5.10
APPLICATIONS OF DERIVATIVES
5.11
Maxima and Minima
Example 1 Find the values of x at which maximum and minimum values of y and points of inflexion occur on the curve y = 12 ln x + x2 − 10x dy d2 y d2 y =0 Turning points; >0 minimum; <0 maximum 2 dx dx dx2 dy 12 = + 2x − 10 dx x dy =0 dx 2 2x − 10x + 12 = 0 2 (x − 2) (x − 3) = 0 x = 2, x = 3 TP @ (2, −7. 68) (3, −7. 82) d dy 12 d2 y = 2 =− 2 +2 dx dx dx x 2 12 d y = − 2 + 2 = −1 < 0 2 dx2 x=2 2 d y <0 maximum TP @ (2, −7. 68) dx2 2 12 2 d y =− 2 +2 = 3 3 dx2 x=3 d2 y >0 minimum TP @ (3, −7. 82) dx2 Inflexion point @
d2 y =0 dx2
√ √ d2 y 12 = − 2 + 2 = 0, Solution is: − 6, 6 2 dx x √ Only x= 6 √ Inflexion point @ 6, −7. 74 y = 12 ln x + x2 − 10x
1 Example 2 xαr ln r 1 2 x = kr ln r dx 1 = 2kr ln − kr dr r 1 2kr ln − kr = 0 r 2
69
1 kr 2 ln −1 =0 r 1 2 ln −1=0 r 1 1 = ln r 2 1 r= = 0.606 53 1 e2 d2 x 1 − 2k − k = 2k ln dr2 r 1 d2 x = 2k ln − 2k − k 2 r dr r=0.606 53 1 − 2k − k = −2k 2k ln 0.606 53
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5.12
Optimization
Example 1 The total area of a sheet of paper is 20 cm2 . The margins at the 3 top and bottom are 1 cm and on the sides cm in length. What must the 4 dimensions of the paper be so that the area between the margins is a maximum.
3 cm 4
y
3 cm 4 1cm
1cm
x Total Area AT = xy = 20
3 Area between margins ABM = (x − 2) y − 2 3 Total area of Margins AM = 2y + (x − 2) 2 Maximize Area between margins xy = 20 20 y= x 3 Area between margins ABM = (x − 2) y − 2 20 3 ABM (x) = (x − 2) − x 2 71
20 3 20 40 − 20x 3 d 20 (ABM ) = − + (x − 2) − 2 = + − dx x 2 x x x2 2 d 20 40 − 20x 3 (ABM ) = + − dx x x2 2 2 d d 20 40 − 20x 3 1 40 + = 3 (40x − 80) − 2 (A ) = − BM 2 2 dx dx x x 2 x x d2 40x − 80 40 (ABM ) = − 2 dx2 x3 x d d2 (ABM ) = 0 Maximum @ and (ABM ) < 0 dx dx2 d (ABM ) = 0 dx 20 40 − 20x 3 d (ABM ) = + − =0 dx x x2 2 4√ 4√ x=− 15, x= 15 3 3 √ 4 x= 15 Only solution because measurement can never be −ve 3 4√ 40 15 − 80 40 3 d2 = − 2 = −0.580 95 < 0 (A ) BM 4 3 √ dx2 4√ 4√ x= 15 15 15 3 3 3 4√ x= 15 = 5. 164 0 3 √ 20 y = √ = 15 = 3. 873 0 4 15 3 The dimensions of the paper
x = 5. 16 y = 3. 87
so that the area between
the margins is a maximum.
Example 2 A piece of wire 12 m long is cut into two lengths. The one length is bent into a circle and the other into a square. if the sum of the areas of the circle and the square are required to be a minimum, find the radius of the circle and the sides of the square. Lc = 2πr Ls = 4l L = 2πr + 4l = 12 πr l =3− 2 Ac = πr2 As = l2 A = πr2 + l2 πr 2 A = πr2 + 3 − 2
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dA πr = 2πr − π 3 − dr 2 dA =0 dr πr =0 π 2r − 3 + 2 3 r= = 0.84 1 π+2 2 d2 A >0 minimum dr2 2 π d A 1 =π 2+ =π π + 2 = 11. 218 > 0 dr2 2 2 3 = 0.840 15 r= 1 π+2 2 π
3 1 π+2 2 = 1. 680 3 2
l =3− Check: L = 2πr + 4l = 20 2π (0.84) + 4 (1. 68) = 11. 998 ≈ 12.0 .
Example 3 A farmer witht a field adjancent to a straight river wishes to fence a rectangular region for grazing. If no fence is needed along the river, and the farmer has available 1600 metres of fence, what should be the dimentions of the field in order that it ha a maximum area?
l
w
SOLUTION l + 2w = 1600 73
(1)
We are to maximize A = lw
(2)
equation (2) is a function of two variables. Using equation (1), thus l = 1600 − 2w, substitute this into equation (2), thus A = w (1600 − 2w) = 1600w − 2w2 . Then
d2 A = −4. dw2
dA = 1600 − 4w, dw
dA = 1600 − 4w = 0 ⇒ w = 400, and the second derivative is negative thus dw maximum area is achieved. Thus l = 800, w = 400. Maximum area is 320 000 m2
Example 4 The strength of a rectangular beam varies directily as the width and the square of the de[th. what are the dimentions of the strngest rectangular beam that can be cut from a cylindrical log of radious r.
y
P (x, y)
(r, 0)
x
Let the strength of the beam be denoted by S : S = 2x (2y)2 = 8xy2 2 2 2 2 2 2 since P lies on the circle x + y = r ⇒ y = r − x S = 8x r2 − x2 = 8x r2 − x2 = 8 r2 x − x3 dS d2 S = 8 r2 − 3x2 , = −48x. dx dx2 dS r = 8 r2 − 3x2 = 0 ⇒ x = ± √ dx 3 74
r r Thus x = √ because x = − √ has no physical meaning. 3 3
2 2 √ r = 0.666 667 r2 = 3 √ 0.666 667 r2 2r 2 the width is therefore √ = 1. 154 7r and the depthis 2r = 1. 632 99r for 3 3 a beam of maximum strength. √ for this x value we have y = r2 − x2 =
r2
r2 − = 3
Example 5 Find two positive numbers whosw product is 100 and whse sum is as small as possible SOLUTION 10 and 10
Example 6 Determine the altitude(height = h) of a cylinder of maximum volume that can be inscribed in a right circular come of height (H) and base radius (R).
H
R
From the notion of similar triangles:
75
E
D
A
B
C
AC CE AE △ACE ≡ △BCD Thus: = = BC CD DB R H H = =⇒ h = (R − r) R−r h R Volume of a cylinder Vc = πr2 h H Vcyl = πr2 (R − r) R dVc For maximum volume: =0 dr dVcyl H = πr (2R − 3r) dr R 2 H πr (2R − 3r) = 0, Solution is: r = 0, or r= R R 3 r=0 has no meaning since we want to have a cylinder inscribed in the cone thus radius can not be zero. 2 1 Thus for a maximum volume r = R and h = H 3 3
Example 7 Show that the right circular cylinder of greatest volume that can 4 be inscribed in a right circular cone has volume that is of the cone 9 From example 6: A right circular cylinder of greatest volume that can be inscribed in a right 2 1 circular cone has r = R and h = H 3 3 1 3 Vcone = πR2 H =⇒ H = Vcone 3 πR2 2 1 Vcyl = πr2 h Since: r = R and h = H 3 3 2 2 1 4 2 Vcyl = π R H = πR H 27 3 3 4 3 2 πR Vcone = 27 πR2 76
Figure 8: 4 Vcone 9 4 = Vcone PROVED 9 =
Vcyl
Example 7 Find two numbers whose sum is 18 and for which the sum of the fourth power of the first and the sqaure of the second is a minumum.
SOLUTION: Let the first number be x and the second y. x + y = 18 =⇒ y = 18 − x A = x4 + y 2 =⇒ A = x4 + (18 − x)2 dA = 4x3 + 2 (18 − x) (−1) = 4x3 + 2x − 36 dx dA For minimum =0 dx 3 4x + 2x − 36 = 0, Solution is: x = 2. y = 16 ∴ x = 2, y = 16
Example 7 Plans for a new supermaerket requires a floor area of 900m2 . The supermarket is to be rectangular in shape with three solid brick walls and a very fancy all-glass front. If the glass costs 1.88 times as much as the brick wall per 77
linear metre, what should be the dimentions of the building so that the cost of meterials for the walls is a minimum. 900 Area: xy = 900 =⇒ y = x Cost: A = xh + 2yh + 1.882 xh 900 A = xh 1 + 1.882 + 2yh x dA minimize the cost: =0 dx dA 900 = h 1 + 1.882 − 2h 2 dx x dA =0 dx 900 h 1 + 1.882 − 2h 2 = 0 x 750 √ x= 1417 = 19. 924 0 1417 900 900 6√ = = y= 1417 = 45. 171 7 750 √ x 5 1417 1417 x = 19.9m, y = 45.2m Check if cost is minumum: d2 A d 900 h 2 h 1 + 1.88 = − 2h = 3600 3 > 0 2 2 dx dx x x Thus cost is indeed minumum. The dimensions of the building so that the cost of meterials for the walls is a minimum are x = 19.9m and y = 45.2m.
Example 7b Suppose that the supermarket has a heat loss across the glass front and the heat loss is seven times as great as the heat loss across the brick walls per square metre. neglecting the heat loss across the roof and through the floor. What should be the dimentions of the building so that the heat loss is a minimum. 900 Area: xy = 900 =⇒ y = x Heat loss: H = xh + 2yh + 7xh dH Minimize heat loss: =0 dx 900 H = 8xh + 2 h x dH 1800 = 8h − 2 h = 0 dx x Solution is: x = ±15 But x = −15, has no physical meaning thus x = 15. 900 y= = 60m 15 Check if heat loss is minumum:
78
d2 A d h 1800 = 8h − h = 3600 3 > 0 dx2 dx x2 x Thus heat loss is indeed minumum. The dimentions of the building so that the heat loss is a minimum are x = 15m and y = 60m.
Exercise 3 An open cylindrical can has to be made out of a sheet metal with a predetermined volume of V litres. If the radius is r cm determine the following: 1. the area of sheet metal required in terms of the radius 2. the value of the radius (r) for a minimum surface area for the cylindrical can. 3. the value for minimum area required to construct the cylinder.
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5.13
Partial Differentiation and Applications
Consider: z = f (x; y) let x change by δx while y is held temporarily constant. Then z changes by δz because z depends on x then: z + δz = f (x + δx; y) δz = f (x + δx; y) − f (x; y) δz f (x + δx; y) − f (x; y) = δx δx ∂z δz f (x + δx; y) − f (x; y) = = lim lim δx→0 δx ∂x δx→0 δx This is denoted by ∂z f (x + δx; y) − f (x; y) = lim ∂x δx→0 δx Similarly ∂z f (x; y + δy) − f (x; y) = lim ∂y δy→0 δy Notations: ∂z = zx ∂x ∂z fy (x, y) = = zy ∂y
fx (x, y) =
The partial derivative of f (x, y) with ∂z respect to x. = zx . ∂x Examples
80
The partial derivative of f (x, y) with ∂z respect to y. = zy . ∂y