 2 x2 = + x + ln x( x − 1)4 ( x − 2)2 + C 2 =

19.

2 x2 − 5x − 2

A B C + + x − 1 x − 2 ( x − 2) 2

=

2

( x − 2) ( x − 1)

2 x 2 − 5 x − 2 = A( x − 2) 2 + B ( x − 1)( x − 2) + C ( x − 1) If x = 1, then –5 = A. If x = 2, then –4 = C. If x = 0, then –2 = 4A + 2B – C, –2 = –20 + 2B + 4, or B = 7. ⎡ −5 −4 ⎤ 2x2 − 5x − 2 7 ∫ ( x − 2)2 ( x − 1) dx = ∫ ⎢⎢ x − 1 + x − 2 + ( x − 2)2 ⎥⎥ dx ⎣ ⎦ = −5ln x − 1 + 7 ln x − 2 +

20.

−3x3 + 2 x − 3 2

2

x ( x − 1)

=

4 4 ( x − 2)7 +C = + ln +C x−2 x−2 ( x − 1)5

−3 x 3 + 2 x − 3 2

x ( x + 1)( x − 1)

=

A B C D + + + 2 x x x +1 x −1

3

−3x + 2 x − 3 = Ax( x + 1)( x − 1) + B( x + 1)( x − 1) + Cx 2 ( x − 1) + Dx 2 ( x + 1)

If x = 0, then −3 = −B, or B = 3. If x = −1, then −2 = −2C, or C = 1. If x = 1, then −4 = 2D, or D = −2. If x = 2, then −23 = 6A + 3B + 4C + 12D, −23 = 6A + 9 + 4 − 24, or A = −2. ⎛ −2 3 −3x3 + 2 x − 3 −2 ⎞ 1 ∫ x 2 ( x 2 − 1) dx = ∫ ⎝⎜ x + x 2 + x + 1 + x − 1 ⎠⎟ dx = −2 ln x −

21.

2( x 2 + 8) 3

x + 4x

=

3 3 x +1 + ln x + 1 − 2 ln x − 1 + C = − + ln +C 2 x x x ( x − 1)2 2 x 2 + 16

(

2

x x +4

(

)

)

=

A Bx + C + x x2 + 4

2 x 2 + 16 = A x 2 + 4 + ( Bx + C ) x 2 x 2 + 16 = ( A + B) x 2 + Cx + 4 A Thus A + B = 2, C = 0, 4A = 16. This gives A = 4, B = –2, C = 0.

∫ 22.

⎡ x4 ⎤ ⎛4 −2 x ⎞ 1 1 2 = − = 4 ln x − ln x + 4 + C = ln 4 dx [2 x dx ] dx = ∫ ⎜ + dx ⎢ 2 ⎥+C ⎟ ∫ x ∫ x2 + 4 x3 + 4 x ⎝ x x2 + 4 ⎠ ⎣⎢ x + 4 ⎦⎥

2( x 2 + 8)

4 x3 − 3 x 2 + 2 x − 3 2

( x + 3)( x + 1)( x − 2)

(

=

Ax + B 2

x +3

+

C D + x +1 x − 2

4 x3 − 3 x 2 + 2 x − 3 = ( Ax + B)( x + 1)( x − 2) + C ( x 2 + 3)( x − 2) + D( x 2 + 3)( x + 1)

If x = −1, then −12 = −12C, or C = 1. If x = 2, then 21 = 21D, or D = 1. If x = 0, then −3 = −2B − 6C + 3D, −3 = −2B − 6 + 3, 0 = −2B, or B = 0. 625

)



If x = 1, then 0 = −2(A + B) − 4C + 8D, 0 = −2A − 4 + 8, −4 = −2A, or A = 2. ⎛ 2x 4 x3 − 3 x 2 + 2 x − 3 1 1 ⎞ ∫ ( x 2 + 3)( x + 1)( x − 2) dx = ∫ ⎝⎜ x2 + 3 + x + 1 + x − 2 ⎠⎟ dx = ln( x 2 + 3) + ln x + 1 + ln x − 2 + C = ln ( x 2 + 3)( x + 1)( x − 2) + C

23.

− x3 + 8 x 2 − 9 x + 2

( x + 1) ( x − 3) 2

2

Ax + B

=

2

x +1

+

C D + x − 3 ( x − 3) 2

( = ( Ax + B) ( x − 6 x + 9 ) +C ( x

) (

)

− x3 + 8 x 2 − 9 x + 2 = ( Ax + B )( x − 3)2 + C ( x − 3) x 2 + 1 + D x 2 + 1 2

3

3

) (

)

− 3x 2 + x − 3 + D x 2 + 1 2

= ( A + C ) x + ( B − 6 A − 3C + D) x +(9 A − 6 B + C ) x + (9 B − 3C + D) Thus A + C = –1, B – 6A – 3C + D = 8, 9A – 6B + C = –9, 9B – 3C + D = 2. This gives A = –1, B = 0, C = 0, D = 2. ⎛ −x − x3 + 8 x 2 − 9 x + 2 1 2 0 2 ⎞ 2 ∫ x2 + 1 ( x − 3)2 dx = ∫ ⎜⎜ x2 + 1 + x − 3 + ( x − 3)2 ⎟⎟ dx = − 2 ln x + 1 − x − 3 + C ⎝ ⎠

(

24.

(

)

5x4 + 9 x2 + 3 2

x( x + 1)

=

2

)

A Bx + C Dx + E + + x x 2 + 1 ( x 2 + 1)2

5 x 4 + 9 x 2 + 3 = A( x 2 + 1) 2 + ( Bx + C ) x( x 2 + 1) + ( Dx + E ) x = A( x 4 + 2 x 2 + 1) + ( Bx + C )( x3 + x) + Dx 2 + Ex = ( A + B) x 4 + Cx3 + (2 A + B + D) x 2 + (C + E ) x + A Thus, A + B = 5, C = 0, 2A + B + D = 9, C + E = 0, and A = 3. This gives A = 3, B = 2, C = 0, D = 1, and E = 0. ⎛3 ⎞ x 5x4 + 9 x2 + 3 2x ∫ x( x2 + 1)2 dx = ∫ ⎜⎜ x + x2 + 1 + ( x2 + 1)2 ⎟⎟ dx ⎝ ⎠ 1 2 = 3ln x + ln x + 1 − +C 2( x 2 + 1) 1 = ln x3 ( x 2 + 1) − +C 2 2( x + 1)

25.

14 x3 + 24 x

( x + 1)( x 2

2

+2

(

)

=

Ax + B 2

x +1

)

+

Cx + D x2 + 2

(

)

14 x3 + 24 x = x 2 + 2 ( Ax + B) + x 2 + 1 (Cx + D) = ( A + C ) x3 + ( B + D) x 2 + (2 A + C ) x + (2 B + D)

Thus A + C = 14, B + D = 0, 2A + C = 24, 2B + D = 0. This gives A = 10, B = 0, C = 4, D = 0.

626


∫

(

14 x3 + 24 x

Section 15.2

⎛ 10 x 4x ⎞ dx = ∫ ⎜ + ⎟ dx 2 2 ⎝ x +1 x + 2 ⎠ x +1 x + 2 2

)(

2

)

= 5∫

1 2

x +1

[2 dx] + 2 ∫

1

[2 dx]

2

x +2

( ) ( ) ⎡ ⎤ = ln ⎢( x + 1) ( x + 2 ) ⎥ + C ⎣ ⎦

= 5ln x 2 + 1 + 2 ln x 2 + 2 + C 5

2

26.

2

2

12 x3 + 20 x 2 + 28 x + 4

1 ⎛ Ax + B Cx + D ⎞ = ⎜ + ⎟ 2 3( x + 2 x + 3)( x + 1) 3 ⎝ x + 2 x + 3 x 2 + 1 ⎠ 2

2

12 x3 + 20 x 2 + 28 x + 4 = ( Ax + B )( x 2 + 1) + ( x 2 + 2 x + 3)(Cx + D) = ( A + C ) x3 + ( B + D + 2C ) x 2 + ( A + 2 D + 3C ) x + ( B + 3D ) Thus, A + C = 12, B + D + 2C = 20, A + 2D + 3C = 28, B + 3D = 4. This gives A = 4, B = 4, C = 8, D = 0. 12 x3 + 20 x 2 + 28 x + 4 1 ⎛ 4x + 4 8x ⎞ ∫ 3 x2 + 2 x + 3 x2 + 1 dx = 3 ∫ ⎜⎝ x2 + 2 x + 3 + x 2 + 1 ⎟⎠ dx

(

)(

)

=

(

⎡ = ln ⎢ x 2 + 2 x + 3 ⎢⎣

(

27.

3 x3 + 8 x 2

( x + 2)

2

=

Ax + B 2

x +2

+

)

(

)

1⎡ 2 ln x 2 + 2 x + 3 + 4 ln x 2 + 1 ⎤ + C ⎦⎥ 3 ⎣⎢

)( 2 3

4 3 ⎤ x2 + 1 ⎥ + C ⎥⎦

)

Cx + D ( x 2 + 2) 2

3x3 + 8 x = ( Ax + B)( x 2 + 2) + Cx + D = Ax3 + Bx 2 + (2 A + C ) x + (2 B + D) Thus, A = 3, B = 0, 2A + C = 8, 2B + D = 0. This gives A = 3, B = 0, C = 2, D = 0. ⎛ 3x ⎞ 3 x3 + 8 x 2x 3 1 2 ∫ ( x 2 + 2)2 dx = ∫ ⎜⎜ x2 + 2 + ( x2 + 2)2 ⎟⎟ dx = 2 ln( x + 2) − x2 + 2 + C ⎝ ⎠

28.

3x2 − 8 x + 4

∫ x3 − 4 x2 + 4 x − 6 dx = ∫ x3 − 4 x 2 + 4 x − 6 ⎡⎣⎢( 3x 1

2

)

− 8 x + 4 dx ⎤ ⎦⎥

⎛ ⎛1⎞ ⎞ ⎜ Form: ∫ ⎜ ⎟ du ⎟ (Partial fractions not required.) ⎝u⎠ ⎠ ⎝

Answer: ln x3 − 4 x 2 + 4 x − 6 + C

627


29.

2 − 2x

ISM: Introductory Mathematical Analysis If x = –2, then 30 = B. If x = 0, then –18 = 2A + B, –18 = 2A + 30, or A = –24.

2 − 2x A B = + 2 ( x + 3)( x + 4) x + 3 x +4 x + 7 x + 12 2 – 2x = A(x + 4) + B(x + 3) If x = –4, then 10 = –B, or B = –10. If x = –3, then 8 = A. 1 1⎛ 8 2 − 2x −10 ⎞ ∫0 x2 + 7 x + 12 dx = ∫0 ⎜⎝ x + 3 + x + 4 ⎟⎠ dx =

= ⎡⎣8ln x + 3 − 10 ln x + 4 ⎤⎦

16

∫0

1

30.

x2 + 4 x + 3

= 3+

= 6 − 24 ln 3 − 10 − (−24 ln 2 − 15) = 11 + 24 ln

3x + 4

x2 + 4 x + 3 3x + 4 = 3+ ( x + 1)( x + 3)

200(q + 3) 2

10

3250 ⎤ ⎡ = ⎢120 ln(16) + 80 ln(11) − − [120 ln(6)] 22 ⎥⎦ ⎣ 8 1625 = 120 ln + 80 ln(11) − ≈ $161.80 3 11

) ≥ 0 on [0, 1].

+ 1)

( x + 2)2

0

6( x 2 + 1)

Problems 15.3

6 x2 + 1

1 6( x 2

= 6+

1. Let u = x, a 2 = 9 . Then du = dx. dx x +C ∫ (9 − x2 )3 / 2 = 9 9 − x2

dx

−24 x − 18

(by long division) ( x + 2) 2 A B = + 2 x + 2 ( x + 2)2 ( x + 2) –24x – 18 = A(x + 2) + B 2

200(q + 3) A B = + (q + 6)(q + 1) q + 6 q + 1

325 ⎤ ⎡ = ⎢120 ln q + 6 − 80 ln q + 1 − q 22 ⎥⎦ 0 ⎣

1 5 1 5 ⎛ ⎞ = 6 + ln 3 + ln 5 − ⎜ 3 + ln 2 + ln 4 ⎟ 2 2 2 2 ⎝ ⎠ 1 5 1 5 = 3 + ln 3 + ln 5 − ln 2 − ln 4 2 2 2 2

Area = ∫

=

q + 7q + 6 200(q + 3) = A(q + 1) + B(q + 6) If q = –1, then 400 = 5B, or B = 80. If q = –6, then –600 = –5A, or A = 120. 10 ⎡ 120 80 325 ⎤ CS = ∫ ⎢ + − ⎥ dq 0 q + 6 q +1 22 ⎦ ⎣

2

( x + 2)

2 sq units. 3

10 ⎡ 200( q + 3) 325 ⎤ − 32. CS = ∫ ⎢ ⎥ dq 0 q 2 + 7q + 6 22 ⎦⎥ ⎣⎢

1 5 ⎛ ⎞ = ⎜ 3x + ln x + 1 + ln x + 3 ⎟ 2 2 ⎝ ⎠1

2

2 3

The area is 11 + 24 ln

1 If x = −1, then 1 = 2A, or A = . If x = −3, then 2 5 −5 = −2B, or B = . 2 2 2 3 x + 15 x + 13 ∫1 x2 + 4 x + 3 dx 2⎛ 1 1 5 1 ⎞ = ∫ ⎜3+ ⋅ + ⋅ dx 1 ⎝ 2 x + 1 2 x + 3 ⎟⎠

(

( x + 2)2 1⎡ −24 30 ⎤ = ∫ ⎢6 + + ⎥ dx 0 x + 2 ( x + 2) 2 ⎥⎦ ⎢⎣ 1

3x + 4 A B = + ( x + 1)( x + 3) x + 1 x + 3 3x + 4 = A(x + 3) + B(x + 1)

31. Note that

2

30 ⎤ ⎡ = ⎢6 x − 24 ln x + 2 − x + 2 ⎥⎦ 0 ⎣

0

= 8 ln 4 − 10 ln 5 − (8 ln 3 − 10 ln 4) = 18 ln(4) – 10 ln(5) – 8 ln(3) 3x 2 + 15 x + 13

( x + 1) dx

( x + 2) −24 x − 18

628


Section 15.3

2. Let u = 2x, a 2 = 25 . Then du = 2dx. dx 1 (2dx) = ∫ ∫ 3 3 2 2 ⎡ 25 − (2 x) 2 ⎤ 2 25 − 4 x 2 ⎣ ⎦ ⎡ ⎤ 1 (2 x) ⎥+C = ⎢ 2 ⎢ 25 25 − (2 x) 2 ⎥ ⎣ ⎦

(

=

8. Formula 32 with u = x, a 2 = 7 . Then du = dx. dx x ∫ ( x2 + 7)3 / 2 = 2 + C 7 x +7

)

x 25 25 − 4 x 2

9. Formula 12 with u = x, a = 2, b = 3, c = 4, k = 5. Then du = dx. x dx ∫ (2 + 3x)(4 + 5 x)

+C

=

3. Let u = 4x, a 2 = 3 . Then du = 4 dx. dx (4 dx) = 4∫ ∫ 2 2 2 x 16 x + 3 (4 x) (4 x)2 + 3

10. Formula 37 with u = 5x, a = 2. Then du = 5 dx. 1 5x 1 25 x 5x ∫ 2 dx = 5 ∫ 2 (5 dx) = 5 ⋅ ln 2 + C

⎡ (4 x)2 + 3 ⎤ ⎥+C = 4 ⎢− ⎢ 3(4 x) ⎥ ⎣ ⎦

11. Formula 45 with u = x, a = 5, b = 2, c = 3. Then dx 1 3 x − ln 5 x + 2e3 x + C du = dx. ∫ = 3x 15 5 + 2e 12. Formula 14 with u = x, a = 1, b = 1. Then du = dx.

4. Let u = x 2 , a 2 = 9. Then du = 2x dx. 3 dx 3 (2 x dx) ∫ 3 4 = 2∫ 2 2 2 2 x x −9 (x ) (x ) − 9

=

∫x

2 2 ⎤ 3 ⎡⎢ − ( x ) − 9 − + C⎥ ⎥ 2⎢ 9 x2 ⎣ ⎦

x4 − 9 6 x2

+C

3

105

+C

du = 11 dx .

∫

dx x 5 − 11x 2

=−

7. Formula 28 with u = x, a = 3. Then du = dx.

∫

1 + x dx =

)

14. Formula 20 with u = 11x , a = 5 . Then

6. Formula 8 with u = x, a = 2, b = 5. Then du = dx. ⎡ x 2 dx ⎤ 3 x 2 dx ∫ (2 + 5 x)2 = 3 ⎢⎢ ∫ (2 + 5 x)2 ⎥⎥ ⎣ ⎦ ⎡ x ⎤ 4 4 ln 2 + 5 x ⎥ + C = 3⎢ − − 25 125(2 5 ) 125 x + ⎣ ⎦

1 = ln 2 x x +9 3

2

(

2 8 − 12 x + 15 x 2 (1 + x) 2

13. Formula 9 with u = x, a = 5, b = 2. Then du = dx. ⎡ ⎤ dx 7 dx ∫ x(5 + 2 x)2 = 7 ⎢⎢ ∫ x(5 + 2 x)2 ⎥⎥ ⎣ ⎦ ⎡ 1 1 x ⎤ = 7⎢ + ln ⎥+C ⎣ 5(5 + 2 x) 25 5 + 2 x ⎦

5. Formula 5 with u = x, a = 6, b = 7. Then du = dx. dx 1 x ∫ x(6 + 7 x) = 6 ln 6 + 7 x + C

dx

)

(

16 x 2 + 3 =− +C 3x

=

1 ⎡4 2 ⎤ ln 4 + 5 x − ln 2 + 3x ⎥ + C 2 ⎢⎣ 5 3 ⎦

x2 + 9 − 3 +C x

629

1 5

ln

=∫

11dx

(

11x

) ( 5) −(

5 + 5 − 11x 2 11x

2

+C

11x

)

2



15. Formula 3 with u = x, a = 2, b = 1. Then du = dx. 1 1 x dx ⎛4⎞ ∫0 2 + x = ( x − 2 ln 2 + x ) 0 = 1 − 2 ln 3 + 2 ln 2 = 1 – ln 9 + ln 4 = 1 + ln ⎜⎝ 9 ⎟⎠ 16. Formula 4 with u = x, a = 3, b = 7. Then du = dx. ⎛ x2 3x ⎞ 2 x 2 dx x 2 dx 9 = = 2 2 ∫ 3 + 7 x ∫ 3 + 7 x ⎜⎜ 14 − 49 + 343 ln 3 + 7 x ⎟⎟ + C ⎝ ⎠ 17. Formula 23 with u = x, a 2 = 3 . Then du = dx. 1⎛ ⎞ 2 2 2 ∫ x − 3 dx = 2 ⎜⎝ x x − 3 − 3ln x + x − 3 ⎟⎠ + C 18. Formula 11 with u = x, a = 1, b = 5, c = 3, k = 2. Then du = dx. dx 1 1 + 5x ∫ (1 + 5 x)(2 x + 3) = 13 ln 2 x + 3 + C 19. Formula 38 with u = x, a = 12. Then du = dx. 1/12

∫0

12 x

xe

e12 x dx = (12 x − 1) 144

1/12

= 0

1 1 [e(0) − 1(−1)] = 144 144

20. Formula 46 with u = 3x, a = 2, b = 5. Then du = 3 dx. 1 2 + 3x 1 2 + 3x ∫ 5 + 3x dx = 3 ∫ 5 + 3x (3 dx) = 3 ⎡⎣ (2 + 3x)(5 + 3x) − 3ln

(

)

2 + 3x + 5 + 3x ⎤ + C ⎦

21. Formula 39 with u = x, n = 2, a = 1. Then du = dx.

∫x

2 x

e dx = x 2 e x − 2 ∫ xe x dx

Applying Formula 38 on

∫x

∫ xe

x

dx with u = x, a = 1 (so du = dx) gives

(

∫ xe

x

dx = e x ( x − 1) + C1 . Thus

)

e dx = x 2 e x − 2 ⎡ e x ( x − 1) ⎤ + C = e x ⎡ x 2 − 2( x − 1) ⎤ + C = e x x 2 − 2 x + 2 + C ⎣ ⎦ ⎣ ⎦

2 x

22. Formula 6 with u = x, a = 1, b = 1. Then du = dx. 2

∫1

4 dx x 2 (1 + x )

= 4∫

2

1

2

⎛ 1 1+ x ⎞ 3⎞ 3 ⎛ 1 = 4 ⎜ − + ln ⎟ = 4 ⎜ − + ln ⎟ − 4(−1 + ln 2) = 2 + 4 ln 2 2 2 4 x x ⎝ ⎠ x (1 + x) ⎝ ⎠1 dx

23. Formula 26 with u = 5 x, a 2 = 1. Then du = 5 dx.

∫

5x2 + 1 2x2

5

5x2 + 1

(

)

5 dx ∫ 5x2 2 5 ⎞ 5 ⎛⎜ 5x2 + 1 = − + ln 5 x + 5 x 2 + 1 ⎟ + C ⎟ 2 ⎜ 5x ⎝ ⎠

dx =

630


Section 15.3

24. Formula 17 with u = x, a = 2, b = –1. Then du = dx.

∫x

dx 2− x

=

1 2

2− x − 2

ln

2− x + 2

+C

25. Formula 7 with u = x, a = 1, b = 3. Then du = dx. x dx 1⎛ 1 ⎞ ∫ (1 + 3x)2 = 9 ⎜⎝ ln 1 + 3x + 1 + 3x ⎟⎠ + C 26. Formula 47 with u = 3x, a = 5, b = 6. Then du = 3 dx. 3 dx 11 ∫ (5 + 3x)(6 + 3x) = ln 2 + 3x + (5 + 3x)(6 + 3x) + C 27. Formula 34 with u = 5 x, a = 7 . Then du = 5dx dx

∫ 7 − 5x2

=

1 5

∫

1

( 7 ) − ( 5x) 2

2(

)

5dx =

1 ⎛ 1 ⎜ ln 5 ⎜⎝ 2 7

7 + 5x ⎞ ⎟+C 7 − 5 x ⎟⎠

28. Formula 24 with u = 3 x, a 2 = 6 . Then du = 3dx .

∫ 7x

2

3x 2 − 6 dx =

=

7

( 3)

3

∫(

3x

) ( 3x ) 2

2

−6

(

3 dx

)

⎤ 7 ⎡ 3x 36 (6 x 2 − 6) 3 x 2 − 6 − ln 3 x + 3 x 2 − 6 ⎥ + C ⎢ 8 3 3⎣ 8 ⎦

29. Formula 42 with u = 3x, n = 5. Then du = 3 dx. 36 5 5 5 ∫ 36 x ln(3x)dx = 36∫ x ln(3x)dx = 36 ∫ (3x) ln(3x)(3 dx) 4 ⎡ (3x)6 ln(3 x) (3 x)6 ⎤ 6 = ⎢ − ⎥ + C = x [6 ln(3x) − 1] + C 81 ⎣⎢ 6 36 ⎦⎥ 30. Formula 10 with u = x, a = 3, b = 2. Then du = dx. ⎡ ⎤ ⎡ 5 dx dx 3 + 4x 4 3 + 2x ∫ x2 (3 + 2 x)2 = 5 ⎢⎢ ∫ x2 (3 + 2 x)2 ⎥⎥ = 5 ⎢⎣− 9 x(3 + 2 x) + 27 ln x ⎣ ⎦

⎤ ⎥+C ⎦

31. Formula 13 with u = x, a = 1, b = 3. Then du = dx. 3 ⎡ 2 2(9 x − 2)(1 + 3 x ) ∫ 270 x 1 + 3xdx = 270∫ x 1 + 3x dx = 270 ⎢⎢ 15 ⋅ 9 ⎣

⎤ ⎥+C ⎥ ⎦

3

= 4(9 x − 2)(1 + 3 x) 2 + C

32. Formula 42 with u = x, n = 2. Then du = dx. ⎛ x3 ln x x3 ⎞ 2 2 9 x ln x dx = 9 x ln x dx = 9 − ⎟ + C = 3x3 (ln x) − x3 + C ⎜ ∫ ∫ ⎜ 3 9 ⎟⎠ ⎝ 631



33. Formula 27 with u = 2x, a 2 = 13 . Then du = 2 dx. dx 1 1 = ∫ (2 dx) ∫ 2 4 x − 13 2 (2 x)2 − 13 =

38. Formula 2 with u = x3 , a = 1, b = 2. Then du = 3 x 2 dx. 1 3x2

1 3 ∫0 1 + 2 x3 2 ln 1 + 2 x 0 1 1 = ln 3 − ln 1 = ln 3 2 2

1 ln 2 x + 4 x 2 − 13 + C 2

34. Formula 44 with u = 2x. Then du = 2 dx. dx (2 dx) ∫ x ln(2 x) = ∫ (2 x) ln(2 x)

x dx

=

∫

(

1 2 x

dx

16 − 9 x 2 +C 8x

( 2 ) − ( 3x )

∫ 6x

x π + 7e

4 x

2 + 2 − 3x2 3x

)

π + 7e

dx = 3 ⋅

u

du.

2 x5 / 2 e 5∫

6 x5 / 2 e +C 5

2 x 2 + 1dx = 2

(

3dx

)

2

3 2

(

)

= 2x2 + 1

+C

⎡ 5 3/ 2 ⎤ ⎢ 2 x dx ⎥ ⎣ ⎦

⎛ 1 ⎞ dx ⎟ ⎜ ⎝2 x ⎠

3 2

(

)

3 2x2 + 1 2∫

1 2

(4 x dx)

3 2

+C

+C

5 x3 − x 1 −1 ⎞ ⎛5 dx = ∫ ⎜ x 2 − x 2 ⎟ dx 2x 2 ⎝2 ⎠ 5 3 = x − x +C 6

42.

∫

43.

∫ x2 − 5 x + 6 dx = ∫ ( x − 3)( x − 2) dx

dx

4 x

5/ 2

3 ( 2 x + 1) = ⋅

2

3x

1

xe x

∫e

41. Can be put in the form ∫ u n du .

2

= 2∫

)

=

37. Formula 45 with u = x , a = π, b = 7, c = 4 . Then du =

(

1 ln x 2 + 1 + C 2

∫ 3x

du = 3dx .

= 2 − 3 x 2 − 2 ln

1

40. Can be put in the form

36. Formula 22 with u = 3 x, a = 2 . Then

2 − 3x 2 dx = ∫ x

1

1

∫ u du .

∫ x2 + 1 = 2 ∫ x2 + 1 (2 x dx)

35. Formula 21 with u = 3x, a 2 = 16. Then du = 3 dx. 2 dx (3 dx) = 2(3) ∫ ∫ 2 x 16 − 9 x 2 (3 x)2 16 − (3 x)2 ⎛ 16 − 9 x 2 ⎞ ⎟+C = 6⎜ − ⎜ 16(3x) ⎟ ⎝ ⎠

∫

=


= ln ln(2 x) + C

=−

1

dx

1

1

Formula 11 with u = x, a = −3, b = 1, c = −2, and k = 1. Then du = dx. 1 1 ∫ x2 − 5 x + 6 dx = ∫ ( x − 3)( x − 2) dx x−3 = ln +C x−2

⎡ 1 ⎛ ⎞⎤ = 2 ⎢ ⎜ 4 x − ln π + 7e4 x ⎟ ⎥ + C π 4 ⎝ ⎠ ⎣ ⎦ 1 ⎛ ⎞ 4 x = ⎜ 4 x − ln π + 7e ⎟+C 2π ⎝ ⎠

632


Section 15.3

50. Formula 41 with u = x 2 . Then du = 2x dx.

44. Can be put in the form ∫ u n du . e2 x

(

3

∫1 3x ln x

)

− 12 1 dx = ∫ e2 x + 3 (2e2 x dx) 2x 2 e +3

∫

2

dx e

45. Formula 42 with u = x and n = 3. Then du = dx. x4 ⎡ 1⎤ 3 ∫ x ln x dx = 4 ⎢⎣ln( x) − 4 ⎥⎦ + C

3 e 3 ln( x 2 )[2 x dx] = [ x 2 ln( x 2 ) − x 2 ] ∫ 1 2 2 1 3 2 2 2 = [(e ln(e ) − e ) − (1 ⋅ ln1 − 1)] 2 3 = (e2 + 1) 2

46. Formula 38 with u = x and a = −1. Then du = dx.

51. Formula 15 with u = x, a = 4 and b = –1. Then du = dx.

=

= e2 x + 3 + C

3

∫0

3

xe− x dx = e− x (− x − 1) = e−3 (−4) − 1(−1)

∫1

0

= 1 − 4e

−3

=

2

47. Formula 38 with u = x and a = 3. Then du = 2x dx. 2

x dx

2

=

4− x

2(− x − 8) 4 − x 3

(

2 9 3 − 10 2 3

⎡ e3 x 2 ⎤ = 2⎢ (3 x 2 − 1) ⎥ + C ⎢ 9 ⎥ ⎣ ⎦ 2 2 3x 2 = e (3x − 1) + C 9

3

= 35 ⋅

2

3 + 2 xdx = 35

(

2 72 − 72 x + 60 x

2

2 2

∫1 x

∫2

) (3 + 2 x)

(

1

2

∫0

840

2 x dx

1

8 − x2

=−

49. Formula 43 and then Formula 41. For Formula 43, let u = x, n = 0, and m = 2. Then du = dx.

(8 − x )

(

Now we apply Formula 41 to the last integral with u = x (so du = dx).

0

= −2

(

633

− 12

(−2 x dx)

1

1 2

(

)

0 1 1 2

= −2 0

7 −2 2

=2 2 2− 7

x dx = x(ln x)2 − 2 x(ln x) + 2 x + C

)

1 2

= −2 8 − x 2

2 2 ∫ ln x dx = x ln x − 2∫ ln x dx

(

= −∫ 8 − x2

2

= 98 7 − 25 5

2

2

)

53. Can be put in the form ∫ u n du .

1

∫ ln

3

2 = [23(11)3 / 2 − 14(8)3 / 2 ] 135 2 = 253 11 − 224 2 135

3 + 2 xdx 3 2

) 2(9 x − 4)(2 + 3 x)3 / 2 135

x 2 + 3 x dx =

48. Formula 14 with u = x, a = 3 and b = 2. Then du = dx. 2

1

52. Formula 13 with u = x, a = 2, and b = 3. Then du = dx.

2

3 3x 2 3x ∫ 4 x e dx = 2∫ x e [2 x dx]

∫1 35 x

2

)

)

(

7− 8

)


ISM: Introductory Mathematical Analysis 58. Formula 6 with u = q, a = 1 and b = –1. Then du = dq. 1 0.1 dq n=− 0.4 ∫0.3 q 2 (1 − q )

54. Formula 39 with u = x, n = 2, a = 3. Then du = dx. x 2 e3 x 2 2 3x 3x ∫ x e dx = 3 − 3 ∫ xe dx For

3x

∫ xe

0.1

dx, use Formula 38 with u = x and

a = 3. Then du = dx. ⎤ x 2 e3 x 2 ⎡ e3 x − ⎢ (3x − 1) ⎥ 3 3 ⎢⎣ 9 ⎦⎥ 3x e [9 x 2 − 6 x + 2] = 27

2 3x ∫ x e dx =

=−

1 ⎡ 1 1− q ⎤ ⎢ − − ln ⎥ 0.4 ⎣⎢ q q ⎥⎦ 0.3

=−

1 ⎧ 7 ⎤⎫ ⎡ 10 ⎨[−10 − ln 9] − ⎢ − − ln ⎥ ⎬ 0.4 ⎩ 3 3 ⎦⎭ ⎣

=−

1 ⎛ 20 7⎞ − − ln 9 + ln ⎟ ≈ 20 0.4 ⎜⎝ 3 3⎠

ln 2

ln 2 2 3 x x e 0

∫

⎛ e3 x ⎞ [9 x 2 − 6 x + 2] ⎟ dx = ⎜ ⎜ 27 ⎟ ⎝ ⎠0 8 1 [9(ln 2) 2 − 6 ln 2 + 2] − [2] = 27 27 2 2 [36(ln 2) − 24 ln 2 + 7] = 27

59. a.

9

∫0 1000e

∫1

2

2

1

b. For 10

∫0

qn 0

= ln

du can

dt

10

∫0

500te−0.06t dt use Formula 38 with

500te−0.06t dt 10

1

qn

= ln

10

⎡ e−0.06t ⎤ (−0.06t − 1) ⎥ = 500 ⎢ ⎣⎢ 0.0036 ⎦⎥ 0 500 [e−0.6 (−1.6) − (−1)] = 0.0036 ≈ $16,930.75

∫ k dx .

q0

−0.04t

0

2

dq q = ln 1− q q (1 − q)

u

= 500 ∫ te−0.06t dt

60.

57. Formula 5 with u = q, a = 1, and b = –1. Then du = dq.

∫q

∫e

t = u and a = −0.06, so du = dt.

dx = ∫ 1 dx = x = 2 − 1 = 1 1

dt , the form

1000 9 −0.04t e (−0.04 dt ) −0.04 ∫0 9 1000 −0.04t e =− 0.04 0 1000 −0.36 (e =− − 1) 0.04 ≈ $7558.09

( )

2

−0.04t

=

1 1 = 2 ln(4) − 1 − ln(2) + 2 4 1 3 = 2 ln 22 − ln(2) − 2 4 1 3 = 4 ln(2) − ln(2) − 2 4 7 3 = (ln 2) − 2 4


9

∫0 1000e

be applied.

55. Integration by parts or Formula 42. For Formula 42, let u = 2x, n = 1. Then du = 2 dx. 2 1 2 ∫1 x ln(2 x)dx = 4 ∫1 (2 x) ln(2 x)[2 dx] 1 ⎡ (2 x) 2 ln(2 x) (2 x)2 ⎤ = ⎢ − ⎥ 4 ⎢⎣ 2 4 ⎥⎦

For

qn q0 − ln 1 − qn 1 − q0

T

∫0 ke =−

qn (1 − q0 )

q0 (1 − qn ) 634

− rt

ke

−ke− rt ⎛ 1⎞ T dt = k ⎜ − ⎟ ∫ e− rt (− r dt ) = r ⎝ r⎠ 0

− rT

r

⎛ 1 − e− rT k + = k⎜ ⎜ r r ⎝

⎞ ⎟ ⎟ ⎠

T

0


10

∫0

61. a.

Section 15.4

10

400e0.06(10−t ) dt = 400 ∫ e0.6−0.06t dt

3.

0

10 0.6 −0.06t e e dt 0 0.6 10 −0.06t

= 400∫

∫0

= 400e

e

=

dt

⎛ 1 ⎞ 10 −0.06t = 400e0.6 ⎜ (−0.06 dt ) ⎟∫ e ⎝ −0.06 ⎠ 0 =

400e0.6 −0.06t e −0.06

≈ $5481

10

= 0

4.

5

=

0.04(5−t )

5.

f =

5

dt = 40∫ te0.2 e−0.04t dt

=

0

⎡ e−0.04t ⎤ (−0.04t − 1) ⎥ ⎢ ⎢⎣ 0.0016 ⎥⎦ 0

6.

0.2

40e ⎡ −0.2 e (−0.2 − 1) − 1(−1) ⎤ ≈ $535 ⎦ 0.0016 ⎣

62. Use Formula 38 with u = t and a = −0.07, so du = dt. 5

∫0 50, 000te

−0.07t

3 2 1 1 x3 x dx = ⋅ ∫ 3 − (−1) −1 4 3

7 = 3

0 5

f =

)

3 1 2t 5 dt ∫ 3 − (−3) −3

1 t6 ⋅ 6 3

3 −3

1 4 2 t t + 9 dt 4 − 0 ∫0 ⎛ 1 ⎞⎛ 1 ⎞ 4 = ⎜ ⎟ ⎜ ⎟ ∫ t 2 + 9[2t dt ] ⎝ 4 ⎠⎝ 2 ⎠ 0 f =

)

3 2

4

⎤ ⎥ 49 ⎥ = 6 ⎥ ⎥⎦ 0

7.

1 9 f = 6 xdx = 9 − 1 ∫1

8.

f =

9

1 ⎛ 32 ⎞ ⎜ 4 x ⎟ = 13 8⎝ ⎠1 3

3

= −1

1 3 5 1 5 1⎛ 5 ⎞ = ⎜ − + 5⎟ dx = ⋅ − ∫ 2 1 3 −1 x 2 x1 2⎝ 3 ⎠ 5 = 3

1⎛ −1 ⎞ 9− ⎟ 4 ⎜⎝ 3 ⎠

9. P = =

2

2.

= −1 −1

(

(

dt = 50, 000∫ te−0.07t dt

Problems 15.4 f =

2

1 3 2 x + x + 1 dx 3 − 1 ∫1

⎡ 2 1 ⎢2 t +9 = ⎢ 8⎢ 3 ⎢⎣

5

⎡ e−0.07t ⎤ = 50, 000 ⎢ (−0.07t − 1) ⎥ ⎢⎣ 0.0049 ⎥⎦ 0 50, 000 −0.35 [e (−1.35) − 1(−1)] = 0.0049 = $496, 640

1.

)

1 = [36 − (−3)6 ] 18 =0

5

=

)

⎞ 1 ⎛ x3 x 2 22 = ⎜ + + x⎟ = ⎜ ⎟ 2⎝ 3 2 3 ⎠1

∫

= 40e

(

1 2 x − x3 3

f =

5 40e0.2 te−0.04t dt 0 0.2

(

2 1 2 − 3 x 2 dx 2 − (−1) ∫−1

3

400e0.6 ⎡ −0.6 ⎤ −1 e ⎦ −0.06 ⎣

b. Use Formula 38 with u = t and a = –0.04, so du = dt.

∫0 40te

f =

⎛ 3x 2 ⎞ 1 2 7 (3x − 1)dx = ⎜ − x⎟ = ∫ ⎜ 2 ⎟ 2 −1 1 ⎝ ⎠1 2

(

)

100 1 369q − 2.1q 2 − 400 dq ∫ 100 − 0 0

(

1 184.5q 2 − 0.7 q3 − 400q 100

)

100 0

1 = (1,845, 000 − 700, 000 − 40, 000) − 0 100 = 11,050 Answer: $11,050 635


10. c =

(


)

500 1 4000 + 10q + 0.1q 2 dq 500 − 100 ∫100

1 ⎛ 0.1q3 ⎞ = ⎜ 4000q + 5q 2 + ⎟ 400 ⎜⎝ 3 ⎟⎠

14.


≈ 17,333.33

1. Separating variables, we have dI = −0.0085I dx dI = −0.0085dx I 1 ∫ I dI = − ∫ 0.0085 dx ln I = −0.0085 x + C1 To solve for I, we convert to exponential Formula

100

1 2 3000e0.05t dt 2 − 0 ∫0 3000 1 2 0.05t = ⋅ e [0.05 dt ] 2 0.05 ∫0 = 30, 000e0.05t

2 0

(

)

T

R (1 + α t ) 2 dt F1

= 30, 000 e0.1 − 1 ≈ 3155.13

Answer: $3155.13 1 T R 1 12. C = dt = ∫ 0 T −0 F (t ) T R 1 ⋅ TF1 α =

R α TF1

T

∫0

∫0

R (1 + α t ) [α dt ] = α TF1 2

I = e−0.0085 x +C1 = Ce−0.0085 x . Since I = I 0

when x = 0, I 0 = Ce0 = C , so I ( x) = I 0 e−0.0085 x .

T

⎡ (1 + α t )3 ⎤ ⎢ ⎥ 3 ⎣⎢ ⎦⎥ 0

Problems 15.5

⎡ (1 + α T ) 1⎤ − ⎥ ⎢ 3 3 ⎥⎦ ⎢⎣ 3

1. y′ = 2 xy 2 dy = 2 xy 2 dx dy = 2 x dx y2

R ⎡ = 1 + 3α T + 3α 2T 2 + α 3T 3 − 1⎤ ⎦ 3α TF1 ⎣ =

=

R 1 ⎛ ⎞ (3α T ) ⎜1 + α T + α 2T 2 ⎟ 3α TF1 3 ⎝ ⎠

(

2 2

R 1 + α T + 13 α T

∫y

)

−

F1

r ( q0 )

r ( q0 ) q0

−2

dy = ∫ 2 x dx

1 = x2 + C y

y=−

1 dr 13. Average value = dq . ∫ q0 − 0 dq 1 ⎡ r ( q0 ) − r (0) ⎤⎦ = q0 ⎣

But r(0) = 0, so avg. value =

1 1 1 dx ≈ 0.32 1 − 0 ∫0 x 2 − 4 x + 5

500

Answer: $17,333.33 11.

f =

1 2

x +C

2. y ′ = x 2 y 2 dy = x2 y 2 dx dy = x 2 dx y2

. Since

= [price per unit when q0 units are sold] ⋅q0 , we have ⎡ price per unit ⎤ ⎢ when q0 units ⎥ ⋅ q0 ⎢ ⎥ are sold ⎦ avg. value = ⎣ q0 = price per unit when ⋅q0 units are sold. 636

ISM: Introductory Mathematical Analysis dy

∫ y2 = ∫ x

2

Section 15.5

6. y ′ = e x y 3 dy = e x y3 dx dy = e x dx 3 y dy x ∫ y3 = ∫ e dx 1 − = ex + C 2 2y 1 2 y =− x 2(e + C )

dx

1 x3 = + C1 y 3 1 1 − = ( x3 + 3C1 ) y 3 1 1 = − ( x3 + C ) 3 y 3 y=− 3 x +C −

3.

dy − 3x x 2 + 1 = 0 dx

(

)

dy = 3x x 2 + 1

∫ dy = 3∫ x ( x

1 2

dx

dy dx dy y

) dx 3 ∫ dy = 2 ∫ ( x + 1) [2 x dx] 3 ( x + 1) y= ⋅ +C 2

+1

1 2

1 2

2

∫

3 2

2

2

3 2

(

)

y = x2 + 1

3 2

dy dx = y ∫ x

ln y = ln x + C1 ln y = ln x + ln C, where C > 0. ln y = ln(Cx) ⇒ y = Cx , where C > 0.

+C

8.

dy x = 4. dx y y dy = x dx ∫ y dy = ∫ x dx

dy + xe x = 0 dx dy = − xe x dx x

∫ dy = ∫ − xe dx y = ∫ − xe x dx

y 2 x2 = + C1 2 2

Using integration by parts or formula 38 gives y = (1 − x)e x + C

y 2 = x 2 + 2C1 y 2 = x2 + C

5.

y , where x, y > 0. x y = x dx = x

7. y′ =

9. y ′ =

dy = y , where y > 0. dx dy = dx y

1

where y(1) = 1. y2 dy 1 = dx y 2

y 2 dy = dx

∫y

dy ∫ y = ∫ dx ln y = x + C1 y = e x +C1 = eC1 e x = Ce x , where C = eC1 . Thus

2

dy = ∫ dx

y3 = x+C 3

Given y(1) = 1, we obtain

y = Ce x , where C > 0. 637

13 = 1 + C , so 3



3 13. (3x 2 + 2)3 y ′ − xy 2 = 0, where y (0) = . 2 dy (3x 2 + 2)3 = xy 2 dx dy x = 2 2 y (3x + 2)3 dy x ∫ y 2 = ∫ (3x2 + 2)3 dx 1 2 −2 −3 ∫ y dy = 6 ∫ (3x + 2) [6 x dx] 1 1 − =− +C 2 y 12(3 x + 2)2

2 2⎞ ⎛ C = − . Thus y 3 = 3 ⎜ x − ⎟ = 3 x − 2, 3 3⎠ ⎝ y = 3 3x − 2.

10. y′ = e x − y , where y(0) = 0 dy e x = dx e y e y dy = e x dx

∫e

y

dy = ∫ e x dx

e y = ex + C

Since y(0) = 0, we have e0 = e0 + C , 1 = 1 + C,

3 we have 2 2 1 1 1 − =− + C , − = − + C , so 3 2 3 48 2(2) 2

Given that y (0) =

C = 0. Thus e y = e x , so y = x. 11. e y y′ − x 2 = 0 , where y = 0 when x = 0. ey

dy = x2 dx

31 . Thus, 48 1 1 31 − =− − 2 2 y 48 12(3 x + 2)

C=−

e y dy = x 2 dx

∫e

y

dy = ∫ x 2 dx

x3 ey = +C 3

=−

Given that y(0) = 0, we have e0 = 0 + C , so 1 = C ⇒ ey =

3

1 y2

3

x x +3 +1 , ey = , so 3 3

.

48(3 x 2 + 2)2 4 + 31(3x 2 + 2) 2

.

14. y ′ + x3 y = 0 and y = e when x = 0. dy = − x3 y dx dy = − x3 dx y dy 3 ∫ y = − ∫ x dx

= 0, where y(1) = 2

dy 1 =− dx y2 dx y 2 dy = − x2 dx 2 ∫ y dy = − ∫ x2 y3 1 = +C 3 x x2

Now, y(1) = 2 implies C =

48(3 x 2 + 2)2

Hence, y =

x3 + 3 y = ln . 3

12. x 2 y′ +

4 + 31(3 x 2 + 2)2

x4 +C 4 Given y(0) = e, ln e = 0 + C, so C = 1. 4 x4 − x +1 Thus ln y = − + 1, so y = e 4 . 4 ln y = −

5 . Thus 3

y3 1 5 3 3 = + , y 3 = + 5, y = 3 + 5. 3 x 3 x x

638


15.

Section 15.5

2 dy 3x 1 + y = , where y > 0 and y (1) = 8 . dx y

y dy 1 + y2

17. 2

)

(

∫ xe

∫(

2

+2

2

+9

) dydx = 3xy

y2 + 9

)

(

)

2 2 y +9 3

1 2

3x2 + 2

3 2

x3 + 2 x + 1

[2 y dy ] = ∫

, where y(0) = 0.

2

e− x = 2( y 2 + 1) −1/ 2 − 1.

dx

1

(

19. (q + 1) 2

)

⎡ 3x 2 + 2 dx ⎤ 3 ⎥⎦ x + 2 x + 1 ⎣⎢

1

dc = cq dq q

∫ c dc = ∫ (q + 1)2 dq

3

= ln x + 2 x + 1 + C

Using partial fractions or Formula 7 for q ∫ (q + 1)2 dq , we obtain 1 ln c = ln(q + 1) + + C . Now, fixed cost is q +1 given to be e, which means that c = e when q = 0. This implies 1 = 0 + 1 + C, so C = 0. Thus ln( q +1) + q1+1 1 ln c = ln(q + 1) + , ⇒c=e q +1

2 Now y(0) = 0 implies that (27) = ln(1) + C , so 3 C = 18. Thus 3 2 2 2 y + 9 = ln x3 + 2 x + 1 + 18 . 3

(

dx = ∫ y ( y 2 + 1)−3 / 2 dy

1 − x2 1 e [−2 x dx] = ∫ ( y 2 + 1) −3 / 2 [2 y dy ] ∫ 2 2 1 − x 2 1 ( y 2 + 1)−1/ 2 − e = +C 2 2 −1/ 2 1 − x2 e = ( y 2 + 1)−1/ 2 + C 2 1 1 1 Now y(0) = 0 gives = + C , so C = − . 2 2 1 1 − x2 1 Thus e = ( y 2 + 1) −1/ 2 − or 2 2

2

16. 2 y x3 + 2 x + 1

− x2

−

⎡ 3x2 3 ⎤ Since y > 0, y = ⎢ + ⎥ −1 . 2 ⎥⎦ ⎣⎢ 2

y + 9 dy = ∫

)

2

⎡ 3x 2 3 ⎤ y2 = ⎢ + ⎥ −1 2 ⎥⎦ ⎣⎢ 2

∫ 2y

(

xe− x dx = y ( y 2 + 1)−3 / 2 dy

2

2

)

18. x( y 2 + 1)3 / 2 dx = e x y dy, where y(0) = 0.

2

(

(

)

2

)

⎡ 3x2 3 ⎤ 1 + y2 = ⎢ + ⎥ 2 ⎥⎦ ⎣⎢ 2

)

(

)

−1 1 2 2 1 y [2 y dy ] = 3∫ x dx + 2∫ 1 3x 2 2 1 + y2 = +C 2 1 3 y (1) = 8 ⇒ (1 + 8) 2 = + C 2 3 C= 2 Thus 1 3x 2 3 2 1 + y2 = + 2 2

(

(

1

−2 1 x x2 + 3 dx 2 − 12 1 1 y 2 ∫ e dy = 2 ⋅ 2 ∫ x + 3 [2 x dx] 1 1 2 2 ey = x +3 +C 2 1 Now, y (1) = 0 ⇒ e0 = (2) + C , so C = 0. Thus 2 1 1 2 ⎛1 2 ⎞ 2 ey = x + 3 ⇒ y = ln ⎜ x +3⎟. 2 ⎝2 ⎠

e y dy =

= 3x dx

(

dy xe− y = , where y(1) = 0. dx x2 + 3

)

1

1

c = eln( q +1) e q +1 , or c = (q + 1)e q +1 . 639


20.

ISM: Introductory Mathematical Analysis 23. Let N be the population at time t, where t = 0 corresponds to 1985. Since N follows

dy xe x = xe x − y = dx ey

exponential growth, N = N 0 ekt . Now, N = 40,000 when t = 0, so N 0 = 40, 000 .

y x ∫ e dy = ∫ xe dx

Using integration by parts or formula 38 gives

Therefore N = 40, 000ekt . Since N = 48,000

e y = e x ( x − 1) + C . Now, f (1) = 0 ⇒ 1 = e(0) + C , 1 = C, so

when t = 10, we have 48, 000 = 40, 000e10k , ln1.2 1.2 = e10k , ln 1.2 = 10k, k = 10 Thus

e y = e x ( x − 1) + 1 , y = ln ⎡ e x ( x − 1) + 1⎤ . Thus ⎣ ⎦

(

)

f (2) = ln e2 + 1 .

21. 1

N = 40, 000e

dy = −0.025 y dt

N = 40, 000e

(*)

( ) t 10

0.18

(First form) t 10

From (*), we have N = 40, 000 ⎡ eln1.2 ⎤ , so ⎣ ⎦

ln y = −0.025t + C

Given that y = 1000 when t = 0, we have ln 1000 = −0 + C = C. Thus ln y = −0.025t + ln1000. To find t when money is 95% new, we note that y would be 5%(1000) = 50. Solving ln 50 = −0.025t + ln 1000 gives ln1000 − ln 50 ≈ 120 weeks. t= 0.025

t

N = 40, 000(1.2) 10 (Second form) At year 2005, t = 20 and so 20

N = 40, 000(1.2) 10 = 40, 000(1.2) 2 = 57, 600 .

24. Exponential growth applies, so N = N 0 ekt . When t = 0, then N = 50,000, So N 0 = 50, 000 . Thus N = 50, 000ekt . When t = 50, then N = 100,000, or 100, 000 = 50, 000e50 k or ln 2 k= . Thus 50

dr −r = (50 − 4q)e 5 dq r

∫ e 5 dr = ∫ (50 − 4q)dq r 5

( 10t )

N = 40, 000e0.018t

∫ y dy = −0.025∫ dt

22.

ln(1.2)

N = 50, 000e

2

5e = 50q − 2q + C Since r = 0 when q = 0, we have 5(1) = C, C = 5.

t ln 2 50

( )t

N = 50, 000e

r

N = 50, 000e0.0138t

5e 5 = 50q − 2q 2 + 5 r 2 e 5 = 10q − q 2 + 1 5 2 r = ln 10q − q 2 + 1 5 5 2 r = 5ln 10q − q 2 + 1 5 1 5 2 Since r = pq, p = r = ln 10q − q 2 + 1 . q q 5

(*)

0.69 50

(First form)

From (*), N = 50, 000 ⎡eln 2 ⎤ ⎣ ⎦ t

N = 50, 000(2) 50 . When t = 100, then

t 50

, so

(Second form)

100

N = 50, 000(2) 50 = 50, 000(2) 2 = 200, 000

25. Let N be the population (in billions) at time t, where t is the number of years past 1930. N follows exponential growth, so N = N 0 ekt . When t = 0, then N = 2, so N 0 = 2 . Thus N = 2ekt . Since N = 3 when t = 30, then

640


Section 15.5 28. N = N 0 e−λt After 100 s, 70% remains.

3 = 2e30k 3 = e30k 2 3 30k = ln 2 3 ln k= 2 30

0.7 N 0 = N 0 e−100λ

t

0.7 = e−100λ −100λ = ln 0.7 ln 0.7 λ=− 100 λ ≈ 0.0035667 The half-life is ln 2 ln 2 ln 2 = = −100 ≈ 194.3 s. 0.7 ln 0.7 λ − ln100

ln 3

Thus N = 2e 30 2 . In 2015, t = 85 and so 85

N = 2e 30

ln 32

≈ 2e1.14882 billion.

26. Let N = population at time t and N 0 = population at t = 0. Then N = N 0 ekt . When t = 100, then N = 3 N 0 , so

29. Let N be the amount of

C present in the scroll

t years after it was made. Then N = N 0 e−λt , where N 0 is amount of 14 C present when t = 0. We must find t when N = 0.7 N 0 .

ln 3 . 100 Setting N = 2 N 0 and solving for t gives 3N 0 = N 0 e100k or k =

2 N0 = N0e

14

0.7 N 0 = N 0 e−λt

t ln 3 100

0.7 = e−λt −λ t = ln 0.7 ln 0.7 so t = − . By Equation 15 in the text,

t ln 3 100

2=e t ln 3 ln 2 = 100 100 ln 2 t= ≈ 63. ln 3 The population will double in approximately 63 years.

λ

ln 2 , so 5730 ln 0.7 5730 ln 0.7 t=− =− ≈ 2900 years. ln 2 ln 2

λ=

5730

27. Let N be amount of sample that remains after t seconds. Then N = N 0 e−λt , where N 0 is the initial amount present. When t = 100, then N = 0.3 N 0 . Thus

30. N = N 0 e−λt 0.1N 0 = N 0 e−λt 0.1 = e−λt −λ t = ln(0.1)

0.3N 0 = N 0 e−100λ 0.3 = e−100λ −100λ = ln 0.3 ln 0.3 λ=− 100 Thus λ ≈ 0.01204 . The half-life is ln 2 ln 2 ln 2 = = −100 ≈ 57.57 s. ln 0.3 ln 0.3 λ −

t=−

ln 0.1

λ

By Equation 15 in the text, λ = t=−

100

641

ln 0.1 ln 2 5730

=−

ln 2 , so 5730

5730 ln 0.1 ≈ 19, 000 years. ln 2


31.


dN = kN dt

By Equation 15 in the text, 8 =

N = Aekt

Thus t = −

N 0 = Aekt0 A=

N0 ekt0

Thus N =

(e ) = N e

N0

kt

ekt0

0

35. kt − kt0

, or

k t −t N = N 0 e ( 0 ) , where t ≥ t0 .

32. a.

From Equation 15 in the text, 140 = Thus λ =

b.

ln 2

λ

.

ln 2 . 140

N = N 0 e−λt = N 0 e

ln 2 − t140

= N0 e

2 − 365ln 140

33. N = N 0 e−λt When t = 2, then N = 10. Thus 10 = N 0 e−2λ , N 0 = 10e2λ . By Equation 15 in the text,

λ=

ln 2 ln 2 6 2 ln62

= 10e

ln 2 3

≈ 12.6 units.

ln 2

ln 2 . 8

≈ 5.9 days.

dA = 200 − 0.50 A dt dA ∫ 200 − 0.50 A = ∫ dt 1 ln(200 − 0.50 A) = t + C1 − 0.50 ln(200 − 0.50 A) = −0.50t − 0.50C1 = −0.50t + C2 Thus

36.

−λt

34. N = N 0 e We want to find t when ⎛3⎞ N = ⎜ ⎟ N0 ⎝5⎠

dP

dP = k (150, 000 − 2 P) dx

∫ 150, 000 − 2 P = ∫ k dx 1 − ln[150, 000 − 2 P ] = kx + C 2 Since P(0) = 15,000, we have 1 − ln[150, 000 − 30, 000] = C , so 2 1 1 − ln[150, 000 − 2 P] = kx − ln[120, 000]. 2 2

⎛3⎞ −λt ⎜ 5 ⎟ N0 = N0e ⎝ ⎠ 3 = e−λt 5 ⎛3⎞ −λ t = ln ⎜ ⎟ ⎝5⎠ t=−

8ln 53

,λ=

−t ⎞ ⎛ Thus A = 400 ⎜ 1 − e 2 ⎟ . If t = 1, ⎝ ⎠ − 12 ⎞ ⎛ A = 400 ⎜ 1 − e ⎟ ≈ 157 grams per square meter. ⎝ ⎠

λ

Thus N 0 = 10e

ln 2 8

=−

λ

200 − 0.50 A = e−0.50t +C2 = e−0.50t eC2 A 200 − = Ce−0.50t 2 Given that A = 0 when t = 0, we have C = 200, A −t so 200 − = 200e 2 2 A −t 200 − 200e 2 = 2 A − 2t ⎞ ⎛ 200 ⎜ 1 − e ⎟ = ⎝ ⎠ 2

N − 365ln 2 = e 140 ≈ 0.164 N0

6=

ln 53

ln 2

ln 53

λ

642


Section 15.6

Since P(1000) = 70,000, 1 − ln[150, 000 − 140, 000] 2

b.

1 = 1000k − ln[120, 000] 2 1 ln[120, 000] − ln[10, 000] ln12 k= ⋅ = 2 1000 2000 Thus 1 ln12 1 − ln[150, 000 − 2 P ] = x − ln[120, 000] 2 2000 2 ln12 ln[150, 000 − 2 P ] = − x + ln[120, 000] 1000 150, 000 − 2 P = e

ln12 x − 1000

1 + be−ct M = 100,000 Since N = 50,000 at t = 0 (1995), we have 100, 000 100, 000 50, 000 = , so 1 + b = = 2, or 1+ b 50, 000 b = 1. 100, 000 Hence, N = . If t = 5, then N = 60,000, 1 + e−ct so 100, 000 60, 000 = 1 + e−5c 100, 000 5 1 + e−5c = = 60, 000 3 5 2 e−5c = − 1 = 3 3 1/ 5 2 ⎛ ⎞ e−c = ⎜ ⎟ ⎝3⎠ 100, 000 Hence, N = . In 2005, t = 10, so t /5 1 + 23

dV = kV dt 1 ∫ V dV = ∫ k dt lnV = kt + C1 V = ekt eC1

or V = Cekt . Now t = 0 corresponds to July 1, 1996 where V = 0.75 ⋅ 80,000 = 60,000, so 60,000 = C(1). Thus V = 60, 000ekt . Also V = 38,900 for t = 9.5, so

( )

9.5k

t

ln

M

1. N =

P = 75, 000 − 60, 000(12−2 ) ≈ $74,583.

Thus V = 60, 000e 9.5

t ln ( 389 ) 7 = e 9.5 600 30 t ⎛ 7 ⎞ ⎛ 389 ⎞ ln ⎜ ⎟ = ln ⎜ ⎟ 30 9.5 ⎝ ⎠ ⎝ 600 ⎠ 7 9.5ln 30 t= ≈ 31.903 ln 389 600

Problems 15.6

e

38,900 = 60, 000e 389 = e9.5k 600 ⎛ 389 ⎞ 9.5k = ln ⎜ ⎟ ⎝ 600 ⎠ 1 ⎛ 389 ⎞ k= ln ⎜ ⎟ 9.5 ⎝ 600 ⎠

( 389 600 )

This corresponds to about 31 years and 11 months after July 1, 1996 ⇒ June 2028.

1⎛ − ln12 x ⎞ P = ⎜ 150, 000 − 120, 000e 1000 ⎟ 2⎝ ⎠ x ⎛ − 1000 ⎞ = 75, 000 − 60, 000 ⎜ 12 ⎟ ⎝ ⎠ If x = 2000, then

37. a.

ln

( ) ( )

ln12 x − 1000 ln[120,000]

150, 000 − 2 P = 120, 000e

t

14, 000 = 60, 000e 9.5

N=

100, 000 1+

2. N =

( ) 2 3

2

≈ 69, 200.

M

1 + be−ct Since M = 500, and N = 200 when t = 0, we have 500 200 = 1+ b 500 5 3 = ⇒b= . 1+ b = 200 2 2 500 Hence N = . When t = 1 we are given 1 + 32 e−ct

( 389 600 ) .

643



N = 300. Thus 500 300 = 1 + 32 e−c 3 500 5 1 + e−c = = 2 300 3 3 −c 2 e = 2 3 4 −c e = 9 500 Hence N = 1 + 32 94

( )

N=

3. N =

500 1 + 32

( 94 )

2

30, 000 1+ b 30, 000 = 75 1+ b = 400 b = 74 30, 000 Hence N = . If t = 1, then N = 1200. 1 + 74e−ct Thus 30, 000 1200 = 1 + 74e−c 30, 000 1 + 74e−c = = 25 1200 400 =

t

. When t = 2, then

74e−c = 24 24 12 e−c = = 74 37 30, 000 Hence N = 1 + 74 12 37

≈ 386 .

M

( )

− ct

1 + be M = 40,000, and N = 20 when t = 0, so 40, 000 20 = 1+ b 40, 000 1+ b = = 2000 20 b = 1999 40, 000 Hence N = . 1 + 1999e−ct 40, 000 Since N = 100 when t = 1, 100 = , 1 + 1999e−c 40, 000 = 400 1 + 1999e−c = 100 399 e−c = 1999 40, 000 Hence N = . t 399 1 + 1999 1999

(

If t = 2, then N =

40, 000

(

399 1999

)

2

.

M

1 + be−ct M = 100,000, and since N = 500 when t = 0, we have 100, 000 500 = 1+ b 100, 000 1+ b = = 200 500 b = 199 100, 000 . If t = 1, then Hence N = 1 + 199e−ct N = 1000. Thus 100, 000 1000 = 1 + 199e−c 100, 000 = 100 1 + 199e−c = 1000

)

1 + 1999

4. N =

5. N =

t

≈ 500.

199e−c = 99 99 e−c = 199

M

Hence N =

1 + be−ct Since M = 30,000, and N = 400 when t = 0, we have

100, 000 1 + 199

N=

100, 000 1 + 199

644

99 ( 199 )

2

( ) 99 199

t

. If t = 2, then

≈ 1990 .


6. a.

Section 15.6

dN = N (1 − N ) dt dN = dt N (1 − N ) dN ∫ N (1 − N ) = ∫ dt Using Formula 5 in the Table of Integrals, dN N for ∫ , we get ln = t + C. 1− N N (1 − N )

c.

= =

b.

e +1

1+ e 2 − (1 + e −t ) 2(1 + e−t ) 1 − e −t 2(1 + e−t ) et − 1

e −t − 1

1 − et

=

=−

et − 1

= − f (t ) 2(e−t + 1) 2(1 + et ) 2(et + 1) Thus, f(t) is symmetric about the origin.

d. The graph of N(t) is the graph of f(t) shifted 1 unit upward. Thus, since f(t) is 2 symmetric about (0, 0), N(t) is symmetric ⎛ 1⎞ about ⎜ 0, ⎟ . ⎝ 2⎠ 1 N (t ) = f (t ) + 2 1 N (−t ) = f (−t ) + 2 1 = − f (t ) + 2 1⎤ ⎡ = − ⎢ f (t ) + ⎥ + 1 2⎦ ⎣ = 1 − N (t )

N (et + 1) = et =

1 2

denominator by et .

Also, N > 0, and since M = 1, N < 1. Thus ⎛ N ⎞ ln ⎜ ⎟ = t. ⎝ 1− N ⎠ N = et 1− N N = (1 − N )et

t

−

−t

2(et + 1) Replace t by −t then multiply numerator and

1

et

1

=

1 Since N (0) = , ln 2 = ln1 = 0 = C. 2 1 − 12

N=

f (t ) =

1 1 + e −t

dN = N (1 − N ) = N − N 2 dt d2N = 1− 2N dt 2 d2N

1 = 0 when N = . 2 1 1 − 2N > 0 when N < and 1 − 2N < 0 2 1 when N > , so there is an inflection point 2 1 when N = . 2 1 1 = 2 1 + e −t dt 2

e.

1.5

N

t 3

1 + e −t = 2 e −t = 1 t =0

7. a.

⎛ 1⎞ Thus the point ⎜ 0, ⎟ is an inflection point ⎝ 2⎠ on the graph.

N= ≈

b.

645

375 5.2 − 2.3t

1+ e 375

=

375 1+ e

5.2 −2.3t

e

1 + 181.27e−2.3t

lim N =

t →∞

375 = 375 1 + 181.27(0)


8. a.

N= = ≈


0.2524 e

−2.128 x

11.

+ 0.005125

0.2524 0.005125 e−2.128 x + 0.005125 0.005125

49.25 e−2.128 x 0.005125

+1

≈

49.25 1 + 195.1e−2.128 x

b. If x = 0, then N ≈

200, 000 − x = e− kt −C = e−C e− kt = Ae− kt , where A = e−C . Thus x = 200, 000 − Ae− kt . If t = 0, then x = 50,000, so 50, 000 = 200, 000 − A ⇒ A = 150, 000 . Thus

49.25 ≈ 0.2511 cm2. 1 + 195.1

x = 200, 000 − 150, 000e− kt . If t = 1, then x = 100,000, so

dT = k (T − a) where a = −10. 9. dt dT = k dt T + 10 dT ∫ T + 10 = ∫ k dt Thus ln(T + 10) = kt + C. At t = 0, we have T = 28, so ln(28 + 10) = 0 + C, C = ln 38, and ln(T + 10) = kt + ln 38. ln(T + 10) − ln 38 = kt ⎛ T + 10 ⎞ Hence ln ⎜ ⎟ = kt. ⎝ 38 ⎠

100, 000 = 200, 000 − 150, 000e− k 150, 000e− k = 100, 000 100, 000 2 e−k = = 150, 000 3 t

⎛2⎞ Thus x = 200, 000 − 150, 000 ⎜ ⎟ . If t = 3, then ⎝3⎠ ⎛ 8 ⎞ x = 200, 000 − 150, 000 ⎜ ⎟ ≈ $155,555.56. ⎝ 27 ⎠

⎛ 20 + 10 ⎞ If t = 1, then T = 20. Thus ln ⎜ ⎟ = k ⋅1, so ⎝ 38 ⎠ 30 15 k = ln = ln . Hence 38 19 T + 10 ⎛ ⎞ ⎛ 15 ⎞ ln ⎜ ⎟ = ⎜ ln ⎟ t. ⎝ 38 ⎠ ⎝ 19 ⎠

12.

dN = kN 2 dt dN ∫ N 2 = ∫ k dt 1 − = kt + C N

If t = 0, then N = N 0 . Thus −

⎛ 47 ⎞ ⎛ 15 ⎞ If T = 37, then ln ⎜ ⎟ = ⎜ ln ⎟ t ⎝ 38 ⎠ ⎝ 19 ⎠ ln 47 t = 38 ≈ −0.90 hr 15 ln 19

−

1 = C , so N0

1 1 = kt − N N0

1 −kN 0 t + 1 = N N0

which corresponds to 54 minutes. Time of murder: 4:15 A.M. − 54 min = 3:21 A.M. 10.

dx = k (200, 000 − x) dt dx ∫ 200, 000 − x = ∫ k dt − ln(200, 000 − x) = kt + C ln(200,000 – x) = –kt – C

N=

dp = kp( I − p ) dt This is logistic growth, so the maximum rate of I formation (growth) occurs when p = , which 2 is when there are equal amounts of both enzymes.

N0 1 − kN 0 t −

⎛ 1 ⎞ + As t → ⎜ ⎟ , then 1 − kN 0t → 0 , so kN 0 ⎝ ⎠ N →∞.

646


13.

Section 15.7

dN = k (M − N ) dt dN ∫ M − N = ∫ k dt –ln(M – N) = kt + C If t = 0, then N = N 0 , so − ln ( M − N 0 ) = C . Thus we have − ln( M − N ) = kt − ln ( M − N 0 )

2.

1

∞

∫1

(3x − 1)

2

dx = lim

M − N0 = kt M −N M −N ln = −kt M − N0

3.

∫

r →∞ 0

(

4.

1 x

3

(

)

r

0

)

5.

∫

r →∞

= 0+

1 3

− 23

dx

r

2

r

= lim − ∫ e− x [− dx] = lim (−e− x ) 1

r →∞

(

)

r →∞

∫0 ( 5 + e ∞

r →∞

3

⎛ 1 1⎞ 1 1⎛ 1⎞ 1 lim ⎜ − ⎟ = − ⎜ 0 − ⎟ = 2 2 r →∞ ⎝ r 9⎠ 2⎝ 9 ⎠ 18

7.

∞

∫1

−x

) dx = lim ∫ (5 + e ) dx r

(

)

r →∞ 0 r

0

−x

(

)

= lim ⎡ 5r − e− r − (0 − 1) ⎤ ⎦⎥ r →∞ ⎣⎢

1 x

dx = lim

(

∫

r −1 2

r →∞ 1

)

x

1

dx = lim 2 x 2 r →∞

= lim 2 r − 2 = ∞ ⇒ diverges r →∞

647

1

⎛ 1 1⎞ = lim ⎜ − + ⎟ e⎠ r →∞ ⎝ e r

⎛ ⎞ 1 = lim ⎜ 5r − + 1⎟ = ∞ ⇒ diverges r r →∞ ⎝ e ⎠

r

r

1 1 = e e

= lim 5 x − e− x

r

r →∞ 3

1 1 lim 2 r →∞ x 2

∞ −x e dx 1

= lim −e− r + e−1

dx = lim ∫ x −3 dx

3

1

= lim 3 ⎡ 3 r + 2 − 3 (4) ⎤ ⎦ r →∞ ⎣ = ∞ ⇒ diverges

6.

=−

( x + 2)

( x + 2) 3

r →∞

)

r

∫

r

r →∞ 2

3e−0.1t − 3e−0.3t dt

x −2 = lim r →∞ −2 =−

( x + 2) 2

= lim

Problems 15.7 ∞

3

)

(

∫3

∫2

dx = lim

− 3e−0.3t dt

⎡ 30 ⎤ 10 = lim ⎢ − + − −30e0 + 10e0 ⎥ 0.3r r →∞ ⎣ e −0.1r e ⎦ ⎡ 30 ⎤ 10 = lim ⎢ − + − (−20) ⎥ 0.3r r →∞ ⎣ e0.1r e ⎦ = 0 + 0 + 20 = 20 The total amount of the drug that is eliminated is approximately 20 milliliters.

1.

1

∞

1

= lim −30e−0.1t + 10e−0.3t r →∞

r

r →∞


r

r1 1 dx = lim ∫ dx = lim ln x 1 x x r →∞ r →∞

r →∞

N = M − ( M − N 0 ) e− kt

= lim

∞

∫1

= lim ln r = ∞ ⇒ diverges

M − N = ( M − N 0 ) e− kt

−0.1t

(3x − 1) −2 [3 dx]

= lim ( ln r − 0 )

M −N = e− kt M − N0

∫0 ( 3e

r

r

ln

1.

∫

⎡ ⎤ 1 = lim ⎢ − ⎥ r →∞ ⎣ 3(3 x − 1) ⎦ 1 ⎡ 1 1⎤ = lim ⎢ − + ⎥ r − 3(3 1) 6⎦ r →∞ ⎣ 1 = 0+ 6 1 = 6

ln ( M − N 0 ) − ln( M − N ) = kt

∞

1

r →∞ 3 1

r

1


8.

x

∞

∫4

( x + 9)

3

2

(


)

− 32 1 r 2 + x 9 [2 x dx] ∫ r →∞ 2 4

dx = lim

12.

0

r

9.

)

1

−3

−3

lim ∫ ∫−∞ ( x + 1)2 dx = r →−∞ r

1

3

( x + 1) −2 dx

3

7−x

dx = lim − ∫ (7 − x) 1

= lim − 2(7 − x) 2 r →−∞

Thus

11.

∞

∫−∞ 2 xe 0

∫−∞

−x

2

dx = ∫

0

−∞

r

k ⎞ ⎛ k lim ⎜ − + ⎟ =1 r →∞ ⎝ r 800 ⎠ k 0+ =1 800 k = 800

[− dx]

0 r

⎡ 1 = lim ⎢ −1 + 2 r →−∞ ⎢ er ⎣ r

2

⎤ ⎥ = −1 + 0 = − 1 ⎥⎦

r →∞

∞

∫−∞ 2 xe

− x2

ke−2 x dx = 1

lim −

r

ke−2 x lim − 2 r →∞

0

⎡ 1 ⎤ = lim ⎢ − 2 + 1⎥ = 0 + 1 = 1 r →∞ ⎢ e r ⎥⎦ ⎣

Thus

r →∞

k r −2 x ∫ e [−2 dx] = 1 r →∞ 2 1

2

2

∞

∫1

14.

0

= lim − e x

dx = lim 800∫

r

1200

x −2 dx

2 2 ⎛ 800 800 ⎞ = lim ⎜ − + = 0+ = ⎟ r 1200 ⎠ 3 3 r →∞ ⎝

2 xe− x dx = lim − ∫ e− x [−2 x dx] r →∞

x

2

800 x 1200 r →∞

r

2

800

= lim −

2

r →−∞

∞

∞

∫1200

r

2 xe− x dx = lim − ∫ e− x [−2 x dx] = lim − e− x

∫0

b.

2

0

r →−∞

x −2 dx = 1

k =1 r →∞ x 800

∞

0

r

800

lim −

2 xe− x dx + ∫ 2 xe− x dx

2

k

∞

r →∞

)

2

diverges

∫800 x2 dx = 1 lim k ∫

= lim −4 + 2 7 − r = ∞ ⇒ diverges r →−∞

∞

∫−∞ (5 − 3x)dx

13. a.

3 r

(

− 12

r

r →−∞

0

0

−3

∫−∞

(5 − 3 x)dx

3 ⎞ ⎛ = lim ⎜ 5 x − x 2 ⎟ 2 ⎠r r →−∞ ⎝ ⎡ 3 ⎞⎤ ⎛ = lim ⎢ (0 − 0) − ⎜ 5r − r 2 ⎟ ⎥ 2 ⎠⎦ r →−∞ ⎣ ⎝ 3 ⎞ ⎛ = lim ⎜ −5r + r 2 ⎟ = ∞ 2 ⎠ r →−∞ ⎝

1 = lim − r →−∞ x + 1 r 1 ⎤ ⎡1 = lim ⎢ + ⎥ r →−∞ ⎣ 2 r + 1 ⎦ 1 = +0 2 1 = 2

10.

∞

0

lim ∫ (5 − 3x )dx ∫−∞ (5 − 3x)dx = r →−∞ r

⎡ − 12 ⎤ ⎡ 1 1⎤ = lim ⎢ − x 2 + 9 + ⎥ ⎥ = lim ⎢ − r →∞ ⎣⎢ ⎥⎦ 4 r →∞ ⎣⎢ r 2 + 9 5 ⎦⎥ 1 1 = 0+ = 5 5

(

∞

∫−∞ (5 − 3x)dx = ∫−∞ (5 − 3x)dx + ∫0

r

=1 1

k k ⎞ ⎛ + lim ⎜ − ⎟ =1 r 2 r →∞ ⎝ 2e 2e2 ⎠ k =1 0+ 2e 2

dx = −1 + 1 = 0.

Thus k = 2e2 . 648


15.

∞

∫0

Chapter 15 Review

∞

18. V = π0 ∫ eθ t e− ρ t dt = π0 lim

240, 000e−0.06t dt

= lim

r →∞

0

240, 000 r −0.06t e [−0.06 dt ] −0.06 ∫0

1 r →∞ θ − ρ

= π0 lim

r

240, 000 −0.06t e 0.06 r →∞

0

⎞ 240, 000 ⎛ 1 − 1⎟ ⎜ 0.06 ⎝ e0.06 r r →∞ ⎠ 240, 000 =− (−1) = 4, 000, 000 0.06

=−

r

16. α = ∫ e− x dx = lim − ∫ e− x [− dx] xc

= lim − e− x r →∞

19.

⎛ 1 ⎞ = lim ⎜ − + e− xc ⎟ = 0 + e− xc = e− xc r →∞ ⎝ e r ⎠ ∞ 1 −x r −x ⎡ 1 ⎤ β =∫ e 8 dx = lim − ∫ e 8 ⎢ − dx ⎥ xc 8 xc r →∞ ⎣ 8 ⎦ = lim − e

− 8x

r →∞

= 0+e =e 0

0

(since θ − ρ < 0 )

π0

π0 π = 0 . θ − ρ ρ −θ

−

(t + 2)

2

r

40, 000

dt ∫ r →∞ 0 (t + 2) 2

dt = lim

⎡ 40, 000 40, 000 ⎤ = lim ⎢ − + 2 ⎥⎦ r+2 r →∞ ⎣ 40, 000 = 0+ = 20, 000 increase 2

r


( 18 ) xc

1. Use Formula 42 with u = x and n = 1. Then du = dx. x 2 ln x x 2 ln x x dx = − +C ∫ 2 4

( 18 ) xc 1 0 3x ∫ e [3 dx] r →−∞ 3 r ⎡1 1 ⎤ = lim ⎢ − er ⎥ r →−∞ ⎣ 3 3 ⎦

−e3 x dx = lim 0

40, 000

r

xc

−

∞

∫0

40, 000 t+2 0 r →∞

2. Use Formula 27 with u = 2x, a 2 = 1 . Then du = 2 dx. 1 1 1 (2 dx) ∫ 2 dx = 2 ∫ 4x +1 (2 x)2 + 1

1 3x ⋅e r →−∞ 3 r 1 1 = − 0 = sq units 3 3 = lim

π0 ⎡ (θ − ρ ) r ⎤ e −1 ⎦ r →∞ θ − ρ ⎣

= lim

= lim −

⎡ 1 −( 1 ) xc ⎤ = lim ⎢ − r + e 8 ⎥ r →∞ ⎢ ⎥⎦ ⎣ e8

−∞

[(θ − ρ )dt ]

r xc

17. Area = − ∫

dt

θ −ρ

Thus V = −

xc

r →∞

r

π = 0 [0 − 1] θ −ρ

= lim −

∞

r (θ − ρ )t

∫0 e

π0 (θ − ρ )t = lim e r →∞ θ − ρ

= lim −

r (θ − ρ )t

∫ e r →∞ 0

=

649

1 ln 2 x + 4 x 2 + 1 + C 2



3. Use Formula 23 with u = 3x, a 2 = 16. Then du = 3 dx. 2 1 2 2 2 ∫0 9 x + 16 dx = 3 ∫0 (3x) + 16 (3 dx) 1 ⎡1 ⎛ = ⎢ ⎜ (3x ) 9 x 2 + 16 + 16 ln 3x + 9 x 2 + 16 3 ⎣2 ⎝ 8 8 ⎛ ⎞ ⎛ ⎞ = ⎜ 2 13 + ln 6 + 2 13 ⎟ − ⎜ 0 + ln 4 ⎟ 3 3 ⎝ ⎠ ⎝ ⎠ 8 ⎛ 6 + 2 13 ⎞ = 2 13 + ln ⎜⎜ ⎟⎟ 3 ⎝ 4 ⎠ 8 ⎛ 3 + 13 ⎞ = 2 13 + ln ⎜⎜ ⎟ 3 ⎝ 2 ⎟⎠

(

4. By long division,

)

⎛

16 x

2

⎞⎤ ⎟⎥ ⎠⎦ 0

12 ⎞

∫ 3 − 4 x dx = ∫ ⎜⎝ −4 + 3 − 4 x ⎟⎠ dx = −4 x − 3ln 3 − 4 x + C

Or, by Formula 3 with u = x, a = 3, and b = –4. Then du = dx. 16 x x 3 ⎡ x ⎤ ∫ 3 − 4 x dx = 16∫ 3 − 4 x dx = 16 ⎢⎣ −4 − 16 ln 3 − 4 x ⎥⎦ + C = −4 x − 3ln 3 − 4 x + C 5.

15 x − 2

⎛

15 x

⎞

2

∫ (3x + 1)( x − 2) dx = ∫ ⎜⎝ (3x + 1)( x − 2) − (3x + 1)( x − 2) ⎟⎠ dx For

15 x

∫ (3x + 1)( x − 2) dx, use Formula 12 with u = x, a = 1, b = 3, c = −2, and k = 1. Then du = dx. 15 x

x

⎡1 ⎛

1

⎞⎤

∫ (3x + 1)( x − 2) dx = 15∫ (3x + 1)( x − 2) dx = 15 ⎢⎣ −7 ⎜⎝ −2 ln x − 2 − 3 ln 3x + 1 ⎟⎠ ⎥⎦ + C For

2

∫ (3x + 1)( x − 2) dx, use Formula 11 with

u = x, a = 1, b = 3, c = −2, and k = 1. Then du = dx. 2 dx ⎛ 1 3x + 1 ⎞ ∫ (3x + 1)( x − 2) dx = 2∫ (3x + 1)( x − 2) = 2 ⎜⎝ −7 ln x − 2 ⎟⎠ + C Thus,

15 x − 2

∫ (3x + 1)( x − 2) dx 30 5 2 3x + 1 ln x − 2 + ln 3x + 1 + ln +C 7 7 7 x−2 30 5 2 2 = ln x − 2 + ln 3x + 1 + ln 3 x + 1 − ln x − 2 + C 7 7 7 7 = 4 ln x − 2 + ln 3 x + 1 + C =

650


6. The integral can be put in the form

Chapter 15 Review

1

∫ u du

12. Use Formula 15 with u = x, a = 2, and b = 5. Then du = dx. x 2(5 x − 4) 2 + 5 x +C ∫ 2 + 5 x dx = 75

with

u = ln x. e2

e2

e2 1 ⎡ 1 1 ⎤ dx = ∫e x ln x ∫e ln x ⎢⎣ x dx ⎥⎦ = ln ln x e

13. Use Formula 38 with u = x and a = 7. Then du = dx. 7x 7x ∫ 49 xe dx = 49∫ xe du

= ln ln e2 − ln ln e = ln 2 − ln1 = ln(2) – 0 = ln 2

⎡ e7 x ⎤ = 49 ⎢ (7 x − 1) ⎥ + C = e7 x (7 x − 1) + C ⎢⎣ 49 ⎥⎦

7. Use Formula 9 with u = x, a = 2, and b = 1. Then du = dx. dx 1 1 x ∫ x( x + 2)2 = 2( x + 2) + 4 ln x + 2 + C

14. Use Formula 45 with u = x, a = 2, b = 3, and c = 4. Then du = dx. dx 1 4x ∫ 2 + 3e4 x = 8 ⎣⎢⎡4 x − ln 2 + 3e ⎦⎥⎤ + C

(

8. Use Formula 35 with u = x and a = 1. Then du = dx. dx 1 x −1 ∫ x2 − 1 = 2 ln x + 1 + C

15. The integral has the form dx

9. Use Formula 21 with u = 4x and a 2 = 9 . Then du = 4 dx. dx (4 dx) = 4∫ ∫ 2 2 2 (4 x) 9 − (4 x) 2 x 9 − 16 x ⎡ 9 − 16 x 2 = 4 ⎢− 9(4 x) ⎢ ⎣ =−

1 ⎡2

1

∫ u du . ⎤

1

∫ 2 x ln x 2 = 4 ∫ ln x2 ⎢⎣ x dx ⎥⎦ = 4 ln ln x

2

+C

16. Use Formula 5 with u = x, a = 2, and b = 1. Then du = dx. dx x 1 ∫ x(2 + x) = 2 ln 2 + x + C

⎤ ⎥+C ⎥ ⎦

17. Long division or Formula 3. For long division, 2x 3 ⎤ ⎡ ∫ 3 + 2 x dx = ∫ ⎢⎣1 − 3 + 2 x ⎥⎦ dx

9 − 16 x 2 +C 9x

10. Use Formula 42 with u = x 2 and n = 1. Then du = 2x dx. 1 3 2 2 2 ∫ x ln x dx = 2 ∫ ( x ) ln( x )[2 x dx] 1 ⎛ ( x 2 )2 ln x 2 ( x 2 )2 ⎞ = ⎜ − ⎟+C 2 ⎜⎝ 2 22 ⎟⎠ =

1

)

= x − 3⋅

1 1 [2 dx] ∫ 2 3 + 2x

3 = x − ln 3 + 2 x + C . 2 For Formula 3, use u = x, a = 3, and b = 2. Then du = dx. x 2x ∫ 3 + 2 x dx = 2∫ 3 + 2 x dx 3 ⎛x 3 ⎞ = 2 ⎜ − ln 3 + 2 x ⎟ + C = x − ln 3 + 2 x + C 2 ⎝2 4 ⎠

1 4 1 x ln x 2 − x 4 + C 4 8

11. Use Formula 35 with u = x and a = 3. Then du = dx. 9 dx dx ⎡1 x − 3 ⎤ ∫ x2 − 9 = 9∫ x2 − 9 = 9 ⎢⎣ 6 ln x + 3 ⎥⎦ + C 3 x −3 = ln +C 2 x+3

18. Use Formula 30 with u = 2x and a 2 = 9. Then du = 2 dx. dx (2 dx) ∫ 2 2 = 2∫ 2 x 4x − 9 (2 x) (2 x)2 − 9 2 ⎛ − 4 x2 − 9 ⎞ ⎟ + C = 4x − 9 + C = 2⎜ − ⎜ 9(2 x ) ⎟ 9 ⎝ ⎠ 651


ISM: Introductory Mathematical Analysis 22. Integration by parts

19. Partial fractions 5x2 + 2 5x2 + 2 A Bx + C = = + 3 2 x +x x( x + 1) x x 2 + 1

u = x2 dv = e x dx

5 x 2 + 2 = A( x 2 + 1) + ( Bx + C ) x

Then du = 2x dx and v = e x .

= ( A + B) x 2 + Cx + A Thus, A + B = 5, C = 0, A = 2. This gives A = 2, B = 3, C = 0. ⎡2 5x2 + 2 3x ⎤ ∫ x3 + x dx = ∫ ⎢⎣ x + x 2 + 1 ⎥⎦ dx

∫x

(

x 4 + x3 + x 2

x

2

= Ax( x + x + 1) + B ( x + x + 1) + (Cx + D) x

23.

24.

ln( x + 1) x +1

+C

)

4

(

)

(

1 4 1 3 3 x 2 + 2 x dx = x + x2 4 − 2 ∫2 2 2

1 [(64 + 16) − (8 + 4)] = 34 2

f =

∫t

1 1 2 t t e dt 1 − 0 ∫0

2 t

e dt = t 2 et − 2 ∫ tet dt

For ∫ tet dt , use Formula 38 with u = t and a = 1. Then du = dt. 2 t 2 t t ∫ t e dt = t e − 2[e (t − 1)] + C

)

= et (t 2 − 2t + 2) + C Thus,

1 1 f = ∫ t 2 et dt = et (t 2 − 2t + 2) = e(1) − 1(2) 0

dx

0

= e − 2.

1 1 dx and v = 2( x + 1) 2 . x +1

25. y′ = 3x 2 y + 2 xy , y > 0

(

1 2

= 2( x + 1) ln( x + 1) − 2 ∫ ( x + 1)

− 12

)

dy = 3 x 2 + 2 x dx y

dx

1

x

and a = 1. Then du = dt.

21. Integration by parts u = ln(x + 1)

∫

x

For ∫ t 2 et dt , use Formula 39 with u = t, n = 2,

3 = ln x − + ln x 2 + x + 1 + C x

Then du =

f = =

2

= ( A + C ) x3 + ( A + B + D ) x 2 + ( A + B ) x + B Thus, A + C = 3, A + B + D = 5, A + B = 4, B = 3. This gives A = 1, B = 3, C = 2, D = 1. 3 x3 + 5 x 2 + 4 x + 3 ∫ x4 + x3 + x2 dx ⎡1 3 2x +1 ⎤ = ∫⎢ + + ⎥ dx 2 2 x x x + x + 1⎦ ⎣

dv = ( x + 1)

x

= e x ( x 2 − 2 x + 2) + C

3 x3 + 5 x 2 + 4 x + 3

− 12

x

∫ 2 xe dx = 2 xe − ∫ 2e dx = 2 xe − 2e 2 x 2 x x x ∫ x e dx = x e − (2 xe − 2e ) + C

x 2 ( x 2 + x + 1) A B Cx + D = + + x x2 x2 + x + 1

(

dx, use integration by parts again.

Then du = 2 dx and v = e x .

3 x3 + 5 x 2 + 4 x + 3

2

x

dv = e x dx

)

=

∫ 2 xe

u = 2x

20. Partial fractions 3 x3 + 5 x 2 + 4 x + 3

e dx = x 2 e x − ∫ 2 xe x dx

For

3 ln x 2 + 1 + C 2

= 2 ln x +

2 x

dx

∫

1

= 2( x + 1) 2 ln( x + 1) − 4( x + 1) 2 + C

(

)

dy = 3 x 2 + 2 x dx y ∫

3 2 ln y = x3 + x 2 + C1 , from which y = e x + x +C1 ,

= 2 x + 1[ln( x + 1) − 2] + C

3 2 y = Ce x + x , where C > 0.

652


26. y′ − 2 xe x

2

− y +3

Chapter 15 Review

= 0 , y(0) = 3

30.

2 dy = 2 xe x +3e− y dx

y

0

y(0) = 3 implies e3 = e3 + C , C = 0. Thus

∞

∫0

2 e y = e x +3 , or y = x 2 + 3 .

27.

∫1

x 2.5

dx =

1− x 2

= lim −

1 0 1− x 2 ∫ e [−2 x dx] r →−∞ 2 r

dx = lim −

1

2

1 r 1− x 2 ∫ e [−2 x dx] r →∞ 2 0

2

xe1− x dx = lim −

r

Thus

3x1.5 1 2 2⎞ ⎛ = lim ⎜ − + ⎟ r →∞ ⎝ 3r1.5 3 ⎠ 2 = 0+ 3 2 = 3

∞

1− x 2

∫−∞ xe

r →∞

∫−∞

0 2x e r →−∞ r

e2 x dx = lim

∫

N 0 = 100, 000 . Thus N = 100, 000ekt . Since N = 120,000 when t = 15, then 120, 000 = 100, 000e15k 1.2 = e15k

e2 x r →−∞ 2

0

ln 1.2 = 15k, or k =

dx = lim

r

N = 100, 000e

t ln1.2 15

ln1.2 . Thus 15

(

= 100, 000 eln1.2

)

t 15

t 15

= 100, 000(1.2) For the year 2015 we have t = 30 and

r

r 1 1 1 dx = lim ln x ∫1 2 x dx = rlim ∫ r →∞ 2 →∞ 1 2 x 1 1 ⎡ ⎤ = lim ⎢ ln r − 0 ⎥ = ∞ ⇒ diverges r →∞ ⎣ 2 ⎦

∞

1 1 = − e+ e = 0 2 2

31. N = N 0 e kt Since N = 100,000 when t = 0 (1985),

1 ⎡1 1 ⎤ 1 = lim ⎢ − e 2r ⎥ = − 0 = 2 r →−∞ ⎣ 2 2 ⎦ 2

29.

2

0

r

∫

x r →∞ −1.5

0

∞

2

xe1− x dx + ∫ xe1− x dx

2 1 = lim − e1− x r →∞ 2 0 2 1 1 ⎤ 1 1 ⎡ = lim ⎢ − e1− r + e ⎥ = 0 + e = e 2 ⎦ 2 2 r →∞ ⎣ 2

r lim x −2.5 dx 1 r →∞ −1.5 r

= lim

28.

0

−∞

0

2 dy = ∫ 2 xe x +3 dx

1

dx = ∫

2 1 = lim − e1− x r →−∞ 2 r 2⎤ 1 1 1 ⎡ 1 = lim ⎢ − e + e1− r ⎥ = − e − 0 = − e 2 2 2 r →−∞ ⎣ 2 ⎦

2 e y = e x +3 + C

∞

1− x 2

∫−∞ xe

2 e y dy = 2 xe x +3 dx

∫e

∞

∫−∞ xe

30

N = 100, 000(1.2) 15 = 100, 000(1.2) 2 = 144, 000

32. N = N 0 e kt When t = 0, then N = 40,000. Thus N 0 = 40, 000 and N = 40, 000e kt . When t = 10, then N = 80,000, so 80, 000 = 40, 000e10k 2 = e10k

10 k = ln 2, or k =

653

t ln 2 ln 2 . Thus N = 40, 000e 10 . 10



33. N = N 0 e−λt , where N 0 is the original amount present. When t = 100, then N = 0.95 N 0 , so we have

N = 300 ⇒ 300 =

1 + 224e −6c 450 3 = 1 + 224e −6c = 300 2 3 1 224e−6c = − 1 = 2 2 1 e −6c = 448

0.95N 0 = N 0 e −100λ

0.95 = e−100λ −100λ = ln 0.95 ln 0.95 λ=− ≈ 0.0005 (decay constant). After 100

e6c = 448 6c = ln 448 ln 448 c= ≈ 1.02 6 450 Thus N ≈ . 1 + 224e−1.02t

200 years, N = N 0 e −200λ . Thus 0.95 ⎤ −200 ⎡⎣ − ln100 N ⎦ = e 2 ln 0.95 = e −200λ = e N0 ≈ 0.90 = 90%

34.

dq = −kq dt dq = −k dt q dq ∫ q = ∫ −k dt ln q = −kt + C When t = 0, q = q0 , so ln q0 = 0 + C = C. Thus ln q = −kt + ln q0

36. N =

35. N =

7 q , = e −7 ≈ 0.09%. k q0

450 1 + be −ct

If t = 0, then N = 2. Thus 2 =

2000

1 + be −ct When t = 0 (last year), then N = 1000. Thus 2000 1000 = 1+ b 2000 =2 1+ b = 1000 b=1 2000 . When t = 1 then N = 1100. So N = 1 + e −ct Thus 2000 1100 = 1 + e−c 2000 20 1 + e−c = = 1100 11 9 e−c = 11 2000 Hence N = . When t = 2, then t 9 1 + 11

q = e− kt eln q0 = q0 e− kt

When t =

450

450 , 1+ b

450 450 = 225 , b = 224, so N = . 2 1 + 224e −ct If t = 6, then 1+ b =

( )

N=

37.

654

2000

( )

2 9 1 + 11

≈ 1200 .

dT = k (T − 25) dt dT = k dt T − 25 dT ∫ T − 25 = ∫ k dt ln(T – 25) = kt + C


Chapter 15 Review

If t = 0, then T = 35. Thus ln 10 = C, so ⎛ T − 25 ⎞ ln(T − 25) = kt + ln 10, or ln ⎜ ⎟ = kt . If ⎝ 10 ⎠

∫t

⎛ 9 ⎞ t = 1, then T = 34 and ln ⎜ ⎟ = k . Thus ⎝ 10 ⎠

=

t2 1

2=

10te

41. a.

∞

⎡ e −0.06t ⎤ (−0.06t − 1) ⎥ dt = 10 ⎢ ⎣⎢ 0.0036 ⎦⎥ 0

f ( x) dx

= lim

∫

r

r →∞ 0

(0.007e −0.01x + 0.00005e−0.0002 x )dx

= lim (−0.7e−0.01x − 0.25e−0.0002 x ) r →∞

t1

∫−∞

A0 ekt dt = lim

∫

t1

r →−∞ r

= lim A0 ⋅ r →−∞

A0 e kt r →−∞ k

r 0

= lim r

r →−∞

⎡ ek ( t2 −t1 ) − 1⎤ . ⎢⎣ ⎥⎦

e kt2 kt1

(1)

k t −t = e ( 2 1 ) . Substituting into (1) gives

Total revenue = r (12) − r (0) = ∫

12

0

TR ≈

A0 ekt dt

t1

)

−1

dr dq . dq

3 (138) = 207 2 Simpson’s f (0) = 25 4 f (3) = 88 2 f (6) = 36 4 f (9) = 52 f (12) = 7 208

1 t1 kt e (k dt ) k ∫r

= lim

)

TR ≈

0.25 ⎡ 0.7 ⎤ = lim ⎢ − − − (−0.7 − 0.25) ⎥ −0.0002 r r →∞ ⎣ e0.01r e ⎦ = 0 − 0 + 0.7 + 0.25 = 0.95

40.

(

dr dq n = 4, a = 0, b = 12 b − a 12 − 0 = =3 h= n 4 Trapezoidal f (0) = 25 2 f (3) = 44 2 f (6) = 36 2 f (9) = 26 f (12) = 7 138

10 [e −0.72 (−0.72 − 1) − (−1)] 0.0036 ≈ $452

∫0

e

t1

A0 kt2 e − ekt1 k

f (q) =

=

39.

kt2 − kt1

=

e A0 kt1 A e [2 − 1] = 0 ekt1 . k k

12

∫0

t2

If A0 e kt2 = 2 A0 ekt1 , then e kt2 = 2ekt1 ,

38. Use Formula 38 with u = t, and a = –0.06, so du = dt. −0.06t

(e

A0 kt1 e k A = 0 e kt1 k

⎛ T − 25 ⎞ ln ⎜ ⎟ = (ln 0.9)t . If ⎝ 10 ⎠ T = 37, 12 ln = (ln 0.9)t 10 ln 1.2 = (ln 0.9)t, ln1.2 t= ≈ −1.73 ln 0.9 Note that 1.73 hr corresponds approximately to 1 hr 44 min. Thus 6:00 P.M. – 1 hr 44 min = 4:16 P.M.

12

A e kt A0 e dt = 0 k kt

3 (208) = 208 3

b. Total variable cost c(12) − c(0) = ∫

12

0

(

A0 kt1 e − ekr k

)

dc dq dq

dc dq a = 0, b = 12 Using as few data values as possible, we choose n = 1 for Trapezoidal and n = 2 for Simpson’s (n must be even). Trapezoidal (n = 1) f (q) =

A = 0 e kt1 since ekr → 0 as r → −∞ for k k > 0.

655



b − a 12 − 0 = = 12 n 1 f (0) = 15

3 VC ≈ (142) = 142 3 Using these results and those of part (a), we have: Trapezoidal P(12) ≈ 207 – 141 – 25 = 41 Simpson’s P(12) ≈ 208 – 142 – 25 = 41

h=

f (12) = 7 22 12 (22) = 132 2 Simpson’s (n = 2) b − a 12 − 0 h= = =6 n 2 f (0) = 15 VC ≈

Mathematical Snapshot Chapter 15 1. C = 2000, w0 = 200 C 2000 = ≈ 114 17.5 17.5 C ⎛ C ⎞ −0.005t w(t ) = + ⎜ w0 − ⎟e 17.5 ⎝ 17.5 ⎠ 2000 ⎛ 2000 ⎞ −0.005t = + ⎜ 200 − ⎟e 17.5 ⎝ 17.5 ⎠ Letting w(t) = 175 and solving for t gives 2000 ⎛ 2000 ⎞ −0.005t 175 = + ⎜ 200 − ⎟e 17.5 ⎝ 17.5 ⎠ 2000 ⎛ 2000 ⎞ −0.005t 175 − = ⎜ 200 − ⎟e 17.5 ⎝ 17.5 ⎠ weq =

4 f (6) = 48 f (12) = 7 70 6 (70) = 140 3 To each of our results we must add on the fixed cost of 25 to obtain total cost. Thus for trapezoidal we get TC ≈ 132 + 25 = 157, and for Simpson’s we have TC ≈ 140 + 25 = 165. VC ≈

c.

We use the relation 12 ⎡ dr dc ⎤ P (12) = ∫ ⎢ − ⎥ dq − 25 . First we 0 dq dq ⎣ ⎦ determine variable cost for each rule with b − a 12 − 0 = = 3. n = 4 and h = n 4 Trapezoidal f (0) = 15 2 f (3) = 28 2 f (6) = 24 2 f (9) = 20 f (12) = 7 94

175 − 2000 17.5 200 − 2000 17.5

= e−0.005t

⎡ 175 − 2000 ⎤ 17.5 ⎥ −0.005t = ln ⎢ ⎢⎣ 200 − 2000 ⎥ 17.5 ⎦ ⎡ 175− 2000 17.5 ⎤ ln ⎢ 2000 ⎥ − 200 17.5 ⎦ t= ⎣ ≈ 69 −0.005 Thus weq = 114 and t = 69 days.

2.

3 VC ≈ (94) = 141 2 Simpson’s f (0) = 15 4 f (3) = 56 2 f (6) = 24 4 f (12) = 40 f (12) = 7 142

656

dw 1 = (C − 17.5w) dt 3500 dw 1 = dt C − 17.5w 3500 dw 1 ∫ C − 17.5w = ∫ 3500 dt 1 1 − ln C − 17.5w = t + C1 17.5 3500


ln C − 17.5w = −


17.5 t − 17.5C1 = −0.005t + C2 3500

C − 17.5w = e−0.005t +C2 = eC2 e−0.005t = C3e−0.005t

Thus C − 17.5w = C4 e−0.005t , where C4 is a constant and C4 = ±C3 . When t = 0, then w = w0 , so C − 17.5w0 = C4 . Thus C − 17.5w = ( C − 17.5w0 ) e−0.005t −17.5w = −C + ( C − 17.5w0 ) e−0.005t C ⎛ C 17.5 ⎞ −0.005t w0 ⎟ e +⎜− + 17.5 ⎝ 17.5 17.5 ⎠ C ⎛ C ⎞ −0.005t w= + ⎜ w0 – ⎟e 17.5 ⎝ 17.5 ⎠ which is Equation 2. w=

C ⎛ C ⎞ −0.005t + ⎜ w0 − ⎟e 17.5 ⎝ 17.5 ⎠ C = weq , we have w(t ) = weq + w0 − weq e−0.005t . Simplifying the equation Since 17.5 1 1 w(t + d ) = w(t ) − ⎡⎣ w(t ) − weq ⎤⎦ gives w(t + d ) = ⎡⎣ w(t ) + weq ⎤⎦ . Thus 2 2 1 weq + w0 − weq e−0.005(t + d ) = ⎡ weq + w0 − weq e−0.005t + weq ⎤ , or ⎦ 2⎣ 1 weq + w0 − weq e−0.005(t + d ) = weq + w0 − weq e−0.005t 2 Solving for d gives 1 e−0.005(t + d ) = e−0.005t 2 1 e−0.005t e−0.005d = e−0.005t 2 1 −0.005d e = 2 1 −0.005d = ln = − ln 2 2 ln 2 d= 0.005 as was to be shown.

3. w(t ) =

(

)

(

)

4. BMI =

w 2

(

(

)

(

)

)

, so w = BMI ⋅ h 2 with w in kilograms and h in meters. 5 feet, 8 inches equals 68 inches, or

h 1.7272 meters. The upper BMI limit then corresponds to a weight of 24.9(1.7272)2 ≈ 74.28 kilograms, or about 163 pounds. So the woman would need to lose 27 pounds. On a 2200 calorie-per-day diet, ≈ 125.71 lb and the weight function is weq = 2200 17.5

w(t) = 125.71 + (190 – 125.71)e–0.005t = 125.71 + 64.29e–0.005t. The solution of the equation 163 = 125.71 + 64.29e−0.005t is t ≈ 109. It would take about 109 days. 5. Answers may vary.

657

Chapter 16 Principles in Practice 16.1 b.

1. The uniform density function is given by ⎪⎧ 1 , if 0 ≤ x ≤ 60 f ( x) = ⎨ 60 otherwise. ⎪⎩0, The probability of waiting between 25 and 45 minutes is P (25 ≤ X ≤ 45) = ∫

45

25

=

45

c.

1 x dx = 60 60 25

d.

1

6

1

3 3/ 2

c1

1

∫1 6 ( x + 1)dx = 2

( x + 1) 12

c

= 1

1 2

(c + 1) 1 1 − = 12 3 2 (c + 1)2 − 4 = 6 (c + 1) 2 = 10

c + 1 = ± 10 c = −1 ± 10 We choose c = −1 + 10 since 1 < c < 3.

2. a.

P (3000 < X < 4000) = ∫

4000 1000

3000

x2

dx

4000

=−

b.

1000 1 ⎛ 1⎞ 1 = − −⎜− ⎟ = 4 ⎝ 3 ⎠ 12 x 3000

P ( X > 2000) = ∫ r

∫ r →∞ 2000

2 2

∞

2000

= lim ( x + 1)dx =

2.5

2

Problems 16.1 21

6

( x + 1) 2 12

3 1 3⎞ ( x + 1)2 ⎛ P⎜ X ≥ ⎟ = ∫ ( x + 1)dx = 2 ⎠ 3/ 2 6 12 ⎝

( x + 1) 2 12

3. The exponential density function is given by ⎪⎧0.2e−0.2 x , for x ≥ 0 f ( x) = ⎨ otherwise. ⎪⎩0, 1 1 = 5. The mean is given by µ = = k 0.2 The standard deviation is given by 1 1 σ= = = 5. k 0.2

=

1

( x + 1)dx =

16 25 13 = − = = 0.8125 12 48 16

45 − 25 20 1 = = . 60 60 3

P (1 < X < 2) = ∫

2.5 1

49 4 11 = − = = 0.6875 48 12 16

2. The exponential density function is given ⎧ 1 − 10x ⎪ e , if x ≥ 0 by f ( x) = ⎨10 ⎪⎩0, if x < 0. The probability that the break pads will break down after the warranty period is P(5 < X) = 1 – P(0 ≤ X ≤ 5) ⎛ −x 5⎞ 5 1 −x = 1− ∫ e 10 dx = 1 − ⎜ −e 10 ⎟ 0 10 ⎜ ⎟ 0⎠ ⎝ −1 −1 ⎛ −5 ⎞ = 1 − ⎜ −e 10 + e0 ⎟ = 1 + e 2 − 1 = e 2 ≈ 0.607 ⎝ ⎠

1. a.

P ( X < 2.5) = ∫

1000

x2 r

1000

x2

dx

dx

1000 ⎛ 1000 1 ⎞ 1 = lim ⎜ − + ⎟= x 2000 r →∞ ⎝ r 2⎠ 2 r →∞

= lim − 1

9 4 5 − = 12 12 12

658


3. a.

Section 16.1

⎧⎪ 1 , if 1 ≤ x ≤ 4 f ( x) = ⎨ 3 ⎪⎩0, otherwise

j.

1

x 1

b.

4

7 3 7⎞ 2 − 2 2 ⎛3 P⎜ < X < ⎟ = = 2 ⎠ 4 −1 3 ⎝2 1

2

0

1

P (0 < X < 1) = ∫ 0 dx = 0

d.

P ( X ≤ 3.5) = P (1 ≤ X ≤ 3.5)

4. a. 4−3 1 = 4 −1 3

e.

P ( X > 3) = P (3 < X ≤ 4) =

f.

P(X = 2) = 0

g.

4 −1 P ( X < 5) = P (1 ≤ X ≤ 4) = =1 4 −1

h.

i.

µ=∫

1

x2 ⎛1⎞ x ⎜ ⎟ dx = 6 ⎝3⎠

4 1

4

x

6

f(x)

1 5

x 5

16 1 5 = − = 6 6 2

σ2 = ∫

4

t x −1 . = 31 3

⎧⎪ 1 , if 0 ≤ x ≤ 5 f ( x) = ⎨ 5 ⎪⎩0, otherwise 1

4 2 ⎛1⎞ x ⎜ ⎟ dx − µ 2 1 ⎝ 3⎠

x3 = 9

3

x

dt =

F(x)

2

3.5 − 1 2.5 5 = = 4 −1 3 6

4

1

3

c.

=

x1

if x < 1 ⎧0, ⎪⎪ x − 1 , if 1 ≤ x ≤ 4 Thus F ( x) = ⎨ ⎪ 3 if x > 4. ⎪⎩1, 2 −1 1 = P ( X < 2) = F (2) = 3 3 2 2 P (1 < X < 3) = F (3) − F (1) = − 0 = 3 3

f(x)

1 3

If 1 ≤ x ≤ 4, F ( x) = ∫

b.

3 −1 2 = 5−0 5

c.

5 − 4.5 0.5 1 = = 5−0 5 10

d. P(X = 4) = 0

2

⎛5⎞ −⎜ ⎟ ⎝2⎠ 1 ⎡ 64 1 ⎤ 25 = ⎢ − ⎥− ⎣ 9 9⎦ 4 25 = 7− 4 3 = 4 3 3 = . Thus, σ = 4 2

e.

P ( X > 2) = P (2 < X ≤ 5) =

5−2 3 = 5−0 5

f.

P ( X < 5) = P (0 ≤ X ≤ 5) =

5−0 =1 5−0

g.

P ( X > 5) = ∫

h.

659

∞

5

µ=∫

5 0

0dx = 0

x2 ⎛1⎞ x ⎜ ⎟ dx = 10 ⎝5⎠

5

= 0

25 5 −0 = 10 2

Chapter 16: Continuous Random Variables

i.

2

σ =∫

5 2⎛1⎞ x ⎜ ⎟ dx − µ 2 0 ⎝5⎠

5

x3 ⎛5⎞ = −⎜ ⎟ 15 ⎝2⎠ 0


2

c.

b

x3 ⎛ a+b⎞ = −⎜ ⎟ 3(b − a ) ⎝ 2 ⎠ a

⎡125 ⎤ 25 25 25 25 =⎢ − 0⎥ − = − = 3 4 12 ⎣ 15 ⎦ 4

Thus σ =

j.

25 5 5 3 = = . 12 6 12

If 0 ≤ x ≤ 5, F ( x) = ∫

x1

0

5

x

dt =

t x = 50 5

=

1 2 1

3.5 1 − 5 5

=

b 2 + ab + a 2 a 2 + 2ab + b 2 − 3 4

b

∫a

k dx = 1

b

kx a = 1 k (b − a ) = 1

F(x)

1 b−a Thus X is uniformly distributed. k=

b. If a ≤ x ≤ b F ( x) = ∫

⎧⎪ 1 , if a ≤ x ≤ b f ( x) = ⎨ b − a otherwise ⎪⎩0,

µ=∫ =

b3 − a 3 ( a + b ) 2 − 3(b − a ) 4

6. a.

x

b.

=

b 2 − 2ab + a 2 (b − a )2 = 12 12 b−a Thus σ = . 12

5

5. a.

2

=

⎧0, if x < 0 ⎪ Thus F ( x) = ⎨ 5x , if 0 ≤ x ≤ 5 ⎪ ⎩ 1, if x > 5 P (1 < X < 3.5) = F (3.5) − F (1) =

⎛ 1 ⎞

b

2 σ 2 = ∫ x2 ⎜ ⎟ dx − µ a − b a ⎝ ⎠

2

b a

x a

x

1 1 x−a dt = t = b−a b−a a b−a if x < a

⎧0, ⎪ Thus F ( x) = ⎨ bx−−aa ⎪ ⎩1

b

x2 ⎛ 1 ⎞ x⎜ ⎟ dx = 2(b − a ) ⎝b−a⎠ a

7.

2

b −a a+b = 2(b − a ) 2

if a ≤ x ≤ b if x > b

⎪⎧3e−3 x , if x ≥ 0 f ( x) = ⎨ if x < 0 ⎪⎩0,

a.

4

P (1 < X < 4) = ∫ 3e−3 x dx 1

= −e −3 x

4 1

= −e −12 + e−3 ≈ 0.04978

b.

4

P ( X < 4) = ∫ 3e−3 x dx 0

= −e −3 x −12

4 0

= −e +1 ≈ 0.99999 660


Section 16.1

P ( X > 6) = 1 − P ( X ≤ 6)

c.

6

= 1 − ∫ 3e −3 x dx 0

= −e

= 1 − (−e−18 + 1) = e −18 ≈ 0.00000 1 1 = . k 3 ⎛ 1 ⎞ P ( µ − 2σ < X < µ + 2σ ) = P ⎜ − < X < 1⎟ ⎝ 3 ⎠

= ∫ 3e

−3 x

0

= −e −3 x

−2.5

e.

c

∫ 0 0.5e

∞

∞

dx

−0.5 x

0

c 0

=

e−0.5c =

1 2 1 2 1 2 1 2 1 2

1 2 −0.5c = − ln 2 c = 2 ln 2 −0.5c = ln

−3 x

dx

r

−3 x

= lim (−e

−3 x

= lim (−e

−3r

r →∞ r →∞

dx

r

9. a.

) 0

F ( x) = P ( X ≤ x) = ∫

x

−∞

4

∫0

kx dx = 1 kx 2 2

+ 1)

= 0 +1 = 1

f.

dx =

−e −0.5c + 1 =

∫ 3e r →∞ 0

= lim

0

+ 1 ≈ 0.918

−e−0.5 x

1

∫−∞ f ( x) dx = ∫ 0 3e

0

P (0 < X < c) =

= −e −3 + 1 ≈ 0.95021

e.

5

d. P(X = 4) = 0

d. From the text, µ = σ =

1

5

P ( X < 5) = ∫ 0.5e−0.5 x dx = −e−0.5 x

4

=1 0

8k = 1 1 k= 8

f (t ) dt

If x ≥ 0, x

x

0

0

F ( x) = ∫ 3e−3t dt = −e−3t

0 if ⎧ Thus F ( x) = ⎨ −3 x + 1 if ⎩ −e

8.

= −e−3 x + 1.

b.

x<0

P (2 < X < 3) = ∫

3x 2

8

dx =

x2 16

3

= 2

9 4 − 16 16

5 = 16

x≥0

⎪⎧0.5e−0.5 x , if x ≥ 0 f ( x) = ⎨ if x < 0 ⎪⎩0,

c.

4 x x2 P ( X > 2.5) = ∫ dx = 2.5 8 16 4

2.5

a.

P ( X > 4) = 1 − P ( X ≤ 4) = 1 − ∫ 0.5e−0.5 x dx 0

(

= 1 − −e −0.5 x

b.

25 39 = 1− = ≈ 0.609 64 64

4

) 0 = 1 − ( −e−2 + 1) = e−2 ≈ 0.135 4

P (0.5 < X < 2.6) = ∫ = −e −0.5 x

2.6 0.5

2.6 0.5

d. P(X > 0) = P(0 ≤ X ≤ 4) = 1

0.5e−0.5 x dx

e.

= −e−1.3 + e−0.25 ≈ 0.506

661

4

⎛ x⎞

x3

4

64

8

µ = ∫ x ⎜ ⎟ dx = = −0 = 0 ⎝8⎠ 24 24 3 0


f.

2

σ =∫ = 8−

4 2⎛ x⎞ x ⎜ ⎟ dx − µ 2 0 ⎝8⎠

2

c.

8 8 2 2 = = 9 3 3

= 0

d.

1 2

P (3 < X < 5) = P (3 < X < 4) = ∫

10. a.

4 3

11. P ( X ≤ 7) = ∫

x ⎞ ⎜ − 1⎟ dx 2 ⎝ ⎠

7

1 x 7 dx = = 0 10 10 0 10 7

10 x2 ⎛1⎞ E ( X ) = ∫ x ⎜ ⎟ dx = = 5 min 0 20 ⎝ 10 ⎠ 0 10

x dx 3 8 4

12 − 11.93 0.07 1 = = 12.07 − 11.93 0.14 2 P(X = 12) = 0 12.07 ⎛ 1 ⎞ E( X ) = ∫ x dx 11.93 ⎜⎝ 12.07 − 11.93 ⎟⎠ 12.07 x =∫ dx 11.93 0.14

12. P ( X < 12) =

16 9 7 = − = 16 16 16

P (2 ≤ X ≤ 4) = 1 4⎛ x ⎞ ∫ 2 ⎜⎝ 2 + k ⎟⎠ dx = 1 4

⎛ x2 ⎞ + kx ⎟ = 1 ⎜ ⎜ 4 ⎟ ⎝ ⎠2 (4 + 4k ) − (1 + 2k ) = 1 2 k = −2 k = −1

=

x2 0.28

12.07

= 12 oz 11.93

13. P(X > 1) = 1 – P(X ≤ 1)

(

1

= 1 − ∫ 3e−3 x dx = 1 − −e−3 x

b. P(X ≥ 2.5) = P(2.5 ≤ X ≤ 4)

(

0

)

= 1 − −e −3 + 1 ≈ 0.050

4

⎛ x2 ⎞ ⎛x ⎞ = ∫ ⎜ − 1⎟ dx = ⎜ − x⎟ ⎜ ⎟ 2.5 ⎝ 2 ⎠ ⎝ 4 ⎠ 2.5 4

10 / 3 ⎛ 2

10 / 3

c2 = 8 c = ±2 2 We choose c = 2 2 since 0 < c < 4.

x2 = 16

P (2 < X < µ ) = ∫

⎛ x2 ⎞ =⎜ − x⎟ ⎜ 4 ⎟ ⎝ ⎠2 ⎡⎛ 25 10 ⎞ ⎤ 4 = ⎢⎜ − ⎟ − (1 − 2) ⎥ = 9 3 ⎠ ⎣⎝ ⎦ 9

c2 1 = 16 2

h.

⎞

⎡⎛ 32 ⎞ ⎛4 ⎞ ⎤ 10 = ⎢⎜ − 8 ⎟ − ⎜ − 2 ⎟ ⎥ = 3 3 ⎠ ⎝ ⎠⎦ 3 ⎣⎝

1 2 cx 1 ∫ 0 8 dx = 2 c

4 ⎛ x2

⎞

4

P ( X < c) =

x2 16

⎛x

4

µ = ∫ x ⎜ − 1⎟ dx = ∫ ⎜ − x ⎟ dx ⎟ 2 ⎝2 2⎜ 2 ⎠ ⎝ ⎠ ⎛ x3 x 2 ⎞ =⎜ − ⎟ ⎜ 6 2 ⎟⎠ ⎝ 2

64 8 = 9 9

Thus σ =

g.

4

x4 ⎛8⎞ = −⎜ ⎟ 32 ⎝ 3⎠ 0


⎡ ⎛ 25 5 ⎞ ⎤ 15 = ⎢(4 − 4) − ⎜ − ⎟ ⎥ = = 0.9375 ⎝ 16 2 ⎠ ⎦ 16 ⎣

662

)0 1


14. P ( X ≤ 3) = ∫

3 2 −2 x 5 0

5

= −e

− 65

e

dx = −e

Section 16.2

3

− 25 x

e.

0

+ 1 ≈ 0.699

P ( X > 5) = 1 − P( X ≤ 5) = 1 − ∫ ⎛ = 1 − ⎜ −e ⎝ ≈ 0.135

− 52 x

5

5 2 −2 x 5 0

5

e

dx f.

⎞ −2 ⎟ = 1 − (−e + 1) ⎠0

Problems 16.2 1. a. b.

c.

P ( Z > −1.23) = 0.5 + A(1.23) = 0.5 + 0.3907 = 0.8907

d.

P ( Z ≤ 2.91) = 0.5 + A(2.91) = 0.5 + 0.4982 = 0.9982

e.

P (−2.51 < Z ≤ 1.3) = A(2.51) + A(1.3) = 0.4940 + 0.4032 = 0.8972

f.

P ( Z > 0.03) = 0.5 − A(0.03) = 0.5 − 0.0120 = 0.4880

2. a.

P (−1.96 < Z < 1.96) = 2 A(1.96) = 2(0.4750) = 0.9500

b.

P (−2.11 < Z < −1.35) = A(2.11) − A(1.35) = 0.4826 − 0.4115 = 0.0711

c.

P ( Z < −1.05) = 0.5 − A(1.05) = 0.5 − 0.3531 = 0.1469

d.

1⎞ 1⎞ ⎛ ⎛ 1 ⎛1⎞ P ⎜ Z < ⎟ = P ⎜ − < Z < ⎟ = 2A⎜ ⎟ 2⎠ 2⎠ ⎝ ⎝ 2 ⎝2⎠ = 2(0.1915) = 0.3830

3.

P ( Z < z0 ) = 0.5517 0.5 + A ( z0 ) = 0.5517 A( z0 ) = 0.0517 z0 = 0.13

4.

P ( Z < z0 ) = 0.0668 0.5 − A ( − z0 ) = 0.0668 A(− z0 ) = 0.4332 − z0 = 1.5 z0 = −1.5

5.

P ( Z > z0 ) = 0.8599 0.5 + A ( − z0 ) = 0.8599 A(− z0 ) = 0.3599 − z0 = 1.08 z0 = −1.08

6.

P ( Z > z0 ) = 0.4129 0.5 − A ( z0 ) = 0.4129 A( z0 ) = 0.0871 z0 = 0.22

P(0 < Z < 1.7) = A(1.7) = 0.4554 P (0.43 < Z < 2.89) = A(2.89) − A(0.43) = 0.4981 − 0.1664 = 0.3317

P ( Z > 2 ) = P ( Z < −2) + P ( Z > 2) = 2[0.5 − A(2)] = 2[0.5 − 0.4772] = 0.0456

7. P ( − z0 < Z < z0 ) = 0.2662 2 A ( z0 ) = 0.2662 A( z0 ) = 0.1331 z0 = 0.34 8. P ( Z > z0 ) = 0.3174 0.3174 = 0.1587 2 0.5 − A( z0 ) = 0.1587 A( z0 ) = 0.3413 z0 = 1.00 P ( Z > z0 ) =

P ( Z > 3σ ) = P ( Z > 3) = 0.5 − A(3) = 0.5 − 0.4987 = 0.0013

663


9. a.

b.

c.

ISM: Introductory Mathematical Analysis 15. P ( X > 54) = P ( Z > z0 ) = 0.0401

27 − 16 ⎞ ⎛ P ( X < 27) = P ⎜ Z < ⎟ 4 ⎠ ⎝ = P ( Z < 2.75) = 0.5 + A(2.75) = 0.5 + 0.4970 = 0.9970

0.5 − A ( z0 ) = 0.0401 A ( z0 ) = 0.4599 z0 = 1.75

10 − 16 ⎞ ⎛ P ( X < 10) = P ⎜ Z < ⎟ 4 ⎠ ⎝ = P ( Z < −1.5) = 0.5 − A(1.5) = 0.5 – 0.4332 = 0.0668

Since 14

σ

b.

σ

= 1.75

= 1.75

so σ =

P (10.8 < X < 12.4) 12.4 − 16 ⎞ ⎛ 10.8 − 16 = P⎜
10. a.

54 − 40

16. Case 1.

14 =8. 1.75

Suppose x0 > 16 . Then x − 16 ⎞ ⎛ P (16 < X < x0 ) = P ⎜ 0 < Z < 0 2.25 ⎟⎠ ⎝

150 − 200 ⎞ ⎛ P ( X > 150) = P ⎜ Z > ⎟ 40 ⎝ ⎠ = P(Z > –1.25) = 0.5 + A(1.25) = 0.5 + 0.3944 = 0.8944

⎛ x − 16 ⎞ = A⎜ 0 ⎟ = 0.4641 . Thus ⎝ 2.25 ⎠ x0 − 16 = 1.8 , so x0 = 20.05 . 2.25

P (210 < X < 250)

Case 2.

250 − 200 ⎞ ⎛ 210 − 200 = P⎜
Suppose x0 < 16 . Then ⎛ x − 16 ⎞ < Z < 0⎟ P ( x0 < X < 16 ) = P ⎜ 0 ⎝ 2.25 ⎠ ⎛ x − 16 ⎞ = A⎜ − 0 ⎟ = 0.4641 . Thus ⎝ 2.25 ⎠ x − 16 − 0 = 1.8 , so x0 = 11.95 . 2.25

−2 − (−3) ⎞ ⎛ 11. P ( X > –2) = P ⎜ Z > ⎟ 2 ⎝ ⎠ 1⎞ ⎛ ⎛1⎞ = P ⎜ Z > ⎟ = 0.5 − A ⎜ ⎟ 2⎠ ⎝ ⎝2⎠ = 0.5 – 0.1915 = 0.3085

Therefore, x0 can be either 11.95 or 20.05. 17. Let X be score on test. Then the probability that X lies within 2σ = 2(100) = 200 points of 500 is 0.95. Thus, 95% of those who took the test had scores between 300 and 700.

3−0 ⎞ ⎛ 12. P ( X < 3) = P ⎜ Z < ⎟ = P( Z < 2) 1.5 ⎠ ⎝ = 0.5 + A(2) = 0.5 + 0.4772 = 0.9772

18. Let X be score on test and let x0 be least score a person could get and yet score in about the top 20 percent. Then x − 65 ⎞ ⎛ = 0.20. Thus P ( X ≥ x0 ) = P ⎜ Z ≥ 0 10 ⎟⎠ ⎝

13. Since σ 2 = 100 , σ = 10. Thus 35 = 65 − 30 = µ − 3σ 95 = 65 + 30 = µ + 3σ Thus P (35 < X ≤ 95) = P( µ − 3σ < X ≤ µ + 3σ ) = 0.997 (99.7%)

⎛ x − 65 ⎞ ⎛ x0 − 65 ⎞ 0.5 − A ⎜ 0 ⎟ = 0.20 or A ⎜ 10 ⎟ = 0.30. 10 ⎝ ⎠ ⎝ ⎠ x0 − 65 ≈ 0.84. Hence 10 so x0 ≈ 73.4 ≈ 74.

7 −8⎞ ⎛ 14. P ( X > µ − σ ) = P( X > 7) = P ⎜ Z > ⎟ 1 ⎠ ⎝ = P(Z > –1) = 0.5 + A(1) = 0.5 + 0.3413 = 0.8413

664


Section 16.3

19. Let X be height of an adult. Then 72 − 68 ⎞ ⎛ P ( X < 72) = P ⎜ Z < ⎟ = P(Z < 1.33) 3 ⎠ ⎝ = 0.5 + A(1.33) = 0.5 + 0.4082 = 0.9082 90.82% are over 6 feet.

Principles in Practice 16.3 1. X is the number of winners and X is binomial 1 with n = 60 and p = . To find 4 P(X = 20), use the normal approximation to the ⎛1⎞ binomial distribution with µ = np = 60 ⎜ ⎟ = 15 ⎝4⎠

20. Let X be the yearly income (in dollars) of a person in the group. a.

b.

45 ⎛ 1 ⎞⎛ 3 ⎞ and σ = npq = 60 ⎜ ⎟ ⎜ ⎟ = ≈ 3.35 . 4 ⎝ 4 ⎠⎝ 4 ⎠ Converting the correct X-values 19.5 and 20.5 to Z-values gives 19.5 − 15 z1 = ≈ 1.34

46, 000 − 60, 000 ⎞ ⎛ P ( X < 46, 000) = P ⎜ Z < ⎟ 5000 ⎝ ⎠ = P( Z < −2.8) = 0.5 − A(2.8) = 0.5 − 0.4974 = 0.0026 0.0026(10,000) = 26 people have yearly incomes less than $46,000.

45 4

z2 =

⎛ 75, 000 − 60, 000 ⎞ P ( X > 75, 000) = P ⎜ ⎟ 5000 ⎝ ⎠ = P ( Z > 3) = 0.5 − A(3) = 0.5 − 0.4987 = 0.0013 0.0013(10,000) = 13 people have yearly incomes over $75,000.

45 4

≈ 1.64

Thus P(X = 20) ≈ P(1.34 ≤ Z ≤ 1.64) = A(1.64) – A(1.34) = 0.4495 – 0.4099 = 0.0396 The probability of 20 winners out of 60 contestants is 0.0396. Problems 16.3 1. n = 150, p = 0.4, q = 0.6, µ = np = 150(0.4) = 60, σ = npq = 150(0.4)(0.6) = 36 = 6 P ( X ≤ 52) = P ( X < 52.5)

21. Let X be IQ of a child in population. a.

20.5 − 15

125 − 100.4 ⎞ ⎛ P ( X > 125) = P ⎜ Z > ⎟ 11.6 ⎝ ⎠ = P(Z > 2.12) = 0.5 – A(2.12) = 0.5 – 0.4830 = 0.0170. Thus 1.7% of the children have IQ’s greater than 125.

52.5 − 60 = −1.25 6 P(X ≤ 52) = P(X ≤ 52.5) ≈ P(Z ≤ –1.25) = 0.5 – A(1.25) = 0.5 – 0.3944 = 0.1056 73.5 − 60 ⎞ ⎛ P(X ≥ 74) = P(X ≥ 73.5) ≈ P ⎜ Z ≥ ⎟ 6 ⎝ ⎠ = P(Z ≥ 2.25) = 0.5 – A(2.25) = 0.5 – 0.4878 = 0.0122

z=

b. If x0 is the value, then P ( X > x0 ) = 0.90 . x − 100.4 ⎞ ⎛ Thus P ⎜ Z > 0 = 0.90 or 11.6 ⎟⎠ ⎝ ⎛ x − 100.4 ⎞ 0.5 + A ⎜ − 0 = 0.90 . Hence 11.6 ⎟⎠ ⎝ ⎛ x − 100.4 ⎞ = 0.4 , so A⎜ − 0 11.6 ⎟⎠ ⎝ x − 100.4 − 0 = 1.28 or x0 = 85.552 ≈ 85.6 . 11.6

22. Since P (4 < X < 16) = P( µ − 3σ < X < µ + 3σ ) = 0.25 ≠ 0.997, X cannot be normally distributed. 665



2. n = 50, p = 0.3, q = 0.7, µ = np = 50(0.3) = 15, σ = npq = 50(0.3)(0.7) = 10.5 ≈ 3.24 P ( X = 19) = P (18.5 ≤ X ≤ 19.5) ⎛ 18.5 − 15 19.5 − 15 ⎞ ≈ P⎜ ≤Z≤ ⎟ 10.5 ⎠ ⎝ 10.5 = P (1.08 ≤ Z ≤ 1.39) = A(1.39) − A(1.08) = 0.4177 − 0.3599 = 0.0578 P ( X ≤ 18) = P ( X ≤ 18.5) ⎛ 18.5 − 15 ⎞ ≈ P⎜Z ≤ ⎟ 10.5 ⎠ ⎝ = P ( Z ≤ 1.08) = 0.5 + A(1.08) = 0.5 + 0.3599 = 0.8599

5. Let X = no. of times 5 occurs. Then X is 1 5 binomial with n = 300, p = , q = , 6 6 125 ≈ 6.45 . 3 P(45 ≤ X ≤ 60) = P(44.5 ≤ X ≤ 60.5) ⎛ ⎞ 44.5 − 50 60.5 − 50 ⎟ ≈ P⎜ ≤Z≤ ⎜⎜ 125 125 ⎟⎟ 3 3 ⎝ ⎠ = P(–0.85 ≤ Z ≤ 1.63) = A(0.85) + A(1.63) = 0.3023 + 0.4484 = 0.7507

µ = np = 50, σ = npq =

6. Let X = no. of heads that occurs. Then X is binomial with n = 200, p = 0.4, q = 0.6, µ = np = 80, σ = npq = 48 ≈ 6.93 . P (90 ≤ X ≤ 100) = P(89.5 ≤ X ≤ 100.5) ⎛ 89.5 − 80 100.5 − 80 ⎞ ≈ P⎜ ≤Z≤ ⎟ 48 48 ⎠ ⎝ = P(1.37 ≤ Z ≤ 2.96) = A(2.96) – A(1.37) = 0.4985 – 0.4147 = 0.0838

3. n = 200, p = 0.6, q = 0.4, µ = np = 200(0.6) = 120 σ = npq = 200(0.6)(0.4) = 48 ≈ 6.93 P ( X = 125) = P(124.5 ≤ X ≤ 125.5) ⎛ 124.5 − 120 125.5 − 120 ⎞ ≈ P⎜ ≤Z≤ ⎟ 48 48 ⎝ ⎠ = P (0.65 ≤ Z ≤ 0.79) = A(0.79) − A(0.65) = 0.2852 − 0.2422 = 0.0430 P (110 ≤ X ≤ 135) = P (109.5 ≤ X ≤ 135.5) ⎛ 109.5 − 120 135.5 − 120 ⎞ ≈ P⎜ ≤Z≤ ⎟ 48 48 ⎝ ⎠ = P (−1.52 ≤ Z ≤ 2.24) = A(1.52) + A(2.24) = 0.4357 + 0.4875 = 0.9232

7. Let X = no. of trucks out of service. Then X can be considered binomial with n = 60, p = 0.1, q = 0.9, µ = np = 6, σ = npq = 5.4 ≈ 2.32 ⎛ 6.5 − 6 ⎞ P ( X ≥ 7) = P( X ≥ 6.5) ≈ P ⎜ Z ≥ ⎟ 5.4 ⎠ ⎝ = P(Z ≥ 0.22) = 0.5 – A(0.22) = 0.5 – 0.0871 = 0.4129

8. Let X = no. of defective items in sample. Then X is binomial with n = 200, p = 0.05, q = 0.95, µ = 200(0.05) = 10, σ = npq = 9.5 ≈ 3.08 P(X ≥ 7) = P(X ≥ 6.5) 6.5 − 10 z= ≈ −1.14 9.5 P ( X ≥ 7) = P ( X ≥ 6.5) ≈ P( Z ≥ −1.14) = 0.5 + A(1.14) = 0.5 + 0.3729 = 0.8729

4. n = 25, p = 0.25, q = 0.75, µ = np = 25(0.25) = 6.25, σ = npq = 25(0.25)(0.75) = 4.6875 ≈ 2.17 P ( X ≥ 7) = P ( X ≥ 6.5) ⎛ 6.5 − 6.25 ⎞ ≈ P⎜ Z ≥ ⎟ 4.6875 ⎠ ⎝ = P ( Z ≥ 0.12) = 0.5 − A(0.12) = 0.5 − 0.0478 = 0.4522

666


Chapter 16 Review 12. Let X = the number of subjects that choose the cola from the sponsoring company. Then X is binomial with n = 35, p = 0.5, q = 0.5, µ = np = 17.5, σ = npq = 8.75 ≈ 2.96 P ( X ≥ 25) = P( X ≥ 24.5) 24.5 − 17.5 ⎞ ⎛ = P⎜Z ≥ ⎟ 8.75 ⎠ ⎝ = P( Z ≥ 2.37) = 0.5 − A(2.37) = 0.5 − 0.4911 = 0.0089

9. Let X = no. of correct answers. Then X is binomial and p = 0.5, q = 0.5. If n = 25, then µ = np = 25(0.5) = 12.5, σ = npq = 25(0.5)(0.5) = 6.25 = 2.5 and P ( X ≥ 13) = P( X ≥ 12.5) 12.5 − 12.5 ⎞ ⎛ ≈ P⎜Z ≥ ⎟ 2.5 ⎝ ⎠ = P( Z ≥ 0.00) = 0.5 − A(0.00) = 0.5 − 0 = 0.5 If n = 100, then µ = 100(0.5) = 50, σ = npq = 100(0.5)(0.5) = 25 = 5 P ( X ≥ 60) = P( X ≥ 59.5) 59.5 − 50 ⎞ ⎛ = P⎜Z ≥ ⎟ 5 ⎝ ⎠ = P( Z ≥ 1.9) = 0.5 − A(1.9) = 0.5 − 0.4713 = 0.0287

Chapter 16 Review Problems 1. a.

P (0 ≤ X ≤ 1) = 1 1⎛ 1

∫ 0 ⎜⎝ 3 + kx

2⎞

⎟ dx = 1 ⎠ 1

⎛ x kx3 ⎞ ⎜ + ⎟ =1 ⎜3 3 ⎟⎠ ⎝ 0 1 k + =1 3 3 k 2 = 3 3 k=2

10. Let X = no. of correct answers on last 20 questions. Then X is binomial with n = 20, p = 0.25, q = 0.75, µ = np = 20(0.25) = 5, σ = npq = 20(0.25)(0.75 = 3.75 ≈ 1.94 P ( X ≥ 10) = P ( X ≥ 9.5) 9.5 − 5 ⎞ ⎛ ≈ P⎜Z ≥ ⎟ 3.75 ⎠ ⎝ = P ( Z ≥ 2.32) = 0.5 − A(2.32) = 0.5 − 0.4898 = 0.0102

b.

3/ 4⎛ 1 3⎞ ⎛1 2⎞ P⎜ < X < ⎟ = ∫ ⎜ + 2 x ⎟ dx 1/ 2 4⎠ ⎝2 ⎝3 ⎠ 3/ 4 3/ 4 ⎛ x 2 x3 ⎞ 1 =⎜ + = x + 2 x3 ⎟ ⎜3 3 ⎟⎠ 3 1/ 2 ⎝ 1/ 2

(

)

1 ⎡⎛ 3 27 ⎞ ⎛ 1 1 ⎞ ⎤ 9 = ⎢⎜ + ⎟ − ⎜ + ⎟ ⎥ = 3 ⎣⎝ 4 32 ⎠ ⎝ 2 4 ⎠ ⎦ 32

11. Let X = no. of deals consisting of three cards of one suit and two cards of another suit. Then X is binomial with n = 100, p = 0.1, q = 0.9, µ = np = 10, σ = npq = 9 = 3 . P ( X ≥ 16) = P ( X ≥ 15.5) 15.5 − 10 ⎞ ⎛ ≈ P⎜Z ≥ ⎟ 3 ⎝ ⎠ = P ( Z ≥ 1.83) = 0.5 − A(1.83) = 0.5 − 0.4664 = 0.0336

c.

1 ⎛1 1⎞ ⎛ ⎞ P ⎜ X ≥ ⎟ = ∫ ⎜ + 2 x 2 ⎟ dx 1/ 2 2 3 ⎝ ⎠ ⎝ ⎠ 1 1 ⎛ x 2 x3 ⎞ 1 =⎜ + = x + 2 x3 ⎟ ⎜3 3 ⎟⎠ 3 1/ 2 ⎝ 1/ 2

(

1⎡ ⎛ 1 1 ⎞⎤ 3 = ⎢ (1 + 2) − ⎜ + ⎟ ⎥ = 3⎣ ⎝ 2 4 ⎠⎦ 4

667

)


ISM: Introductory Mathematical Analysis 18 − 20 ⎞ ⎛ 14 − 20
x⎛ 1 ⎞ d. If 0 ≤ x ≤ 1, F ( x ) = ∫ ⎜ + 2t 2 ⎟ dt 0 ⎝3 ⎠ x

⎛ t 2t 3 ⎞ x 2 x3 =⎜ + ⎟ = + ⎜3 3 ⎟ 3 ⎝ ⎠0 3

10 − 20 ⎞ ⎛ 8. P ( X > 10) = P ⎜ Z > ⎟ 4 ⎠ ⎝ = P ( Z > −2.5) = 0.5 + A(2.5) = 0.5 + 0.4938 = 0.9938

if x < 0 ⎧0, ⎪⎪ 3 Therefore, F ( x) = ⎨ 3x + 23x , if 0 ≤ x ≤ 1 ⎪ if x > 1 ⎪⎩1,

2.

23 − 20 ⎞ ⎛ 9. P ( X < 23) = P ⎜ Z < ⎟ 4 ⎠ ⎝ = P ( Z < 0.75) = 0.5 + A(0.75) = 0.5 + 0.2734 = 0.7734

⎧⎪ 1 e−(1/ 3) x , if x ≥ 0 f ( x) = ⎨ 3 if x < 0 ⎪⎩0, P ( X > 2) = 1 − P( X ≤ 2) = 1 − ∫

2 1 −(1/ 3) x

0

3

e

dx

33 − 20 ⎞ ⎛ 23 − 20
⎡ − 1/ 3 x 2 ⎤ = 1 − ⎢ −e ( ) ⎥ = 1 − ⎡ −e −2 / 3 + 1⎤ ⎣ ⎦ 0⎦ ⎣ = e −2 / 3 ≈ 0.513

3. a.

b.

µ=∫

2 x3 ⎛ 2 ⎞ x ⎜ x ⎟ dx = 75 ⎝ 25 ⎠

5 0

5

= 0

10 3

11. n = 100, p = 0.35, q = 0.65, µ = np = 35, σ = npq = 22.75 ≈ 4.77 P(25 ≤ X ≤ 47) = P(24.5 ≤ X ≤ 47.5) ⎛ 24.5 − 35 47.5 − 35 ⎞ ≈ P⎜ ≤Z≤ ⎟ 22.75 ⎠ ⎝ 22.75 = P(–2.20 ≤ Z ≤ 2.62) = A(2.20) + A(2.62) = 0.4861 + 0.4956 = 0.9817

⎛ 2 ⎞ x ⎟ dx − µ 2 ⎝ 25 ⎠

5

σ 2 = ∫ x2 ⎜ 0

5

x4 ⎛ 10 ⎞ = −⎜ ⎟ 50 ⎝ 3⎠ 0 =

2

625 100 25 − = 50 9 18

12. n = 100, p = 0.35, q = 0.65, µ = np = 35, σ = npq = 22.75 ≈ 4.77 P(X = 48) = P(47.5 ≤ X ≤ 48.5) ⎛ 47.5 − 35 48.5 − 35 ⎞ ≈ P⎜ ≤Z≤ ⎟ 22.75 ⎠ ⎝ 22.75 = P(2.62 ≤ Z ≤ 2.83) = A(2.83) – A(2.62) = 0.4977 – 0.4956 = 0.0021

25 Thus σ = ≈ 1.18. 18

4. P ( X < 5) = ∫ =

5

51 1 x dx = ∫ dx = 2 6−2 2 4 42 5

5 2 3 − = 4 4 4

13. Let X = height of an individual. X is normally distributed with µ = 68 and σ = 2. 72 − 68 ⎞ ⎛ P ( X > 72) = P ⎜ Z > ⎟ 2 ⎠ ⎝ = P ( Z > 2) = 0.5 − A(2) = 0.5 – 0.4772 = 0.0228

22 − 20 ⎞ ⎛ 5. P ( X > 22) = P ⎜ Z > ⎟ 4 ⎠ ⎝ = P ( Z > 0.5) = 0.5 − A(0.5) = 0.5 – 0.1915 = 0.3085 21 − 20 ⎞ ⎛ 6. P ( X < 21) = P ⎜ Z < ⎟ 4 ⎠ ⎝ = P ( Z < 0.25) = 0.5 + A(0.25) = 0.5 + 0.0987 = 0.5987 668


Mathematical Snapshot Chapter 16 Mathematical Snapshot Chapter 16

14. Let X = number of heads that occurs. X is binomial with n = 500, p = 0.5, q = 0.5, µ = np = 250, σ = npq = 125 ≈ 11.18. P ( X ≥ 215) = P ( X ≥ 214.5) 214.5 − 250 ⎞ ⎛ ≈ P⎜Z ≥ ⎟ 125 ⎠ ⎝ = P ( Z ≥ −3.18) = 0.5 + A(3.18) = 0.5 + 0.4993 = 0.9993

1. The result should correspond to the known distribution function. 2. The derivative of the logistic function should roughly coincide with the normal probability density function used to generate the values. 3. The list of earthquake magnitudes will appear to have a normal density function. This is surprising, since one would expect something more like an exponential density function, with most earthquakes being very low-magnitude. Presumably, the explanation for the normal density function is that as magnitude declines, the likelihood of a quake’s being reported and thus appearing on the list also goes down.

669

Chapter 17 Principles in Practice 17.1 1. a.

14. P (3, 4) =

c(500, 700) = 160 + 2(500) + 3(700) = 160 + 1000 + 2100 = 3260 The cost of manufacturing 500 12-ounce and 700 20-ounce mugs is $3260.

f (2, − 1) = 3(2) 2 (−1) − 4(−1) = −12 + 4 = −8

= (3) 2 (1)(−2) + 3(1) 2 (−2) + 3(1)(−2)2 = −18 − 6 + 12 = −12

2

5 −4

2

=

−9 −9 = = −1 25 − 16 9

1

1

)

7. g (4, 8) = 2(4) 42 − 5 = 2(4)(11) = 88

1

20. 2x + y + 2z = 6 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. The intercepts are (3, 0, 0), (0, 6, 0), and (0, 0, 3).

9. F(2, 0, –1) = 3

z

2(1) 2 = (0 + 1)(3) 3

11.

f ( x0 + h, y0 ) = e( x0 + h ) + y0 = e x0 + h + y0

12.

f (r + t , r ) = (r + t ) 2 r − 3r 3 = r t 2 + 2rt − 2r 2

13.

(

f (400, 400, 80) =

y

x

8. g (8, 4) = 8 4 + 10 = 8(2) + 10 = 16 + 10 = 26

10. F (1, 0, 3) =

3 64

z

6. h(1, 5, 3, 1) = ln(1 · 1) = ln 1 = 0

(

3!⋅1!

19. x + y + z = 1 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. The intercepts are (1, 0, 0), (0, 1, 0), and (0, 0, 1).

4. g (3, 1, − 2)

−3(3)

=

3

18. A plane parallel to the y,z-plane has the form x = constant. Because (–4, –2, 7) lies on the plane, the equation is x = –4.

3. g (0, 3, − 1) = e2⋅0 [3(3) + (−1)] = e0 (8) = 8

5. h(−3, 3, 5, 4) =

3!(4 − 3)!

( 14 ) ( 43 ) =

4!

17. A plane parallel to the x,y-plane has the form z = constant. Because (2, 7, 6) lies on the plane, the equation is z = 6.

2

2.

4 −3

16. A plane parallel to the y,z-plane has the form x = constant. Because (−2, 0, 0) lies on the plane, the equation is x = −2.

Problems 17.1 f (1, 2) = 4(1) − (2) + 3 = 4 − 4 + 3 = 3

3

15. A plane parallel to the x,z-plane has the form y = constant. Because (0, 2, 0) lies on the plane, the equation is y = 2.

b. c(1000, 750) = 160 + 2(1000) + 3(750) = 160 + 2000 + 2250 = 4410 The cost of manufacturing 1000 12-ounce mugs and 750 20-ounce mugs is $4410.

1.

( 14 ) ( 43 )

4!

3

)

6 3

x

400(400) = 2000 80

670

y


Section 17.1

21. 3x + 6y + 2z = 12 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. The intercepts are (4, 0, 0), (0, 2, 0), and (0, 0, 6).

z

y

z 6

x

25. z = 4 − x 2 . The x,z-trace is z = 4 − x 2 , which is a parabola. For any fixed value of y, we obtain

y

2

the parabola z = 4 − x 2 .

4

x

z 4

22. 2x + 3y + 5z = 1 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. ⎛1 ⎞ ⎛ 1 ⎞ The intercepts are ⎜ , 0, 0 ⎟ , ⎜ 0, , 0 ⎟ , and ⎝2 ⎠ ⎝ 3 ⎠ 1⎞ ⎛ ⎜ 0, 0, ⎟ . 5⎠ ⎝

x

z

26. y = z 2 . The y,z-trace is y = z 2 , which is a parabola. For any fixed value of x, we obtain the

1 5

x

y

2

1 2

parabola y = z 2 .

y

1 3

z

23. x + 2y = 2 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. There are only two intercepts: (2, 0, 0) and (0, 1, 0). The x,y-trace is x + 2y = 2, which is a line. For any fixed value of z, we obtain the line x + 2y = 2.

y x

z

27. x 2 + y 2 + z 2 = 9. The x,y-trace is x 2 + y 2 = 9, which is a circle. The x,z-trace is x 2 + z 2 = 9,

1

x

which is a circle. The y,z-trace is y 2 + z 2 = 9, which is a circle. The surface is a sphere.

y

z

2

3

24. y = 3z + 2 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. There are only two intercepts: (0, 2, 0) and 2⎞ ⎛ ⎜ 0, 0, − ⎟ . The y,z-trace is y − 3z = 2, which is 3⎠ ⎝ a line. For any fixed value of x, we obtain the line y − 3z = 2.

3

x

671

3

y

Chapter 17: Multivariable Calculus


28. 3x 2 + 2 y 2 = 1

1

7. g ( p, q ) =

The x,y-trace is 3x 2 + 2 y 2 = 1, which is an ellipse. For any fixed value of z, we obtain the ellipse 3x 2 + 2 y 2 = 1. z

pq = ( pq) 2

g p ( p, q ) =

1 q −1 ( pq ) 2 ⋅ q = 2 2 pq

g q ( p, q ) =

1 p −1 ( pq) 2 ⋅ p = 2 2 pq

( − 1 g w ( w, z ) = ( w2 + z 2 ) 3 3

8. g ( w, z ) = w2 + z 2 = w2 + z 2 y x

g z ( w, z ) =

Problems 17.2 1.

2.

3.

4.

f ( x, y ) = 4 x 2 + 3 y 2 − 7 f x ( x, y ) = 4(2 x) + 0 + 0 = 8 x f y ( x, y ) = 0 + 3(2 y ) + 0 = 6 y

)

− 23

s2 + 4 t −3 1 2s hs ( s, t ) = (2 s ) = t −3 t −3

1 3

2w

(2w) =

(

3 w2 + z 2 (2 z ) =

)

2 3

2z

(

3 w2 + z 2

)

2 3

9. h( s, t ) =

f ( x, y ) = 2 x 2 + 3xy f x ( x, y ) = 2(2 x) + 3(1) y = 4 x + 3 y f y ( x, y ) = 0 + 3x(1) = 3 x

(

)

Rewriting h(s, t) as s 2 + 4 (t − 3) −1 , we have s2 + 4 ht ( s, t ) = s 2 + 4 ⎡(−1)(t − 3) −2 (1) ⎤ = − ⎣ ⎦ (t − 3)2

(

f ( x, y ) = 2 y + 1 f x ( x, y ) = 0 + 0 = 0 f y ( x, y ) = 2(1) + 0 = 2

10. h(u , v) =

)

8uv 2 u 2 + v2

(u 2 + v2 ) (1) − u(2u) 2 (u 2 + v2 ) 8v 2 ( v 2 − u 2 ) = 2 ( u 2 + v2 ) u 2 + v 2 ) (2v) − v 2 (2v) ( hv (u, v) = 8u 2 ( u 2 + v2 )

f ( x, y ) = ln 2 f x ( x, y ) = 0 f y ( x, y ) = 0

hu (u, v) = 8v 2

5. g ( x, y ) = 3 x 4 y + 2 xy 2 − 5 xy + 8 x − 9 y g x ( x, y ) = 3(4) x3 y + 2(1) y 2 − 5(1) y + 8(1) = 12 x3 y + 2 y 2 − 5 y + 8

g y ( x, y ) = 3 x 4 (1) + 2 x(2) y − 5 x(1) − 9(1) = 3 x 4 + 4 xy − 5 x − 9

=

6. g ( x, y ) = ( x + 1) 2 + ( y − 3)3 + 5 xy 3 − 2 g x ( x, y ) = 2( x + 1) + 0 + 5(1) y 3 − 0 = 2( x + 1) + 5 y 3

(

1 2 w + z2 3

2 3

)

( )

g y ( x, y ) = 0 + 3( y − 3)2 + 5 x 3 y 2 − 0 = 3( y − 3)2 + 15 xy 2 672

16u 3v

( u 2 + v2 )

2


Section 17.2

1 1 ln(q1 + 2) + ln(q2 + 5) 2 3 1 1 1 uq1 ( q1 , q2 ) = ⋅ +0 = 2 q1 + 2 2(q1 + 2)

11. u ( q1 , q2 ) =

1 1 1 uq2 ( q1 , q2 ) = 0 + ⋅ = 3 q2 + 5 3(q2 + 5)

12. Q(l , k ) = 2l 0.38 k1.79 − 3l1.03 + 2k 0.13 Ql (l , k ) = 2(0.38)l 0.38−1k1.79 − 3(1.03)l1.03−1 + 0 = 0.76l −0.62 k1.79 − 3.09l 0.03 Qk (l , k ) = 2l 0.38 (1.79)k1.79−1 − 0 + 2(0.13)k 0.13−1 = 3.58l 0.38 k 0.79 + 0.26k −0.87

13. h( x, y ) =

x 2 + 3 xy + y 2

hx ( x, y ) =

( = =

x2 + y2

x2 + y 2

(

x2 + y2

)

1 2

−1 ⎡ ⎤ [2 x + 3 y ] − x 2 + 3xy + y 2 ⎢ 12 x 2 + y 2 2 (2 x) ⎥ ⎣ ⎦

(

) (

⎡ 2 2 ⎢ x +y ⎣

(

)

1 2

2

) ⎡⎢⎣( x2 + y 2 ) (2 x + 3 y) − ( x2 + 3xy + y 2 ) x⎤⎥⎦ − 12

x2 + y 2

2 x3 + 3 x 2 y + 2 xy 2 + 3 y 3 − x3 − 3 x 2 y − xy 2

(

h y ( x, y ) =

⎤ ⎥ ⎦

)

(

x2 + y2

x2 + y 2

)

1 2

)

3 2

=

x3 + xy 2 + 3 y3

(

x2 + y2

)

3 2

−1 ⎡ ⎤ [3 x + 2 y ] − x 2 + 3 xy + y 2 ⎢ 12 x 2 + y 2 2 (2 y ) ⎥ ⎣ ⎦

(

) (

⎡ 2 2 ⎢ x +y ⎣

(

)

1 2

⎤ ⎥ ⎦

)

2

x 2 + y 2 ) ⎡⎢( x 2 + y 2 ) (3x + 2 y ) − ( x 2 + 3 xy + y 2 ) y ⎤⎥ ( ⎣ ⎦ = − 12

x2 + y2

=

3x3 + 2 x 2 y + 3 xy 2 + 2 y 3 − x 2 y − 3xy 2 − y 3

( x2 + y 2 )

3 2

=

3 x3 + x 2 y + y 3

( x2 + y2 )

673

3 2


14. h( x, y ) =


x+9 2

x y + y2 x

− x 2 y + y 2 x ) 12 ( x + 9) − ( x + 9) ( 2 xy + y 2 ) ( hx ( x, y ) = 2 ( x2 y + y 2 x ) 1 ( x + 9) − ⎡ x 2 y + y 2 x − 2( x + 9) 2 xy + y 2 ⎤ ( )⎦⎥ 2 ⎣⎢ = 2 ( x2 y + y 2 x ) 1 2

1 2

1 2

y ⎡ x 2 + xy − 2( x + 9)(2 x + y ) ⎤ ⎦ = ⎣ 1 2 2 2( x + 9) [ xy ( x + y ) ]

(

y ⎡ x 2 + xy − 4 x 2 − 36 x − 2 xy − 18 y ⎤ − 3 x 2 + xy + 36 x + 18 y ⎦= = ⎣ 1 2 x 2 y x + 9( x + y )2 2( x + 9) 2 x 2 y 2 ( x + y ) 2

)

( ) , then −2 hy ( x, y ) = x + 9(−1) ( x 2 y + y 2 x ) ( x 2 + 2 xy ) − x + 9 ( x 2 + 2 xy ) − x x + 9( x + 2 y ) − x + 9( x + 2 y ) = = = 2 2 2 x 2 y 2 ( x + y )2 xy 2 ( x + y )2 + x y y x ( )

Since h( x, y ) = x + 9 x 2 y + y 2 x

−1

15. z = e5 xy ∂z ∂z = e5 xy (5 y ) = 5 ye5 xy ; = e5 xy (5 x) = 5 xe5 xy ∂x ∂y 16. z = ( x 2 + y 2 )e2 x +3 y +1

∂z = ( x 2 + y 2 )[e2 x +3 y +1 (2)] + e2 x +3 y +1[2 x] = (2 x 2 + 2 y 2 + 2 x)e2 x + 3 y +1 ∂x ∂z = ( x 2 + y 2 )[e2 x +3 y +1 (3)] + e2 x +3 y +1[2 y ] = (3 x 2 + 3 y 2 + 2 y )e2 x + 3 y +1 ∂y

(

17. z = 5 x ln x 2 + y

)

⎧⎪ ⎡ 1 ⎫⎪ ⎡ 2 x2 ⎤ ⎤ ∂z (2 x) ⎥ + ln x 2 + y [1]⎬ = 5 ⎢ = 5 ⎨x ⎢ + ln x 2 + y ⎥ 2 2 ∂x ⎦⎥ ⎣⎢ x + y ⎦⎥ ⎩⎪ ⎣⎢ x + y ⎭⎪

(

)

(

⎛ 1 ⎞ ∂z 5x [1] ⎟ = = 5x ⎜ ⎜ x2 + y ⎟ x2 + y ∂y ⎝ ⎠

674

)


Section 17.2

18. z = ln(5 x3 y 2 + 2 y 4 ) 4 = 4 ln(5 x3 y 2 + 2 y 4 ) ∂z 1 60 x 2 y 2 60 x 2 y 2 60 x 2 = 4⋅ = = [5(3x 2 ) y 2 + 0] = ∂x 5 x3 y 2 + 2 y 4 5 x3 y 2 + 2 y 4 y 2 (5 x3 + 2 y 2 ) 5 x3 + 2 y 2 ∂z 1 4(10 x3 y + 8 y 3 ) 8 y (5 x3 + 4 y 2 ) 8(5 x3 + 4 y 2 ) = 4⋅ = = [5 x3 (2 y ) + 2(4 y 3 )] = ∂y y (5 x3 y + 2 y 3 ) 5 x3 y 2 + 2 y 4 5 x3 y 2 + 2 y 4 5 x3 y + 2 y 3

19.

1

(

) ⎡3r 2 − 2s ⎤ + ( r 3 − 2rs + s 2 ) ⎡ 1 (r + 2s )− ⎢2 ⎣ ⎦ ⎣

f (r , s ) = (r + 2 s ) 2 r 3 − 2rs + s 2 1

f r (r , s) = (r + 2s) 2

(

)

= r + 2 s 3r 2 − 2s + f s (r , s ) = (r + 2s)

1 2

⎤ (1) ⎥ ⎦

r 3 − 2rs + s 2 2 r + 2s

[ −2r + 2s ] + ( r 3 − 2rs + s 2 ) ⎡⎢

= 2( s − r ) r + 2 s +

20.

1 2

1 −1 ⎤ (r + 2s ) 2 (2) ⎥ ⎣2 ⎦

r 3 − 2rs + s 2 r + 2s

1

f (r , s ) = (rs ) 2 e 2+ r 1

f r (r , s ) = (rs ) 2 ⎡ e2+ r (1) ⎤ + e2+ r ⎣ ⎦

s ⎤ 2+ r − 12 ⎡1 ⎤ ⎡ ⎢ 2 (rs ) ( s ) ⎥ = ⎢ rs + 2 rs ⎥ e ⎣ ⎦ ⎣ ⎦

2+ r −1 ⎡1 ⎤ re f s (r , s ) = e 2+ r ⎢ (rs ) 2 (r ) ⎥ = ⎣2 ⎦ 2 rs

21.

f (r , s ) = e3− r ln(7 − s ) f r (r , s ) = ln(7 − s ) ⎡e3− r (−1) ⎤ = −e3− r ln(7 − s ) ⎣ ⎦ 3− r ⎡ 1 ⎤ e f s ( r , s ) = e 3− r ⎢ (−1) ⎥ = ⎣7 − s ⎦ s−7

22.

( ) f r (r , s ) = ( 5r 2 + 3s3 ) [2] + (2r − 5s )[10r ] = 2 ( 5r 2 + 3s3 ) + 10r (2r − 5s ) f s (r , s ) = ( 5r 2 + 3s3 ) [−5] + (2r − 5s ) ⎡9 s 2 ⎤ = −5 ( 5r 2 + 3s3 ) + 9s 2 (2r − 5s ) ⎣ ⎦ f (r , s ) = 5r 2 + 3s3 (2r − 5s )

23. g ( x, y, z ) = 2 x3 y 2 + 2 xy 3 z + 4 z 2 g x ( x, y, z ) = 2 y 2 (3 x 2 ) + 2 y 3 z (1) + 0 = 6 x 2 y 2 + 2 y 3 z

g y ( x, y, z ) = 2 x3 (2 y ) + 2 xz (3 y 2 ) + 0 = 4 x3 y + 6 xy 2 z g z ( x, y, z ) = 0 + 2 xy 3 (1) + 4(2 z ) = 2 xy3 + 8 z

675



24. g ( x, y, z ) = 2 xy 2 z 6 − 4 x 2 y 3 z 2 + 3 xyz g x ( x, y, z ) = 2(1) y 2 z 6 − 4(2 x) y 3 z 2 + 3(1) yz = 2 y 2 z 6 − 8 xy 3 z 2 + 3 yz

g y ( x, y, z ) = 2 x(2 y ) z 6 − 4 x 2 (3 y 2 ) z 2 + 3 x(1) z = 4 xyz 6 − 12 x 2 y 2 z 2 + 3 xz g z ( x, y, z ) = 2 xy 2 (6 z 5 ) − 4 x 2 y3 (2 z ) + 3 xy (1) = 12 xy 2 z 5 − 8 x 2 y 3 z + 3xy

25. g (r , s, t ) = e s +t (r 2 + 7 s3 ) g r (r , s, t ) = e s +t [2r + 0] = 2re s +t

( ) = ( 7 s3 + 21s 2 + r 2 ) e s +t gt (r , s, t ) = ( r 2 + 7 s3 ) ⎡ e s +t (1) ⎤ = e s + t ( r 2 + 7 s3 ) ⎣ ⎦

g s (r , s, t ) = e s +t ⎡0 + 21s 2 ⎤ + r 2 + 7 s3 ⎡ e s +t (1) ⎤ ⎣ ⎦ ⎣ ⎦

26. g(r, s, t, u) = rs ln(2t + 5u) g r (r , s, t , u ) = (1) s ln(2t + 5u ) = s ln(2t + 5u ) g s (r , s, t , u ) = r (1) ln(2t + 5u ) = r ln(2t + 5u ) 2rs ⎡ 1 ⎤ gt (r , s, t , u ) = rs ⎢ (2) ⎥ = ⎣ 2t + 5u ⎦ 2t + 5u 5rs ⎡ 1 ⎤ gu (r , s, t , u ) = rs ⎢ (5) ⎥ = ⎣ 2t + 5u ⎦ 2t + 5u

27.

f ( x, y ) = x 3 y + 7 x 2 y 2 f x ( x, y ) = 3 x 2 y + 14 xy 2 f x (1, − 2) = 3(1)2 (−2) + 14(1)(−2)2 = 50

28. z = 2 x3 + 5 xy + 2 y 2

∂z 6 x2 + 5 y = ∂ x 2 2 x3 + 5 xy + 2 y 2 ∂z 5 = ∂ x (0, 1) 2 2 29. g ( x, y, z ) = e x y + 2 z −1 ⎡1 ⎤ g z ( x, y, z ) = e x ⎢ ( y + 2 z ) 2 (2) ⎥ = ⎣2 ⎦

g z (0, 6, 4) =

1 6+8

=

ex y + 2z

1 14

676


Section 17.2

3 x 2 y 2 + 2 xy + x − y xy − yz + xz

30. g ( x, y , z ) =

g y ( x, y , z ) =

( xy − yz + xz )(6 x 2 y + 2 x − 1) − (3 x 2 y 2 + 2 xy + x − y )( x − z ) ( xy − yz + xz )2

(1 − 5 + 5)(6 + 2 − 1) − (3 + 2 + 1 − 1)(1 − 5)

g y (1, 1, 5) =

(1 − 5 + 5) 2

(

)

= 27

31. h(r , s, t , u ) = s 2 + tu ln(2r + 7 st ) hs (r , s, t , u ) =

(

7t s 2 + tu 2r + 7 st

) + 2s ln(2r + 7st )

hs (1, 0, 0, 1) = 0 7 r + 3s 2 u 2 s ht (r , s, t , u ) = 0

32. h(r , s, t , u ) =

ht (4, 3, 2, 1) = 0

33.

(

)

f (r , s, t ) = rst r 2 + s3 + t 4 = r 3 st + rs 4t + rst 5

( )

f s (r , s, t ) = r 3 (1)t + r 4s3 t + r (1)t 5 = r 3t + 4rs 3t + rt 5 f s (1, − 1, 2) = 2 + (−8) + 32 = 26

34. z =

x2 + y2 ex

2

+y

2

= ( x 2 + y 2 )e − ( x

2

+ y2 )

2 2 2 2 ∂z = (2 x)e −( x + y ) + ( x 2 + y 2 )e− ( x + y ) (−2 x) ∂x

= 2 xe−( x ∂z ∂x

x =0 y =0

2

+ y2 )

= 2(0)e0 [1 − (0)] = 0

By symmetry, ∂z ∂y

x =1 y =1

[1 − ( x 2 + y 2 )]

2 2 ∂z = 2 ye −( x + y ) [1 − ( x 2 + y 2 )]. ∂y

= 2(1)e−2 [1 − (2)] = −

2 e2

35. z = xe x − y + ye y − x

∂ z ⎡ x− y x− y ⎤ ⎡ y− x = xe +e + ye (−1) ⎤ ⎦ ⎣ ⎦ ∂x ⎣ ∂ z ⎡ x− y = xe (−1) ⎤ + ⎡ ye y − x + e y − x ⎤ ⎦ ⎣ ⎦ ∂y ⎣ ∂z ∂z Thus + = e x − y + e y − x , as was to be shown. ∂x ∂y 677



(1 + r )1− z ln(1 + r )

36. u = f (t , r , z ) =

(1 + r )1− z − t

∂u ∂ ⎡ (1 + r )1− z ⎤ = ln(1 + r ) ⎢ ⎥ ∂z ∂ z ⎣⎢ (1 + r )1− z − t ⎦⎥ = ln(1 + r )

{

}

⎡ (1 + r )1− z − t ⎤ ∂ ⎡(1 + r )1− z ⎤ − (1 + r )1− z ∂ ⎡(1 + r )1− z ⎤ − 0 ∂z ⎣ ⎣ ⎦∂z ⎣ ⎦ ⎦ ⎡ (1 + r )1− z − t ⎤ ⎣ ⎦

2

−t ∂∂z ⎡ (1 + r )1− z ⎤ ⎣ ⎦ = ln(1 + r ) ⎡ (1 + r )1− z − t ⎤ ⎣ ⎦

= ln(1 + r )

=

2

{

}

−t (1 + r )1− z ln(1 + r )[−1] ⎡ (1 + r )1− z − t ⎤ ⎣ ⎦

t (1 + r )1− z ln 2 (1 + r ) ⎡ (1 + r )1− z − t ⎤ ⎣ ⎦

2

2

, as was to be shown.

bT iC + C 2 ∂F ∂ ⎡ bT ⎤ ∂ ⎡ iC ⎤ bT i = + =− + ⎢ ⎥ ⎢ ⎥ ∂C ∂C ⎣ C ⎦ ∂C ⎣ 2 ⎦ C2 2

37. F (b, C , T , i ) =

38. From η =

r D ∂r ∂D

rL = r + D ⋅ rL = r +

r

η

, we have

∂r r = . Substituting into Equation (3) gives ∂ D Dη

r dC + Dη dD

+

dC dD

⎡ 1 ⎤ dC rL = r ⎢1 + ⎥ + ⎣ η ⎦ dD ⎡η + 1 ⎤ dC rL = r ⎢ ⎥+ ⎣ η ⎦ dD ⎡1 + η ⎤ dC rL = r ⎢ ⎥+ ⎣ η ⎦ dD which is Equation (4).

39. R = f (r , a, n) =

r

( )

1 + a n2−1

∂R ⎡ ⎛ n − 1 ⎞⎤ = r (−1) ⎢1 + a ⎜ ⎟⎥ ∂n ⎝ 2 ⎠⎦ ⎣

⎡ ⎛ n − 1 ⎞⎤ = r ⎢1 + a ⎜ ⎟⎥ ⎝ 2 ⎠⎦ ⎣

−2

⋅

−1

a ra =− 2 2 2 ⎡1 + a n2−1 ⎤ ⎣ ⎦

( )

678


Section 17.3

Problems 17.3

b.

1. c = 7 x + 0.3 y 2 + 2 y + 900 c.

∂c = 0.6 y + 2 ∂y When x = 20 and y = 30, then ∂c = 0.6(30) + 2 = 20 . ∂y 7.

2. c = x x + y + 5000

∂c x = + x+ y ∂ x 2 x+ y

From parts (a) and (b), ∂P ∂ P ⎛αP ⎞ ⎛ β P ⎞ l +k = l⎜ ⎟+k⎜ ⎟ ∂l ∂k ⎝ l ⎠ ⎝ k ⎠ = α P + β P = P (α + β ) = P(1) = P

∂ qA ∂ qA ∂ qB ∂ qB = −50 , =2, = 4, = −20 ∂ pA ∂ pB ∂ qA ∂ pB ∂ qA ∂ qB Since > 0 and > 0 the products are ∂ pB ∂ pA competitive.

When x = 40 and y = 60, then ∂c 40 = + 100 = 12 . ∂ x 2 100

8.

3. c = 0.03( x + y )3 − 0.6( x + y ) 2 + 9.5( x + y ) + 7700

∂c = 0.09( x + y )2 − 1.2( x + y ) + 9.5 ∂x

∂ qA ∂ qA ∂ qB ∂ qB = −1 , = −2 , = −2 , = −3 ∂ pA ∂ pB ∂ pA ∂ pB ∂ qA ∂ qB Since < 0 and < 0 the products are ∂ pB ∂ pA complementary.

When x = 50 and y = 80, then ∂c = 0.09(130)2 − 1.2(130) + 9.5 = 1374.5 . ∂x

1

9. qA = 100 pA −1 pB− 2 −1

qB = 500 pB−1 pA3 1 ∂ qA −100 −2 − 2 = 100(−1) pA pB = 1 ∂ pA 2 2 pA pB

4. P = 15lk − 3l 2 + 5k 2 + 500 ∂P = 15l + 10k ∂k ∂P = 15k − 6l ∂l

∂ qA −50 ⎛ 1 ⎞ −1 − 32 pB = = 100 ⎜ − ⎟ pA 3 ∂ pB ⎝ 2⎠ 2

pA pB

∂ qB −500 ⎛ 1⎞ = 500 ⎜ − ⎟ pB−1 pA = 4 ∂ pA ⎝ 3⎠ 3 − 43

0.357 0.643

5. P = 2.314l k ∂P = 2.314(0.357)l −0.643 k 0.643 ∂l 0.643 ⎛k⎞ = 0.826098 ⎜ ⎟ ⎝l⎠ ∂P = 2.314(0.643)l 0.357 k −0.357 ∂k 0.357 ⎛l⎞ = 1.487902 ⎜ ⎟ ⎝k⎠

3 pB pA

∂ qB 500 = 500(−1) pB−2 pA = − 1 ∂ pB 2 3 − 13

pB pA

Since

∂ qA ∂ qB < 0 and < 0 , the products are ∂ pB ∂ pA

complementary. 10.

6. P = Alα k β a.

∂P βP ⎛β⎞ = Aβ lα k β −1 = ⎜ ⎟ Alα k β = k k ∂k ⎝ ⎠

∂P αP ⎛α ⎞ = Aα lα −1k β = ⎜ ⎟ Alα k β = l ∂l ⎝l ⎠

∂P = 15.18l −0.54 k 0.52 ∂l ∂P = 17.16l 0.46 k −0.48 ∂k If l = 1 and k = 1, then

∂P = 17.16 ∂k 679

∂P = 15.18 and ∂l


11.

ISM: Introductory Mathematical Analysis b. One for which w = w0 and s = s0 since increasing w by 1 while holding s fixed decreases the reading ease score.

∂P = 0.01A0.27 B −0.99C 0.01 D 0.23 E 0.09 F 0.27 ∂B ∂P = 0.01A0.27 B 0.01C −0.99 D 0.23 E 0.09 F 0.27 ∂C

12. P =

a.

16. ω = b −1L−1

kl 2k + 3l

∂ω τ 1 τ = (−1)b −2 L−1 =− 2 ∂b πρ b L πρ

∂P l (2k + 3l ) − kl (2) 3l 2 = = ∂k (2k + 3l )2 (2k + 3l )2

∂ω τ 1 τ = b −1 (−1) L−2 =− 2 ∂L πρ bL πρ

∂P k (2k + 3l ) − kl (3) 2k 2 = = ∂l (2k + 3l )2 (2k + 3l )2

∂ω 1 τ ⎛ 1 ⎞ − 32 1 = ⎜− ⎟ ρ = − 3 ∂ρ bL π ⎝ 2 ⎠ 2bL ρ 2

b. When k = l, then 3l 2 2l 2 ∂P ∂P + = + ∂k ∂l (2l + 3l )2 (2l + 3l )2 17.

25l 2 1 = 5

18. qA = e−( pA + pB ) and qB =

degree had an extra year of work experience before the degree, the manager would receive $4480 more per year in extra compensation.

a.

∂ Sg

⎛ 1 ⎞ − 2 1 ⎛ 7 ⎞ Si = 7 ⎜ ⎟ Se 3 Si2 = ⎜ ⎟ ∂ Se ⎝3⎠ ⎝ 3 ⎠ 3 S2 e

∂ Si

⎛1⎞ = 7 ⎜ ⎟ S e Si ⎝2⎠ 1 3

− 12

−2 −2 pB = 16 pA

∂qA = −e −( pA + pB ) < 0 ∂pB

b. Note that pA and pB are in units of thousands of dollars. When pA = 1 and

3

⎛7⎞ S =⎜ ⎟ e ⎝ 2 ⎠ Si

∂qA 1 = −e −3 = − . ∂pB e3 A decrease in the price of B of $20 is a 20 = 0.01. Thus the decrease in pB of 2000 change in qB is approximately pA = 2 , then

If Se = 125 and Si = 100 , then ∂ S g ⎛ 7 ⎞ 10 14 ∂ Sg ⎛ 7 ⎞ 5 7 and =⎜ ⎟ = = = . ∂ Se ⎝ 3 ⎠ 52 15 ∂ Si ⎜⎝ 2 ⎟⎠ 10 4 Thus if Se increases to 126 and Si remains at 14 ; if 15 Si increases to 101 and Se remains at 125, then

100, then S g increases by approximately

−

1 e

3

(−0.01) =

0.01 e3

by approximately

7 S g increases by approximately . 4

15. a.

16 2 2 pA pB

∂qB −3 −2 = −32 pA pB < 0 ∂pA Since both are < 0, A and B are complementary.

1 2

14. S g = 7 Se Si

∂ Sg

∂g 1 = > 0 for VF > 0 . Thus if x increases ∂ x VF and VF and VS are fixed, then g increases.

∂z 13. = 4480 . If a staff manager with an M.B.A. ∂x

1 3

τ π

∂ω 1 1 ⎛ 1 ⎞ − 12 1 1 = ⎜ ⎟τ = ∂τ bL πρ ⎝ 2 ⎠ 2bL πρτ

5l 2

=

τ 1 τ − 12 1 1 12 = ρ = τ bL πρ πρ bL π

∂R ∂R = −1.015 ; = −0.846 ∂w ∂s 680

. So demand increases

0.01 e3

unit.


19. a.

Section 17.3

∂ qA ⎛ 1 −3 ⎞ = 10 pB ⎜ − pA2 ⎟ ∂ pA ⎝ 2 ⎠ ∂ qA 10 ⎛ 1 − 12 ⎞ = ⎜ pB ⎟ ∂ pB pA ⎝ 2 ⎠ When pA = 9 and pB = 16, then

∂ qA 20 ⎛ 1 1 ⎞ = 10(4) ⎜ − ⋅ ⎟ = − and 27 ∂ pA ⎝ 2 27 ⎠

∂ qA 10 ⎛ 1 1 ⎞ 5 = ⋅ = . ∂ pB 3 ⎜⎝ 2 4 ⎟⎠ 12 ∂ qA 5 = . Hence each $1 reduction in pB decreases qA by ∂ pB 12

b. From (a), when pA = 9 and pB = 16, then

5 unit. Thus a $2 reduction in pB (from $16 to $14) decreases the demand for A by 12 5 5 (2) = unit. approximately 12 6

approximately

20. c =

a.

(

2 3 qA qB + qA

17

)

1 2

1

+ qA qB3 + 600

∂c 1 ⎡ 2 1 3 = ⎢ qA ⋅ qB + qA ∂ qA 17 ⎣ 2

(

) ( − 12

3 + qB + qA

)

1 2

⎤

1

( 2qA )⎥ + qB3

⎦ 1 1 1 − ⎡ ⎤ 1 1 2 3 3 = ⎢ qA + qA 2 ⎥ + qB3 qB + qA 2 + 2qA qB 17 ⎣ 2 ⎦ 1 − ⎤ ∂c 1 ⎡ 2 1 3 1 −2 = ⎢ qA ⋅ qB + qA 2 3qB2 ⎥ + qA ⋅ qB 3 ∂ qB 17 ⎣ 2 3 ⎦ −1 ⎤ 1 2 1 ⎡3 2 2 3 = ⎢ qA qB qB + qA 2 ⎥ + qA qB− 3 17 ⎣ 2 ⎦ 3

(

)

(

) ( )

(

(

)

)

b. When qA = 17 and qB = 8 , then

∂c 1 ⎡1 ⎤ 1 ⎛ 1 ⎞ ⎡1 ⎤ = ⎢ (17)2 ⎜ ⎟ + 2(17)(23) ⎥ + 2 = ⎢ (17) + 2(23) ⎥ + 2 ≈ 48.37 . 23 ∂ qA 17 ⎣ 2 ⎣2 ⎦ ⎝ 23 ⎠ ⎦ c.

21. a.

From (b), if qA is reduced by one unit (from 17 to 16) while qB remains at 8, then the cost will decrease by approximately $48.37.

∂R = 2.5945 − 0.1608 Er − 0.0277 I r ∂ Er ∂R ∂R If Er = 18.8 and I r = 10 , then = −0.70564 . Since < 0, ∂ Er ∂ Er such a candidate should not be so advised.

681


b.

22. S =

a.


∂R = 0.8579 − 0.0122 N ∂N ∂R < 0 , then N > 70.3 ≈ 70% If ∂N AT + 450 A+T2

∂S = ∂T

. Note: A is expressed in hundreds of dollars.

(A+T2 )

( =

1 2

−1 ⎡ ⎤ ( A) − ( AT + 450) ⎢ 12 A + T 2 2 (2T ) ⎥ ⎣ ⎦

(

⎛ A+T2 ⎞ ⎜ ⎟ ⎝ ⎠ A+T2

)

2

) ⎡⎢⎣( A + T 2 ) A − ( AT + 450)T ⎤⎥⎦ = A2 − 450T A+T2 (A+T2 ) − 12

3 2

as was to be shown. b. We want to find when 902 − 450T

(90 + T 2 )

3 2

∂S 9000 ∂S < 0 and A = = 90 . First we find when = 0 and A = 90: 100 ∂T ∂T

= 0 ⇒ 902 − 450T = 0 ⇒ T =

902 = 18 . 450

∂S ∂S > 0 for T < 18, and < 0 for T > 18. Thus 18 months elapse before the sales volume begins to ∂T ∂T decrease. 23. qA = 1000 − 50 pA + 2 pB ⎛ pA ⎞ ∂ qA ⎛ pA ⎞ =⎜ ⎟ ⎟ (−50) ⎝ qA ⎠ ∂ pA ⎝ qA ⎠

η pA = ⎜

⎛ pB ⎞ ∂ qA ⎛ pB ⎞ =⎜ ⎟ ⎟ (2) ⎝ qA ⎠ ∂ pB ⎝ qA ⎠

η pB = ⎜

When pA = 2 and pB = 10 , then qA = 920 , from which η pA = −

5 1 and η pB = 46 46

24. qA = 60 − 3 pA − 2 pB ⎛ pA ⎞ ∂qA ⎛ pA ⎞ =⎜ ⎟ ⎟ (−3) ⎝ qA ⎠ ∂pA ⎝ qA ⎠

η pA = ⎜

⎛ pB ⎞ ∂qA ⎛ pB ⎞ =⎜ ⎟ ⎟ (−2) ⎝ qA ⎠ ∂pB ⎝ qA ⎠

η pB = ⎜

When pA = 5 and pB = 3, then qA = 39, from which η pA = −

682

5 2 and η pB = − . 13 13


25. qA =

Section 17.4 ∂z ⎛ ∂z ⎞ + 2 x 2 ⎜ 2 z ⎟ + 2 z 2 (2 x) − y = 0 ∂x ⎝ ∂x ⎠ ∂z (3z 2 + 4 x 2 z ) = y − 4 xz 2 ∂x ∂z y − 4 xz 2 = ∂x 3z 2 + 4 x 2 z

100

6. 3z 2

pA pB

⎞ ⎟ ⎟ ⎠ ⎛ p ⎞∂q ⎛ p ⎞ ⎛ −50 ⎞⎟ η pB = ⎜ B ⎟ A = ⎜ B ⎟ ⎜ 3 ⎟ ⎝ qA ⎠ ∂ pB ⎝ qA ⎠ ⎜⎝ pA pB ⎠ When pA = 1 and pB = 4 , then qA = 50 . This ⎛ pA ⎝ qA

η pA = ⎜

⎞ ∂ qA ⎛ pA ⎞ ⎛ −100 =⎜ ⎟ ⎟⎜ 2 ⎠ ∂ pA ⎝ qA ⎠ ⎜⎝ pA pB

gives η pA = −1 and η pB = −

7. 0 + e y + e z

∂z ey =− = −e y − z z ∂y e

1 . 2

Problems 17.4

8. xyz + 3 y 3 x 2 − ln z 3 = 0 so

∂z =0 ∂x ∂z 4x 2x =− =− ∂x 10 z 5z

xyz + 3 y3 x 2 − 3ln z = 0.

1. 4 x + 0 + 10 z

2. 2 z

xz + xy

∂ z 10 x 5 x = = z ∂ x 2z

∂z z (− xz − 9 y 2 x 2 ) = ∂y xyz − 3

∂z − 0 − 8y = 0 ∂y

9.

∂ z 8y 4y = = 2 ∂ y 6z 3z 2 4. 0 + 2 y + 6 z 2

∂z =0 ∂y

∂ z −2 y y = =− ∂ y 6z2 3z 2

(

)

5. x 2 − 2 y − z 2 + y x 2 z 2 = 20

10.

∂z ∂z 2 ⎡ ⎤ + y ⎢ x2 ⋅ 2 z + z ⋅ 2x⎥ = 0 ∂x ∂x ⎣ ⎦ ∂z 2 x 2 yz − 2 z = −2 x − 2 xyz 2 ∂x

2x − 0 − 2z

(

)

( (

∂z 1 ∂z + 9 y 2 x2 − 3 ⋅ =0 ∂y z ∂y

3 ⎞ ∂z ⎛ 2 2 ⎜ xy − ⎟ = − xz − 9 y x z ⎠ ∂y ⎝ ⎛ xyz − 3 ⎞ ∂z 2 2 ⎜ ⎟ = − xz − 9 y x ⎝ z ⎠ ∂y

∂z − 10 x + 0 = 0 ∂x

3. 6 z 2

∂z =0 ∂y

) ( ) (

) )

2 x yz 2 + 1 ∂ z −2 x 1 + yz = = ∂ x 2 z x2 y − 1 z 1 − x2 y

683

1∂z ∂z +9 −y=0 ∂x z∂x ⎛1 ⎞∂ z =y ⎜ + 9⎟ z ⎝ ⎠∂ x ⎛ 1 + 9z ⎞ ∂ z =y ⎜ ⎟ ⎝ z ⎠∂ x yz ∂z = ∂ x 9+ z 1 1∂z +0− =0 x z∂x 1∂z 1 − =− z∂x x ∂z z = ∂x x



⎛ ∂z ⎞ 11. ⎜ 2 z + 6 x ⎟ x3 + 5 = 0 y ∂ ⎝ ⎠ ∂z 2z + 6x = 0 ∂y

15. e yz ⋅ y

( ye yz

(

If x = 1, y = 4, z = 1, then

e2 , y = 1, z = 2, then 2 ∂z 2 2 4 =− =− 2 =− . 2 e ∂x e2 e 2 + −2e 2

If x = −

∂z 1 = − = −1 . ∂x 1

16.

(

1 xz + y 2 2

)

− 12

⎡ ∂z ⎤ + 2 y⎥ − x = 0 ⎢x ∂ y ⎣ ⎦

x xz + y 2 − y y ∂z = x− = 2 xz + y 2 ∂ y xz + y 2 xz + y 2 x

⎛ ∂z ⎞ ⎛ ∂z ⎞ 13. 3x ⎜ 2 z ⎟ + 3z 2 + 2 y ⎜ 2 z ⎟ − 7 y (4 x3 ) = 0 ⎝ ∂x ⎠ ⎝ ∂x ⎠ ∂z (6 xz + 4 yz ) = 28 x3 y − 3 z 2 ∂x ∂z 28 x3 y − 3 z 2 = 6 xz + 4 yz ∂x

2 ⎛⎜ x xz + y 2 − y ⎞⎟ ∂z ⎠ = ⎝ ∂y x

If x = 2, y = 2, z = 6, then

28(1)3 (0) − 3(1) −3 1 ∂z = = =− 2 ∂x (1, 0, 1) 6(1)(1) − 4(0)(1) 6

17.

⎡ ∂z ⎤ ∂z = x⎢y + z ⋅1⎥ ∂y ∂ y ⎣ ⎦ ∂ z xe zx − xy = xz ∂y ∂z xz = ∂ y x e zx − y

14. e zx ⋅ x

)

∂ z 2(2 ⋅ 4 − 2) = =6. ∂y 2

1∂z = 4+0 z∂x ∂z = 4z ∂x

If x = 5, y = –20, z = 1, then

)

18.

∂z z = ∂ y e zx − y If x = 1, y = −e−1 , z = –1, then

∂z −1 e = =− . ∂ y e−1 − −e−1 2

(

)

∂z z =− yz ∂x e +x

12. xz(1 + y) – 5 = 0 ⎡ ∂z ⎤ ⎢ x ∂ x + z ⋅1⎥ (1 + y ) − 0 = 0 ⎣ ⎦ ∂z x +z=0 ∂x ∂z z =− x ∂x

(

)

∂z yz =− yz ∂x y e +x

∂ z −6 x 3x = =− ∂ y 2z z

(

∂z ⎡ ∂z ⎤ = −y ⎢x + z ⋅1⎥ . ∂x ∂ x ⎣ ⎦ ∂z + xy = − yz ∂x

)

( s2 + t 2 ) ⎡⎣4rs2 ∂∂ rt ⎤⎦ − 2r 2 s2[2t ] = 1 2 ( s2 + t 2 ) 2 ∂r 4rs 2 ( s 2 + t 2 ) = 4r 2 s 2 t + ( s 2 + t 2 ) ∂t 2 2 2 2 2 ∂ r 4r s t + ( s + t ) = ∂t 4rs 2 ( s 2 + t 2 ) If r = 1, s = 1, t = 1, then

684

∂z = 4. ∂x

∂ r 4 + 22 = = 1. ∂t 4(2)


19.

(

)

(rs ) ⎡⎣ 2t ∂∂ rt ⎤⎦ − s 2 + t 2 [ s ]

b. Differentiating with respect to qA :

=0

(rs ) 2

( (

Section 17.5

∂c 1 ∂c + ⋅ = 9 + qB2 . ∂ qA 2 c ∂ qA ⎛ 1 ⎞ ∂c = 9 + qB2 . ⎜1 + ⎟ ⎝ 2 c ⎠ ∂ qA

) )

∂t − s s2 + t 2 = 0 ∂r ∂t 2rst = s s2 + t 2 ∂r 2rst

(

When qA = 6 and qB = 4 , then c = 36 and

)

1 ⎞ ∂c 13 ∂ c ⎛ =5, ⋅ = 5 , or ⎜1 + ⎟ 12 ∂ qA ⎝ 12 ⎠ ∂ qA ∂ c 60 = . ∂ qA 13

2 2 s2 + t 2 ∂t s s +t = = 2rst 2rt ∂r

If r = 1, s = 2, t = 4, then 20.

∂ t 4 + 16 20 5 = = = . ∂ r 2 ⋅1 ⋅ 4 8 2

Differentiating with respect to qB :

1 ∂z ⎛ ∂z ⎞ ⎜1 + ⎟ + yz + xy ∂x x + y + z ⎝ ∂x ⎠ ∂z ⎛ ∂z ⎞ = e x + y + z + ze x + y + z ⎜ 1 + ⎟ ∂x ⎝ ∂x ⎠ When x = 0, y = 1, and z = 0, then 1 ⎛ ∂z ⎞ ∂z ⎜ 1 + ⎟ + (1)(0) + (0)(1) ∂x 1 ⎝ ∂x ⎠ ∂z 1 ∂z ⎞ 1 ⎛ = e + 0(e ) ⎜1 + ⎟ ∂x ∂ x⎠ ⎝ ∂z ∂z ∂z ∂z 1 = e , 1 = (e − 1), = 1+ ∂x ∂x ∂x ∂x e − 1

qB ∂c 1 ∂c + ⋅ = qA ⋅ ∂ qB 2 c ∂ qB 9 + qB2 q q ⎛ 1 ⎞ ∂c = A B ⎜1 + ⎟ ⎝ 2 c ⎠ ∂ qB 9 + qB2

When qA = 6 and qB = 4 , then c = 36 and 1 ⎞ ∂c 24 13 ∂ c 24 ⎛ = , ⋅ = , or ⎜1 + ⎟ ∂ ∂ 12 q 5 12 q 5 ⎝ ⎠ B B ∂ c 288 = . ∂ qB 65

Problems 17.5

21. c + c = 12 + qA 9 + qB2 a.

1.

f x ( x, y ) = 6(1) y 2 = 6 y 2 f xy ( x, y ) = 6(2 y ) = 12 y f y ( x, y ) = 6 x(2 y ) = 12 xy f yx ( x, y ) = 12(1) y = 12 y

2.

f x ( x, y ) = 6 x 2 y 2 + 12 xy 3 − 3 y

If qA = 6 and qB = 4 , then c + c = 12 + 6(5) = 42 , 2

c = 42 − c ,

2

c = (42 − c) = 42 − 84c + c 2 , c 2 − 85c + 1764 = 0 , 2

85 ± (−85) − 4(1)(1764) 2 85 ± 169 85 ± 13 . Thus c = 49 or = = 2 2 c = 36. However c = 49 is extraneous but c = 36 is not. Thus c = 36. c=

f xx ( x, y ) = 12 xy 2 + 12 y3

3.

f y ( x, y ) = 3 f yy ( x, y ) = 0 f yyx ( x, y ) = 0

685


4.


( ) ( ) = (2 x + y ) ⎡⎢( x 2 + xy + y 2 ) + ( x 2 + xy + 1) ⎤⎥ ⎣ ⎦ 2 2 = (2 x + y ) ( 2 x + 2 xy + y + 1) f xy ( x, y ) = (2 x + y )[2 x + 2 y ] + ( 2 x 2 + 2 xy + y 2 + 1) [1] = ( 4 x 2 + 6 xy + 2 y 2 ) + ( 2 x 2 + 2 xy + y 2 + 1)

f x ( x, y ) = x 2 + xy + y 2 [2 x + y ] + x 2 + xy + 1 [2 x + y ]

= 6 x 2 + 8 xy + 3 y 2 + 1

5.

f y ( x, y ) = 9 ⎡ e2 xy (2 x) ⎤ = 18 xe2 xy ⎣ ⎦

(

)

f yx ( x, y ) = 18 ⎡⎢ x e2 xy ⋅ 2 y + e2 xy (1) ⎤⎥ = 18e2 xy (2 xy + 1) ⎣ ⎦

(

)

f yxy ( x, y ) = 18 ⎡⎢e2 xy (2 x) + (2 xy + 1) e2 xy ⋅ 2 x ⎤⎥ ⎣ ⎦ = 18e 2 xy (2 x)[1 + (2 xy + 1)] = 18e2 xy (2 x)[2 + 2 xy ] = 72 x(1 + xy )e 2 xy

6.

f x ( x, y ) =

1 2

x +y

2

2x

⋅ 2x =

2

x + y2

x 2 + y 2 ) [2] − (2 x)[2 x] 2 ( y 2 − x 2 ) ( f xx ( x, y ) = = 2 2 2 2 x + y ( ) ( x2 + y 2 ) −2 4 xy f xy ( x, y ) = (2 x)(−1) ( x 2 + y 2 ) [2 y ] = − 2 ( x2 + y2 )

7.

(

)

f ( x, y ) = ( x + y )2 ( xy ) = x 2 + 2 xy + y 2 ( xy ) = x3 y + 2 x 2 y 2 + xy 3 2

2

f x ( x, y ) = 3x y + 4 xy + y

3

f y ( x, y ) = x3 + 4 x 2 y + 3xy 2 f xx ( x, y ) = 6 xy + 4 y 2 f yy ( x, y ) = 4 x 2 + 6 xy

8.

f x ( x, y, z ) = 2 xy3 z 4 f xz ( x, y, z ) = 8 xy 3 z 3 f z ( x, y , z ) = 4 x 2 y 3 z 3 f zx ( x, y, z ) = 8 xy 3 z 3

686


9. z = e

x2 + y 2

∂z =e ∂y 2

∂ z ∂y 2

Section 17.5

=

x2 + y 2

⋅

2y 2 x2 + y 2

=

x2 + y 2

ye

x2 + y2

=

zy x2 + y2

2 2 2 2 2 2 ⎛ ⎛ ⎞ ⎞ y y x 2 + y 2 ⎜ e x + y + y ⋅ e x + y ⋅ 2 2 ⎟ − ye x + y ⎜ 2 2 ⎟ x +y ⎠ ⎝ ⎝ x +y ⎠

⎛ x2 + y2 ⎞ ⎜ ⎟ ⎝ ⎠

2

2 2 2 2 2 2 ⎛ ⎞ y2 ( x2 + y 2 ) ⎜ e x + y + e x + y ⋅ 2 2 ⎟ − y 2e x + y x +y ⎠ ⎝ = 2 2 3/ 2 (x + y ) 2 2 x2 + y 2 + y 2 x2 + y 2 − y 2 = e x +y ( x 2 + y 2 )3 / 2

=e = z⋅

10.

x2 + y 2

x2 + y 2 x2 + y 2 ( x 2 + y 2 )3 / 2

x2 + y 2 x2 + y 2 ( x 2 + y 2 )3 / 2

∂z 1 1 2x = ⋅ (2 x) = ∂ x y x2 + 5 y x2 + 5

(

)

∂ 2z 2x ⎛ 1 ⎞ 2x = ⎜⎜ − 2 ⎟⎟ = − 2 2 ∂ y∂ x x + 5 ⎝ y ⎠ y x2 + 5

(

11.

)

f y ( x, y , z ) = 0 f yx ( x, y, z ) = 0 f yxx ( x, y, z ) = 0

Thus f yxx (4, 3, − 2) = 0 . 12.

(

f x ( x, y , z ) = z 2 6 x − 4 y 3

(

)

)

f xy ( x, y, z ) = z 2 −12 y 2 = −12 y 2 z 2 2

f xyz ( x, y, z ) = −24 y z . Thus f xyz (1, 2, 3) = −24(4)(3) = −288 .

13.

f k (l , k ) = 18l 3 k 5 − 14l 2 k 6 f kl (l , k ) = 54l 2 k 5 − 28lk 6

f klk (l , k ) = 270l 2 k 4 − 168lk 5

Thus f klk (2, 1) = 270(4)(1) − 168(2)(1) = 744.

687


14.


f x ( x, y ) = 9 x 2 y 2 + y − 2 xy 2

19.

f xx ( x, y ) = 18 xy 2 − 2 y 2

f y ( x, y ) = 4 x 2 y + 20 y3

f xxy ( x, y ) = 36 xy − 4 y

f xy ( x, y ) = 8 xy

2

f yx ( x, y ) = 8 xy

f xy = 18 x y + 1 − 4 xy

Thus f xy ( x, y ) = f yx ( x, y ) .

f xyx = 36 xy − 4 y = f xxy ( x, y )

Thus f xxy (5, 1) = f xyx (5, 1) = 36(5)(1) − 4 = 176. 15.

f x ( x, y ) = y 2 e x + f xy ( x, y ) = 2 ye

20.

f x ( x, y ) = ye xy

f xx ( x, y ) = y 2 e xy

1 x

f xy ( x, y ) = y ( xe xy ) + e xy (1) = e xy ( xy + 1)

x

f y ( x, y ) = xe xy

f xyy ( x, y ) = 2e x

f yy ( x, y ) = x 2 e xy

Thus f xyy (1, 1) = 2e . 16.

f x ( x, y ) = 24 x 2 + 4 xy 2

f yx ( x, y ) = x( ye xy ) + e xy (1) = e xy ( xy + 1)

Thus, f xx ( x, y ) + f xy ( x, y ) + f yx ( x, y ) + f yy ( x, y )

f x ( x, y ) = 3 x 2 − 6 y 2 + 2 x

= y 2 e xy + e xy ( xy + 1) + e xy ( xy + 1) + x 2 e xy

f xy ( x, y ) = −12 y

= e xy ( x 2 + 2 xy + y 2 + 2)

Thus f xy (1, − 1) = 12 . 17.

(

∂c 1 2 = 3qA + qB3 + 4 ∂ qB 3

(

= f ( x, y )(( x + y )2 + 2)

) (3qB2 ) − 23

2 3 = qB2 3qA + qB +4

)

21.

x 2 + y 2 ) (2) − (2 x)(2 x) 2 ( y 2 − x 2 ) ( = = 2 2 2 2 ∂ x2 x y + ( ) ( x2 + y 2 )

− 23

∂ 2c 2 2 3 = − qB2 3qA + qB +4 ∂ qA ∂ qB 3

(

(

∂ 2z

)

− 53

2 = −4qA qB2 3qA + qB3 + 4

)

( 6qA )

∂z 2y = ∂ y x2 + y2

− 53

x 2 + y 2 ) (2) − (2 y )(2 y ) 2 ( x 2 − y 2 ) ( = = 2 2 2 2 ∂ y2 x y + ( ) ( x2 + y 2 ) 2 2 2 2 ∂ 2z ∂ 2z 2( y − x ) 2( x − y ) + = + =0 2 2 ∂ x2 ∂ y 2 ( x2 + y 2 ) ( x2 + y 2 )

∂ 2z

When pA = 25 and pB = 4 , then qA = 10 − 25 + 16 = 1 and qB = 20 + 25 − 44 = 1 , and 18.

∂z 2x = ∂ x x2 + y 2

∂ 2c 4 1 −5 = −4(8) 3 = − =− . ∂ qA ∂ qB 32 8

f x ( x, y ) = 4 x 3 y 4 + 9 x 2 y 2 − 7 f xy ( x, y ) = 16 x3 y 3 + 18 x 2 y

22. 6 z

f xx ( x, y ) = 12 x 2 y 4 + 18 xy 2

∂z − 6 x2 = 0 ∂x

∂z 6 x 2 x 2 = = ∂x 6 z z

f xyx ( x, y ) = 48 x 2 y 3 + 36 xy f xxy ( x, y ) = 48 x 2 y 3 + 36 xy

∂2 z

Thus f xyx ( x, y ) = f xxy ( xy ) .

∂x 2

688

=

z (2 x) − x 2 z2

∂z ∂x

=

2 xz − x 2 z2

( ) = 2xz x2 z

2

− x4

z3


23. 2 z

Section 17.6

∂z + 2y = 0 ∂y

Problems 17.6

∂z 2y y =− =− ∂y 2z z

∂ 2z

=−

∂ y2

z (1) − y ⋅ ∂∂ zy z2

( )=−z y

=−

z−y − z

1. z = 5x + 3y , x = 2r + 3s, y = r – 2s ∂z ∂z∂x ∂z ∂y = + = (5)(2) + (3)(1) = 13 ∂r ∂x ∂r ∂ y ∂r 2

z2

∂z ∂z∂x ∂z ∂y = + = (5)(3) + (3)(−2) = 9 ∂s ∂x ∂s ∂ y ∂s

+ y2 z3

From the original equation, z 2 + y 2 = 3 x 2 . Thus

∂ 2z ∂y

=−

2

3x 2 z

3

2. z = 2 x 2 + 3xy + 2 y 2 , x = r 2 − s 2 , y = r 2 + s 2 ∂z ∂z ∂x ∂z ∂y = + ∂r ∂x ∂r ∂y ∂r = (4 x + 3 y )(2r ) + (3x + 4 y )(2r ) = 14r ( x + y ) ∂z ∂z ∂x ∂z ∂y = + ∂s ∂x ∂s ∂y ∂s = (4 x + 3 y )(−2 s ) + (3 x + 4 y )(2 s ) = −2 s ( x − y )

.

24. 2 z 2 = x 2 + 2 xy + xz (Eq. 1). Differentiating both sides of Eq. 1 with respect to y: ∂z ∂z ∂z 4z = 2x + x , (4 z − x) = 2x , ∂y ∂y ∂y

∂z 2x = . ∂ y 4z − x

3. z = e x + y , x = t 2 + 3, y = t 3 dz ∂t dx ∂z dy = + dt ∂x dt ∂y dt

Differentiating both sides of Eq. 1 with respect to x: ∂z ⎡ ∂z ⎤ = 2x + 2 y + ⎢ x + z (1) ⎥ , 4z ∂x ⎣ ∂x ⎦ ∂z ∂ z 2x + 2 y + z (4 z − x) = 2x + 2 y + z , . = ∂x ∂x 4z − x ∂z Differentiating with respect to x: ∂y

⎛3 ⎞ = e x + y (2t ) + e x + y ⎜ t1/ 2 ⎟ ⎝2 ⎠ 3 ⎞ x+ y ⎛ t⎟ =e ⎜ 2t + 2 ⎠ ⎝

4. z = 8 x + y , x = t 2 + 3t + 4 , y = t 3 + 4

∂z

dz ∂ z dx ∂ z dy = + dt ∂ x dt ∂ y dt

− 1⎤⎥ ⎦

=

(4 z − x)[1] − x ⎡⎣ 4 ∂ x − 1⎤⎦ ∂ 2z = 2⋅ ∂ x∂ y (4 z − x)2 4(2 x + 2 y + z )

(4 z − x) − x ⎡⎢ 4 z − x ⎣ = 2⋅ (4 z − x )2 = 2⋅ = 2⋅

(4 z − x) 2 − x[4(2 x + 2 y + z ) − (4 z − x)] (4 z − x)

3

16 z − 8 xz − 8 x 2 − 8 xy

= 16 ⋅

= 16 ⋅ =

=

2

(4 z − x)3 2 z 2 − xz − x 2 − xy (4 z − x )3

( x2 + 2 xy + xz ) − xz − x2 − xy

16 xy (4 z − x)3

(4 z − x)3

.

689

4 8x + y

(2t + 3) +

3t 2 + 16t + 24 2 8x + y

1 2 8x + y

(3t 2 )



5. w = x 2 z 2 + xyz + yz 2 , x = 5t, y = 2t + 3, z = 6 – t dw ∂ w dx ∂ w dy ∂ w dz = + + dt ∂ x dt ∂ y dt ∂ t dt

(

)

( ) ( ) = 5 ( 2 xz 2 + yz ) + 2 ( xz + z 2 ) − ( 2 x 2 z + xy + 2 yz )

= 2 xz 2 + yz (5) + xz + z 2 (2) + 2 x 2 z + xy + 2 yz (−1)

(

)

6. w = ln x 2 + y 2 + z 2 , x = 2 – 3t, y = t 2 + 3 , z = 4 – t dw ∂ w dx ∂ w dy ∂ w dz = + + dt ∂ x dt ∂ y dt ∂ z dt = =

(

2x 2

2

2

(−3) +

x +y +z −2(3 x − 2 yt + z )

2y 2

2

x +y +z

2

(2t ) +

2z 2

x + y2 + z2

(−1)

x2 + y2 + z 2

7. z = x 2 + xy 2

) , x = r + s + t, y = 2r – 3s + 8t 3

∂z ∂z∂x ∂z ∂y = + ∂t ∂ x ∂t ∂ y ∂t

( ) ( 2x + y 2 )[1] + 3( x2 + xy2 ) 2 = 3 ( x 2 + xy 2 ) ( 2 x + y 2 + 16 xy ) = 3 x 2 + xy 2

2

2

(2 xy )[8]

8. z = x 2 + y 2 , x = r 2 + s − t , y = r – s + t x y 2 xr + y ∂z ∂z∂x ∂z ∂y = (2r ) + (1) = = + ∂r ∂x ∂r ∂ y ∂r x2 + y 2 x2 + y2 x2 + y 2

9. w = x 2 + xyz + z 2 , x = r 2 − s 2 , y = rs, z = r 2 + s 2

∂w ∂w∂x ∂w∂ y ∂w∂z = + + ∂s ∂x ∂s ∂ y ∂s ∂z ∂s = (2 x + yz )(−2s ) + ( xz )(r ) + ( xy + 2 z ) (2s ) = −2 s (2 x + yz ) + r ( xz ) + 2s ( xy + 2 z )

10. w = e xyz , x = r 2 s3 , y = ln(r − s), z = rs 2 ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r ⎛ s2 ⎞ ⎛ 1 ⎞ xyz ⎜ ⎟ = yze xyz (2rs3 ) + xze xyz ⎜ + xye ⎟ ⎜ 2 ⎟ ⎝r−s⎠ ⎝ 2 rs ⎠ ⎛ xz xys 2 ⎞ ⎟ = e xyz ⎜ 2rs3 yz + + ⎜ r − s 2 rs 2 ⎟⎠ ⎝ 690


Section 17.6

11. y = x 2 − 7 x + 5 , x = 19rs + 2s 2t 2

∂ y dy ∂ x = = (2 x − 7)(19s ) = 19s (2 x − 7) ∂ r dx ∂ r 12. y = 4 − x 2 , x = 2r + 3s – 4t

∂ y dy ∂ x = = (−2 x)(−4) = 8 x ∂ t dx ∂ t 13. z = (4 x + 3 y )3 , x = r 2 s , y = r – 2s; r = 0, s = 1

∂z ∂z∂x ∂z ∂y = + ∂r ∂x ∂r ∂ y ∂r = 12(4 x + 3 y )2 (2rs ) + 9(4 x + 3 y ) 2 (1) = 3(4 x + 3 y ) 2 (8rs + 3)

When r = 0, s = 1, then x = 0, y = –2, and

∂z = 324 . ∂r

14. z = 2 x + 3 y , x = 3t + 5, y = t 2 + 2t + 1; t = 1 dz ∂ z dx ∂ z dy = + dt ∂ x dt ∂ y dt = =

2 2 2x + 3y 3(t + 2)

(3) +

3 2 2x + 3y

(2t + 2)

2x + 3y

When t = 1, then x = 8, y = 4 and

dz 9 9 = = . dt 28 2 7

15. w = e2 x +3 y ( x 2 + 4 z 2 ), x = rs, y = 2s − 3r, and z = r + s; r = 1, s = 0. ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s = [2e2 x +3 y ( x 2 + 4 z 2 ) + e2 x +3 y (2 x)](r ) + 3e2 x +3 y ( x 2 + 4 z 2 )(2) + e2 x +3 y (8 z )(1)

When r = 1, s = 0, then x = 0, y = −3 and z = 1. ∂w = [2e −9 (4) + e−9 (0)](1) + 3e−9 (4)(2) + e−9 (8)(1) ∂s 40 = e−9 [8 + 24 + 8] = e9 x , x = 2t 2 − 3rs − r 2 t ; r = 0, s = 2, t = –1 x−5 ∂ y dy ∂ x −5 = = 4t − r 2 ∂ t dx ∂ t ( x − 5)2

16. y =

(

)

When r = 0, s = 2, and t = –1, then x = 2 and

∂ y 20 = ∂t 9

691


17.


∂c ∂ c ∂ qA ∂ c ∂ qB = + ∂ pA ∂ qA ∂ pA ∂ qB ∂ pA −2 −2 ⎡1 ⎤ ⎡1 ⎤ 2 3 2 3 = ⎢ 3qA + qB + 4 3 ( 6qA ) ⎥ (−1) + ⎢ 3qA + qB + 4 3 3qB2 ⎥ (1) ⎣3 ⎦ ⎣3 ⎦

(

(

2 = 3qA + qB3 + 4

)

(

) ( )

) ( −2qA + qB2 ) − 23

∂c ∂ c ∂ qA ∂ c ∂ qB = + ∂ pB ∂ qA ∂ pB ∂ qB ∂ pB −2 −2 ⎡1 ⎤ ⎡1 ⎤ 2 2 3 = ⎢ 3qA + qB3 + 4 3 ( 6qA ) ⎥ ( 2 pB ) + ⎢ 3qA + qB + 4 3 3qB2 ⎥ (−11) ⎣3 ⎦ ⎣3 ⎦

(

(

2 3 = 3qA + qB +4

)

(

) ( )

) ( 4qA pB − 11qB2 ) − 23

When pA = 25 and pB = 4 , then qA = 10 − 25 + 16 = 1 , qB = 20 + 25 − 44 = 1 , and

18. a.

∂c 1 ∂c 5 −2 −2 = (8) 3 (−1) = − and = (8) 3 (5) = . 4 4 ∂ pA ∂ pB dw ∂ w dx ∂ w dy = + dt ∂ x dt ∂ y dt

b. Since

19. a.

b.

dy dw ∂ w dx ∂ w = 1 , from (a), = + dt dt ∂ x dt ∂ y

∂w ∂w ∂x ∂w ∂y = + ∂t ∂x ∂t ∂y ∂t

w = 2 x 2 ln 3x − 5 y , x = s t 2 + 2 and y = t − 3e2− s . ⎡ 2 x2 ⎤ ∂w ⎡ 2 x 2 (3) ⎤ s (2t ) = ⎢ 4 x ln 3 x − 5 y + +⎢ (−5) ⎥ (1) ⎥ ∂t ⎢⎣ 3x − 5 y ⎦⎥ 2 t 2 + 2 ⎣⎢ 3x − 5 y ⎦⎥ When s = 1 and t = 0, then x = 2 and y = −3e. ⎡ ⎤ ∂w ⎡ 2(2)(3) ⎤ 2(2) = ⎢ 4 2 ln 3 2 − 5(−3e) + (−5) ⎥ ⎥ (0) + ⎢ ∂t ⎣ 3 2 − 5(−3e) ⎦ ⎣ 3 2 − 5(−3e) ⎦ 20 =− 3 2 + 15e

692


Section 17.7

20. p = aP – whL, where P = f(l, k) and l = Lg(h). ∂p ∂P ⎡∂ P ∂ l ∂ P ∂ k ⎤ =a − wh = a ⎢ + ⎥ − wh ∂L ∂L ⎣ ∂ l ∂ L ∂ k ∂ L⎦ ∂P ⎤ ∂P ⎡∂ P = a⎢ ⋅ 0 ⎥ − wh = a g ( h) + g (h) − wh ∂ ∂ ∂l l k ⎣ ⎦ ∂p ∂P ⎡∂ P ∂ l ∂ P ∂ k ⎤ =a − wL = a ⎢ + ⎥ − wL ∂h ∂h ⎣ ∂l ∂ h ∂ k ∂ h⎦

6.

⎧ f x ( x, y, z , w) = 2 x − w = 0 ⎪⎪ f y ( x, y, z , w) = 2 y + w = 0 ⎨ ⎪ f z ( x, y, z , w) = 2 z − 2 w = 0 ⎪⎩ f w ( x, y, z , w) = − x + y − 2 z + 6 = 0

Solving the system gives the critical point (1, –1, 2, 2).

∂P ⎤ ⎡∂ P Lg ′(h) + = a⎢ ⋅ 0 − wL ∂ k ⎥⎦ ⎣ ∂l ∂P Lg ′(h) − wL =a ∂l

7.

f ( x, y ) = x 2 + y 2 − 5 x + 4 y + xy ⎧ f x ( x, y ) = 2 x + y − 5 = 0 ⎨ f ( x, y ) = x + 2 y + 4 = 0 ⎩ y Solving the system gives the critical point ⎛ 14 13 ⎞ ⎜ , − ⎟. 3⎠ ⎝ 3

2.

3⎞ ⎛ 2 ⎜ −2, ⎟ , D = (2)(6) − 0 = 12 > 0 and 2⎠ ⎝ 3⎞ ⎛ f xx ( x, y ) = 2 > 0 . Thus at ⎜ −2, ⎟ there is a 2⎠ ⎝ relative minimum.

f ( x, y ) = x 2 + 4 y 2 − 6 x + 16 y

8.

⎧ f x ( x, y ) = 2 x − 6 = 0 ⎨ f ( x, y ) = 8 y + 16 = 0 ⎩ y Critical point: (3, –2)

3.

f ( x, y ) =

f ( x, y ) = −2 x 2 + 8 x − 3 y 2 + 24 y + 7 ⎧ f x ( x, y ) = −4 x + 8 = 0 ⎨ ⎩ f y ( x, y ) = −6 y + 24 = 0 Critical point: (2, 4) Second-Derivative Test f xx ( x, y ) = −4 , f yy ( x, y ) = −6 ,

5 3 2 3 15 2 x + y − x + y2 − 4 y + 7 3 3 2

f xy ( x, y ) = 0 . At (2, 4),

⎧⎪ f x ( x, y ) = 5 x 2 − 15 x = 0 ⎨ 2 ⎪⎩ f y ( x, y ) = 2 y + 2 y − 4 = 0 Both equations are easily solved by factoring. Critical points: (0, −2), (0, 1), (3, −2), (3, 1)

D = (−4)(−6) − 02 = 24 > 0 and f xx ( x, y ) = −4 < 0 ; thus there is a relative maximum at (2, 4).

4. f(x, y) = xy − x + y f x ( x, y ) = y − 1 f y ( x, y ) = x + 1

9.

f ( x, y ) = y − y 2 − 3 x − 6 x 2 ⎧ f x ( x, y ) = −3 − 12 x = 0 ⎨ f ( x, y ) = 1 − 2 y = 0 ⎩ y ⎛ 1 1⎞ Critical point ⎜ − , ⎟ ⎝ 4 2⎠ Second-Derivative Test f xx ( x, y ) = −12 , f yy ( x, y ) = −2 , f xy ( x, y ) = 0

Critical point: (−1, 1) 5.

f ( x, y ) = x 2 + 3 y 2 + 4 x − 9 y + 3 ⎧ f x ( x, y ) = 2 x + 4 = 0 ⎨ f ( x, y ) = 6 y − 9 = 0 ⎩ y 3⎞ ⎛ Critical point ⎜ −2, ⎟ 2⎠ ⎝ Second-Derivative Test f xx ( x, y ) = 2 , f yy ( x, y ) = 6 , f xy ( x, y ) = 0 . At

Problems 17.7 1.

f ( x, y, z , w) = x 2 + y 2 + z 2 − w( x − y + 2 z − 6)

f ( x, y, z ) = 2 x 2 + xy + y 2 + 100 − z ( x + y − 200) ⎧ f x ( x, y , z ) = 4 x + y − z = 0 ⎪ ⎨ f y ( x, y , z ) = x + 2 y − z = 0 ⎪ f ( x, y, z ) = − x − y + 200 = 0 ⎩ z Solving the system gives the critical point (50, 150, 350).

⎛ 1 1⎞ At ⎜ − , ⎟ , D = (−12)(−2) − 02 = 24 > 0 and ⎝ 4 2⎠ ⎛ 1 1⎞ f xx ( x, y ) = −12 < 0 . Thus at ⎜ − , ⎟ there is a ⎝ 4 2⎠ relative maximum. 693


10.

ISM: Introductory Mathematical Analysis Second-Derivative Test f xx ( x, y ) = 2 x − 4 , f yy ( x, y ) = 16 y − 4 ,

3 2 y + 3 xy − 10 x − 9 y + 2 2 ⎧ f x ( x, y ) = 4 x + 3 y − 10 = 0 ⎨ f ( x, y ) = 3 y + 3 x − 9 = 0 ⎩ y Critical point: (1, 2) Second-Derivative Test f xx ( x, y ) = 4, f yy ( x, y ) = 3, f xy ( x, y ) = 3. f ( x, y ) = 2 x 2 +

f xy ( x, y ) = 0 . At (0, 0), D = (−4)(−4) − 02 = 16 > 0 and f xx ( x, y ) = −4 < 0 ; thus a relative maximum. ⎛ 1⎞ At ⎜ 4, ⎟ , D = (4)(4) − 02 = 16 > 0 and ⎝ 2⎠ f xx ( x, y ) = 4 > 0 ; thus a relative minimum.

At (1, 2), D = (4)(3) − 32 = 3 > 0 and f xx ( x, y ) = 4 > 0; thus there is a relative minimum at (1, 2). 11.

⎛ 1⎞ At ⎜ 0, ⎟ , D = (−4)(4) − 02 = −16 < 0 ; thus ⎝ 2⎠ neither.

f ( x, y ) = x 2 + 3xy + y 2 + x + 3

At (4, 0), D = (4)(−4) − 02 = −16 < 0 , thus neither.

⎧ f x ( x, y ) = 2 x + 3 y + 1 = 0 ⎨ f ( x, y ) = 3 x + 2 y = 0 ⎩ y 3⎞ ⎛2 Critical point: ⎜ , − ⎟ 5 5 ⎝ ⎠ Second-Derivative Test

14.

⎧⎪ f ( x, y ) = 2 x − y + 3 x 2 = 0 ⎨ x ⎪⎩ f y ( x, y ) = 2 y − x = 0 1⎞ ⎛ 1 Critical points: (0, 0), ⎜ − , − ⎟ 2 4⎠ ⎝ Second-Derivative Test f xx ( x, y ) = 2 + 6 x , f yy ( x, y ) = 2 ,

3⎞ ⎛2 f xx ( x, y ) = 2, f yy = 2, f xy = 3. At ⎜ , − ⎟ , 5 5 ⎝ ⎠ D = (2)(2) − (3)2 = −5 < 0, so there is no 3⎞ ⎛2 relative extremum at ⎜ , − ⎟ . 5⎠ ⎝5

12.

f ( x, y ) =

f xy ( x, y ) = −1 . At (0, 0),

x3 + y 2 − 2 x + 2 y − 2 xy 3

D = (2)(2) − (−1) 2 = 3 > 0 and f xx ( x, y ) = 2 > 0 ; thus relative minimum.

⎧⎪ f x ( x, y ) = x 2 − 2 − 2 y = 0 ⎨ ⎪⎩ f y ( x, y ) = 2 y + 2 − 2 x = 0 Critical points: (2, 1), (0, –1) Second-Derivative Test f xx ( x, y ) = 2 x , f yy ( x, y ) = 2 , f xy ( x, y ) = −2 .

1⎞ ⎛ 1 At ⎜ − , − ⎟ , D = (−1)(2) − (−1)2 = −3 < 0 ; 4⎠ ⎝ 2 thus neither.

15.

At (2, 1), D = (4)(2) − (−2) 2 = 4 > 0 and f xx ( x, y ) = 4 > 0 , so a relative minimum at (2, 1). At (0, –1), D = (0)(2) − (−2)2 = −4 < 0 ; thus neither at (0, –1). 13.

f ( x, y ) =

(

) (

f ( x, y ) = x 2 + y 2 − xy + x3

)

1 3 x + 8 y3 − 2 x2 + y 2 + 1 3

l2 + 2lk + 3k 2 − 69l − 164k + 17 2 ⎧ fl (l , k ) = l + 2k − 69 = 0 ⎨ f (l , k ) = 2l + 6k − 164 = 0 ⎩ k Critical point: (43, 13) Second-Derivative Test fll (l , k ) = 1 , f kk (l , k ) = 6 , flk (l , k ) = 2 f (l , k ) =

At (43, 13), D = (1)(6) − 22 = 2 > 0 and fll (l , k ) = 1 > 0 ; thus there is a relative minimum at (43, 13).

⎧⎪ f x ( x, y ) = x 2 − 4 x = 0 ⎨ 2 ⎪⎩ f y ( x, y ) = 8 y − 4 y = 0 ⎛ 1⎞ ⎛ 1⎞ Critical points: (0, 0), ⎜ 4, ⎟ , ⎜ 0, ⎟ , (4, 0) ⎝ 2⎠ ⎝ 2⎠

694


16.

Section 17.7 Critical points: (0, −2), (0, 2) [Note that y = 0 does not give rise to a common solution of (1) and (2).] Second-Derivative Test

f (l , k ) = l 2 + k 2 − 2lk ⎧ fl (l , k ) = 2l − 2k ⎨ f (l , k ) = 2k − 2l ⎩ k Critical points: (r, r) where r is any real number. Second Derivative Test fll (l , k ) = 2, f kk (l , k ) = 2, and flk (l , k ) = −2.

(

f ( p, q ) = pq −

(0, 2), D = (0)(0) − (4)2 = −16 < 0 ; thus neither.

1 1 − p q

20.

thus there is a relative minimum at (–1, –1). f ( x, y ) = ( x − 3)( y − 3)( x + y − 3) = ( y − 3)( x 2 + xy − 6 x − 3 y + 9)

21. P = f (l , k )

2

= 1.08l 2 − 0.03l 3 + 1.68k 2 − 0.08k 3 ⎪⎧ Pl = 2.16l − 0.09l 2 = 0 ⎨ 2 ⎪⎩ Pk = 3.36k − 0.24k = 0 Critical points: (0, 0), (0, 14), (24, 0), (24, 14) Second-Derivative Test Pll = 2.16 − 0.18l , Pkk = 3.36 − 0.48k , Plk = 0 .

= ( x − 3)( xy − 3 x + y − 6 y + 9) ⎧ f x ( x, y ) = ( y − 3)(2 x + y − 6) = 0 ⎨ f ( x, y ) = ( x − 3)( x + 2 y − 6) = 0 ⎩ y Critical points: (2, 2), (3, 3), (3, 0), (0, 3) Second-Derivative Test f xx ( x, y ) = 2( y − 3) , f yy ( x, y ) = 2( x − 3) , f xy ( x, y ) = 2 x + 2 y − 9 . At (2, 2),

At (0, 0), D = (2.16)(3.36) − 02 > 0 and Pll = 2.16 > 0 ; thus relative minimum.

D = (−2)(−2) − (−1)2 = 3 > 0 and f xx ( x, y ) = −2 < 0 ; thus relative maximum.

At (0, 14), D = (2.16)(−3.36) − 02 < 0 ; thus no extremum. At (24, 0),

At (3, 3), D = (0)(0) − 32 = −9 < 0 ; thus neither.

D = (−2.16)(3.36) − 02 < 0 ; thus no extremum.

At (3, 0), D = (−6)(0) − (−3)2 = −9 < 0 ; thus

At (24, 14), D = (−2.16)(−3.36) − 02 > 0 and Pll = −2.16 < 0 ; thus l = 24, k = 14 gives a relative maximum.

neither. At (0, 3), D = (0)(−6) − (−3)2 = −9 < 0 ; thus neither.

(

)(

)

f ( x, y ) = y 2 − 4 e x − 1

( (

) )

⎧ f ( x, y ) = e x y 2 − 4 = 0 ⎪ x ⎨ x ⎪ f y ( x, y ) = 2 y e − 1 = 0 ⎩

f ( x, y ) = ln( xy ) + 2 x 2 − xy − 6 x 1 ⎧ ⎪⎪ f x ( x, y ) = x + 4 x − y − 6 = 0 ⎨ 1 ⎪ f y ( x, y ) = − x = 0 y ⎪⎩ ⎛3 2⎞ The only critical point is ⎜ , ⎟ . ⎝2 3⎠ 1 1 f xx ( x, y ) = − + 4 , f yy ( x, y ) = − , 2 x y2 ⎛3 2⎞ f xy ( x, y ) = −1 . At ⎜ , ⎟ , ⎝2 3⎠ ⎛ 32 ⎞⎛ −9 ⎞ D = ⎜ ⎟⎜ ⎟ − (−1)2 = −9 < 0 ; thus neither. ⎝ 9 ⎠⎝ 4 ⎠

D = (2)(2) − 12 = 3 > 0 and f pp ( p, q ) = 2 > 0 ;

19.

)

D = (0)(0) − (−4) 2 = −16 < 0 ; thus neither. At

⎧ f ( p, q ) = q + 1 = 0 ⎪⎪ p p2 ⎨ ⎪ f q ( p, q ) = p + 12 = 0 q ⎪⎩ Critical point: (–1, –1) Second-Derivative Test 2 2 f pp ( p, q ) = − , f qq ( p, q) = − , 3 p q3 f pq ( p, q) = 1 . At (–1, –1),

18.

(

f xy ( x, y ) = 2 ye x . At (0, –2),

At (r, r), D = (2)(2) − (−2) 2 = 0, thus we cannot make a conclusion. 17.

)

f xx ( x, y ) = e x y 2 − 4 , f yy ( x, y ) = 2 e x − 1 ,

(1) (2) 695



22. Q = 18c + 20d − 2c 2 − 4d 2 − cd ⎧Qc = 18 − 4c − d = 0 ⎨Q = 20 − 8d − c = 0 ⎩ d Critical point: c = 4, d = 2 Qcc = −4, Qdd = −8, Qcd = −1

When c = 4 and d = 2, then D = (−4)(−8) − (−1) 2 > 0 and Qcc = −4 < 0 ; thus relative maximum at c = 4, d = 2. 23. Profit per lb for A = pA − 60 . Profit per lb for B = pB − 70 .

Total Profit = P = ( pA − 60 ) qA + ( pB − 70 ) qB P = ( pA − 60 ) ⎡⎣5 ( pB − pA ) ⎤⎦ + ( pB − 70 ) ⎡⎣500 + 5 ( pA − 2 pB ) ⎤⎦

Thus ⎧ ∂ P = −10 ( p − p + 5 ) = 0 A B ⎪ ∂ pA ⎨ ∂P = 10 ( pA − 2 pB + 90 ) = 0 ⎪ ⎩ ∂ pB Critical point: pA = 80 , pB = 85

∂ 2P ∂ pA2

= −10 ,

∂ 2P ∂ pB2

= −20 ,

∂ 2P = 10 . When pA = 80 and pB = 85 , then ∂ pB∂ pA

D = (−10)(−20) − (10)2 = 100 > 0 and

∂ 2P ∂ pA2

= −10 < 0 ; thus relative maximum at pA = 80 , pB = 85 .

24. Profit per lb for A = pA − a . Profit per lb for B = pB − b .

Total Profit = P = ( pA − a ) qA + ( pB − b ) qB P = ( pA − a ) ⎡⎣5 ( pB − pA ) ⎤⎦ + ( pB − b ) ⎡⎣500 + 5 ( pA − 2 pB ) ⎤⎦

⎧ ∂ P = −5 ( 2 p − 2 p + b − a ) = 0 A B ⎪ ∂ pA ⎨ ∂P = 5 ( 2 pA − 4 pB + 2b − a + 100 ) = 0 ⎪ ⎩ ∂ pB a b Critical point: pA = 50 + , pB = 50 + 2 2 2 2 2 ∂ P ∂ P ∂ P = −10 , = −20 , = 10 2 2 p ∂ ∂ pA ∂ pB B∂ pA

When pA = 50 +

a b ∂ 2P and pB = 50 + , then D = (−10)(−20) − (10)2 = 100 > 0 and = −10 < 0 ; thus a relative 2 2 ∂ pA2

maximum at pA = 50 +

a b , pB = 50 + . 2 2

696


Section 17.7

25. pA = 100 − qA , pB = 84 − qB , c = 600 + 4 ( qA + qB ) . Revenue from market A = rA = pA qA = (100 − qA ) qA . Revenue from market B = rB = pB qB = ( 84 − qB ) qB . Total Profit = Total Revenue – Total Cost P = (100 − qA ) qA + ( 84 − qB ) qB − ⎡⎣ 600 + 4 ( qA + qB ) ⎤⎦ ⎧ ∂ P = 96 − 2q = 0 A ⎪ ∂ qA ⎨∂P = 80 − 2qB = 0 ⎪ ⎩ ∂ qB Critical point: qA = 48 , qB = 40

∂ 2P ∂ qA2

= −2 ,

∂ 2P ∂ qB2

= −2 ,

∂ 2P =0. ∂ qB∂ qA

At qA = 48 and qB = 40 , then D = (−2)(−2) − 02 = 4 > 0 and

∂ 2P ∂ qA2

= −2 < 0 ; thus relative maximum at

qA = 48 , qB = 40 . When qA = 48 and qB = 40 , then selling prices are pA = 52, pB = 44 , and profit = 3304.

26. qA = 3 − pA + 2 pB , qB = 5 + 5 pA − 2 pB Revenue from A = pA qA . Revenue from B = pB qB . Total cost of producing qA units of A and qB units of B is 3qA + 2qB . Total Profit = Total Revenue − Total Cost P = pA qA + pB qB − (3qA + 2qB )

2 P = − pA − 2 pB2 + 7 pA pB − 4 pA + 3 pB − 19

⎧ ∂P ⎪⎪ ∂p = −2 pA + 7 pB − 4 A ⎨ ∂P ⎪ = 7 pA − 4 pB + 3 ⎩⎪ ∂pB 5 22 , pB = 41 41 2 2 2 ∂ P ∂ P ∂ P = −2, = −4, =7 2 2 ∂pB ∂pA ∂pA ∂pB

Critical point: pA = −

5 22 , pB = , then D = (−2)(−4) − 7 2 = −41 < 0, so this is not a relative extremum. Thus the profit 41 41 cannot be maximized.

At pA = −

3 2 2 , pB = 72 − 2qB2 qA + 3qB2 , pA = 60 − qA 2 Total Profit = Total Revenue – Total Cost P = ( pA qA + pB qB ) − c

27. c =

⎞ 3 3 ⎛3 2 P = 60qA − qA + 72qB − 2qB − ⎜ qA + 3qB2 ⎟ 2 ⎝ ⎠ ⎧ ∂ P = 60 − 3q − 3q 2 = 3 ( 5 + q )( 4 − q ) A A A A ⎪ ∂ qA ⎨∂P = 72 − 6qB − 6qB2 = 6 ( 4 + qB )( 3 − qB ) ⎪ ⎩ ∂ qB

Since we want qA ≥ 0 and qB ≥ 0 , the critical point occurs when qA = 4 and qB = 3. 697


∂ 2P ∂ qA2 and

= −3 − 6qA ,

∂ 2P ∂ qA2

∂ 2P ∂ qB2

= −6 − 12qB ,


∂ 2P = 0 . When qA = 4 and qB = 3, then D = (−27)(−42) − 02 > 0 ∂ qB∂ qA

= −27 < 0 ; thus relative maximum at qA = 4, qB = 3.

28. c = 2(qA + qB + qA qB ), Total Profit = Total Revenue − Total Cost P = ( pA qA + pB qB ) − c = pA (20 − 2 pA ) + pB (10 − pB ) − [20 − 2 pA + 10 − pB + (20 − 2 pA )(10 − pB )] 2 = −2 pA − pB2 − 2 pA pB + 42 pA + 31 pB + 230

⎧ ∂P ⎪⎪ ∂p = −4 pA − 2 pB + 42 A ⎨ ∂P ⎪ = −2 pA − 2 pB + 31 ⎪⎩ ∂pB 11 , pB = 10 2 ∂2 P ∂2 P ∂2 P = −4, = −2, = −2 2 ∂pB ∂pA ∂pA ∂pB2

Critical point: pA =

When pA =

∂2 p 11 = −4 < 0, so the maximum profit occurs , pB = 10, then D = (−4)(−2) − (−2)2 = 4 > 0, and 2 2 ∂pA

when pA = 5.5 and pB = 10. At these prices, qA = 9, qB = 0, and the total profit is 40.5. 29. Refer to the diagram in the text. xyz = 6 C = 3xy + 2[1(xz)] + 2[0.5(yz)] 6 . Thus Note that z = xy ⎛ 6 ⎞ ⎛ 6 ⎞ 12 6 C = 3xy + 2 xz + yz = 3xy + 2 x ⎜ ⎟ + y ⎜ ⎟ = 3xy + + y x ⎝ xy ⎠ ⎝ xy ⎠ 6 ⎧∂ C ⎪⎪ ∂ x = 3 y − 2 = 0 x ⎨ dC 12 ⎪ = 3x − =0 ⎪⎩ dy y2

A critical point occurs at x = 1 and y = 2. Thus z = 3. ∂ 2 C 12 ∂ 2 C 24 ∂ 2C = , = , = 3. ∂ x 2 x3 ∂ y 2 y 3 ∂ x∂ y When x = 1 and y = 2, then d = (12)(3) − (3) 2 = 27 > 0 and dimensions should be 1 ft by 2 ft by 3 ft.

698

∂ 2C ∂ x2

= 12 > 0 . Thus we have a minimum. The


Section 17.7

30. p = 92 − qA − qB , cA = 10qA , cB = 0.5qB2 Since Profit = Total Revenue – Total Cost, then Profit of A = pqA − cA and Profit of B = pqB − cB . Thus profit P of monopoly is P = pqA − cA + pqB − cB

32. y =

2

⎛ x − 10 ⎞ f ( x, y ) = x 2 + 4 ⎜ ⎟ +6 ⎝ 4 ⎠ Setting the derivative equal to 0 gives ⎛ x − 10 ⎞ ⎛ 1 ⎞ 2 x + 4(2) ⎜ ⎟ ⎜ ⎟ = 0 , from which x = 2. ⎝ 4 ⎠⎝ 4 ⎠ 5 The second-derivative is > 0 ,so we have a 2 relative minimum If x = 2, then y = –2. Thus at (2, –2) there is a relative minimum

= p ( qA + qB ) − cA − cB

= ( 92 − qA − qB )( qA + qB ) − 10qA − 0.5qB2 2 = 82qA + 92qB − qA − 2qA qB − 1.5qB2

⎧ ∂ P = 82 − 2q − 2q = 0 A B ⎪ ∂ qA ⎨∂P = 92 − 2qA − 3qB = 0 ⎪ ⎩ ∂ qB Critical point: qA = 31, qB = 10

∂ 2P ∂ qA2

= −2,

∂ 2P ∂ qB2

= −3 ,

2 + 3qB2 + 2qA qB + aqA + bqB + d 33. c = qA

We are given that ( qA , qB ) = ( 3, 1) is a critical point. ⎧ ∂ c = 2q + 2q + a = 0 A B ⎪ ∂ qA ⎨ ∂c = 6qB + 2qA + b = 0 ⎪ ⎩ ∂ qB Substituting the given values for qA and qB into both equations gives a = –8 and b = –12. Since c = 15 when qA = 3 and qB = 1 , from the jointcost function we have

∂ 2P = −2 ∂ qB∂ qA

When qA = 31 and qB = 10 , then D = (−2)(−3) − (−2) 2 = 2 > 0 and

∂ 2P ∂ qA2

= −2 < 0 ; thus relative maximum at

( )

qA = 31, qB = 10 .

31. y =

x − 10 4

15 = 32 + 3 12 + 2(3)(1) + (−8)(3) + (−12) + d ,

15 = –18 + d, 33 = d. Thus a = –8, b = –12, d = 33.

3x − 7 2 2

⎛ 3x − 7 ⎞ f ( x, y ) = −2 x 2 + 5 ⎜ ⎟ +7 ⎝ 2 ⎠ Setting the derivative equal to 0 gives ⎛ 3x − 7 ⎞ ⎛ 3 ⎞ −4 x + 5(2) ⎜ ⎟⎜ ⎟ = 0 , ⎝ 2 ⎠⎝ 2 ⎠ 15 −4 x + (3x − 7) = 0 , −8 x + 15(3 x − 7) = 0 , 2 105 . The second-derivative is 37x = 105, or x = 37 37 > 0 , so we have a relative minimum. If 2 105 28 , then y = . Thus there is a relative x= 37 37 ⎛ 105 28 ⎞ minimum at ⎜ , ⎟. ⎝ 37 37 ⎠

34. D(a, b) > 0 f xx (a, b) f yy (a, b) − ( f xy (a, b))2 > 0 f xx (a, b) f yy (a, b) > ( f xy (a, b))2 ≥ 0

a.

Since the product f xx (a, b) f yy (a, b) is positive, f xx (a, b) and f yy (a, b) must have the same sign. That is f xx (a, b) < 0 if and only if f yy (a, b) < 0.

b. Since the product f xx (a, b) f yy (a, b) is positive, f xx (a, b) and f yy (a, b) must have the same sign. That is f xx (a, b) > 0 if and only if f yy (a, b) > 0.

699

Chapter 17: Multivariable Calculus 35. a.


Profit = Total Revenue – Total Cost P = pA qA + pB qB − total cost

(

)

1 2⎞ ⎛ 2 3 = 35 − 2qA + qB qA + ( 20 − qB + qA ) qB − ⎜ −8 − 2qA + 3qA qB + 30qA + 12qB + qA ⎟ 2 ⎝ ⎠ 1 2 P = 5qA − qA − qA qB + 8qB − qB2 + 8 2 ⎧ ∂ P = 5−q −q = 0 A B ⎪ ∂ qA ⎨∂P = −qA + 8 − 2qB = 0 ⎪ ⎩ ∂ qB Critical point: qA = 2, qB = 3

∂ 2P ∂ qA2

= −1 ,

∂ 2P ∂ qB2

= −2 ,

∂ 2P = −1 ∂ qB∂ qA

At qA = 2 and qB = 3 , then D = (−1)(−2) − (−1) 2 = 1 > 0 and

∂ 2P ∂ qA2

= −1 < 0 ; thus there is a relative

maximum profit for 2 units of A and 3 units of B. b. Substituting qA = 2 and qB = 3 into the formulas for pA , pB , and P gives a selling price for A of 30, a selling price for B of 19, and a relative maximum profit of 25. ⎡ 5x 2y ⎤ 36. P = 250 ⎢ + ⎥−x− y ⎣2 + x 5+ y⎦ 2500 ⎧ ∂P −1 ⎪ ∂x = (2 + x) 2 ⎪ ⎨ ∂P 2500 ⎪ = −1 ⎪⎩ ∂y (5 + y )2

Critical point: x = 48, y = 45 5000 5000 ∂2 P ∂2 P ∂2 P , , =− =− =0 (2 + x)3 ∂y 2 (5 + y )3 ∂y∂x ∂x 2 ∂2 P 1 ⎛ 1 ⎞⎛ 1 ⎞ At x = 48 and y = 45, then D = ⎜ − ⎟ ⎜ − ⎟ − 02 > 0 and =− < 0. 2 25 25 25 ⎝ ⎠⎝ ⎠ ∂x Thus relative maximum profit at x = 48, y = 45.

37. a.

(

)

P = 5T 1 − e− x − 20 x − 0.1T 2

700


b.

(

Section 17.8

)

∂P = 5 1 − e− x − 0.2T ∂T ∂P = 5Te − x − 20 ∂x At the point (T, x) = (20, ln 5), ∂P = 5 1 − e− ln 5 − 0.2(20) ∂P ⎛1⎞ ∂T = 5(20)e− ln 5 − 20 = 100 ⎜ ⎟ − 20 = 0 ⎛ 1⎞ ∂ x ⎝5⎠ = 5 ⎜1 − ⎟ − 4 = 0 5 ⎝ ⎠

(

)

5⎞ ⎛ Thus (20, ln 5) is a critical point. In a similar fashion we verify that ⎜ 5, ln ⎟ is a critical point. 4⎠ ⎝

c.

∂ 2P ∂T2

= −0.2 ,

At (20, ln 5),

∂ 2P ∂ x2

= −5Te− x ,

∂ 2P = 5e− x ∂ T∂ x

(

D = (−0.2) ⎡ −5(20)e − ln 5 ⎤ − 5e− ln 5 ⎣ ⎦

and

∂ 2P ∂T2

)

2

⎛ 1 ⎞ ⎡ ⎛ 1 ⎞⎤ = 20 ⎜ ⎟ − ⎢5 ⎜ ⎟ ⎥ = 3 > 0 , ⎝ 5 ⎠ ⎣ ⎝ 5 ⎠⎦

2

= −0.2 < 0 . Thus we get a relative maximum at (20, ln 5).

5⎞ ⎛ At ⎜ 5, ln ⎟ , 4⎠ ⎝ 2

2

− ln ( 54 ) ⎤ ⎡ − ln ( 54 ) ⎤ ⎡ ⎛ 4 ⎞ ⎡ ⎛ 4 ⎞⎤ D = (−0.2) ⎢ −5(5)e ⎥ − ⎢5e ⎥ = 5 ⎜ ⎟ − ⎢5 ⎜ ⎟ ⎥ = −12 < 0 , so there is no relative extremum at ⎝ 5 ⎠ ⎣ ⎝ 5 ⎠⎦ ⎣ ⎦ ⎣ ⎦ 5⎞ ⎛ ⎜ 5, ln ⎟ . 4⎠ ⎝

Problems 17.8 1.

f ( x, y ) = x 2 + 4 y 2 + 6, 2 x − 8 y = 20 F ( x, y, λ ) = x 2 + 4 y 2 + 6 − λ (2 x − 8 y − 20) ⎧ Fx = 2 x − 2λ = 0 (1) ⎪ (2) ⎨ Fy = 8 y + 8λ = 0 ⎪ F = −2 x + 8 y + 20 = 0 (3) ⎩ λ From (1), x = λ ; from (2), y = −λ . Substituting x = λ and y = −λ into (3) gives −2λ − 8λ + 20 = 0 , −10λ = −20 , so λ = 2 . Thus x = 2 and y = –2. Critical point of F: (2, –2, 2). Critical point of f: (2, –2).

2.

f ( x, y ) = −2 x 2 + 5 y 2 + 7 , 3x – 2y = 7 F ( x, y, λ ) = −2 x 2 + 5 y 2 + 7 − λ (3 x − 2 y − 7) ⎧ Fx = −4 x − 3λ = 0 ⎪ ⎨ Fy = 10 y + 2λ = 0 ⎪ F = −3 x + 2 y + 7 = 0 ⎩ λ

(1) (2) (3)

701



λ 3λ ; from (2), y = − . 5 4 λ 3λ and y = − into (3) Substituting x = − 5 4 9λ 2λ 140 gives − + 7 = 0 , from which λ = − . 4 5 37 105 28 Thus x = and y = . Critical point of F: 37 37 ⎛ 105 28 140 ⎞ , ,− ⎜ ⎟ . Critical point of f: 37 ⎠ ⎝ 37 37 ⎛ 105 28 ⎞ , ⎜ ⎟. ⎝ 37 37 ⎠

5.

From (1), x = −

3.

= 2 x 2 + xy + y 2 + z − λ ( x + 2 y + 4 z − 3) ⎧ Fx ⎪⎪ F y ⎨ F ⎪ z ⎪⎩ Fλ

F ( x, y , z , λ ) = x 2 + y 2 + z 2 − λ (2 x + y − z − 9) = 2 x − 2λ = 0 = 2y − λ = 0 = 2z + λ = 0 = −2 x − y + z + 9 = 0

(1) (2) (3) (4)

From (1), x = λ ; from (2), y = z=−

λ 2

λ 2

; from (3),

6.

2

= 1 − λ yz = 0 = 1 − λ xz = 0 = 1 − λ xy = 0 = − xzy + 8 = 0

)

⎧ Fx = yz 2 − λ = 0 (1) ⎪ ⎪ Fy = xz 2 + λ = 0 (2) ⎨ 2 0 (3) λ F xyz = − = ⎪ z ⎪⎩ Fλ = − x + y − z + 20 = 0 (4) From (1) and (2), y = –x, From (1) and (3), z = 2x. Hence from (4), x = 5, so y = –5 and z = 10. Critical point of f is (5, –5, 10). Note that it is not necessary to determine λ .

λ ⎛ −λ ⎞

7. f(x, y, z) = xyz, x + 2y + 3z = 18 (xyz ≠ 0) f ( x, y, z , λ ) = xyz − λ ( x + 2 y + 3 z − 18)

4. f(x, y, z) = x + y + z, xyz = 8 F ( x, y , z , λ ) = x + y + z − λ ( xyz − 8) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ

(

f ( x, y, z ) = xyz 2 , x – y + z = 20 xyz 2 ≠ 0 f ( x, y, z , λ ) = xyz 2 − λ ( x − y + z − 20)

. Substituting into (4) gives

+⎜ ⎟ + 9 = 0 , −6λ + 18 = 0 , so ⎝ 2 ⎠ 3 3 λ = 3 . Thus x = 3, y = , z = − . Critical point 2 2 3 3 ⎛ ⎞ of F: ⎜ 3, , − , 3 ⎟ . Critical point of f: 2 ⎠ ⎝ 2 3⎞ ⎛ 3 ⎜ 3, , − ⎟ . 2 2⎠ ⎝ −2λ −

= 4x + y − λ = 0 = x + 2 y − 2λ = 0 = 1 − 4λ = 0 = −x − 2 y − 4z − 3 = 0

1 From the third equation we have λ = . 4 Substituting this value into the first two equations 1 and then eliminating y gives x = 0 and y = . 4 Finally, solving for z in the last equation gives 7 z=− . 8 7 1⎞ ⎛ 1 Critical point of F: ⎜ 0, , − , ⎟ 4 8 4⎠ ⎝ 7⎞ ⎛ 1 Critical point of f: ⎜ 0, , − ⎟ 8⎠ ⎝ 4

f ( x, y , z ) = x 2 + y 2 + z 2 , 2 x + y − z = 9 ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎩⎪ Fλ

f ( x, y, z ) = 2 x 2 + xy + y 2 + z , x + 2y + 4z = 3 F ( x, y , z , λ )

⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ

(1) (2) (3) (4)

= yz − λ = 0 = xz − 2λ = 0 = xy − 3λ = 0 = − x − 2 y − 3z + 18 = 0

From (1) and (2), y =

From (1) and (2), λ yz = λ xz , so y = x. From (2) and (3), λ xz = λ xy , so y = z. Therefore x = y = z, so from (4), x = y = z = 2. Hence, Critical point of f is (2, 2, 2). Note that it is not necessary to determine λ .

(1) (2) (3) (4)

x . From (1) and (3), 2

x . Hence from (4), x = 6, so y = 3 and z = 2. 3 Critical point of f is (6, 3, 2). Note that it is not necessary to determine λ . z=

702


8.

Section 17.8

f ( x, y , z ) = x 2 + y 2 + z 2 , x + y + z = 3 F ( x, y , z , λ ) = x 2 + y 2 + z 2 − λ ( x + y + z − 3) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎩⎪ Fλ

= 2x − λ = 0 = 2y − λ = 0 = 2z − λ = 0 = −x − y − z + 3 = 0

(1) (2) (3) (4)

λ

λ

λ

λ

. Substituting into (4), − − − + 3 = 0 , so λ = 2 . Thus x = 1, y = 1, z = 1 . 2 2 2 2 Critical point of F: (1, 1, 1, 2 ) . Critical point of f: (1, 1, 1) .

From (1)–(3), x = y = z =

9.

f ( x, y , z ) = x 2 + 2 y − z 2 , 2x – y = 0, y + z = 0 Since there are two constraints, two Lagrange multipliers are used. F ( x, y, z , λ1 , λ2 ) = x 2 + 2 y − z 2 − λ1 (2 x − y ) − λ2 ( y + z )

⎧ Fx = 2 x − 2λ1 = 0 ⎪ Fy = 2 + λ1 − λ2 = 0 ⎪⎪ ⎨ Fz = −2 z − λ2 = 0 ⎪ Fλ1 = −2 x + y = 0 ⎪ ⎪⎩ Fλ2 = − y − z = 0

(1) (2) (3) (4) (5)

From (1), x = λ1 . From (3), z = −

λ2 2

. From (4) and (5), 2x = –z, so λ1 =

λ2 4

. Substituting λ1 =

λ2 4

into (2)

8 2 2 4 4 . Thus λ1 = , x = , and z = − . From (5), y = –z and hence y = . Critical point of f: 3 3 3 3 3 4⎞ ⎛2 4 ⎜ , ,− ⎟ 3⎠ ⎝3 3

yields λ2 =

10.

f ( x, y, z ) = x 2 + y 2 + z 2 , x + y + z = 4, x – y + z = 4 Since there are two constraints, two Lagrange multipliers are used. F ( x, y, z , λ1 , λ2 ) = x 2 + y 2 + z 2 −λ1 ( x + y + z − 4) − λ2 ( x − y + z − 4) ⎧ Fx = 2 x − λ1 − λ2 = 0 ⎪ Fy = 2 y − λ1 + λ2 = 0 ⎪⎪ ⎨ Fz = 2 z − λ1 − λ2 = 0 ⎪ Fλ1 = − x − y − z + 4 = 0 ⎪ ⎪⎩ Fλ2 = − x + y − z + 4 = 0

(1) (2) (3) (4) (5)

From (4) and (5), y = 0. From (1) and (3), z = x. Substituting into (5) gives x = 2 . Thus z = 2 . Critical point of f: ( 2, 0, 2 )

703


11.


f ( x, y , z ) = xy 2 z , x + y + z = 1, x − y + z = 0 (xyz ≠ 0) Since there are two constraints, two Lagrange multipliers are used. F ( x, y, z , λ1 , λ2 ) = xy 2 z − λ1 ( x + y + z − 1) − λ2 ( x − y + z ) ⎧ F = y 2 z − λ1 − λ2 = 0 ⎪ x ⎪ Fy = 2 xyz − λ1 + λ2 = 0 ⎪ ⎨ Fz = xy 2 − λ1 − λ2 = 0 ⎪F = − x − y − z + 1 = 0 ⎪ λ1 ⎪⎩ Fλ2 = − x + y − z = 0

(1) (2) (3) (4) (5)

Subtracting (3) from (1) gives y 2 z − xy 2 = 0, so x = z (since xy 2 z ≠ 0). Subtracting (5) from (4) gives 1 1 1 1 1 −2y + 1 = 0, so y = . Substituting z = x and y = in (5) gives −2 x + = 0, so x = . Thus, z = . Critical 2 2 2 4 4 ⎛1 1 1⎞ point of f: ⎜ , , ⎟ . ⎝4 2 4⎠

12.

f ( x, y, z , w) = 3 x 2 + y 2 + 2 z 2 − 5w2 , x + 6y + 3z + 2w = 4 F ( x, y , z , w, λ ) = 3x 2 + y 2 + 2 z 2 − 5w2 − λ ( x + 6 y + 3 z + 2 w − 4) ⎧ Fx ⎪F ⎪ y ⎨ Fz ⎪ Fw ⎪F ⎪⎩ λ

= 6x − λ = 0 = 2 y − 6λ = 0 = 4 z − 3λ = 0 = −10 w − 2λ = 0 = − x − 6 y − 3z − 2w + 4 = 0

3λ λ and w = − . , y = 3λ , z = 5 6 4 240 40 720 180 48 Substituting into the last equation gives λ = and w = − . Thus x = , y= ,z= . 1201 1201 1201 1201 1201 48 240 ⎞ ⎛ 40 720 180 Critical point of F: ⎜ , , ,− , ⎟ 1201 1201 1201 1201 1201 ⎠ ⎝ 48 ⎞ ⎛ 40 720 180 Critical point of f: ⎜ , , ,− ⎟ 1201 1201 1201 1201 ⎝ ⎠

Solving the first four equations for x, y, z, and w in terms of λ gives x =

λ

13. We minimize c = f ( q1 , q2 ) = 0.1q12 + 7q1 + 15q2 + 1000 subject to the constraint q1 + q2 = 100 . F ( q1 , q2 , λ ) = 0.1q12 + 7 q1 + 15q2 +1000 − λ ( q1 + q2 − 100 )

⎧ Fq1 = 0.2q1 + 7 − λ = 0 ⎪ ⎨ Fq2 = 15 − λ = 0 ⎪ F = − q − q + 100 = 0 1 2 ⎩ λ

(1) (2) (3)

From (2), λ = 15 . Substituting λ = 15 into (1) gives 0.2q1 + 7 − 15 = 0 , so q1 = 40 . Substituting q1 = 40 into (3) gives −40 − q2 + 100 = 0 , so q2 = 60 . Thus λ = 15 , q1 = 40, and q = 60. Thus plant 1 should produce 40 units and plant 2 should produce 60 units.

704


Section 17.8

14. We minimize c = 3q12 + q1q2 + 2q22 subject to the constraint q1 + q2 = 200 . F ( q1 , q2 , λ ) = 3q12 + q1q2 + 2q22 − λ ( q1 + q2 − 200 ) ⎧ Fq1 = 6q1 + q2 − λ = 0 ⎪ ⎨ Fq2 = q1 + 4q2 − λ = 0 ⎪ F = −q − q + 200 = 0 1 2 ⎩ λ

(1) (2) (3)

3 3 Eliminating λ from (1) and (2) yields q1 = q2 . Substituting q1 = q2 into (3) yields q2 = 125 and thus 5 5 q1 = 75 . Thus plant 1 should produce 75 units and plant 2 should produce 125 units.

15. We maximize f (l , k ) = 12l + 20k − l 2 − 2k 2 subject to the constraint 4l + 8k = 88. F (l , k , λ ) = 12l + 20k − l 2 − 2k 2 −λ (4l + 8k − 88) (1) ⎧ Fl = 12 − 2l − 4λ = 0 ⎪ (2) ⎨ Fk = 20 − 4k − 8λ = 0 ⎪⎩ Fλ = −4l − 8k + 88 = 0 (3) Eliminating λ from (1) and (2) yields k = l – 1. Substituting k = l – 1 into (3) yields l = 8, so k = 7. Therefore the greatest output is f(8, 7) = 74 units (when l = 8, k = 7).

16. We maximize f (l , k ) = 20l + 25k − l 2 − 3k 2 subject to the constraint 2l + 4k = 50. F (l , k , λ ) = 20l + 25k − l 2 − 3k 2 − λ (2l + 4k − 50) ⎧ Fl = 20 − 2l − 2λ = 0 ⎪ ⎨ Fk = 25 − 6k − 4λ = 0 ⎪⎩ Fλ = −2l − 4k + 50 = 0

(1) (2) (3)

25 2 − λ . Substituting these expressions for l and k into (3) yields 6 3 20 90 85 ⎛ 90 85 ⎞ 3725 ≈ 133 units (when λ = − . Thus l = and k = . Therefore the greatest output is f ⎜ , ⎟= 28 7 7 14 ⎝ 7 14 ⎠ 90 85 ⎞ l = , k = ⎟. 7 14 ⎠

From (1), l = 10 − λ and from (2), k =

1

3

17. We maximize P( x, y ) = 9 x 4 y 4 − x − y subject to the constraint x + y = 60,000. 1

3

F ( x, y, λ ) = 9 x 4 y 4 − x − y − λ ( x + y − 60, 000) 9 − 34 34 ⎧ ⎪ Fx = 4 x y − 1 − λ = 0 ⎪ 27 14 − 14 ⎪ x y −1− λ = 0 ⎨ Fy = 4 ⎪ ⎪ Fλ = − x − y + 60, 000 = 0 ⎪⎩

(1) (2) (3)

9 − 34 34 27 14 − 14 9 − 3 3 27 14 − 14 x y − x y = 0 , x 4 y4 = x y , y = 3x. 4 4 4 4 Substituting for y in (3) gives –4x + 60,000 = 0, so x = 15,000, from which y = 45,000. Thus each month $15,000 should be spent on newspaper advertising and $45,000 on TV advertising.

Solving (2) for λ and substituting in (1) gives

705


2


3

18. We maximize f (l , k ) = 6l 5 k 5 subject to the constraint 25l + 69k = 25,875. 2

3

F (l , k , λ ) = 6l 5 k 5 − λ (25l + 69k − 25,875) 12 − 35 35 ⎧ ⎪ Fl = 5 l k − 25λ = 0 ⎪ 18 2 − 2 ⎪ ⎨ Fk = l 5 k 5 − 69λ = 0 5 ⎪ ⎪ Fλ = −25l − 69k + 25,875 = 0 ⎪⎩ 3

12 − 53 53 18 52 − 52 From the first two equations, l k = 25λ and l k = 69λ . Thus, 5 5

3

12 l − 5 k 5 5 2 2 18 l 5 k − 5 5

=

25λ 25 , from which = 69λ 29

25 l. Substituting this for k in the third equation and solving for l gives l = 414 so k = 225. 46 414 units of labor and 225 units of capital should be invested. k=

19. We minimize B( x, y, z ) = x 2 + y 2 + 2 z 2 subject to x + y = 20 and y + z = 20. Since there are two constraints, two Lagrange multipliers are used. F ( x, y, z , λ1 , λ2 ) = x 2 + y 2 + 2 z 2 −λ1 ( x + y − 20) − λ2 ( y + z − 20) ⎧ Fx = 2 x − λ1 = 0 ⎪ Fy = 2 y − λ1 − λ2 = 0 ⎪⎪ ⎨ Fz = 4 z − λ2 = 0 ⎪ Fλ1 = − x − y + 20 = 0 ⎪ ⎪⎩ Fλ2 = − y − z + 20 = 0

(1) (2) (3) (4) (5)

Eliminating y from (4) and (5) gives x = z. From (1) and (3), λ1 = 2x and λ2 = 4z . Substituting in (2) we have 2y – 2x – 4z = 0, 2y – 2x – 4x = 0, 2y – 6x = 0, y = 3x. Substituting in (5) gives –(3x) – x + 20 = 0, so x = 5. Thus z = 5 and y = 15. Therefore, x = 5, y = 15, z = 5. 20. a.

P = TR – TC = 64q – (8l + 16k) ⎡ 65 − 4(l − 4)2 − 2(k − 5) 2 ⎤ = 64 ⎢ ⎥ − 8l − 16k 16 ⎢⎣ ⎥⎦ P = −196 − 16l 2 + 120l − 8k 2 + 64k

b.

15 4 Pk = −16k + 64 = 0 ⇒ k = 4 Pl = −32l + 120 = 0 ⇒ l =

⎛ 15 ⎞ Thus there is one critical point: (l , k ) = ⎜ , 4 ⎟ ⎝ 4 ⎠ Second-Derivative Test: Pll = −32, Pkk = −16, Plk = 0 . ⎛ 15 ⎞ ⎛ 15 ⎞ 2 Thus D(l , k ) = Pll Pkk − [ Plk ] = (−32)(−16) − 02 = 512. At ⎜ , 4 ⎟ , D ⎜ , 4 ⎟ = 512 > 0 and Pll = −32 < 0 . ⎝ 4 ⎠ ⎝ 4 ⎠ 15 Thus there is a relative maximum at l = , k = 4 . Substituting these values into the profit function gives a 4 relative maximum profit of $157.00.

706


c.

Section 17.8

F (l , k , q, λ ) = 64q − 8l − 16k − λ ⎡16q − 65 + 4(l − 4)2 + 2(k − 5)2 ⎤ ⎣ ⎦ F = − 8 − 8 λ ( l − 4) = 0 (1) ⎧ l ⎪ Fk = −16 − 4λ (k − 5) = 0 (2) ⎪ ⎨ Fq = 64 − 16λ = 0 (3) ⎪ ⎪⎩ Fλ = −16q + 65 − 4(l − 4)2 − 2(k − 5)2 = 0 (4) 15 . Similarly, from (2) 4 251 251 ⎞ ⎛ 15 we get k = 4. Substituting for l and k in (4) gives q = . Thus (l , k , q ) = ⎜ , 4, ⎟. 4 64 ⎠ 64 ⎝

From (3), λ = 4 . Substituting λ = 4 into (1) gives –8 – 32(l – 4) = 0, so l =

(

21. U = x3 y 3 , px = 2 , p y = 3 , I = 48 x3 y 3 ≠ 0

)

We want to maximize U = x3 y 3 subject to 2x + 3y = 48. F ( x, y, λ ) = x3 y 3 − λ (2 x + 3 y − 48) ⎧ Fx = 3x 2 y3 − 2λ = 0 (1) ⎪⎪ 3 2 (2) ⎨ Fy = 3 x y − 3λ = 0 ⎪ F = −2 x − 3 y + 48 = 0 (3) ⎪⎩ λ 3 3 3 From (1), λ = x 2 y3 and from (2), λ = x3 y 2 . Thus x 2 y 3 = x3 y 2 , so x = y . 2 2 2 3 ⎛ ⎞ Substituting this expression for x into (3) yields y = 8. Hence x = ⎜ ⎟ 8 = 12 . ⎝2⎠

22. U = 40 x − 5 x 2 + 4 y − 2 y 2 , px = 2, p y = 3, I = 10 We want to maximize U = 40 x − 5 x 2 + 4 y − 2 y 2 subject to 2x + 3y = 10. ⎧ F = 40 − 10 x − 2λ = 0 ⎪ x F ( x, y, λ ) = 40 x − 5 x 2 + 4 y − 2 y 2 − λ (2 x + 3 y − 10) ⎨ Fy = 4 − 4 y − 3λ = 0 ⎪ F = −2 x − 3 y + 10 = 0 ⎩ λ

λ

3λ . 4 20 208 38 Substituting these values into the third equation gives λ = . Thus x = and y = . 53 53 53

From the first equation, x = 4 −

5

and from the second equation y = 1 −

23. U = f(x, y, z) = xyz px = p y = p z = 1, I = 100 (xyz ≠ 0) We want to maximize U = xyz subject to x + y + z = 100. F ( x, y, z, λ ) = xyz − λ ( x + y + z − 100) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎩⎪ Fλ

= yz − λ = 0 = xz − λ = 0 = xy − λ = 0 = − x − y − z + 100 = 0

(1) (2) (3) (4)

707



From (1) and (2), yz = xz , so y = x. Similarly, from (1) and (3), z = x. Substituting 100 . Thus y = x and z = x into (4) yields x = 3 100 100 y= . and z = 3 3

Problems 17.9 1. n = 6, Σxi = 21 , Σyi = 18.6 , Σxi yi = 75.7 , Σxi2 = 91 . a = 0.98 b = 0.61 Thus yˆ = 0.98 + 0.61x . When x = 3.5, then yˆ = 3.12 .

24. To maximize U = f(x, y) subject to the constraint xp x + yp y = I , we consider

(

5

)

F ( x, y, λ ) = f ( x, y ) − λ xp x + yp y − I .

For maximum satisfaction, Fx = f x ( x, y ) − λ px = 0 and Fy = f y ( x, y ) − λ p y = 0

y

(1) (2) x

f x ( x, y ) and from (2), px f y ( x, y ) f ( x, y ) f y ( x , y ) λ= . Thus λ = x = py px py

From (1), λ =

8

2. n = 7, Σxi = 28 , Σyi = 29.3 , Σxi yi = 154.1 , Σxi2 = 140 . a = −1.09 , b = 1.32 . Thus yˆ = −1.09 + 1.32 x . When x = 3.5, then yˆ = 3.53 .

Since f x ( x, y ) represents change in total utility from a one unit change in X (which costs px ),

10

y

f x ( x, y ) is the marginal utility of a dollar’s px f y ( x, y ) worth of X. Likewise is the marginal py

then

utility of a dollar’s worth of Y. Thus maximum satisfaction is obtained when the consumer allocates the budget so that the marginal utility of a dollar’s worth of X is equal to the marginal utility of a dollars worth of Y. Similarly, for U = f(x, y, z, w) subject to the constraint xp x + yp y + zpz + wpw = I , U is maximized

x 8

3. n = 5, Σxi = 22 , Σyi = 37 , Σxi yi = 189 , Σxi2 = 112.5 . a = 0.057 , b = 1.67 . Thus yˆ = 0.057 + 1.67 x . When x = 3.5, then yˆ = 5.90 .

when f ( x, y, z , w) f y ( x, y, z , w) λ= x = px py

15

f z ( x, y, z, w) f w ( x, y, z , w) = . pz pw That is, U is maximized when the marginal utility of a dollar’s worth of each of the products is the same.

y

=

x 10

708


Section 17.10

4. n = 6, Σxi = 27, Σyi = 21.6, Σxi yi = 105.8 ,

10.

Σxi2

= 139 . a = 1.39 , b = 0.49 . Thus yˆ = 1.39 + 0.49 x . When x = 3.5, then yˆ = 3.12 .

5

Year ( x) 1 3 5 7 Index ( y ) 77 100 126 134

n = 4, Σxi = 16, Σyi = 437, Σxi yi = 1945 , Σxi2 = 84 . a = 69.85, b = 9.85. Thus yˆ = 69.85 + 9.85 x .

y

11. a.

Year ( x) 1 2 3 4 5 Quantity ( y ) 35 31 26 24 26

n = 5, Σxi = 15, Σyi = 142, Σxi yi = 401 , x

Σxi2 = 55 . a = 35.9, b = −2.5. Thus yˆ = 35.9 − 2.5 x .

8

5. n = 6, Σpi = 250, Σqi = 322, Σpi qi = 11, 690 , Σpi2 = 13,100. a = 80.5 b = −0.643 Thus qˆ = 80.5 − 0.643 p.

6. n = 4,

∑

xi2

b.

n = 5, Σxi = 0, Σyi = 142, Σxi yi = −25 , Σxi2 = 10 . a =

∑ xi = 80, ∑ yi = 23.9, ∑ xi yi = 498.4,

b=

= 1920, a = 4.7, b = 0.06.

Thus yˆ = 4.7 + 0.06 x. When x = 20, then yˆ = 5.9.

12.

Σxi2 = 7040 . a = 100, b = 0.13. Thus yˆ = 100 + 0.13x . When x = 40, then yˆ = 105.2 .

∑ xi = 539,

a = 1.95,

Year ( x) 1 2 3 4 5 Production ( y ) 10 15 16 18 21

n = 5, Σxi = 15, Σyi = 80, Σxi yi = 265, a = 8.5 b = 2.5 Thus yˆ = 8.5 + 2.5 x

Σxi yi Σxi2

Σyi = 407.2 and n

= 26.4 . Thus yˆ = 407.2 + 26.4 x .

Problems 17.10 1.

Σxi2

= −2.5 . Thus yˆ = 28.4 − 2.5 x .

Year ( x) − 2 −1 0 1 2 Index ( y ) 357 380 403 434 462

b=

b = 1.04. Thus yˆ = 1.95 + 1.04 x. 9.

Σxi2

Σxi2 = 10 . a =

∑ yi = 569,

∑ xi yi = 76, 736, ∑ xi2 = 72, 691,

Σxi yi

Σyi = 28.4 and n

n = 5, Σxi = 0, Σyi = 2036, Σxi yi = 264 ,

7. n = 4, Σxi = 160, Σyi = 420.8, Σxi yi = 16,915.2 ,

8. n = 4,

Year ( x) −2 −1 0 1 2 Quantity ( y ) 35 31 26 24 26

2.

= 55 .

3.

4.

709

3 4

∫0 ∫0

4 3

∫1 ∫0

3

4

3

3

0

0

0

0

x dy dx = ∫ xy dx = ∫ 4 x dx = 2 x 2

y dy dx = ∫

1 1

4

1

1 x2 y

∫0 ∫0 xy dx dy = ∫0 2 3 2 x dy dx 0 0

∫ ∫

y2 2

2

3 0

dx = ∫

1 0

49

1

2

4

dx =

9x 27 = 2 1 2

y y2 dy = ∫ dy = 02 4 1

= 18

1

= 0

2

3

2

2

0

0

0

0

= ∫ x 2 y dx = ∫ 3 x 2 dx = x3

1 4

=8


5.

∫1 ∫1 ( 3 2

=∫


2

⎞ x − y dx dy = ∫ ⎜ − xy ⎟ dy ⎟ 1⎜ 3 ⎝ ⎠1

3 ⎛ x3

)

2

11.

3 ⎡⎛ 8

1

3⎛ 7 ⎞ ⎛1 ⎞⎤ ⎞ ⎢⎜ 3 − 2 y ⎟ − ⎜ 3 − y ⎟ ⎥ dy = ∫1 ⎜ 3 − y ⎟ dy ⎠ ⎝ ⎠⎦ ⎝ ⎠ ⎣⎝

1 x2

∫0 ∫3x 1

1

0

(

6

= ∫ 7 x − 63 x 0

4

)

3

⎛7 y2 ⎞ 9⎞ ⎛7 1⎞ 2 ⎛ =⎜ y− ⎟ = ⎜7 − ⎟ −⎜ − ⎟ = ⎜3 ⎟ 2 ⎠ 2⎠ ⎝3 2⎠ 3 ⎝ ⎝ 1

12. 2

⎛ y3 ⎞ 6. ∫ ∫ ( y − 2 xy )dy dx = ∫ ⎜ − xy 2 ⎟ dx ⎟ −2 0 −2 ⎜ 3 ⎝ ⎠0 3 ⎡⎛ 8 3 8 ⎤ ⎞ ⎛ ⎞ = ∫ ⎢⎜ − 4 x ⎟ − 0 ⎥ dx = ∫ ⎜ − 4 x ⎟ dx −2 ⎣⎝ 3 −2 ⎝ 3 ⎠ ⎦ ⎠ 3

2

3

2

2 x2

∫0 ∫0 =∫

3

13.

⎛8 ⎞ ⎛ 16 ⎞ = ⎜ x − 2 x 2 ⎟ = (8 − 18) − ⎜ − − 8 ⎟ ⎝3 ⎠ −2 ⎝ 3 ⎠ 10 = 3

2

3

∫0 ∫0

9− x 2

x6 12

2

2

= 0

) 3x dx 1

0

dx =

x2

⎛ 63 x5 ⎞ 58 dx = ⎜ x7 − ⎟ =− ⎜ ⎟ 5 ⎠ 5 ⎝ 0

xy dy dx = ∫

2 x5

0

(

14 x 2 y dy dx = ∫ 7 x 2 y 2

xy 2 2

x2

dx 0

16 3

y dy dx = ∫

9− x 2

3 y2

0

dx

2

0

⎞ 3 ⎛ 9 − x2 1 3 =∫ ⎜ − 0 ⎟ dx = ∫ (9 − x 2 )dx ⎟ 0⎜ 2 2 0 ⎝ ⎠ 3

7.

1⎛

1 2

⎝

(

1

0

∫0 ∫0 ( 3 x

=∫

∫1 ∫0

⎞ ⎟ dx ⎟ ⎠0

0

1

)0 =3

14.

3 4 x3 x3 ⎞ x4 + ⎟ dx = ∫ dx = 0 3 3 ⎟⎠ 3

y dy dx = ∫

4

1

y2 2

4

5x 0

dx = ∫

2

1

= 27 0

x dx

25 3 525 = x = 6 2 1

10.

2 x −1

∫1 ∫0 2

1

y

1

2

− y 3 ) dy

1− x

1

1

2

2 y dy dx = ∫ y 2

= ∫ ( x − 1)2 dx =

1

1− x ⎛ y2 ⎞ 15. ∫ ∫ 3( x + y )dy dx = ∫ 3 ⎜ xy + dx ⎟ −1 x −1 ⎜ 2 ⎟⎠ ⎝ x 2 2 ⎡ ⎤ ⎛ ⎞ 1 (1 − x) x = ∫ 3 ⎢ x(1 − x) + − ⎜ x2 + ⎟ ⎥ dx ⎜ ⎟⎥ −1 ⎢ 2 2 ⎝ ⎠⎦ ⎣ 1 ⎡ 5 x 2 (1 − x) 2 ⎤ = ∫ 3⎢x − + ⎥ dx −1 2 2 ⎦⎥ ⎣⎢

3

4 25 2

1

1 y

∫0 ∫ y 2 y dx dy = ∫0 xy y 2 dy = ∫0 ( y 1 ⎛ y3 y 4 ⎞ 1 1 1 =⎜ − ⎟ = − = ⎜ 3 ⎟ 4 ⎝ ⎠ 0 3 4 12

x

)

⎝

4 5x

x3 ⎞ 1⎛ 1 = ⎜ 9 x − ⎟ = (27 − 9) − 0 = 9 2 ⎜⎝ 3 ⎟⎠ 2

2

3⎛ y3 ⎞ x 2 + y 2 dy dx = ∫ ⎜ x 2 y + ⎟ dx 0⎜ 3 ⎟⎠ ⎝ 0

3⎛ 3 ⎜x 0⎜

9.

y 2

∫0 ∫0 ( x + y)dy dx = ∫0 ⎜⎜ xy + = ∫ (2 x + 2)dx = x 2 + 2 x

8.

2

1

( x − 1) 3

⎡ x 2 5 x3 (1 − x)3 ⎤ = 3⎢ − − ⎥ 6 6 ⎦⎥ ⎣⎢ 2 −1

x −1

dx 0

3 2 1

=

⎡1 5 ⎤ ⎡1 5 4⎤ = 3 ⎢ − − 0 ⎥ − 3 ⎢ + − ⎥ = −1 ⎣2 6 ⎦ ⎣2 6 3⎦

1 3

710


16.

3 3y

∫0 ∫ y

2

5 x dx dy = ∫

3 ⎛ 45 y 2

=∫ ⎜ 0⎜ ⎝

3y

3 5x2

0

Section 17.10

21.

dy

2

y

5y ⎞ − ⎟ dy 2 ⎟⎠

y 1 y x+ y 1 e dx dy = e x + y dy 0 0 0 0 1 ⎡ e2 y ⎤ e2 ⎛1 y

∫

1 1 y− x

∫0 ∫0 e

1

(

=∫ e 0

1

1

0

0

2y

−e

y

) dy

22.

1

0

0

2

2

9y −1 ∫−1

2 3

z dx dy dz = ∫

=∫

∫

0 0

2

0

2

−1

−1

2 3

z dy dz = ∫ 3 y 3 z 3 0 −1

1 x x+ y 2 1 x

0

3x −1 ∫−1

27 z 4 = ∫ 27 z 3 dz = −1 4

∫0 ∫0 ∫0

1 3x

0

1

y

e x

y

e x

∫

y dy dx = ∫

e

1

y2 2

x

ln x

dx = ∫

e x2

1

2

−

(ln x)2 dx 2

e

e3 e ⎛ 1 ⎞ e3 e 5 − − ⎜ − 1⎟ = − + 6 2 ⎝6 ⎠ 6 2 6

2 2 −( x + y )

∫

e

dx dy

2 2 3

2 2 = ∫ −e− ( x + y ) dy = ∫ ⎡ −e−(2+ y ) + e− y ⎤ dy ⎣ ⎦ 1 1 0

2

y z

dy dz

2

1

= ⎡e − (2+ y ) − e− y ⎤ = e−4 − e−2 − e−3 + e−1 ⎣ ⎦1

dz

dx =

51

3x 10

∫

∫

0

12e−4 x −3 y dx dy

4

−3e−4 x −3 y ) dy 2( 3 6 −16 −3 y = ∫ ( −3 + 3e−12−3 y ) dy 2 6 = ( e −16−3 y − e−12−3 y ) 2 =∫

0 0

=

∫

2 3

1 x 2 x+ y x z dy dx 0 1 x 3 2

x

6 4

24. P (3 ≤ x ≤ 4, 2 ≤ y ≤ 6) = ∫

0 0

[ x ( x + y ) − 0]dy dx = ∫

2

dx x2

23. P (0 ≤ x ≤ 2, 1 ≤ y ≤ 2) = ∫

( x + x y )dy dx

1⎛ 1 ⎡⎛ x2 y 2 ⎞ x4 ⎞ ⎤ = ∫ ⎜ x3 y + ⎟ dx = ∫ ⎢⎜ x 4 + ⎟ − 0 ⎥ dx 0⎜ 0 ⎢⎜ 2 ⎟⎠ 2 ⎟⎠ ⎦⎥ ⎝ ⎝ ⎣ 0

=∫

2

⎡ x3 1 ⎤ = ⎢ − ( x ln 2 x − 2 x ln x + 2 x) ⎥ ⎢⎣ 6 2 ⎥⎦ 1

27 =− 4

x dz dy dx = ∫

2

4

0

x

1 0

0

20.

1 xy 2

dy dx 0

2

2

0

xy dy dx = ∫

1 ln x

−1

∫−1 ∫−1 ∫1 6 xy =∫

∫

e x

=∫

= (−e0 + e1 ) − (−e−1 + e0 ) = −1 + e + e−1 − 1

0

0 x2

∫1 ∫ln x ∫0 dz dy dx = ∫1 ∫ln x z 0 dy dx

=

1

= ∫ (−e y −1 + e y )dy = (−e y −1 + e y )

19.

∫

z

⎡ x 4 x6 ⎤ x5 ⎤ 1 = ∫ ⎢ − ⎥ dx = ⎢ − ⎥ = 0 2 2 ⎦⎥ ⎢⎣ ⎣⎢ 8 12 ⎥⎦ 0 24

dx dy = ∫ −e y − x dy

= −2 + e + e

1 x

xy

1 x

1 ⎡ x3

2 1 ⎞ e =⎢ −e ⎥ = − e − ⎜ − 1⎟ = −e+ 2 ⎝2 ⎠ 2 ⎣⎢ 2 ⎦⎥ 0 2

18.

dz dy dx = ∫

2

0 x2

3

∫∫

∫0 ∫x ∫0 =∫

⎛ 15 y 3 y 5 ⎞ 405 243 =⎜ − − = 81 ⎟ = ⎜ 2 ⎟ 2 ⎠ 2 2 ⎝ 0

17.

xy

2

4

2

1 x

6

= e −34 − e −30 − e−22 + e−18

1 1 1 1⎞ ⎛ 25. P ⎜ x ≥ , y ≥ ⎟ = ∫ ∫ 1 dx dy 1/ 3 1/ 2 2 3⎠ ⎝ 1 1 ⎛ 1⎞ 1 = ∫ x 1/ 2 dy = ∫ ⎜ 1 − ⎟ dy 1/ 3 1/ 3 ⎝ 2⎠ 1 1 1 1 1⎛ 1⎞ 1⎛ 2⎞ 1 dy = y =∫ = ⎜1 − ⎟ = ⎜ ⎟ = 1/ 3 2 2 1/ 3 2 ⎝ 3 ⎠ 2 ⎝ 3 ⎠ 3

5

3(1) 3 −0 = 10 10

26.

711

1 11

∫0 ∫0 8

dx dy = ∫

1x

08

1 0

dy = ∫

11

08

1

dy =

y 1 = 80 8




4. x 2 + z 2 = 1 . The x,z-trace is x 2 + z 2 = 1 , which is a circle. For any fixed value of y, we obtain

1. x + y + z = 1 can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. The intercepts are (1, 0, 0), (0, 1, 0), and (0, 0, 1).

the circle x 2 + z 2 = 1 . z 1

z

y

1

1

x

y

1 1

5.

x

f x ( x, y ) = 4(2 x) + 6(1) y + 0 − 0 = 8 x + 6 y f y ( x, y ) = 0 + 6 x(1) + 2 y − 0 = 6 x + 2 y

2. z = x can be put in the form Ax + By + Cz + D = 0, so the graph is a plane. Every point on the y-axis is an intercept. The x,z-trace is z = x, which is a line. For any fixed value of y, we obtain the line z = x.

6.

z

7. y

∂P = 3l 2 + 0 − (1)k = 3l 2 − k ∂l ∂P = 0 + 3k 2 − l (1) = 3k 2 − l ∂k ∂ z ( x + y )(1) − x(1) y = = 2 ∂x ( x + y) ( x + y )2 Because z = x( x + y )−1 ,

∂z x = x ⎡(−1)( x + y ) −2 (1) ⎤ = − . ⎣ ⎦ ∂y ( x + y )2

x

3. z = y 2 The y,z-trace is z = y 2 , which is a parabola. For any fixed value of x, we obtain the curve

8.

z = y2.

9. z

f pB ( pA , pB ) = 0 + 5(1 − 0) = 5

y

10. w = x

(

)

1 ln x 2 + y 2 2 y ∂ 1 1 [ f ( x, y )] = ⋅ (2 y ) = 2 2 x2 + y 2 ∂y x + y2 f ( x, y ) = ln x 2 + y 2 =

x x2 + y 2

= x( x 2 + y 2 )

− 12

∂w xy −3 ⎡ 1 ⎤ = x ⎢ − ( x 2 + y 2 ) 2 (2 y ) ⎥ = − 3 ∂y 2 ⎣ 2 ⎦ (x + y2 ) 2

11. wx ( x, y, z ) = 2 xyze x

2

yz

(

)

2 2 wxy ( x, y, z ) = 2 xz ⎡⎢ y e x yz ⋅ x 2 z + e x yz ⋅1⎤⎥ ⎣ ⎦

(

)

2 = 2 xze x yz x 2 yz + 1

712


12.

Chapter 17 Review

⎡ ⎛ 1 ⎤ ⎞ f x ( x, y ) = y ⎢ x ⎜ ⋅ y ⎟ + ln( xy ) ⋅1⎥ = y[1 + ln( xy )] ⎣ ⎝ xy ⎠ ⎦ ⎡1 ⎤ f xy ( x, y ) = y ⎢ ⋅ x ⎥ + [1 + ln( xy )] ⋅1 = 1 + 1 + ln( xy ) = 2 + ln(xy) ⎣ xy ⎦

13.

∂ [ f ( x, y, z )] = ( x + y + z )(2 z ) + ( x 2 + y 2 + z 2 )(1) = 3z 2 + 2 z ( x + y ) + x 2 + y 2 ∂z ∂2 ∂ z2

(

[ f ( x, y, z )] = 6 z + 2( x + y ) = 2 x + 2 y + 6 z

14. z = x 2 − y

)( y 2 − 2xy ) = x2 y 2 − 2 x3 y − y3 + 2 xy2

∂z = 2 x 2 y − 2 x3 − 3 y 2 + 4 xy ∂y ∂ 2z ∂ y2 15.

= 2x2 − 6 y + 4 x

w = e x + y + z ln xyz = e x + y + z (ln x + ln y + ln z ) ⎛1⎞ ∂w = e x + y + z (ln x + ln y + ln z ) + e x + y + z ⎜ ⎟ ∂y ⎝ y⎠ ⎡ 1⎤ = e x + y + z ⎢ ln xyz + ⎥ y⎦ ⎣

By symmetry,

∂w = ex+ y+ z ∂x

∂2 w = e x+ y+ z ∂z∂x

1⎤ ⎡ ⎢ ln xyz + x ⎥ . ⎣ ⎦

1 ⎤ x+ y + z ⎡ ⎢ ln xyz + x ⎥ + e ⎣ ⎦ 1 1⎤ ⎡ = e x + y + z ⎢ ln xyz + + ⎥ . x z⎦ ⎣

16.

17.

⎡1⎤ ⎢z⎥ ⎣ ⎦

∂P = 100 ⎡ (0.11)l 0.11−1 ⎤ k 0.89 = 11l −0.89 k 0.89 ⎣ ⎦ ∂l 2 ∂ P = 11l −0.89 ⎡ (0.89)k 0.89−1 ⎤ = 9.79l −0.89 k −0.11 ⎣ ⎦ ∂ k∂ l x+ y 1 y = + xz z xz y f x ( x, y , z ) = − x2 z 1 f xy ( x, y, z ) = − x2 z 1 f xyz ( x, y, z ) = 2 2 x z 1 1 f xyz (2, 7, 4) = = 2 2 64 2 ⋅4 f ( x, y , z ) =

713


18.

f x ( x , y , z ) = 6e y

2


ln( z +1)

f xy ( x, y, z ) = 12 y ln( z + 1)e y

2

ln( z +1)

⎡ y 2 ⎪⎫ y 2 ln( z +1) 1 ⎤ ⎪⎧ 2 ⋅ f xyz ( x, y, z ) = 12 y ⎢ ln( z + 1) ⎨e y ln( z +1) ⋅ ⎥ ⎬+e z + 1 ⎪⎭ z + 1 ⎥⎦ ⎢⎣ ⎪⎩ f xyz (0, 1, 0) = 12[0 + 1] = 12

19.

( )

∂w ∂w∂x ∂w∂ y ⎛ 1 ⎞ = + = (2 x + 2 y ) er + (2 x + 6 y ) ⎜ ⎟ ∂r ∂x ∂r ∂ y ∂r ⎝r+s⎠ 2( x + 3 y ) r+s ∂w ∂w∂x ∂w∂ y ⎛ 1 ⎞ = + = (2 x + 2 y )(0) + (2 x + 6 y ) ⎜ ⎟ ∂s ∂x ∂s ∂ y ∂s ⎝r+s⎠

= 2( x + y )er +

=

20.

2( x + 3 y ) r+s

⎛ 1 ⎞ ∂z ∂z ∂x ∂z ∂ y ⎛1 ⎞ = + = + ye xy − y ⎟ ( 2 s ) + ⎜ − + xe xy − x ⎟ (2(r + s)) ∂ s ∂ x ∂ s ∂ y ∂ s ⎜⎝ x y ⎠ ⎝ ⎠

21. 2 x + 2 y − 4 z (−4 z + x)

∂z ⎡ ∂z ⎤ + x + z (1) ⎥ + 0 = 0 ∂ x ⎢⎣ ∂ x ⎦

∂z = −(2 x + 2 y + z ) ∂x

∂ z −(2 x + 2 y + z ) 2 x + 2 y + z = = ∂x −4 z + x 4z − x 22. z 2 + ln( yz ) + ln z + x + z = 0 or z 2 + ln y + ln z + ln z + x + z = 0, thus z 2 + ln y + 2 ln z + x + z = 0. ∂z 1 1 ∂z ∂z + =0 2z + + 2 ⋅ ∂y y z ∂y ∂y ∂z ⎡ 2 ⎤ 1 2 z + + 1⎥ = − ∂y ⎢⎣ z ⎦ y 2 ∂z ⎡ 2 z + 2 + z ⎤ 1 ⎢ ⎥=− ∂y ⎣⎢ z y ⎦⎥ ∂z z =− 2 ∂y y (2 z + 2 + z )

714


Chapter 17 Review

23. P = 20l 0.7 k 0.3 . Marginal productivity functions ∂P = 20(0.7)l −0.3 k 0.3 and are given by ∂l ∂P ∂P = 20(0.3)l 0.7 k −0.7 . Thus = 14l −0.3 k 0.3 ∂k ∂l ∂P and = 6l 0.7 k −0.7 . ∂k

29. z

xyz = 32 (xyz ≠ 0). Let S be the amount of cardboard used. S = xy + 2yz + 2xz ⎡ 32 ⎤ ⎡ 32 ⎤ = xy + 2 y ⎢ ⎥ + 2 x ⎢ ⎥ xy ⎣ ⎦ ⎣ xy ⎦

24. c = 3x + 0.05xy + 9y + 500 Marginal cost with respect to x is ∂c = 3 + 0.05 y. When x = 50 and y = 100, then ∂x ∂c = 8. ∂x

= xy +

The critical point occurs when x = 4, y = 4, and z = 2, which gives a minimum. The dimensions are 4 ft by 4 ft by 2 ft.

∂ qA ∂ qB = 2 > 0 and = 3 > 0 , A and B ∂ pB ∂ pA

30.

∂α ∂α = 0.530; = −0.027 ∂P ∂S

27.

f ( x, y ) = x 2 + 2 y 2 − 2 xy − 4 y + 3

f y ( x, y ) = 2by + cx − 20

At (1, 2), f x ( x, y ) = 0 and f y ( x, y ) = 0 . Thus 2a + 2c − 10 = 0 (1) and 4b + c − 20 = 0 (2) From Eq. (1), a = 5 – c; from Eq. (2), 20 − c . b= 4 f xx ( x, y ) = 2a, f yy ( x, y ) = 2b, f xy ( x, y ) = c

⎧ f x ( x, y ) = 2 x − 2 y = 0 ⎨ f ( x, y ) = 4 y − 2 x − 4 = 0 ⎩ y Critical point: (2, 2) f xx ( x, y ) = 2, f yy ( x, y ) = 4, f xy ( x, y ) = −2

At (2, 2), D = (2)(4) − (−2) 2 = 4 > 0 and f xx ( x, y ) = 2 > 0 ; thus relative minimum at (2, 2). 28.

3

f ( x, y ) = ax 2 + by 2 + cxy − 10 x − 20 y f x ( x, y ) = 2ax + cy − 10 ;

are competitive products. 26.

64 64 + x y

∂S 64 ∂ S 64 = y− , = x− 2 ∂y ∂x x y2

25. qA = 100 − pA + 2 pB , qB = 150 + 3 pA − 2 pB . Since

y

x

At (1, 2), D = (2a )(2b) − c 2 = 0 . ⎛ 20 − c ⎞ 2 Thus 4ab − c 2 = 0 , 4(5 − c) ⎜ ⎟−c = 0 , ⎝ 4 ⎠

3

f ( w, z ) = 2w + 2 z − 6wz + 7

100 − 25c + c 2 − c 2 = 0 , 100 = 25c, or c = 4. So a = 5 – c = 5 – 4 = 1 and 20 − c 20 − 4 b= = = 4. 4 4 Thus a = 1, b = 4, c = 4.

⎧⎪ f w ( w, z ) = 6 w2 − 6 z = 0 ⎨ 2 ⎪⎩ f z ( w, z ) = 6 z − 6 w = 0 Critical points: (0, 0), (1, 1) f ww ( w, z ) = 12 w, f zz ( w, z ) = 12 z , f wz ( w, z ) = −6

At (0, 0), D = (0)(0) − (−6) 2 = −36 < 0 ; thus neither relative maximum nor relative minimum. At (1, 1), D = (12)(12) − (−6) 2 = 108 > 0 and f ww ( w, z ) = 12 > 0 ; thus relative minimum at (1, 1).

715



31. Profit = P = ( pA − 50 ) qA + ( pB − 60 ) qB P = ( pA − 50 ) ⎡⎣ 250 ( pB − pA ) ⎤⎦ + ( pB − 60 ) ⎡⎣32, 000 + 250 ( pA − 2 pB ) ⎤⎦ .

∂P = ( pA − 50 ) (−250) + ⎡⎣ 250 ( pB − pA ) ⎤⎦ (1) + 250 ( pB − 60 ) ∂ pA

= −500 pA + 500 pB − 250(10) = 500 ( − pA + pB − 5 ) ∂P Also, = ( pA − 50 ) (250) + ( pB − 60 ) (−500) + ⎡⎣32, 000 + 250 ( pA − 2 pB ) ⎤⎦ (1) ∂ pB = 500 pA − 1000 pB + 49,500 = 500 ( pA − 2 pB + 99 )

Setting

∂P ∂P = 0 and = 0 gives ∂ pA ∂ pB

− pA + pB − 5 = 0 (1) and pA − 2 pB + 99 = 0 (2) Adding Equations (1) and (2) gives − pB + 94 = 0 . So pB = 94 . From Equation (1), pA = pB − 5 , so pA = 94 − 5 = 89 . At pA = 89 and pB = 94 , D =

∂ 2P ∂ pA2

∂ 2P ∂ 2P ∂ pA2 ∂ pB2

−

∂ 2P = (−500)(−1000) − (500)2 > 0 and ∂ pB∂ pA

= −500 < 0 .

Thus there is a relative maximum profit when the price of A is 89 cents per pound and the price of B is 94 cents per pound. 32.

f ( x, y, z ) = xy 2 z , x + y + z − 1 = 0 F ( x, y, z , λ ) = xy 2 z − λ ( x + y + z − 1) ⎧ Fx ⎪ ⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ

= y2 z − λ = 0 = 2 xyz − λ = 0 = xy 2 − λ = 0 = −x − y − z +1 = 0

From the first and third equations, we have λ = y 2 z = xy 2 , so x = z (since xyz ≠ 0). With x = z in the second equation, we have λ = 2 x 2 y. Combining this with λ = xy 2 , we get y = 2x. Substituting y = 2x and z = x in the 1 1 1 fourth equation gives −x − 2x − x + 1 = 0 so 4x = 1 and x = . Thus y = and z = . The critical point is 4 4 2 ⎛1 1 1⎞ ⎜ , , ⎟. ⎝4 2 4⎠

716


33.


f ( x, y, z ) = x 2 + y 2 + z 2 , 3x + 2y + z = 14 F ( x, y , z , λ )

37.

= x 2 + y 2 + z 2 − λ (3x + 2 y + z − 14) ⎧ Fx ⎪⎪ Fy ⎨ ⎪ Fz ⎪⎩ Fλ

= 2 x − 3λ = 0 = 2 y − 2λ = 0 = 2z − λ = 0 = −3x − 2 y − z + 14 = 0

From (1), x =

λ

z=

1 y2

∫0 ∫0

xy dx dy = ∫

0

(1) (2) (3) (4)

1 y6 = ⋅ 2 6

38.

1

= 0

y dy dx = ∫

∫1 ∫x2

4

2

= −0.43

39.

1 x2

∫0 ∫

x 1

Year ( x) 1 2 3 4 5 6 Expenditures ( y ) 15 22 21 26 27 34

2

nΣxi yi − ( Σxi )( Σyi ) 2

nΣxi2 − ( Σxi ) Thus yˆ = 12.67 + 3.29 x

∫ ∫ =

6 2

y 18

1

=

=∫

= 3.29

dx x

1

⎤ ⎥ ⎥ ⎦0


y 2 x3 y 2

1

)

)

x2

7 5 ⎡ 5 x x 6 x 7 2 x 2 x3 2 x 2 ⎢ =7 + − − − + ⎢5 6 7 7 3 5 ⎣ 1 ⎡1 1 1 2 1 2 ⎤ = 7⎢ + − − − + ⎥−0 = 5 6 7 7 3 5 30 ⎣ ⎦

Σxi2 ) ( Σyi ) − ( Σxi )( Σxi yi ) ( a= = 12.67

2 y 2 2 x y dx dy 1 0

(

(

Σxi2 = 91

36.

)

3 1⎡ ⎛ 5 ⎞⎤ = 7 ∫ ⎢ x 4 + x5 − x6 − ⎜ x 2 + x 2 − x 2 ⎟ ⎥dx 0⎣ ⎝ ⎠⎦

n = 6, Σxi = 21, Σyi = 145, Σxi yi = 565,

b=

(

7 x 2 + 2 xy − 3 y 2 dy dx

= 7 ∫ x 2 y + xy 2 − y3 0

nΣxi2 − ( Σxi )

x

⎛ 2 x3 x5 ⎞ =⎜ − ⎟ ⎜ 3 10 ⎟⎠ ⎝ 1 ⎛ 128 512 ⎞ ⎛ 2 1 ⎞ =⎜ − ⎟−⎜ − ⎟ 5 ⎠ ⎝ 3 10 ⎠ ⎝ 3 603 =− 10

Thus pˆ = 85.15 − 0.43t 35.

dx 2

x4 ⎞ = ∫ ⎜ 2 x 2 − ⎟ dx 1 ⎜ 2 ⎟⎠ ⎝

Σti2 ) ( Σpi ) − ( Σti )( Σti pi ) ( a= = 85.15

2

2x

4

4⎛

Σti = 104, Σpi = 381, Σti pi = 7482, Σti2 = 3192

nΣti2 − ( Σti )

0

1 1 1 ⋅ −0 = 2 6 12 y2 1 2

4 2x

34. n = 5,

nΣti pi − ( Σti )( Σpi )

dy

⎞ y4 ⋅ y 1 1 =∫ ⎜ − 0 ⎟ dy = ∫ y 5 dy ⎟ 0⎜ 2 2 0 ⎝ ⎠

. Substituting into (4) gives 2 λ ⎛ 3λ ⎞ −3 ⎜ ⎟ − 2λ − + 14 = 0 , from which λ = 2 . 2 ⎝ 2 ⎠ Thus x = 3, y = 2, and z = 1. Critical point of F: (3, 2, 1, 2), so the critical point of f is (3, 2, 1).

b=

2

y2

1⎛

3λ ; from (2), y = λ , from (3), 2

nΣti2 − ( Σti )

1 x2 y

3

0

dy = ∫

2 y5

1

3

20

y

dy

64 1 63 7 − = = 18 18 18 2

x 10 ax

y = Ceax + 5, y − 5 = Ce , ln( y − 5) = ax + ln C

717


x y 0 15 1 12 4 9 7 7 10 6

y −5 10 7 4 2 1


ln( y − 5) 2.30259 1.94591 1.38629 0.69315 0.00000

dT = k (T − a ) , where dt dT dT = k dt , = k (T − 45) , a = 45. Thus T − 45 dt dT ∫ T − 45 = ∫ k dt , ln T − 45 = kt + C . Because T – 45 > 0, ln(T – 45) = kt + C. Thus

3. Newton’s law of cooling:

n = 5, Σxi = 22, Σ ln ( yi − 5 ) = 6.32794 , Σ ⎡⎣ xi ln ( yi − 5 ) ⎤⎦ = 12.34312, Σxi2 = 166 a= =

T − 45 = ekt +C , or T = e kt +C + 45 = eC e kt + 45 = C1e kt + 45 , where

nΣ ⎡⎣ xi ln ( yi − 5 ) ⎤⎦ − ( Σxi ) ⎡⎣Σ ln ( yi − 5 ) ⎤⎦

( )

n Σxi2 − ( Σxi )

5(12.34312) − 22(6.32794) 5(166) − (22) 2

C1 = eC . So T = C1e kt + 45 . When t = 0, then

2

T = 124. Hence 124 = C1 + 45, or C1 = 79 . Thus

≈ −0.22399

Σxi2 ) ⎡⎣Σ ln ( yi − 5 ) ⎤⎦ − ( Σxi ) {Σ ⎡⎣ xi ln ( yi − 5 ) ⎤⎦} ( ln C = 2 n ( Σxi2 ) − ( Σxi ) =

166(6.32794) − 22(12.34312) 5(166) − (22)2

≈ 2.25112

C

, ln y = ln C − ln x r , x xr ln y = ln(C) – r ln x. Thus r(ln x) + (ln y) – ln(C) = 0. Since ln C and r are constants, ln x and ln y are linearly related. r

. ln y = ln

( )

ln 19 19 79 ,k= ≈ −0.01113 . Thus 79 128

4. Using a graphics calculator on the points displayed in Fig. 17.25 produces the same result as in the snapshot. Performing an exponential regression on the points shown in Fig. 17.24, however, does not produce the right curve, because the exponential model lacks a constant term. This difficulty can be overcome by subtracting 45 (the long-term temperature) from every temperature value, running the exponential regression, and then adding a constant term of 45 to the resulting model.

Thus y = 9.50e−0.22399 x + 5 . C

128k = ln

T = 79e −0.01113t + 45 .

C ≈ e 2.25112 ≈ 9.50

2. y =

T = 79ekt + 45 . When t = 128, then T = 64, so 19 64 = 79e128k + 45 , 19 = 79e128k , e128k = , 79

718

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