All Solutions 1078

Instructor’s Manual Essential Mathematics for Economic Analysis nd

2

edition

Knut Sydsæter Peter Hammond

For further instructor material please visit:

www.pearsoned.co.uk/sydsaeter ISBN 13: 978-0-273-68185-4/ISBN 10: 0-273-68185-0 © Knut Sydsæter and Peter Hammond 2006 Lecturers adopting the main text are permitted to download the manual as required

Pearson Education Limited

Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout throughout the world Visit us on the World Wide Web at:

www.pearsoned.co.uk First published 2006 © Knut Sydsæter and Peter Hammond 2006 The rights of Knut Sydsæter and Peter Hammond to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. ISBN 13: 978-0-273-68185978-0-273-68185-4/ISBN 4/ISBN 10: 0-273-68185-0 0-273-68185-0 All rights reserved. Permission is hereby given for the material in this publication to be reproduced for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London W1T 4LP. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the publishers.

© Knut Sydsæter and Peter Hammond 2006

Pearson Education Limited

Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout throughout the world Visit us on the World Wide Web at:

www.pearsoned.co.uk First published 2006 © Knut Sydsæter and Peter Hammond 2006 The rights of Knut Sydsæter and Peter Hammond to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988. ISBN 13: 978-0-273-68185978-0-273-68185-4/ISBN 4/ISBN 10: 0-273-68185-0 0-273-68185-0 All rights reserved. Permission is hereby given for the material in this publication to be reproduced for OHP transparencies and student handouts, without express permission of the Publishers, for educational purposes only. In all other cases, no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Copyright Licensing Agency Ltd, 90 Tottenham Court Road, London W1T 4LP. This book may not be lent, resold, hired out or otherwise disposed of by way of trade in any form of binding or cover other than that in which it is published, without the prior consent of the publishers.


Preface This instructor’s and solutions manual accompanies Essential Mathematics for Economic Analysis (2nd edition, FT Prentice Hall, 2006). Its main purpose is to provide solutions to the even-numbered problems. (Answers to the odd-numbered problems are given in the main text.) For many of the more interesting and/or difficu dif ficult lt pro proble blems, ms, det detail ailed ed sol soluti utions ons are pro provid vided— ed—mor moree det detail ailed ed tha than n ins instru truct ctors ors are li likel kely y to nee need d for the themmselves. Nevertheless, the extra detail may be useful if answers are supplied to the students, or as a guide to marking answers. Sometimes only an outline of the solution is given. For some of the simpler problems, only the final answer is presented. In addition, for each chapter we have a section with comments on the content. Sometimes we explain why certain topics are included and others are excluded. There are also occasional useful hints based on our experience of teaching the material. In some cases, we also comment on alternative approaches, sometimes with mild criticism of other ways of dealing with the material that we believe to be less suitable. Chapters 1 and 2 in the main text review elementary algebra. This manual includes a Test I, designed for the students themselves to see if they need to review particular sections of Chapters 1 and 2. Many students using our text will probably have some background in calculus. The accompanying Test II is designed to give information to both the students and the instructors about what students actually know about single variable calculus, and about what needs to be studied more closely, perhaps in Chapters 6 to 9 of the text. Oslo and Stanford, Stanford, Augus Augustt 2005 Knut Sydsæter, Peter Hammond Contact addresses: [email protected] [email protected]


Contents 1

Introd Int roduct uctory ory Topi opics cs I: Alg Algebr ebraa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2

Introd Int roduct uctory ory Topi opics cs II: II: Equat Equation ionss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3

Introducto Intro ductory ry Topic Topicss III: Miscel Miscellane laneous ous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4

Funct Fun ction ionss of One One Vari ariabl ablee . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

5

Proper Pro pertie tiess of Funct Function ionss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6

Differ Dif ferent entia iatio tion n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

7

Deriv Der ivati ative vess in Use Use . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

8

Single Sin gle-V -Vari ariabl ablee Optimi Optimizat zation ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

9

Integ Int egrat ration ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

10

Interest Inter est Rates Rates and and Present Present Value aluess . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

11

Funct Fun ction ionss of Many Many Vari ariabl ables es . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

12

Tool oolss for Compa Comparat rativ ivee Stati Statics cs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

13

Multi Mul tiva varia riable ble Opti Optimiz mizati ation on . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

14

Constr Con strain ained ed Opti Optimiz mizati ation on . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

15

Matrix Mat rix and and Vec ector tor Alg Algebr ebraa . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67


16

Determinants and Inverse Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

17

Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Test I (Elementary Algebra) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Test II (Elementary Mathematics) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Answers to Test I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Answers to Test II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87


6

C H A P T ER 2

I N T R O DU C T O R Y T O P I C S I I : E Q U A T I ON S

Chapter 1 Introductory Topics I: Algebra The main purpose of Chapter 1 is to help those students who need to revise elementary algebra. (Those who never learned it will need more help than we can provide here.) Section 1.1 briefly discusses the real number system and distinguishes between natural numbers, integers, rational numbers, and irrational numbers. Section 1.2 concerns rules for manipulating powers, starting with an explanation of the notation, and the rules that a 0 1, a −n 1 /a n (if a 0). Then fundamental properties of exponents are described. Next, it is shown how powers can be used to calculate compound interest. Finally, negative powers are applied to find how much money should have been deposited in an account a given number of years ago in order to yield a target amount today. Section 1.3 concerns algebraic rules. Twelve of the most common rules are presented first, and illustrated in examples. One is even illustrated geometrically. (A surprisingly large number of students seem unaware of how algebraic rules can be illustrated in this way.) There follow three identities concerning squares. The student is told what to do with a minus sign outside parentheses. The anatomy of algebraic expressions is defined—terms in particular, and numerical coefficients. Fractions may have frightened some students in the past. Section 1.4 may help to overcome the induced amnesia. Notation is explained. Common factors are cancelled. Denominators are made rational. “Stupid” errors are described. Rules for signs are explained. Algebraic fractions are added, and the rules illustrated numerically. The use of LCDs (lowest common denominators) is illustrated. Parentheses are suggested where they help avoid errors. Finally come rules for multiplying and dividing fractions. All very elementary, but crucially important, and sources of common errors among students. Section 1.5 explains fractional powers and begins with the well-known case of square roots. Fractional powers are important in economics as will be shown in later chapters. (See Section 4.8.) The next Section 1.6 discusses inequalities, which are often not much discussed in elementary algebra courses. Notation and some elementary properties start the section. It is shown how to “solve” linear inequalities. Among the properties presented is the often neglected (5)—if each side of an inequality is multiplied by a negative number, the inequality gets reversed. Then sign diagrams are introduced as a useful device for seeing when certain products or quotients are positive, and when they are negative. The section concludes with a brief discussion of double inequalities. Intervals on the real line, absolute values, and the usual measure of distance are discussed in Section 1.7. All the problems in Chapter 1 are supplied with answers in the text itself to make this chapter more suitable for self study.

=

=

=

Chapter 2 Introductory Topics II: Equations Mathematics for economic analysis often involves solving equations, which is the topic of this chapter. In Section 2.1 even the simplest linear equations receive the attention that some students may well need. Nonlinear equations can lead to attempts to divide by zero, which must be carefully avoided. Often, the hardest part of solving a problem is in formulating an appropriate equation. This skill is hard to teach, except by numerous examples. Some practice can come from the problems. © Knut Sydsæter and Peter Hammond 2006

C H A P T ER 3

I N T R O D UC T O R Y T O P I C S I I I : MI S C E L LA N E O U S

7

Section 2.2 discusses equations with parameters which arise in many economic applications, yet are often neglected in textbooks. The problems for this section are aimed at training students to solve such equations. Then Section 2.3 covers quadratic equations (in a single variable). It describes the crucially important technique of completing the square, as well as how to use the roots of a quadratic equation to factor the left-hand side. Only after several examples is the technique described in general. Then the expressions for the sum and product of the two roots (when they are real) are given. Section 2.4 deals with the solution of two equations in two unknowns. First, the nature of the problem is explained. Then two methods of solution are proposed. The first method involves using one equation to solve for one variable in terms of the other. After this, the solution for both variables is easily derived by substitution, of course. The second method is to eliminate one of the variables by adding or subtracting multiples of one equation from the other. An appropriate tableau for doing this is described. The section ends with a formula giving the solution to a general pair of two linear equations in two unknowns. More general simultaneous linear equations, together with techniques for solving them, receive extended coverage in Chapters 15 and 16. Section 2.5 gives some examples of how to solve some types of nonlinear equation which frequently occur in optimization problems. All the problems in this chapter are supplied with answers in the text itself to make this material more suitable for self study.

Chapter 3 Introductory Topics III: Miscellaneous The chapter discusses some miscellaneous mathematical fundamentals. Section 3.1 explains summation notation, a topic that often causes difficulties to the untutored. Simple examples illustrate the general notation. It is important to understand that the index of summation is a “dummy variable”, and also to understand what to do with indices that are not indices of summation. Then Section 3.2 presents some rules for manipulating sums, including sums of sums and common factors. An important example in which most terms cancel is presented. So are formulas connected with the (arithmetic) mean and variance of a set of numbers. Some common errors are also noted. Then the formula for the sum of an arithmetic progression is presented. The formula is generalized in Problem 3.2.5. We also give the formulas for the sums of the first n squares and first n cubes. Next comes the binomial formula. Pascal’s triangle is displayed (and attributed more appropriately). Section 3.3 gives a brief coverage of double sums. There is an analogy with a standard spreadsheet. The main point is that summing each row and then summing the row totals is equivalent to summing each column and then summing the column totals. In other words, the order of summation does not matter. Section 3.4 discusses some aspects of logic. Propositions are explained, as are implications. Also, necessary and sufficient conditions are clearly distinguished from each other. The use of logic to solve equations and inequalities is illustrated. Next, Section 3.5 discusses proofs, including both direct proofs (that P Q) and indirect proofs (that not Q not P ). Also mentioned is the distinction between deductive and inductive reasoning. A brief exposition of set theory comes in Section 3.6. Sometimes sets can be defined by listing their members. More generally, they can be defined by the common property or properties each of their members must satisfy. Standard notation for set membership, unions, intersections, differences, the empty set, and complements is introduced. So are Venn diagrams, which are then applied to show some easy results. Finally, the last Section 3.7 of this chapter treats proof by (mathematical) induction. It is an important idea to grasp.

⇒


⇒

8

C H A P T ER 3

I N T R O DU C T O R Y T O P I C S I I I : M I S C E LL A N E O US

Answers to Even-Numbered Problems 3.1

√ + √ + √ + √ + √ = + √ + √ + + + + + + + = + + + + +···+ + + f (x2 )x2 + · · · + f (x 2·3+3·5+4·7 6 + 15 + 28 49 · · 100 = 100 = · 100 ≈ 144 .12 1·3+2·5+3·7 3 + 10 + 21 34

2. (a) 2 0 2 1 2 2 2 3 2 4 2 (3 2 3) (b) (x 0)2 (x 2)2 (x 4)2 (x 6)2 4 (x 2 6x 14) (c) a1i b2 a2i b3 a3i b4 ani bn+1 (d) f (x0 )x0 f (x1 )x1 4.

6. (a) The total number of people moving from region i .

m )xm

(b) The total number of people moving to region j .

3.2

+ b)6 = a6 + 6a5 b + 15a4 b2 + 20a3 b3 + 15a2 b4 + 6ab5 + b6 . (The coefficients are those in the

2. (a

seventh row of Pascal’s triangle in the text.)

  =



· · = 5 · 4 · 3 · 2 · 1 = 5! = 5! . In general, m = m(m − 1) ··· (m − k + 1) · · 1 · 2 · 3 · 2 · 1 3!2! 2!3! k k! − k + 1) · (m − k)! = m! . = m(m − 1) ···k(m !(m − k)! (m − k) ! k ! 8 8·7·6 8 8 8·7·6·5·4 8 8 = = = = = + = (b) 56. Also, 56; 3 1·2·3 8−3 5 1·2·3·4·5 3 3+1 8·7·6·5 8+1 9 9·8·7·6 = = = = 126. 56 + 126 and 1·2·3·4 3+1 4 1·2·3·4

5 4. (a) 3

5 4 3 1 2 3



(c)

 =

m!

m

m

− k)!k! = (m −+k)1!+(km+−1)k)! k m!(k

(m

            = − + + =   + + = = m

and

m k (m 1)!

m

k

m

− k)!(k + 1)!

(m

1

k

k

+

1 1

  

m!

m! + (m − k) !k ! (m − k − 1)!(k + 1)!

3.3 2. (a) The total number of units of good i .

(b) The total number of units of all goods owned by person j . (c) The total number of units of goods owned by the group as a whole.

− ¯ − ¯ − ¯ = − ¯

4. Because arj a is independent of the summation index s , it is a common factor when we sum over s , so m m a)(asj a) (a rj a) s =1 (asj a) for each r . Next, summing over r gives s =1 (arj



m

m





− ¯

m

− a¯ )(a

(arj

r 1 s 1

= =

  − ¯ =

sj

a)

   − ¯ − ¯ m

(arj

r 1

=

a)

(asj

∗∗)

a)

(

s 1

=

¯

Using the properties of sums and the definition of aj , we have m

 r 1

=

m

m

 −  ¯ = − ¯ =

(arj

a)

arj

r 1

=

a

r 1

=

¯ − ma¯ = m(a¯ − a¯ )

m aj

Similarly, replacing r with s as the index of summation, one also has Substituting these values into ( ) then confirms ( ).

∗∗


∗

j



m s 1 (asj

=

− a¯ ) = m(a¯ − a¯). j

C H A P T ER 3

I N T R O D UC T O R Y T O P I C S I I I : MI S C E L LA N E O U S

9

3.4

= 2. (x = −1, 0, and 1 make the equation meaningless. Multiplying each term by the common denominator x (x − 1)(x + 1) yields 2 x(x 2 − 3x + 2) = 0, or 2x(x − 1)(x − 2) = 0. Hence, x = 2 is

2. x

the only solution.) 4. (a) x (c) x

≥ 0 is necessary, but not sufficient. ≥ 4 is necessary and sufficient.

(b) x

≥ 50 is sufficient, but not necessary.

− = + − √ + +

6. (a) No solutions. (Squaring each side yields x 4 x 5 18 x 5 81, which reduces to with solution x 20. But x 20 does not satisfy the given equation.) (b) x 20

=

=

√ x + 5 = 5,

= √ √ (i) (ii) 8. (a) x + x + 4 = 2 ⇒ x + 4 = 2 − x ⇒ x + 4 = 4 − 4x + x 2 ⇒ x 2 − 5x = 0 ⇒ x − 5 = 0 ⇐ x = 5. Here implication (i) is incorrect ( x 2 − 5x = 0 ⇒ x − 5 = 0 or x = 0.) Implication (ii) is correct, but it breaks the chain of implications. (b) x = 0. (After correcting implication (i), we see that the given √ equation implies x = 5 or x = 0. But only x = 0 is a solution; x = 5 solves x − x + 4 = 2.) 3.5 2. (a) Logically the two statements are equivalent.

(b)Appending the second statement is still an expressive

poetic reinforcement.

3.6

∩ B ∩ C is the set of all female biology students in the university choir; M ∩ F the female mathematics students; (M ∩ B) \ C \ T the students who study both mathematics and biology but neither play tennis

2. F





nor belong to the university choir. 4. (a) B

⊂ M

(b) F

∩ B ∩ C = ∅

(c) T

∩ B = ∅

(d) F (T

\ ∪ C) ⊂ B = { } = { }

6. (b) and (c) are true, the others are wrong. (Counter example for (a), (d), and (f): A 1, 2 , B 1, C 1, 3 . As for (e), note in particular that A B A C A whenever B and C are subsets of A, even if B C .)

= { } =

∪ = ∪ =

=

8. (a) Consider Fig. M3.6.8, and let nk denote the number of people in the set marked S k , for k 1 , 2, . . . , 8. Obviously n1 n2 n8 1000. The responses imply that: n1 n3 n4 n7 420; n1 n2 n5 n7 316; n2 n3 n6 n7 160; n 1 n7 116; n3 n7 100; n 2 n7 30; n 7 16, and n 8 334. From these equations we easily find n 1 100, n 2 14, n 3 84, n 4 220, n 5 186, n6 46, n 7 16. (i) n 3 n4 304 had read A but not B ; (ii) n6 46; (iii) n 8 334. (b) We find n(A B) n 3 n4 304, n(C (A B)) n 6 46, and n( (A B C)) n 8 334. The last equality is a special case of n( D) n() n(D). (The number of persons who are in  , but not in D , is the number of persons in all of  minus the number of those who are in D .)

+ + · · · + = + = + + + = = = = + = \ = + =


+ + + = + = + = + = = = = = = = \ ∪ = = \ ∪ ∪ = \ = −

+ + = = =

10

C H A P T ER 3

I N T R O D UC T O R Y T O P I C S I I I : M I S C E LL A N E O U S

B

A S 5

S 1 S 4 S 7

S 2

S 3 S 6

S 8

C

Figure M3.6.8

3.7

= 1 the LHS and the RHS of (3.2.6) are both equal to 1. As the induction hypothesis, suppose (3.2.6) is true for n = k , so that

2. We prove only (3.2.6); the proof of (3.2.5) is very similar, but slightly easier. For n

k

 i 1

=

Then

k 1

+

i3

= 13 + 23 + · · · + k3 = [ 12 k(k + 1)]2

k

 = i

3

i 1

i3

+ (k + 1)3 = [ 12 k(k + 1)]2 + (k + 1)3 = (k + 1)2 ( 14 k2 + k + 1)

i 1

=

=

But this last expression is equal to 14 (k 1 )2 (k2 4 k 4 ) [ 12 (k (3.2.6) is true for n k 1. By induction, we have proved (3.2.6).

+

+ 1)(k + 2)]2 , which proves that

+ + =

= + 4. The claim is true for n = 1. As the induction hypothesis, suppose k 3 + (k + 1)3 + (k + 2)3 is divisible by 9. Note that (k + 1 )3 + (k + 2 )3 + (k + 3 )3 = (k + 1 )3 + (k + 2 )3 + k 3 + 9 k 2 + 27 k + 27 = k 3 + (k + 1)3 + (k + 2)3 + 9(k 2 + 3k + 3). This is divisible by 9 because the induction hypothesis implies that the sum of the first three terms is divisible by 9, whereas the last term is also obviously divisible by 9.

Review Problems for Chapter 3 2. (a)12 4 22 5 32 6 42 7 4 20 54 112 1 1/2 1 4 25 216 495 /2

· + · + · + · = + + + = 190 + + + + + = 38

4. (a)

 i 4

=

 n

i(i

+ 2)

(b)

=

i 1

1 xi

16

(c)

 j 0

=

x 2j

(d)

+ 1

2j

(b)1

− 16 = 56 (c)1−2 +2−1 +30 +41 +52 +63 =

81

−

( 1)k −1

k 1

=

1 k

⇒ false (because x 2 = 16 also has the solution x = −4), ⇐ true (c) ⇒ true, ⇐ ⇒ and ⇐ both true 8. A ∩ B = ∅; A ∪ B = {1, 2, 4, 6, 11, 12};  \ B = {1, 3, 4, 5, 6, 7, 8, 9, 10}; C A =  \ A = {2, 3, 5, 7, 8, 9, 10, 11, 12} 10. For n = 1, the inequality is correct: 1 + x ≥ 1 + x . Suppose (1 + x) ≥ 1 + kx for an arbitrary natural number k . We have to prove that then ( 1 + x) +1 ≥ 1 + (k + 1)x . But because x ≥ −1 and k > 0, we have (1 + x) +1 = ( 1 + x) (1 + x) ≥ ( 1 + kx)(1 + x) = 1 + (k + 1)x + kx 2 ≥ 1 + (k + 1)x . Thus, by 6. (a)

⇒ true, ⇐ false

(b) false (x can be 3) (d)

k

k

k

k

induction, Bernoulli’s inequality is true for all natural numbers n .


C H A PT E R 4

F U N CT I O NS O F O N E V A RI A B LE

11

Chapter 4 Functions of One Variable Functions of one real variable are the main subject of Chapters 4 to 9. Chapter 4 begins with a gentle introduction. Section 4.1 presents a few simple examples, mostly drawn from economics, before Section 4.2 presents a definition, and introduces functional notation systematically. The domain and range of a function are defined, as are increasing and decreasing functions. In our experience, training students to master functional notation is very important, yet sometimes neglected in introductory texts. The students should be encouraged to do many of the numerous problems in this section, and throughout the chapter. Next, Section 4.3 begins with a review of how functions can be represented by graphs in a two-dimensional coordinate system. It displays the graphs of a number of important simple functions which students need to learn as soon as possible. The special case of linear functions is the subject of Section 4.4. The intercept is defined, and the formula for the slope provided. Next come the “point–slope” formula for the straight line with a given slope through a specified point, and the “point–point” formula for the line through two specified points with different x coordinates. The section concludes with the general equation for a straight line that includes the case when the line is vertical, along with the graphical method of solving linear equations, and the graphical representation of linear inequalities. Section 4.5 considers a (dubious) linear model of population growth and the linear consumption function as well as a simple model of market equilibrium of supply and demand. Section 4.6 treats quadratic functions, and the possible shapes of their graphs. The solution of quadratic equations is reviewed (see also Section 2.3) and the method of completing the square is used to identify the maximum or the minimum point of a quadratic function. The Cauchy–Schwarz inequality is discussed as a harder problem. In order to have students see mathematics applied as soon as possible to economic optimization problems, we consider a monopolist with a quadratic cost function who faces a linear (inverse) demand function. The perfectly competitive firm is treated as a special case. It is shown that, in order that the profit maximum for a monopolist should coincide with that for a competitive firm, its output should be subsidized (rather than taxed). (Alternatively, it should have its price regulated.) Next, Section 4.7 covers more general polynomials, starting with cubic functions. An example shows how they can represent a richer family of cost curves. The solutions of polynomial equations are discussed. Techniques for finding integer roots are emphasized. Then division of polynomials is considered, especially the remainder theorem. The correspondence between the roots and linear factors of a polynomial is pointed out. Then a technique for dividing polynomials is presented. Finally, rational functions—both proper and improper—are defined. Power functions are the subject of Section 4.8. The need for fractional powers is illustrated, and rational powers are then defined. The graphs of power functions are illustrated. Section 4.9 treats exponential functions. The natural exponential function is introduced already at this stage, although the reason for its importance will only be clear once students have mastered some calculus. Chapter 4 concludes with a discussion of logarithmic functions, emphasizing the natural logarithmic function which is now the most commonly used.


12

C H A PT E R 4

F U NC T I ON S O F O N E V A R IA B LE


= F (−3) = 10, F (a + h) − F(a) = 10 − 10 = 0 √ √ √ √ 4. (a) f (−1/10) = −10/101, f (0) = 0, f (1/ 2) = 2/3, f ( π ) = π /(1 + π), f (2) = 2 /5 (b) f (−x) = −x/( 1 + (−x) 2 ) = −x/( 1 + x 2 ) = −f(x) and f (1/x) = (1/x)/[1 + (1/x) 2 ] = (1/x) · x 2 /[1 + (1/x) 2 ] · x 2 = x /(1 + x 2 ) = f (x) 6. (a) C (0) = 1000, C (100) = 41 000, and C (101) − C( 100) = 501 (b) C (x + 1) − C(x) = 2 x + 301 = incremental cost of increasing production from x to x + 1. 8. (a) f (tx) = 100 (tx)2 = 100 t 2 x 2 = t 2 f(x) (b) P (tx) = (tx)1 2 = t 1 2 x 1 2 = t 1 2 P(x) 10. (a) No. f (2 + 1) = f (3) = 18, whereas f (2) + f (1) = 10. (b) Yes. f (2 + 1) = f (2) + f (1) = −9 + = = √ ≈ + = √ + ≈ 2. F (0)

/

(c) No. f (2

1)

f (3)

3

1 .73, whereas f (2)

/

f (1)

2

/

1

/

2 .41.

12. See Figs. M4.2.12a and M4.2.12b. x 1

1

·

1 1

1

·

1 x

x2

1 x

x

1

x

·

x

Figure M4.2.12a The area is (x 1)2 x 2 2x 1

+

Figure M4.2.12b The area is x 2 1

= + +

+

14. (a) Defined for x 2, i.e. Df ( 3x 6 (c) f (x) 3 3x x 2

= = −∞, 2) ∪ (2, ∞) (b) f (8) = 5 + = − = ⇐⇒ + 6 = 3(x − 2) ⇐⇒ 6 = −6, which is impossible.

4.3 2. (a) f ( 5) 0, f ( 3) 3, f ( 2) (b) Df [ 5, 4], Rf [ 3, 4]

− = = −

x 4. h(x)

− = 0, f (0) = 2, f (3) = 4, f (4) = 0

− = − = −

= x 2 − 2x − 3

−2

−1

0

1

2

3

4

5

0

−3

−4

−3

0

5

See Fig. M4.3.4. x 6. G(x)

= 1 − 2−

x

−2

−1

0

1

2

3

−3

−1

0

1/2

3/4

7/8

See Fig. M4.3.6. © Knut Sydsæter and Peter Hammond 2006

C H A PT E R 4

13


y

6 5 3 2 1

−2

1

2

3

y

6 5 4 3 2 1

4

4 x

-2 -1 -1 -2 -3 -4

−3 −4 Figure M4.3.4

1 2 3 x

Figure M4.3.6

4.4 2. See Figs. M4.4.2a, M4.4.2b, M4.4.2c y

y

y

1

4 3

−1 −2 −3 −4 −5

2 1 1

2

3

4

x

Figure M4.4.2a

4 1

2

3

4

5

6 7

8

10 x

3 2 1 1

−1

Figure M4.4.2b

2

3

4

5 x

Figure M4.4.2c

= x + 2, with slope 1; L2 is y = − 35 x + 3, with slope −3/5; L3 is y = 1, with slope 0; = 3x − 14, with slope 3; L5 is y = 19 x + 2, with slope 1/9. −1400 (Q − 100) 6. If P is the price of Q copies, then applying the point–point formula gives P − 1400 = 3000 500−100 or P = 1000 + 4Q. The price of printing 300 copies is therefore P = 1000 + 4 · 300 = 2200. 4. L1 is y L4 is y

8. (a) See Figs. M4.4.8a, M4.4.8b, and M4.4.8c. y

y

y

1

1 1

x

Figure M4.4.8a


−1 Figure M4.4.8b

1

x

1

Figure M4.4.8c

x

14

C H A PT E R 4


10. See Fig. M4.4.10. Each arrow points to the side of the line where the relevant inequality is satisfied. The

shaded triangle is the required set. y

y x

3

− y = 1

3 2 1

x

-4 -3 -2 -1 -1 -2 -3 -4

2 1 2 3x

3

4 3x

+ 4y = 12

+ y = 3

Figure M4.4.10

f(x)

= − 12 x 2 − x + 23

1 2 3 x

Figure M4.6.2

4.5 e

e

e

e

− 3P = 20 + 2P , and hence P = 11. (b) P = 90 4. C = 0 .8y + 100. (The equation is C = ay + b. We are told that 900 = 1000a + b and a = 80 /100 = 0 .8. So b = 100.) 6. (a) April 1960 corresponds to t = 9 /4, when N (9/4) = −17 400 · (9/4) + 151 000 = 111 850. (b) −17 400 t + 151 000 = 0 implies t = 8 .68, which corresponds roughly to September 1966. 2. (a) 75

4.6

−4

x 2. (a)

−3

−2

−1

0

1

2

= − 12 x 2 − x + 32

−2.5 −2.5 0 1.5 2 1.5 0 (b) See Fig. M4.6.2. (c) f (x) = − 12 (x + 1)2 + 2. Maximum point ( −1, 2). (d) x = −3 and x = 1. (e) f (x) > 0 in ( −3, 1), f (x) < 0 for x < −3 and for x > 1. 4. (a) x (x + 4). Zeros 0 and −4. (b) No factoring is possible. No zeros. √ √ (c) −3(x − x1 )(x − x2 ), where the zeros are x 1 = 5 + √ 15 and x 2 = 5 − √ 15. (d) 9(x − x1 )(x − x2 ), where the zeros are x 1 = 1 /3 + 5 and x 2 = 1 /3 − 5. (e) −(x + 300)(x − 100). Zeros −300 and 100. (f) (x + 200)(x − 100). Zeros −200 and 100. 6. U(x) has maximum for x = 4 (r − 1)/(1 + r 2 ). (Use (5).) 8. (a) If P = α1 − 13 Q denotes the selling price in England, and P = α2 + 16 Q denotes the buying price in Ghana, π(Q) = (P − P − γ )Q = − 12 Q2 + (α1 − α2 − γ )Q. (b) Using (5), we see that Q∗ = α 1 − α2 − γ maximizes profit if α 1 − α2 − γ > 0. If α 1 − α2 − γ ≤ 0, then Q∗ = 0. (c) π(Q) = − 12 Q2 + (α1 − α2 − γ − t)Q and Q∗ = α 1 − α2 − γ − t if α 1 − α2 − γ − t > 0. (d) T = t Q∗ = t (α1 − α2 − γ − t). Here t = 12 (α1 − α2 − γ ) maximizes export tax revenue. f(x)

E

G

E

G

4.7 2. (a) 1 and

−2 (b) 1, 5, and −5 (c) −1 4. (a) y = 21 (x + 1)(x − 3) (b) y = −2(x + 3)(x − 1)(x − 2) (c) y = 12 (x + 3)(x − 2)2 6. c4 + 3c2 + 5 = (c 2 + 3/2)2 + 11/4 is not 0 for any choice of c , so the division has to leave a remainder. © Knut Sydsæter and Peter Hammond 2006

C H A PT E R 4

15


4.8 2. (a): C

(b): D

(c): E

(d): B

(e): A

x

(f) F: y

= 2 − (1/2) 4. (a) 23 = 8, so x = 3/2 (b) 3x + 1 = −4, so x = −5/3 (c) x 2 − 2 x + 2 = 2, so x 2 − 2 x = 0, implying that x = 0 or x = 2. 2 3 = 6. V = (4/3)πr 3 implies r 3 = 3V /4π and so r = (3V /4π )1 3 . Hence, S = 4π r 2 = 4π 3V /4π √

 

/

3

36π V 2/3 .

/

4.9 2. P(t)

= 1.22 · 1.034 . The doubling time t ∗ is given by the equation (1.034) ∗ = 2, and we find t ∗ ≈ 20.7 t

t

(years).

4. The graphs are drawn in Fig. M4.9.4. x

−3

−2

−1

0

1

2

3

2x

1/8

1/4

1/2

1

2

4

8

2−x

8

4

2

1

1/2

1/4

1/8

y

8

= 2−x

y

y

6

= 2x

4 y

f(x)

2

2

= 1 + x 4+x 4 2

1

−3 −2 −1

1

2

3 x

−4 −3 −2 −1

Figure M4.9.4

1

2

3

4

5

x

Figure M4.R.2

= 3.91 · 105 /5.1 ≈ 0 .7667 · 105 , and (taking ln of each side) we find t ≈ 327. So the year is 1969 + 327 = 2296. This is when every Zimbabwean would have only 1 m 2 of land on average. 8. (b) and (d) do not define exponential functions. (In (f): y = ( 1/2) .) 10. Use the two points on each graph to determine A and b from the equation y = Ab . You then get: (a) y = 2 · 2 (b) y = 2 · 3 (c) y = 4 (1/2) . 6. We find ( 1.035)t

x

x

x

x

x

4.10

= = (b) x = e 3 (c)√ x 2 − 4x + 5 = 1 so (x − 2)2 = 0. Hence, − √ − = = − = − = 0, so x = 1 ± 2. (e) x = 0 or ln(x + 3) = 0, so x = 0

2. (a) ln 3x x ln 3 ln 8, so x ln 8/ ln 3. x 2. (d) x (x 2) 1 or x 2 2x 1 or x 2. (f) x 5 1 so x 36.

=

= = −


16

C H A PT E R 4

4. (a) t


= r −1 s ln BA

(b) t

=

countries have the same GNP. 6. (a) Wrong. (Let A

1 5 .6 ln 0.07 1.2

≈ 22. It will take approximately 22 years before the two

= e .) (b) and (c) are right.

Review Problems for Chapter 4 2. (a) F (0)

= 1, F (−2) = 0, F (2) = 2, and F (3) = 25/13

becomes large positive or negative.

(b) F (x)

(c) See Fig. M4.R.2.

= 1 + x +44/x tends to 1 as x

| | ≥ 1, i.e. x ≤ −1 or x ≥ 1 (b) x > 4 (c) 3 ≤ x ≤ 5 6. (a) Slope −4 (b) Slope −3/4 (c) Slope −b/a 8. f (2) = 3 and f (−1) = −3 give 2a + b = 3 and −a + b = −3, so a = 2, b = −1. Hence f(x) = 2 x − 1 and f (−3) = −7. (Or use the point–point formula.) 10. y = 2 x 2 + x − 6. (−3 = a + b + c, −6 = c , and 15 = 9 a + 3b + c) 12. The new profit is π = 100 Q − 52 Q2 − tQ, which is maximized at Q = 15 (100 − t). 14. (a) p(x) = x (x − 3)(x + 4) (b) q(x) = 2 (x − 2)(x + 4)(x − 1/2) 16. (a) k = 4 (b) k = −3/2 and k = 1 (c) k = 26 (d) k = −1 and k = 4 18. (1 + p/100)15 = 2 gives p = 100 (21 15 − 1) ≈ 4 .7 as the percentage rate. √ 20. (a) t = ( ln x − b)/a (b) t = ( ln 2)/a (c) t = ± ln 8/ 2π 4. (a) x

t

t

/

  


C H A P T ER 5

P R O P ER T I E S O F F U N C T I ON S

17

Chapter 5 Properties of Functions This chapter begins in Section 5.1 by examining how changes in a function relate to shifts in its graph. An economic application is to shifting demand and supply curves. Section 5.2 considers how to construct new functions from old ones, and includes a discussion of symmetry. Next, Section 5.3 introduces inverse functions. By definition, if a function is one-to-one, there is an inverse from its range or image space back to its domain. The obvious geometric properties of the graphs of a function and its inverse are explained. Any equation in two variables can be represented by a curve (or a set of points) in the xy -plane. Some examples are given in Section 5.4. Also included is the “vertical line test” for a graph to represent a function, and the effect on a graph of changing units. The section ends with some examples of compound functions. The distance between points in a plane is defined in Section 5.5, and used to derive the equations whose graph is a circle with a given centre and radius. Ellipses and other conics are briefly discussed. Section 5.6 concludes the chapter with a discussion of the general concept of a function and the associated terminology—domain, target, range, image. Definitions of one-to-one functions and of inverse functions are also given.


= f (x) is moved 2 units to the right. See Fig. M5.1.2a. (b) The graph of y = f (x) is moved downwards by 2 units. See Fig. M5.1.2b. (c) The graph of y = f(x) is reflected about the

2. (a) The graph of y

y -axis. See Fig. M5.1.2c. y

y

1

1 1

1 1 x

x

Figure M5.1.2a

4. Move y

y

1

Figure M5.1.2b

x

Figure M5.1.2c

= |x | two units to the left. Then reflect the graph about the x -axis. Finally, move the graph up 2 y

units. See Fig. M5.1.4.

1

−1 Figure M5.1.4 2 Ad + c = Bd −2Bd

6. y ∗


x

18

C H A P T ER 5


5.2 2. See Figs. M5.2.2a to M5.2.2c. (Note that the graph of e −x y

2

+ x is not symmetric about (0, 1).)

y

y

x

x

x

Figure M5.2.2a

Figure M5.2.2b

Figure M5.2.2c

= f (3x + 7) = 3(3x + 7) + 7 = 9x + 28. Then 9x + 28 = 100 for x = 8.

4. f(f(x))

5.3

= ( 157.8/D)10 3 /

2. p

{−4, −2, 0, 2, 4, 6, 8}. f −1 (2) = −1 (b) f (x) = 2x + 4, f −1 (x) = 12 x − 2 6. (a) f (x) = x/ 2 and g(x) = 2x are inverse functions. (b) f (x) = 3x − 2 and g(x) = 13 (x + 2) are inverse functions. (c) C = 59 (F − 32) and F = 95 C + 32 are inverse functions. 4. (a) The domain of f −1 is

8. (a) See Fig. M5.3.8a.

(b) See Fig. M5.3.8b. Triangles O BA and O BC are congruent.

y

y

= x

y

D

C

= x

y

= (b,a)

y

3

(3, 5) B

2

(1, 3)

(5, 3)

1

= (a,b)

A

(3, 1)

1 E

x

Figure M5.3.8a

10. (a) x

5.4

= ln y − 4, y > 0

Figure M5.3.8b

(b) x

= e +4 , y ∈ (−∞, ∞) y

2

3

x

x

Figure M5.4.2

(c) x

y

= 3 + ln(e − 2), y > ln 2

√ − = − = − √ − = − √ √ − − − √ − + = = − − + − = − − = = = =

2. (a) See Fig. M5.4.2. (b) If a > 1, then g(a) a 1 and g(a) < 1 (see the figure), so that 1 g(g(a)) g(1 a 1) ( 1 a 1)2 2(1 a 1) 2 a . If a < 1, then g(a) a 2 2a 2, and g(a) > 1, so that g(g(a)) 1 a 2 2a 2 1 1 (a 1)2 1 (a 1) a . Moreover, g(g(1)) 1. It follows that g(g(a)) a for all a , so g is its own inverse. (Alternatively, show that one can solve y g(x) for x in terms of y , in which case x g(y).)




= − + = + − =

C H A P T ER 5

4. F (100 000)


19

= 4070. The graph is the thick line sketched in Fig. M5.4.4. y y

Y

4070

= ( 3, 2)

A

2

(2, 4)

2

x

2 P RN

7500

2

100000

Figure M5.4.4

Figure M5.5.2

5

= (5, −4)

B

x

Figure M5.R.10

5.5

− 2)2 + (y − 4)2 = 13, or y 2 − 8y + 12 = √ 0, with solutions y = 2 =and y = 6. Geometric explanation: The circle with centre at (2, 4) and radius 13 intersects the line x 5 at two points. See Fig. M5.5.2. 4. (a) (x − 2)2 + (y − 3)2 = 16 (b) (x − 2)2 + (y − 5)2 = 13 6. The condition is that (x + 2)2 + y 2 = 2 (x − 4)2 + y 2 , which reduces to (x − 6)2 + y 2 = 4 2 . 2. (5

5.6





2. The function in (b) is one-to-one and has an inverse: the rule mapping each youngest child alive today to

his/her mother. (Though the youngest child of a mother with several children will have been different at different dates.) The function in (d) is one-to-one and has an inverse: the rule mapping the surface area to the volume. The function in (e) is one-to-one and has an inverse: the rule that maps (u, v) to (u 3, v). The other functions are many-to-one, in general, and so have no inverses.

−

Review Problems for Chapter 5

+ g)(x) = x 2 − 2, (f − g)(x) = 2x 3 − x 2 − 2, (fg)(x) = x 2 (1 − x)(x 3 − 2 ), (f/g)(x) = − 2)/x2 (1 − x), f (g(1)) = f (0) = −2, and g(f (1)) = g (−1) = 2. 4. p = ( 64 − 10D)/3 √ 6. (a) x = 50 − 12 y (b) x = y/ 2 (c) x = 13 [2 + ln(y/ 5)] √ √ 8. (a) 13 (b) 17 (c) (2 − 3a) 2 = |2 − 3a | (Note that 2 − 3a is the correct answer only if 2 − 3a ≥ 0, i.e. a ≤ 2 /3. Ask students with the wrong answer to put, say, a = 3.) 10. (x − 3)2 + (y − 2)2 = (x − 5)2 + (y + 4)2 , which reduces to x − 3y = 7. See Fig. M5.R.10. 2. (f (x 3

5




20

C H A PT E R 6

D I F FE R EN T I AT I O N

Chapter 6 Differentiation Our treatment of the single variable calculus begins in Chapter 6. Section 6.1 is devoted to the slopes of curves and their tangents, as well as the tangent definition of derivative. Section 6.2 discusses how the tangent is the limit of a sequence of lines joining nearby points on the graph of a function. It proceeds to present the formula for the Newton (or differential) quotient, whose limit is the derivative. The derivative is then used in the general equation for the tangent to the graph of a function at any point where it is differentiable. A general recipe for finding derivatives is provided, and Leibniz’s differential notation is introduced. Section 6.3 discusses increasing and decreasing functions and explains how the sign of the derivative can help us to find the intervals of increase/decrease of a differentiable function. (Note 8.4.2 shows how the mean-value theorem can be used to prove the main results of this section.) Section 6.4 is devoted to the economic significance of the derivative of a function. The instantaneous and proportional rates of change are distinguished. Marginal cost, revenue, profit, propensity to consume, product, are all defined in terms of derivatives. Marginal cost is distinguished from incremental cost. It is crucially important for economics students to understand the definition of the derivative. In fact, this is more important than knowing how to differentiate very complicated functions. Example 6.4.3 and problems like 6.4.5 should be emphasized. The procedure of smoothing a function to fit a set of discrete data points is illustrated at the end of this section. Limits were used to define derivatives informally in Section 6.2. The more careful discussion that is really needed is the topic of Section 6.5, though even this remains rather informal. The discussion proceeds mostly through examples. Some general rules for limits of sums, differences, products, ratios, and powers are presented. After these preliminaries, Section 6.6 is devoted to some simple rules for differentiation, including the rule for differentiating a power function. Section 6.7 presents the rules for differentiating sums, differences, products, and quotients. The product rule in particular is illustrated economically. The quotient rule is used to show that average cost is increasing iff it is less than marginal cost. Section 6.8 states the chain rule, first using Leibniz’s notation in a suggestive way, then more carefully. An incomplete proof of the chain rule is provided in small print; we judge that a complete proof belongs to more rigorous courses in mathematical analysis. The next section 6.9 introduces second-order derivatives. It also defines convex and concave functions using the sign of the second derivative. In fact, much modern economic analysis has been built on specific assumptions to the effect that certain functions are either concave or convex. The examples illustrated in Figures 4 and 5, or similar examples, should be carefully discussed. In Example 5 it is shown that an increasing concave transformation of a concave function is concave. Higher-order derivatives, found by repeated differentiation, are discussed next. Sections 4.9 and 4.10 introduced exponential and logarithmic functions, which are often used in economics, of course, as well as in statistics. Section 6.10 shows that if the special number e 2 .718281828 . . . is used as the base for the exponential function, then dex /dx e x for all x , so the derivative of the function is the function itself. It also shows how to differentiate eg(x) using the chain rule. Finally, the rule for differentiating the general exponential function a x is derived. Section 6.11 shows how to differentiate the natural logarithmic function, d ln x/dx 1/x , and also the derivative d ln[h(x)]/dx for any positive valued differentiable function h(x). The section includes a discussion of the useful technique of logarithmic differentiation—i.e., taking the logarithm of an expression before differentiating. As shown later in Section 7.7, this is often a natural way for economists to calculate elasticities.

=

=

=


C H A PT E R 6

D I F FE R EN T IA T I ON

21

A subsection characterizes the number e as lim h→0 (1 h)1/ h . Finally, when a is any real number and x > 0, the power x a can be defined as exp (a ln x) , and using the chain rule, the power rule then follows for all real exponents a . In fact, Chapter 6 almost completes the inventory of functions of a single variable used in this book, and in most mathematical work in economics. The major omission is the family of trigonometric functions discussed in FMEA. In economics, they are used almost exclusively to solve difference and differential equations, which are topics in FMEA.

+


= 1, g  (5) = 1

2. g( 5)

6.2 2. (a) f  (x)

= 6x + 2 (b) f  (0) = 2, f (−2) = −10, f  (3) = 20. The tangent equation is y = 2x − 1. −h = −1 −→ − 1 f (x + h) − f(x) x − (x + h) 1/(x + h) − 1/x 4. = = = h h hx(x + h) hx(x + h) x(x + h) →0 x 2 6. (a) f (x + h) − f(x) = a(x + h)2 + b(x + h) + c − (ax 2 + bx + c) = 2ahx + bh + ah 2 , so [f (x + h) − f(x)]/ h = 2 ax + b + ah → 2 ax + b as h → 0. Thus f  (x) = 2 ax + b. (b) f  (x) = 0 for x = −b/2a . The tangent is parallel to the x -axis at the minimum/maximum point. √ x + h − √ x)/h. The identity 8. (a) Use the difference of squares formula. (b) [f (x + h) − f(x)]/ h = in (a) yields the result. (c) Letting h → 0, the formula follows. (The students may need reminding that √ x = x 1 2 and 1/√ x = x −1 2 .) f (x + h) − f(x) (x + h)1 3 − x 1 3 1 1 = = → 10. as h → 0, h h (x + h)2 3 + (x + h)1 3 x 1 3 + x 2 3 3x 2 3 and 1/(3x 2 3 ) = 13 x −2 3 . h

/



/

/

/

/

/

/

/

/

/

/

6.3

= 13 (4 − √ 13) ≈ 0.13 and x1 = 13 (4 + √ 13) ≈ −∞, x0 ] and in [x1 , −∞).

2. f  (x) 3x 2 8x 1 3(x x0 )(x x1 ), where x0 2.54. Then f (x) is increasing in [ x0 , x1 ], decreasing in (

= − + − = − −

−

6.4 2. I is the fixed cost, whereas k is the marginal cost, and also the (constant) incremental cost of producing

each additional unit. 4. T  (y)

= t , so the marginal tax rate is constant. 6. (a) C  (x) = 3 x 2 − 180x + 7500 (b) x = 30 (C  (x) has a minimum at x = 180 /6 = 30, using (4.6.5).) 6.5 2. (a) 0.6931

(b) 1.0986 (c) 0.4055 (Actually, using the result in Example 7.12.2, the precise values of these limits are ln 2, ln 3, and ln (3/2), respectively.)


22

C H A PT E R 6

4. (a) 5

(b) 1/5

D I F FE R EN T I AT I O N

(c) 1

(d)

−2

(e) 3x 2

(f) h2

6. (a) 4 ( f  (1)) (b) 5 ( f (2) f (1)) (c) 6 ( f  (2)) (d) 2a 2 ( f  (a) ) (e) 2a 2 ( f  (a) ) (f)[f (a h) f (a h)]/ h [ (a h)2 2(a h) (a h)2 2(a h)]/ h 4 a 4 4 a 4 as h 0,so f  (a) 4 a 4. (Alternatively, [f (a h) f (a h)]/ h [ f (a h) f(a)]/ h [f(a) f (a h)]/ h f  (a) f  (a) 4 a 4 as h 0.)

= = − = + = + = + − − = + + + − − − − = + → + → = + + − − = + − + − − → + = + →

6.6

− 16 g (x) (c) 13 g (x) 4. (a) 8π r (b) A(b + 1)y (c) (−5/2)A−7 2 6. (a) F(x) = 13 x 3 + C (b) F(x) = x 2 + 3x + C 2. (a) 2g  (x)

(b)

b

/

(c) F (x)

an arbitrary constant.

x a +1

= a + 1 + C , where in all three cases C is

6.7 2. (a) 65 x

− 14x 6 − 12 x −1 2

4. (a)

3

(b) 4x( 3x 4

√ √ + 1)2 − + x + 1) + 1)2 −√ , , ∞)

2 x( x 2(x 3 x 2 (f) 3x 3 (x

6. (a) [2

(b)

+ 5x 4 + 4x 3 − x −2 −2(1 + 2x) (d) 4x (c) x3 (x 2 + 1)2

− x 2 − 1) − x 4 + 5x 2 + 18x + 2 (b) (x 2 + 2)2 (x + 3)2 /

(c) 10x 9

− 2x 2 + 2 (e) 2 (x − x + 1)2

  √ ∞ −√ √ √  3 0 and in

3,

(c)

2,

2

x1 )(x − x2 ) −∞, x1 ] and in [0, x2 ], where x1 2 = − 12 ∓ 12 5. (f  (x) = −x(x − ) (x + 1)2 √ √ ∞ 3(x − 1)(x + 1) 8. f  (x) = −∞ − + + . f(x) is increasing in ( , 1], in 1 , 2 3 , and in 2 3, . (−x 2 + 4x − 1)2 10. Differentiating f (x) · f(x) = x gives f  (x) · f(x) + f(x) · f  (x) = 1, so 2f  (x) · f(x) = 1. Hence, 1 1 = √ f  (x) = . 2f(x) 2 x

(d) (

,



 



6.8 2. (a) dY/dt (dY/dV )(dV /dt) ( 3)5(V 1)4 t 2 15t 2 (t 3 /3 (b) dK/dt (dK/dL)(dL/dt ) Aa La−1 b Aab(bt c)a −1

= = − + =− = = = + 4. (dY/dt) = = (dY/dK) = · (dK/dt) = = Y  (K(t 0 ))K  (t 0 ) t t 0

t t 0

t t 0

8. b(t) is the total fuel consumption after t hours.

b (t)

 

+ 1)4 6.

dx

a = − √ dp 2 ap − c

= B  s(t) s (t). So the rate of fuel consumption

per hour is equal to the rate per kilometre multiplied by the speed in kph. 10. (a) y 

= 5(x4 )4 · 4x 3 = 20x 19 (b) y  = 3(1 − x)2 (−1) = −3 + 6x − 3x 2 3 2 12. (a) 1 + f  (x) (b) 2f(x)f  (x) − 1 (c) 4 f(x) f  (x) (d) 2xf (x) + x 2 f  (x) + 3 f(x) f  (x) f  (x) 2xf (x) − x 2 f  (x) 2xf (x)f  (x) − 3[f(x)]2  + √ (e) f (x) xf (x) (f) (g) (h) [f(x)]2 x4 2 f(x)

 


 

C H A PT E R 6

D I F FE R EN T IA T I ON

23

6.9

= (1 + x 2 )−1 2 − x 2 (1 + x 2 )−3 2 = ( 1 + x 2 )−3 2 4. 6. d 2 L/dt 2 = 3 (2t − 1)−5 2 8. Because g(u) is not concave. 2. d 2 y/dx 2

/

/

g  (2)

/

= 2.

(g  (t)

= 2(t − 1)−3 )

10. d 3 L/dt 3 > 0

/

6.10 2.

4

−qt (b) x  =

(a) x  = e t [a + b + t (b + 2c + ct)]

+ 2qt 3 − pt − p

t 2 et t (a bt)( bt 2

2 − (at + bt 2 )2 e = + − + (4b − a)t + 2a) = 2(at + bt )(a + 2(ebt)e )2 e √ √ 4. (a) (−∞, ∞) (b) [0, 1/2] (c) (−∞, − 2/2] and in [0 , 2/2] e − e− 6. (a) e e = e + (b) 12 (e 2 − e− 2 ) (c) − (d) z 2 e (e − 1)−2 3 2 − (e + e ) t

(c) x 

t

t

ex x

ex x

t

t/

t

t/

t

t

z3

z3

/

t

6.11 2

2. (a) x ln x( 3 ln x

+ 2)

4. (a) x >

−1

(b)

− 1)

x( 2 ln x

(ln x) 2

(b) 1/3 < x < 1

(c) x

(c)

10(ln x) 9 x

2 ln x x

+ 6 ln x + 18x + 6

= 0

6. (a) ( 2, 0] ( y is defined only in ( 2, 2).) (b) [e−1/3 , (c) [e, e3 ] ( y  ( 1 ln x)( ln x 3)/2x 2 , x > 0)

−

(d)

− −

= − − f  (x) 2 f  (x) = = 2 ln x + 2 8. (a) (b) f(x) 3(x 2 − 1) f(x) 10. ln y = v ln u, so y  /y = v  ln u + vu  /u.

(c)

∞) (y  = x 2 (3 ln x + 1), x > 0) f  (x) f(x)

3

1 2x 4x = 2x − + + 4 x2 + 1 x4 + 6


+ h) − f(x)]/ h = −6x 2 + 2x − 6xh − 2h2 + h → −6x 2 + 2x as h → 0, so f  (x) = −6x 2 + 2x . 4. Because C  (1000) ≈ C( 1001) − C( 1000), if C  (1000) = 25, the additional cost of producing 1 more 2. [f (x

than 1000 units is approximately 25. It is profitable to increase production if each unit is sold for 30.

≈ A(101) − A(100), the additional cost of increasing the area from 100 to 101 m2 is approximately 250 dollars. √ 8. (a) 2at (b) a 2 − 2t (c) 2xφ − 1/2 φ 6. Because A  (100)

10. (a) dZ/dt (dZ/du)(du/dt) 3 (u2 1)2 2u3t 2 18 t 5 (t 6 1)2 (b) dK/dt (dK/dL)(dL/dt ) ( 1/2 L)( 1/t 2 ) 1/2t 2 1

= =

= =

−√

= − √ − = − + 1/t 12. dR/dt = (dR/dS)(dS/dK)(dK/dt) = α S −1 βγ K −1 Apt −1 = Aαβγpt −1 S −1 K −1 14. (a) y  = − 7e (b) y  = − 6xe −3 (c) y  = xe − (2 − x) (d) y  = e [ln(x 2 + 2) + 2x/(x 2 + 2)] (e) y  = 15 x 2 e5 (f) y  = x 3 e− (x − 4) (g) y  = 10 (e + 2x)(e + x 2 )9 √ √ (h) y  = 1 /2 x( x + 1) 16. (a) dπ/dQ = P(Q) + QP  (Q) − c (b) dπ/dL = P F  (L) − w α

x2

x

x

3


x

γ

p

p

x

x

x

x

α

γ

24

C H A PT E R 7

D E RI V A TI V E S I N U S E

Chapter 7 Derivatives in Use Implicit differentiation, often used in economic analysis, is the subject of the first three sections. Section 7.1 concentrates on theory, with purely mathematical examples and problems, while in Section 7.2 several important economic examples are discussed. In Section 7.3 the formula for the derivative of an inverse function is derived. Linear approximations and differentials are treated in Section 7.4. It is important for students to learn how differentials are manipulated, since economists often make use of them. Then quadratic and general polynomial approximations are discussed in Section 7.5, and the n th order Taylor polynomial is introduced. Section 7.6 extends the discussion of Section 7.5 and introduces the Lagrange remainder formula. This leads to the Taylor formula which is an important result in mathematical analysis. Section 7.7 introduces elasticities, and their definition in terms of derivatives. The alternative characterization of elasticities as logarithmic derivatives is explained. Continuity is the subject of Section 7.8. After some preliminary discussion, the need to consider continuity is explained, then a formal definition in terms of limits is provided. It is pointed out that sums, differences, products, powers, and composites of continuous functions are continuous. So are quotients, where the denominator is not zero. Section 7.9 continues the discussion of limits that was started in Section 6.5. In some sense, limits and continuity are logically prior to the first results on differentiation contained in Chapter 6. Nevertheless, we have chosen this order of presentation because the topics treated in Section 7.9 may seem rather abstract unless one has first seen limits being used in connection with derivatives. One-sided limits are discussed, as are one-sided derivatives for functions whose graphs may have corners. Limits at infinity are briefly described, while Problem 7.9.4 introduces the general definition of asymptotes. In this section we also point out that differentiable functions are continuous. A rigorous definition of limits and continuity is discussed in small print. We feel that at least some students can already benefit at this stage from seeing the δ -definition of limits and continuity, although a real understanding of these concepts is beyond this text. Section 7.10 has a discussion of the intermediate value theorem. Newton’s method for finding the roots of equations also receives brief treatment. Section 7.11 deals with infinite sequences and has a short description of irrational numbers as limits of sequences. The chapter ends with Section 7.12 discussing l’Hôpital’s rule for evaluating limits, which is quite often used in economics.

Answers to Even-Numbered Problems 7.1 dv dv dv dv + v + u du − 3v2 du = 0, so du = 32vu2 +−vu . Hence du = 0 when v = −2u (provided 3v2 − u = 0). Substituting for v in the original equation yields 8 u3 − u2 = 0. So the only point on the curve where = 0 is (u, v) = ( 1/8, −1/4). dv/du = 0 and u  √ 4. (a) y  = −x/y (b) y  = − y/x (c) y  = 2 x( 2x 2 − y 3 )/y 2 (3x 2 + 4y) 2. 2u

1 . (Differentiation w.r.t. x yields 3x 2 F(xy) x 3 F  (xy)(y xy  ) exy (y  F (0) 1 Then put x 1, y 0. Note that F is a function of only one variable, with argument xy .)

6. y 

=

+ = =


+

+

+

+ xy  ) = 1.

C H A PT E R 7


25

7.2 (b) d Q∗ /dP > 0, which is reasonable because if the price received by the producer increases, the optimal production should increase.

2. (a) 1

= C  (Q∗)(dQ∗/dP), so d Q∗/dP = 1/C (Q∗)

= f (Y )+I + ¯X−g(Y). Differentiating w.r.t. I yields dY/dI = f  (Y)(dY/dI) +1−g(Y)(dY/dI). Thus, dY/dI = 1 / 1 − f  (Y ) + g  (Y ) . Imports should increase when income increases, so g  (Y ) > 0. We find that dY/dI > 0. (b) d 2 Y/dI 2 = (f  − g  )(dY/dI)/(1 − f  + g  )2 = (f  − g  )/(1 − f  + g  )3 .

4. (a) Y





7.3 2 2 = 4x√ (3 − x2 ) . So f increases in [ −√ 3, √ 3 ], and decreases in [−2, −√ 3 ] and in [√ 3, 2]. 3 4−x √ f has an inverse in the interval [0 , 3] because f is strictly increasing there. See Figure M7.3.2. (b) √ ) = /f  ( ) = √ / g ( 1

2. (a) f  (x)

3

3

1

1

3 3 8.

y

2 1

+ dr

r

−2 −1

1

2

x

r

−1 −2 Figure M7.3.2

Figure M7.4.10

7.4 1

1 10 ≈ − x . (f (0) = 1 /9, f  (x) = −10(5x + 3)−3 , so f  (0) = −10/27.) 2 (5x + 3) 9 27 4. AK ≈ A 1 + α(K − 1) 6. (a) (px −1 + qx −1 ) dx (b) (p + q)x + −1 dx (c) rp(px + q) −1 dx (d) (pe + qe )dx 8. (a) (i) y = 0 .61, dy = 0 .6 (ii) y = 0 .0601, dy = 0 .06 (b) (i) y = 0 .011494, dy = 0 .011111 (ii) y = 0 .001115, dy = 0 .001111 (c) (i) y = 0 .012461, dy = 0 .0125 (ii) y = 0 .002498, dy = 0 .0025 10. (a) A(r + dr) − A(r) is the shaded area in Fig. M7.4.10. It is approximately equal to the length of the 2.

α

p



q



p q

r

inner circle, 2 π r , times d r . (b) V (r dr) V(r) is the volume of the shell between the ball with radius r radius r . It is approximately equal to the area of the inner ball 4 πr 2 times d r .

+

−

7.5

= x − 21 x 2 + 13 x 3 − 14 x 4 + 15 x 5 4. Follows from formula (1) with f = U , a = y , x = y + M − s . 2. p(x)


px

qx

+ dr and the ball with

26

C H A PT E R 7


6. We find x( 0) 2[x( 0)]2 2. Differentiating the expression for x (t) yields x (t) x(t) t x (t) 4[x(t)]x (t),andso x( 0) x (0) 4[x( 0)]x( 0) 1 4 1 2 9. Hence, x(t) x (0) x( 0)t 12 x( 0)t 2 1 2t 92 t 2 .

˙ + +

˙ =

¨ =

=

= ( 1+x)

8. Use (2) with f(x)

n

+

˙

˙ = + · · =

¨ = + ˙ + ≈ + ˙ + ¨ =

= p/100. Then f  (x) = n(1+x) −1 and f (x) = n(n −1)(1+x) −2 . n

and x

The approximation follows.

n

7.6 2. (a)

(b)

√ 2 − 19 274 ) ≈ 2.924 3 (1 − 2/27)1 3 ≈ 3 (1 − 13 27 √ 25 = 1 5 = + ≈ + 1 − 2 1 ≈ 3

/

5

/

33

2 (1

1/32)

2

2 (1

5 32

·



2 .0125

25 322 )





1  + c)−8 3 x 3 ≤ 815 x 3 1 + 13 x − 19 x 2 (b) |R3 (x) | = 3! g (c)x 3 = 16 10 27 (1 √ = = (c) 1003 10 (1 + 3 · 10−3 )1 3 ≈ 10 .0099900, using part √ (a) to approximate (1 + 3 · 10−3 )1 3 . The error 5 in (b) is |R3 (x) | ≤ 81 (3 · 10−3 )3 = 53 10−9 . So the error in 1003 is ≤ 10 |R3 (x) | = 50 10 −9 < 2 · 10−8 , 3

4. (a) p(x) 3

/

/

3

/

and the answer is correct to 7 decimal places.

7.7 2. ElK T

= 1 .06. A 1% increase in expenditure on road building leads to an increase in the traffic volume

of approx. 1.06 %. 4. (a) Elx eax

= (x/e

ax

)ae ax

= ax

(b) 1/ ln x

(c) p

+ ax

(d) p

+ 1/ ln x

6. Elr D

= 1.23. This means that a 1% increase in income leads to an increase in the demand for apples of

approx. 1.23 %

8. (a) See Fig. M7.7.8. (The straight line has been drawn through the lowest and highest points.) T

36.3

35.0

33.9

32.4

24.7

24.2

ln n

5.04

4.89

4.70

4.54

3.64

3.58

= 1.99e0 12

(b) f ( T )

.

T

(c) The fall in temperature that halves the pulse rate is (ln 2)/0.12

≈ 5.8 degrees.

ln n 5 4 3

20 25 Figure M7.7.8

30

35

40

T

10. (a) Elx A

= (x/A)(dA/dx ) = 0 x x xf  xg     + g = El f + El g (b) El (fg) = (fg) = (f g + f g ) = fg fg f x


x

x

C H A PT E R 7

 =  



− f g xf  xg  = f g = f − g = El f − El g (c) El g f g2 xf  xg  + f g x(f  + g  ) f El f + g El g f g = = (d) El (f + g) = f + g f + g f + g (e) Is like (d), but with +g replaced by −g , and +g  by −g  . x dz x u dz du = = El f(u) El u (f) z = f (g(u)), u = g(x) ⇒ El z = z dx u z du dx f

xg

x

f

xg

gf 

27


x

x

x

x

x

x

u

x

7.8

= 0. g is continuous at x = 2. The graphs of f and g are shown in Figs. M7.8.2a

2. f is discontinuous at x

and M7.8.2b.

y

y

4

4

3

3

2

2

1

1

−2

1

2

3 x

−1 −2

f

−2 −3 Figure M7.8.2a

g

1

2

3

4

5 x

Figure M7.8.2b

4. See Fig. M7.8.4; y is discontinuous at x

= a .

6. See Fig. M7.8.6. (This example shows that the commonly seen statement: “if the inverse function ex-

ists, the original and the inverse function must both be monotonic” is wrong. This claim is correct for continuous functions, however.) y

y

y

f(x)

1 x a

x

x

Figure M7.8.4

1

Figure M7.8.6

Figure M7.9.6

7.9 1

3 − x − 3 x x 2 → 0 as x → ∞. = 2. (a) 2 1 x +1 1+ 2 x © Knut Sydsæter and Peter Hammond 2006

(b)

 + =  + 2

3x 1

−

x

3 1

−

2/x 1/x

→

√

3 as x

→ ∞.

(c) a 2

28

C H A PT E R 7


4. (a) y

= x − 1 (x = −1 is a vertical asymptote). (b) y = 2 x − 3 asymptote). (d) y = 5 x ( x = 1 is a vertical asymptote). 6. f  (0+ ) = 1 and f  (0− ) = 0. See Fig. M7.9.6.

(c) y

= 3x + 5 (x = 1 is a vertical

7.10 2. A person’s height is a continuous function of time (even if growth occurs in intermittent spurts, often

overnight). The intermediate value theorem (and common sense) give the conclusion. 4. Integer root: x

= −3. Newton’s method gives −1.880, 0.346, and 1.533 for the three other roots.

6. If f (x0 ) and f  (x0 ) have the same sign (as in Fig. 2), then (1) implies that x1 < x0 . But if they have opposite signs, then x 1 > x0 .

7.11 2. (a) Converges to 5.

(b) Diverges to

→ 3/√ 2 = 3√ 2/2 as n → ∞)

√

∞.

(c) Converges to 3 2/2. (sn

=

3n

 −

n 2

1/n

= 2

3

 − 2

1/n 2

7.12 x2

− a 2 “0 ” x2 − a2 (x − a)(x + a) 2x 2. (a) lim = = = = = x + a → 2a as x → a .) lim 2 a (or → x − a → 1 0 x − a x − a (1 + x) −1 2 − 1 2(1 + x) 1 2 − 2 − x “0 ” “0 ” = = = = (b) lim lim →0 2(1 + x + x 2 )1 2 − 2 − x →0 (1 + 2x)( 1 + x + x 2 )−1 2 − 1 0 0 −1/2 = − 1 (−1/2)(1 + x) −3 2 = lim →0 2(1 + x + x 2 )−1 2 + (−1/2)(1 + 2x) 2 (1 + x + x 2 )−3 2 2 − 1/2 3 x

a

x

a

/

/

/

x

/

x

/

/

x

/

4. The second fraction is not “0/0”. The correct limit is 5/2. 6. lim x

f(x)

→∞ g(x)

=

lim

f (1/t )

→0+ g(1/t )

t

=

“0 ” 0

f  (1/t)( 1/t 2 )

= →lim0+ g (1/t)(−−1/t 2 ) = t

f  (1/t )

lim →0+ g (1/t )

t

=

lim

x

f  (x)

→∞ g (x)

Review Problems for Chapter 7 2. 5y

4

y  − y 2 − 2xyy  = 0, so y  =

y2

y = = = for y  0 and 2x  5 y 3 . 4 3 5y − 2xy 5y − 2x

Because no point

= 0 can satisfy the given equation, y  is never 0. 4. (a) y  = −4/13 (Implicit differentiation yields ( ∗) 2xy + x 2 y  + 9y 2 y  = 0.) (b) Differentiating (∗) w.r.t. x yields: 2 y + 2xy  + 2xy  + x 2 y  + 18yy  y  + 9y 2 y  = 0. Inserting x = 2, y = 1, and y  = −4/13 gives the answer. − (2/x)(1 + 15 ln x)  = 6. y = 0 for x = e −5 . + 1 1/y 8. (a) f (e2 ) = 2 and f (x) = ln x( ln x − 1)2 = 0 for ln x = 0 and ln x = 1, so x = 1 or x = e . (b) f  (x) = (3/x)(ln x − 1 )(ln x − 1 /3) > 0 for x > e, and so f is strictly increasing in [ e, ∞). It therefore has an inverse h . According to (7.3.3), because f (e2 ) = 2, we have h  (2) = 1 /f  (e2 ) = e 2 /5. √ 10. (a) x dx/ 1 + x 2 (b) 8πr dr (c) 400K 3 dK (d) −3x 2 dx/(1 − x 3 ) with y


C H A PT E R 7

12. Letting x

1 2

1 2

= 1 + 1! +

where c is some number between 0 and

( 12 )2

2! 1 2. 1

+

( 12 )3

3!

+

Now, R 6 ( 12 )

( 12 )4

4!

=

+

( 12 )5

5!

( 12 )6 c

e <

6!

+

( 12 )6

6!

we used the fact that since c < 12 , ec < e 2 < 2. Thus it follows that e

29

= 12 and n = 5, formula (7.6.6) yields, e

1 2


1 2

≈ 1 + 1! +

( 12 )2

2!

+

( 12 )3

3!

+

( 21 )4

4!

+

( 12 )5

5!

( 12 )6

6!

ec

1 = 23040 ≈ 0 .0004340, where

2

1 1 = 1 + 12 + 18 + 481 + 384 + 3840 ≈ 1.6486979

1

≈ 1.649 correct to 3 decimals. 1 2 14. We find y  = 1 /2 and y  = 1 /8, so y(x) ≈ 1 + 12 x + 16 x . 16. El (Dmarg ) = −0.165 and El (Dmah ) = 2.39. When income increased by 1%, the demand for marThe error is less than 0.000043, and e 2

r

r

garine decreased by approximately 0.165 %, while the demand for meals away from home increased by approximately 2.39 %.

18. Put f(x) x 3 x 5. Then f  (x) 3x 2 x1 2 f (2)/f  (2) 1 1/11 1 .909.

= − − = − = −

= − 1. Taking x0 = 2, formula (7.10.1) with n = 1 gives ≈ √ 3 − x + 17 20. (a) 2 (b) Approaches +∞. (c) approaches +∞ as x → −1− , but −∞ as x → −1+ , x + 1 so there is no limit as x → −1. 2e +1 − x 2 − 4x − 5 “0 ” 2e +1 − 2x − 4 “0 ” = 0 = lim = 0 = 22. (a) 1 (b) −1/16 (c) lim →−1 →−1 3(x + 1)2 (x + 1)3 2e +1 − 2 “0 ” 2 e +1 1 = = = lim lim . →−1 6(x + 1) →−1 6 0 3 24. Does not exist if b  = d . If b = d , the limit is “0/0” and√ by l’Hôpital’s rule, it is √ √ 1 1 − − 1 2 1 2 − 2 c(cx + d) /1 = a /2 b − c/2 d = (a − c)/2 b. lim →0 2 a(ax + b) 26. x1 = 0 .9 − f (0.9)/f  (0.9) ≈ 0 .9247924, x2 = x 1 − f (x1 )/f  (x1 ) ≈ 0 .9279565, x3 = x 2 − f (x2 )/f  (x2 ) ≈ 0.9280338, and x4 = x3 −f (x3)/f  (x3 ) ≈ 0.9280339. So it seems that theanswer correct to 3 decimals x

x

x

x

x

x

x

x

x



/

is 0 .928.


/



30

C H A P T ER 8

S I N G L E -V A R I A B LE O P T I M IZ A T I O N

Chapter 8 Single-Variable Optimization The previous chapters have presented most of the mathematical functions used in economic analysis, as well as rules for differentiating them, etc. After these essential preliminaries, Chapter 8 turns to the systematic study of single variable optimization problems. In Section 8.1 we begin by giving algebraic (non-calculus) definitions of maxima and minima. Then we define a stationary point, and discuss the role played by stationary points in optimization theory. Students should be encouraged even more strongly than usual to do at least some of the problems in this section in order to understand that sometimes extreme points can be found by merely examining the formula for the function, without any need for differentiation. Section 8.2 gives an extremely useful and important test for (global) maxima and minima, based only on the variation in the sign of the first derivative (Theorem 8.2.1). When this test is passed, there is no need whatsoever to consider the second derivative. (We emphasize this point because it is often neglected by economists who, in our experience, are routinely taught to consider second-derivative tests even when there is no need to do so.) Theorem 8.2.2 claims that, at an interior stationary point of an interval in which a function is concave (or convex), that function has a maximum (or minimum) over that interval. This is a result of great significance in many economic applications. It is also a result that extends to functions of n variables which are concave or convex on some convex set. The case n 2 is covered in Theorem 13.2.1. Section 8.3 presents some interesting economic examples. The economic interpretation of condition ( ) in Example 1 should be noted. Example 3 illustrates an important type of sensitivity analysis of an optimization problem, and illustrates the envelope theorem in a special case. Next, in Section 8.4, the extreme-value theorem is formulated. It is then pointed out that extreme points, if they are not stationary points, must be either end points of an interval on which the function is defined, or else points where the function is not even differentiable. This is the justification for the usual procedure of searching for extrema by considering stationary points. But it is important to realize when the procedure fails because the relevant extreme point is not an interior point of the domain at which the function is differentiable. Otherwise, one would be restricted to interior and “smooth” maxima for the rest of one’s work in economic analysis—and, of course, corner solutions are often important. Finally, the mean-value theorem is formulated and used to confirm the tests for increasing/decreasing functions discussed in Section 6.3. In Section 8.5 further economic examples are studied. Again, the economic interpretations of the firstorder conditions should be stressed. Section 8.6 moves on to local rather than global extreme points. First, these are defined algebraically (using the concept of a neighbouring interval about a local extreme point). Theorem 8.6.1 gives conditions for a stationary point to be a local maximum or a local minimum, or neither, based on the variation in the sign of the derivative near the stationary point. (These tests are often erroneously claimed to be not only sufficient but also necessary for local optimality.) ThenTheorem 8.6.2 proceeds to the rather standard second-derivative test. Thisgives a sufficient condition for a (strict) local maximum (or minimum), based on the function being twice differentiable around the stationary point, with its second derivative being negative (or positive) at that point. (If f  (c) 0, the test fails. Then Theorem 8.6.1 can usually be applied. Tests based on evaluating higher order derivatives at c , we believe, are only of marginal interest to economists. Again, these nth order tests are often erroneously claimed to be necessary and sufficient.) As a simple exercise in comparative statics, Example 5 shows how the second derivative at an optimum can be used to infer the sign of a response to an exogenous shock.

=

∗

=


C H A P T ER 8

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31

Section 8.7 considers inflection points, for which Theorem 8.7.1 gives a simple test. This section also provides an alternative definition of concavity/convexity based on line segments joining pairs of points on the graph. This leads next to standard definitions of strict concavity or convexity, for which there are obvious second-derivative sufficient conditions. These ideas will be generalized and extensively discussed in FMEA.


−1 at x = 0. (F (0) = −1 and 2 −+2x 2 ≥ −1 for all x because 2 + x 2 ≥ 2 and so 2 ≤ 1.) No maximum. (b) Maximum 2 at x = 1. No minimum. 2 + x2 (c) Minimum 99 at x = 0. No maximum. (When x → ±∞ , H (x) → 100.)

2. (a) Minimum

8.2

√ + √ √ / − 3x)( 2 3x) 2. h (x) = = x . The function has a maximum at 2 3 3 and a minimum at 2 + 4)2 ( 3 x √ √ √ √ √ x = −2 3/3, with h(2 3/3) = 2 3/3 and h(−2 3/3) = −2 3/3. 8(2

4. f  (x) [4x(x 4 1) 2x 2 4x 3 ]/(x 4 1)2 , then simplify and factor. f on [0, x 1, because f (x) increases in [0 , 1] and decreases in [1 , ).

=

=

+ −

+

∞

∞) has maximum 1 at

1 6. f  (x) 3e3x 6 ex 3ex (e2x 2 ). So f  (x) 0 when e2x 2 or 2x ln 2, so x 2 ln 2. If 1 1 1 x < 2 ln 2 then f  (x) < 0, and if x > 2 ln 2 then f  (x) > 0, so x ln 2 is a minimum point. 2 x 2x f(x) e (e 6) tends to as x , so f has no maximum.

=

−

=

−

=

=

=

=

=

= − +∞ → ∞ 8. (a) x = 13 ln 2 is a minimum point (y is convex). (b) x = 13 (a + 2b) is a maximum point ( y is concave). (c) x = 15 is a maximum point ( y is concave). 10. (a) f  (x) = k − Aαe − = 0 when x0 = (1/α) ln(Aα/k). Note that x0 > 0 iff Aα > k . Moreover, αx

f  (x) < 0 if x < x0 and f  (x) > 0 if x > x0 , so x 0 solves the minimization problem. (b) Substituting for A in the answer to (a) gives the expression for the optimal height x0 . Its value increases as p 0 (probability of flooding) or V (cost of flooding) increases, but decreases as δ (interest rate) or k

(marginal construction costs) increases. The signs of these responses are obviously what an economist would expect.

8.3

= (102 − 2Q)Q − (2Q + 21 Q2 ) = 100Q − 52 Q2 , which is maximized at Q = 20. = 100Q − 52 Q2 − tQ, which is maximized at Q = 20 − t/5. = kce − , p (x) = −kc 2 e− . No maximum exists, and p(x) → a + k as x → ∞. See

2. (a) π(Q) (b) π(Q)



4. p  (x)

cx

cx

Fig. M8.3.2.

8.4 2. (a) Maximum

−1 at x = 0. Minimum −7 at x = 3. (b) Maximum 10 at x = −1 and x = 2. Minimum 6 at x = 1. (c) Maximum 5/2 at x = 1 /2 and x = 2. Minimum 2√ at x = 1. √ (d) Maximum 4 at x = −1. Minimum −6 3 at x = 3. (e) Maximum 4.5 · 109 at x = 3000. Minimum 0 at x = 0.


32

C H A P T ER 8

S I N G L E -V A R I A B LE O P T I M IZ A T I O N

p a

= 80Q

R(Q)

4000 3000

+ k

p(x)

= a + k(1 − e−cx )

a

= Q2 + 10Q + 900

1000 x

10

Figure M8.3.2

2

C(Q)

2000

Q0

30 Q∗ 40

50

Q

Figure M8.5.2

2

2 x 4. g  (x) e2−2x ). Stationary points: x 0 and x 5 xe (1 1 4 x 1 and x 1 are minimum points. (Note that g(2) 5 (e

=

=

= −

−

=

= ±1. Here x = 2 is a maximum point, = + e−2 ) > g(0) = 15 (1 + e2 ).) ∗ = √

= √ 2/2 (c) x ∗ = √ 12 (d) x 3 8. f is not continuous at x = −1 and x = 1. It has no maximum because f (x) is arbitrarily close to 1 for x sufficiently close to 1. But there is no value of x for which f (x) = 1. Similarly, there is no minimum. 6. (a) x ∗

= 3/2

(b) x ∗

8.5 2. (a) See Fig. M8.5.2. (b) (i) The requirement is π(Q) 0 and Q [0 , 50], that is Q2 70Q 900 0 and Q [0 , 50]. The firm must produce at least Q0 35 5 13 17 units. (ii) Profits are maximized at Q∗ 35.

∈ = 4. (i) Q∗ = 450

≥ √ ∈ = − ≈

− +

− ≥

(ii) Q∗

= 550 (iii) Q∗ = 0 6. π(Q) is stationary at Q = (P/ab)1 −1 . Moreover, π  (Q) = −ab(b − 1)Q −2 < 0 for all Q > 0, so /(b

)

b

this is a maximum point.

8.6

= − =√ − = √ = − =

2. (a) No local extreme points. (b) Local maximum 10 at x 1. Local minimum 6 at x (c) Local maximum 2 at x 1. Local minimum 2 at x 1. (d) Local maximum 6 3 at x 3. Local minimum 6 3 at x 3. (e) No local maximum point. Local minimum 1 /2 at x 3. (f) Local maximum 2 at x 2. Local minimum 2 at x 0.

−√ = − √ = − = −

= 1.

4. a and d are local minimum points, whereas c is a local maximum point for f . Points b and e are neither.

= −3 is a local (and global) minimum point. No local maximum points. (b) x = 0 is a local minimum point and x = −2/ ln 2 is a local maximum point. (Note that f  (x) = x 2 (2 + x ln 2).)

6. (a) x

x

8.7

−√

√ ∞

2. (a) f  (x) 2 x(x 2 3)(1 x 2 )−3 , so f is convex in [ 3, 0] and in [ 3, ). Inflection points are at x 3, 0, and 3. (b) g  (x) 4 (1 x) −3 > 0 when x > 1, so g is (strictly) convex in ( 1, ). No inflection points. (c) h  (x) ( 2 x)e x , so h is convex in [ 2, ) and x 2 is an inflection point.

= √ − √ = − = + = +

+


− − ∞

= −

− ∞

C H A P T ER 8

S I N G L E- V A R I A BL E O P T I M I ZA T I O N

33

√ = + √ − − = = √ − = √ = = = = − ∞ = = − − = − ≤ = 6. a = −2/5, b = 3/5 (f (−1) = 1 gives −a + b = 1. Moreover, f  (x) = 3ax 2 + 2 bx and f  (x) = 6ax + 2b, so f  (1/2) = 0 yields 3 a + 2b = 0.) 4. (a) For x > 0 one has R p x , C wx F , and π(x) p x wx F . (b) π  (x) p(1/2 x) w 0, or p/ 2 x w . (Marginal cost = price times marginal product.) Then 1 −3/2 < 0 for all x > 0, so profit is maximized over the interval x p 2 /4w2 . Moreover, π  (x) 4 px (0, ). When x p 2 /4w2 , then π p 2 /2w p2 /4w F p 2 /4w F . So this is a profit maximum if F p 2 /4w; otherwise, the firm does better not to start up and to choose x 0.

y y

2 f

g

1

−1 −1 −2

x

Figure M8.7.8

1

2

3

4

5

x

Figure M8.R.4

8. See Fig. M8.7.8. Use definition (2). Review Problems for Chapter 8 2. (a) L∗ 160, L∗∗ 120 (b) Q (120) Q(120)/120 720. In general (see Example 6.7.6),  (d/dL)(Q(L)/L) ( 1/L)(Q (L) Q(L)/L). If L > 0 maximizes output per worker, one must have Q (L) Q(L)/L.

=

=

=

=

−

=

= 4. (a) f has a (global) maximum 27 e−3 at x = 3 and no minimum. f has inflection points at x = 0, √ √ x = 3 − 3, and x = 3 + 3. (b) lim f(x) = 0, lim f(x) = −∞. See Fig. M8.R.4. →∞ →−∞  −  −  6. (a) f (x) = −2x + 1 − e , f (x) = −2 + e . f (x) is strictly increasing in [ −3, − ln 2] and strictly decreasing in [− ln 2, 3]. (b) Note that f  (−3) = 7 − e3 < 0, f  (− ln 2) = 2 ln 2 − 1 > 0, and f  (3) = −5 − e−3 < 0. By the intermediate-value theorem and the strict monotonicity of f  (x) on both sides of x = − ln 2, it follows that f  (x) = 0 has just one solution (say, x 0 ) in (−3, − ln 2), and another (actually x = 0) in (− ln 2, 3). (c) Because f decreases in [ −3, x0 ], increases in [ x0 , 0], then decreases again in [0 , 3], the candidate maxima are x = −3 and x = 0. Because f (−3) = e 3 − 12 > f (0) = 1, the maximum is at x = −3. 8. (a) f  (x) = 4 e4 + 8e − 32e−2 , f  (x) = 16 e4 + 8e + 64e−2 (b) f  (x) = 4 e−2 (e3 + 4)(e3 − 2), so f(x) is increasing in [ 13 ln 2, ∞), decreasing in (−∞, 13 ln 2]. f  (x) > 0 for all x so f is strictly convex. (c) 13 ln 2 is a (global) minimum. No maximum exists because f (x) → ∞ as x → ∞. 10. (a) g  (x) = (a − 1 )(1 − c x − ) and g  (x) = a(a − 1 )c x − −1 . Moreover, lim →0+ g(x) = ∞ and lim →∞ g(x) = ∞. (b) gmin = g(c) = a (c − 1). (Look at the sign of the derivative.) (c) Because g(c) = a (c − 1) < 0, the intermediate-value theorem and strict monotonicity of g(x) on each side of x = c imply that there is one root in each of the intervals (0, c) and (c, ∞). But g(1) = c − 1 < 0 and g(a/(a − 1)) = c (a/(a − 1))1− > 0, so the root that is greater than c must lie in ( 1, a/(a − 1)). x

x

x

x

x

a

x

x

x

a

x

x

a

x

a

x

x

x

x

a

a


a

34

C H A PT E R 9

I N T EG R AT I ON

Chapter 9 Integration Integration is an important mathematical technique which features occasionally in economic analysis. Indeed, after this chapter, integrals appear only rarely in the rest of the book. However, when we study differential equations and control theory in FMEA, integrals become important. Despite this, undergraduate economics students do need a thorough understanding of elementary ideas in the theory of integration, not so much for the occasional economic application they may see, but because integrals play such an enormous role in the statistical foundations of econometrics. Section 9.1 introduces an indefinite integral as, in effect, an antiderivative—i.e., a function whose derivative is the original function. Indefinite integrals are defined only up to an arbitrary additive “constant of integration”. General rules for integration are presented. Section 9.2 shows how integrals are used to find the area of certain plane regions. In fact, it is shown that if A(x) denotes the area under the graph of the non-negative valued function f(x) over the interval [a, x ], then A(x) is an indefinite integral of f(x). Students should be made aware that this discovery is not another boring fact, but historically a fundamental breakthrough in the history of science. On this basis the area under the graphs of complicated functions can often be calculated with ease. We next introduce definite integrals. They can be found, of course, as differences in the value of any indefinite integral (or antiderivative), provided that such an indefinite integral is known. The section concludes with a brief consideration of the areas associated with graphs of functions that may have negative values. In Section 9.3 standard properties of definite integrals are explained—e.g., the integral of a sum is the sum of the integrals, etc. The rules for differentiating a definite integral w.r.t. both its upper and lower limits of integration are given. It is pointed out that any continuous function, at least, can be integrated, and a list of “insoluble” integrals is given. Also, it is briefly explained how the “Newton–Leibniz” integral presented in the chapter can be extended to Riemann integrals (at least for bounded functions). After three sections of almost pure mathematics comes Section 9.4, with economic applications. The first concerns extraction from an oil well. It brings out the elementary but important idea that changes in stocks are integrals of flows. There follows a rather extended discussion of the cumulative distribution function of income, defined as the integral of a density function. It is shown how one can find total and mean income for all individuals with income in a certain interval. There is also a discussion of how, when all individuals’ demands for a commodity are a known function of their incomes, it is possible to find total demand for that commodity once the income distribution is also known. The section concludes with a brief discussion of consumer and producer surplus. Next, Section 9.5 presents the method of integration by parts. This is the counterpart for integrals of the product rule for differentiation. It is a tricky technique to master, however, because it is not very obvious which “part” of the integral needs to be integrated first. So students should be encouraged to attempt as many problems as possible. The second important technique for economics students to learn is that of integration by substitution, as discussed in Section 9.6. This is the counterpart for integration of the chain rule for differentiating a function of a function. In Section 9.7 we study integrals over infinite intervals, partly because some important economic models involve an infinite planning horizon, partly because this generalization is useful in statistics. The chapter ends with a short introduction to differential equations. The first-order equations for proportional growth, growth towards an upper limit, and logistic growth are presented and solved. Of course, FMEA provides a much more extensive discussion of differential equations. © Knut Sydsæter and Peter Hammond 2006

C H A PT E R 9

35

I N T EG R AT I ON

Answers to Even-Numbered Problems 9.1 1 4x

−e− + C (b) 4e + C (c) − 32 e−2 + C (d) (1/ ln 2)2 + C 4. (a) 14 t 4 + t 2 − 3t + C (b) 13 (x − 1)3 + C (c) 13 x 3 + 12 x 2 − 2x + C (d) 14 (x + 2)4 + C (e) 13 e3 − 12 e2 + e + C (f) 13 x 3 − 3x + 4 ln |x | + C x

2. (a)

x

x

x

x

x

6. (a) and (b): Differentiate the right-hand side and check that you get the integrand. (For (a) see also

Problem 9.5.4.) 8. The graph of f  (x) in Fig. 2 can be that of a cubic function, with roots at 3, 1 1 3 1 f  (0) (x 3)(x 1)(x 1) x x2 x 1. If f (0) 1. So f  (x) 3 3 3 1 4 1 3 1 2 f(x) x . Figure M9.1.8 is the graph of this f . 12 x 3x 6x

= − = +

= + − −

+

− =

+ − −

− −1, and 1, and with = 0, integrating gives

10. (a) Differentiate the right-hand side. (Once we have learned integration by substitution in Section 9.6, 1 this is an easy problem.) (b) (i) 10 (2x 1)5 C (ii) 23 (x 2)3/2 C (iii) 2 4 x C

+ + + + − √ − + 1 4 12. The general form for f  is f  (x) = 13 x 3 + A, so that for f is f (x) = 12 x + Ax + B . If we require that 1 4  f (0) = 1 and f (0) = −1, then B = 1 and A = −1, so f (x) = 12 x − x + 1. y

y

y

f

2

2

1

1

= e x

f(x)

−4 −3 −2 −1

1

2

x

−1

−1

Figure M9.1.8

−1 1

x

Figure M9.2.2

1

2

x

−1 Figure M9.2.6

9.2



 

= 20 x 3 = 8 (b) 1/7 (c) e − 1/e. (See the shaded area in Fig. M9.2.2.) (d) 9/10 1 1 4. A = 12 −1 (e + e− ) dx = 12 −1 (e − e− ) = e − e−1 √ √ 6. (a) f  (x) = 3 x 2 − 6x + 2. With x 0 = 1 − 13 3 and x 1 = 1 + 13 3, f (x) increases in ( −∞, x0 ) and in 1 (x1 , ∞). (b) See Fig. M9.2.6. The shaded area is 0 f (x ) d x = 10 ( 14 x 4 − x 3 + x 2 ) = 14 . 8. (a) 6/5 (b) 26/3 (c) α(e − 1)/β (d) − ln 2 2. (a)

2 0

3x 2 dx

x



x

β

x

x





9.3 1 + (b) f(x) = 4 x 3 − 3x 2 + 5. (f  (1) = 6 gives a + b = 6, f  (1) = 18 gives p + q + 1 p + r + 1 2a + b = 18. Then a = 12 and b = −6, so f  (x) = 12 x 2 − 6x . By integration, f (x) = 4 x 3 − 3x 2 + C . 2 Requiring 0 ( 4x 3 − 3x 2 + C ) d x = 18 gives C = 5.)

2. (a)

1




36

C H A PT E R 9

I N T EG R AT I ON

4. (a) Use formula (6) to obtain F  (x) x 2 2. To find G (x) , put u x 2 and use formula (8). Then G (x) [ (x 2 )2 2]2x 2 x 5 4x . (One can also evaluate the integrals and then differentiate: G(x)

= +

=

= + = + = ( 1 t 3 + 2t) = 13 x 6 + 2x 2 , so G (x) = 2 x 5 + 4x .) (b) H  (t) = 2 tK(t 2 )e− (use formula (8)). 0 3 3 √ 6. The parabolas intersect at (0, 0) and (3, 3), so A = 0 ( 3x − x 2 + 2x ) d x = 6. See Fig. M9.3.6.



x

2

ρt 2



= √ x + 4 (√ 2x + 4 − 2) > 0 for x > 0 and f has range (−∞, ∞), so f has an inverse g defined on ( −∞, ∞). We find that the inverse is g(x) = e 2 + 4e 4 . (b) See Fig. M9.3.8. (c) In Fig. M9.3.8 the graphs of f and g are symmetric about the line y = x , so = 10 a + 18 − 2e 2 − 16e 4 . Because a = f (10) = area A B = 10a − 0 (e 2 + 4e 4 ) dx √ = area √ − + − ≈

8. (a) f  (x)

x/

 a

4 ln( 14

x/

x/

2), this simplifies to 10 a

a/

14

8 14

y

10

y

a

4 3

= x

y

5

+ 1 = (x − 1)2 y 2 = 3 x

150

g

−5

1

−1 −1

1

3

4

Figure M9.3.6

f

10 x

5

50 f

x

6

5

g

100

a

A

2

a/

6 .26.

B y

x/

−5

2

Figure M9.3.8

4

6

8

10

t

Figure M9.4.4

9.4

 

= 2b ln 2. (The relevant number of individuals is n 2 Br −2 dr = nB /2b, whose total income is 2 2 M = n Br −1 dr = nB ln 2.) (b) x(p) = nAp r Br −2 dr = nABp b −1 (2 −1 − 1)/(δ − 1) 4. (a) See Fig. M9.4.4. ( f and g both have maximum value 4000 /27 ≈ 148 at t = 20 /3 and at t = 10 /3, 1 g(τ) − f ( τ ) dτ = t 2 (t − 10)2 ≥ 0 for all t . respectively.) (b) 2 2. (a) m

b



b

b

b

b

γ δ

γ δ

δ

b

    =  − + + − + +  =  =   − + + − +  = t

0

10

(c)

10

p(t)f(t)dt

10

0

10

p(t)g(t)dt

0

t 3

9t 2

11t

11

11/(t

1) dt

0

t 3

19t 2

79t

0

121

121/(t

1) dt

940

+ 11ln11 ≈ 966.38,

3980/3

− 121ln11 ≈ 1036.52.

Profile g should be chosen.

  =  = − 50

CS

0 50

PS

(50

0

6000 Q∗ 50

= Q∗ + 10. The only positive solution is Q∗ = 50, and then P ∗ = 60. + 50 6000 − 60 dQ = [6000 ln(Q + 50) − 60Q] = 6000ln 2 − 3000, Q + 50

6. Equilibrium when



Q)dQ

= 1250


0

C H A PT PT E R 9

9.5



1

2. (a)

1

x ln (x

−1

1 2 ln

+ 2) d x =

−1

   − − + 1

4

ln(x

+ 2)

1 2 2x

1

1 2 ln

1 2



1

37

x2

+ 2 d x = 3 − −1 x + 2 d x = (b) 8/(ln 2) − 3/(ln 2)2 (c) e − 2

−1

x

= 2 − 32 ln 3 x + 2 −1 4. Use (1) with f (x) = ln x and g  (x) = x . (Alternatively, simply differentiate the right-hand side.) 3

1 2

 −

1

1 2 2x

I N T EG E G R AT A T I ON ON

2

x

dx

ρ

9.6 1 ( x 2 3)6 C . (Substitute u 2 x 2 3, so d u 24 2 3 1 x3 2 e C . (Substitute u e x 2 .) (c) 14 ln(x 3 2 2 ( x) 5/2 ( x) 3/2 C . (Substitute u 5 1 3 1

2. (a)

+

+ +

+

=

+



= 4x d x. The integral becomes 14 u5 d u.) 2 + 2) + C . (Substitute u = ln (x + 2).) = √ 1 + x .)





= + + − + + (d) − 1 1 + + C . (Substitute u = 1 + x 2 , or u = (1 + x 2 )−1 .) (e) 2 2(1 + x ) 4(1 + x 2 )2 √ 2 (f) 15 (4 − x 3 )5 2 − 89 (4 − x 3 )3 2 + C . (Substitute u = 4 − x 3 .) − 2 d t = 2t − = 4. ln(t 2 − 2t ) = ln (x 2 − 2x) − ln 3 = ln 13 (x 2 − 2x) , so the given equation reduces to 2 3 t − 2t 3 1 2 2 − = − 2x) 1. Hence, x 2 − 4x + 3 = 0, with solutions x = 1 and x = 3. But only x = 3 is in 3 (x 3x the specified domain. So the solution is x = 3. 6. (a) 1/70 (The integrand can be written as −x 4 (x 5 − 1)13 . Then put u = x 5 − 1.) √ √ √ (b) 2 x ln x − 4 x + C . (Let u = x . Then u2 = x , 2 ud udu u = d x , etc. Alternati Alternatively vely,, use integrati integration on by √ part pa rts. s.)) (c (c)) 8/3. (Let u = 1 + x .) (b)

/

/





x

x



9.7 2. (a)

 

+∞

−∞ +∞

(b)

b

a

3(b

b

b

1

− a)

a

a

x3

a

x

xf(x)dx b

 = −  = −  − = −  = − = + + −  + = 1

f(x)dx

−∞

(c)

 = −  = b

3

= 13 bb

a

a

1

dx

b

3

1 2 (a 3

a



On the other hand,

−b

≤ x 2 for x ≥ 1, and 1 + x2

the given integral converges. τ

1



1

2(b

b2 )

1 x2

= r τ (1 − e−

−b

 = − b

dx

1 ln (1 2

1

( 1/x)

− a) = 1 1

b

[x/( 1 x )] d x b

(b

0

− a)

1 ( 2 ln 1

(b2

− a2 ) = 21 (a + b)

+ x 2 ) = 12 ln (1 + b2 ) → ∞ as b → ∞.

+ x 2 ) = 0 for all b , so the limit as b → ∞ is 0. = 1 − 1/b → 1 as b → ∞, so by Theorem 9.7.1

 =

τ

− s) e− −

2(τ

2

−

1



= (1 − e − ) 1 r τ r τ 0 τ 0 1 √ √ √ √ ln x 1 10. Usi √ x d x = (2 x ln x − 4 x) = −4 − (2 h ln h − 4 h) → Using ng the ans answer wer to Pro Proble blem m 9.6 9.6.6( .6(b), b), −4 as h → 0+, so the given integral converges to −4. (√ h ln h = ln h/ h−1 2 → 0, by l’Hôpital’s rule.) 8. (a) z

e−rs d s

1

a

a

x2

2

+ x 2 )] d x = b

1

 =

a)

ab

0

[x/( 1

b

b

b

b

6.

a

2(b

4. The first inte integral gral div diverge ergess beca because use

1

a

x

1

dx

1

b

rτ

(b) z

)

 h


τ 2



rs

ds

h

/

rτ

38

C H A PT PT E R 9


12. (a) The suggested substitution gives



+∞



+∞

+∞−∞

 = √ 1

 = √

f(x)dx

π

2σu)e−

u2

2

e −u d u

−∞

= 1, by (8).

= µ, using part (a) and Example 3. −∞ −∞ +∞ +∞ √ 1 2 = (c) Here I = x f ( x ) d x = √ (2σ 2 u2 + 2 2σ µu + µ2 )e− d u −∞ −∞ √ π +∞ +∞ +∞ 2 2σ 2 2σ µ µ2 − 2 − = √ π u e d u + √ ue d u + √ e− d u = σ 2 + 0 + µ2 . (Note how π π −∞ −∞ −∞ +∞ √ 1 − 1 − 2 − + e d u, so integration by parts gives u e d u = − ue u2 e− d u = 1 π .) (b) Here



xf(x)dx



(µ

π

u2



u2

+

√

+∞

1

 

du

u2



u2



u2

2

u2



u2

2

u2

2

−∞

9.8 2. (a) K(t) (K0 I/δ)e−δt I /δ (b) (i) (i) K(t) 200 50e−0.05t and K(t) tends to 200 from below as t . (ii) K(t) 200 50e−0.05t , and K(t) tends to 200 from above as t .

= − + = − → ∞ → = + → ∞ → ˙ = 0.02N(t) + 4 · 104 . The solution with N (0) = 2 · 106 is N(t) = 2 · 106 (2e0 02 − 1). 4. N(t) 6. The percentage surviving after t seconds satisfies p(t) = 100e− , where p(7) = 70.5 and so δ = − ln 0.705/7 ≈ 0.05. Thus p(30) = 100e−30 ≈ 22.3% are still alive after 30 seconds. Because ≈ l lnn 20/0.05 ≈ 60, it takes about 60 seconds to kill 95%. 100e− = 5 when t ≈ .

t

δt

δ

δt

8. (a) In 1950 there were about 276 thousand. In the next 10 years the number increased by 155 thousand. (b) y 479 .36 as t . See Fig. M9.8.8 M9.8.8 for the graph.

→

→ ∞ →

Tractors (in 1000) 500 400 300 200 100 5

(1950)

10 (1960)

20 (1970)

15

t

Figure M9.8.8

˙ =

10. At abo about ut 11: 11:26. 26. (Me (Measu asurin ring g tim timee in hou hours, rs, wit with h t 0 be bein ing g 12 no noon on,, on onee ha hass T k k((20 T ) with T (0) − kt 35 and T (1) 32. So the body temperature at time t is is T(t) 20 15e with k ln (5/4). Assuming ∗ ∗ that the temperature was the normal 37 degrees at the time of death t , then t ln(17/15)/ ln(5/4) 0.56 hours, or about 34 minutes before 12:00.)

= =

=

= +

= = −

−

Review Problems for Chapter 9 2. (a) e 2x

+C

(b) 21 x 2

2 5

− 252 e + C


(c)

− 13 e−3 + 13 e3 + C x

x

(d)) 2 ln x (d

| + 5| + C

−

= ≈

C H A PT PT E R 9

4. (a) 5/4, 5/4, acco according rding to Exam Example ple 9.7.2.

(c) (d) (e)

 −   =   + = ∞ 0

0

e

2

1 (ln x) d x

2

2 (x 3 9 0

1

1)3/2

0

e

2

1

0 20

∞ ∞ 5e−t d t = = 5 e−t = −5

5t e−t

e

 

(b)



(1

+ x 4 )5 = 31/20

 = −  

e

− 2 1 ln x d x e 2 1(x ln x − x) = e − 2 1 0 2 3 2 1 − = ( 9 1 ) 52 / 9 ( f ) ln (e3 + 5) = 13 (ln 6 − ln 5) = 13 ln (6/5) 9 3 −∞ x( ln x) /

√ 6. F  (x) = 4 ( x − 1). ( Hint:



x

4

1/2

(u

+ xu −1/2 ) d u =

z

x

4

2 3/2 3u

+ 2xu1 2 = 83 x 3 2 − 163 − 4x .) /

/

= βα (e − 1) + γ x + C0 10. P ∗ = Q ∗ = 5, CS = 5 50 0 ln 2 − 25, PS = 1 .25 −3 12. (a) x = Ae −3 (b) x = Ae −4 + 3 (c) x = (d) x = Ae − 12 − Ae−3 1 − 1 2 = (f) x = (B = 2 A) 1 − Ae− Be − − 2 ¯ ¯ ˙ = 14. (a) Y = α (a − 1)Y + + α(b + ¯I ) (b) Y = = Y 0 − 1b −+ aI e− 1− + 1b −+ aI I 8. C(x)


βx

t

1 5 t

t

t

1 2 t

1 2 t






α(

a)t

(e) x

= Ae −2 + 5/3 t

39

40

C H A P T ER ER 1 0

I N T E R ES E S T R A T E S A N D P R E S EN EN T V A L U E S

Chapterr 10 Chapte 10 Intere Interest st Rates Rates and Prese Present nt Values Values This chapter essentially deals with the calculation of interest and present values. Section 10.1 considers different interest periods, and defines the effective yearly interest rate. Section 10.2 considers the case when interest is compounded so frequently that one may as well treat time as continuous and have “continuous compounding” of interest. The idea of present values is important for economists and it is discussed in Section 10.3. Section 10.4 is concerned with geometric series and their sums. The condition for an infinite geometric series to have a convergent sum is presented. Of course, this topic is not specific to finance, but we include it here because it can be motivated by the need to discount payment streams extending into the indefinite future. When the rate of discount is fixed, this is treated in Section 10.5, where the present and future values of annuities are also derived. Mortgage repayments are the topic in Section 10.6, which also includes a short subsection on deposits within an interest period. The final Section 10.7 gives a brief discussion of internal rates of return.

Answers to Even-Numbered Problems 10.1 2. (a) 500 5000 0(1 0.03)10

+

≈ 6719.58

(b)) 37 (b 37.1 .17 7 ye year ars. s. (5 (500 000 0(1.03)t

= 3 ·50 = ln 3/ ln 1.03 ≈ 37.17.) 5000, 00, so t = (ii)) Aft (ii After er 10 yea years: rs: 200 2000 0(1.07)10 ≈ 3934.30

4. (a) ((i) i) Afte Afterr 2 years: years: 2000(1.07)2 2289.80 (b) t ln 3/ ln 1.07 16 .2 years.

= =

=

≈

+ 0 .2/4)4 − 1 = 1.054 − 1 ≈ 0.2155 > 0.215, so = + 4 − = √ − ≈

6. The effective yearly rate for alternative (ii) is (1

alternative alternati ve (i) is (slightly) cheaper cheaper..

8. Let the nominal yearly rate be r . By (2), 0.28

( 1

r/ 4)

1, so r

4 (

4

1.28

1)

0 .25, or 25%.

10.2

+ 0.05)10 ≈ 16 1629 29 (i (ii) i) 10 1000 00(1 + 0.05/12)120 ≈ 16 1647 47 (i (iii ii)) 10 1000 00e0 05·10 ≈ 1649 + 0.05)50 ≈ 11 1146 467 7 (i (ii) i) 10 1000 00(1 + 0.05/12)600 ≈ 12 1211 119 9 (i (iii ii)) 10 1000 00e0 05·50 ≈ 12182 ∗ 4. e−0 1 = 1 /10, so −0.1t ∗ = − ln 10, hence hence t ∗ = l ln n 10/0.1 ≈ 23. 6. h (u) = u/(1 + u)2 > 0 for u > 0, so h(u) > 0 for u > 0, implying that g  (x)/g(x) = h(r/x) > 0 for all x > 0. So g(x) is strictly increasing for x > 0. Because g(x) → e as x → ∞ , it follows that .

2. (a) (i) 1000(1 (b) (i) 1000(1

.

. t

r

g(x) < er for all x > 0. Continuous compounding of interest is best for the lender.

10.3 2. (i) The present present value value is 50 000 1.0575−5

·

≈ 37 80 806 6.64 64..

(iii) 5000 (i 50000 0 e−0.0575·5

·

≈ 37 50 506 6.83

10.4

− = (b) 0.1/(1 − 0.1) = 1 /9 (c) 517/[1 − (1.1)−1 ] = 5687 − + = + a (e) 5/(1 − 3/7) = 35/4

2. (a) ( 1/5)/(1 1/5) 1 /4 (d) a a/ /[1 (1 a) −1 ] 1


C H A P T ER 1 0

41

I N T E R ES T RA T E S A N D PR E S E NT V A LU E S

4. (a) Quotient k (b) Quotient k (c) Quotient k

= 1/p√ . Converges to 1/(p√ − 1√ ) for |p | > 1. = 1/ x . Converges to x x/( x − 1) for √ x > 1, that is, for x > 1. = x 2 . Converges to x 2 /(1 − x 2 ) for |x | < 1.

6. Let x denote the number of years beyond 1971 that the Earth’s extractable resources of iron will last. Then weget794 794 1.05 794 (1.05)x 249 103 . Using (3), 794[1 (1.05)x +1 ]/(1 1.05) 249 103 or (1.05)x +1 249 103 0.05/794 16 .68. Using a calculator, we find x ( ln 16.68/ ln 1.05) 1 56.68, so the resources will be exhausted part way through the year 2028.

+ · +···+ · = = · · ≈

·

−

−

≈

=

·

− ≈

−  P(t)  (t) = P (t)(e − 1) − P(t)re , and t ∗ > 0 can only = P(t)e = (b) Here f 1 − e− e −1 (e − 1)2 ∗ ∗ maximize f (t) if f  (t ∗ ) = 0, that is, if P  (t ∗ )(e − 1 ) = rP(t ∗ )e , which implies that P  (t ∗ ) = ∗ r P(t ∗ )/(1 − e− ). (c) Using l’Hôpital’s rule, P  (t ∗ )/P(t ∗ ) → 1 /t ∗ as r → 0. rt

8. (a) f (t)

rt

rt

rt

rt

rt

rt

rt

rt

10.5 2. (a) 10 years ago the amount was: 100 000(1.04)−10 67556.42 1.064 1 (b) 10000(1.063 1.062 1.06 1) 10 000 43746.16 1.06 1

+

+

≈ − ≈ −

+ =

− (1.06)−5 ≈ 20 540. − (1.06)−1 6. If the largest amount is a , then according to formula (4), a/r = K , so that a = r K . 15 15 1 −0 06 = 050006 1 − e−0 9 ≈ 4945.25. 8. PDV = 0 500e−0 06 dt = 500 0 0−06 e FDV = e 0 06·15 PDV = e 0 9 PDV ≈ 2 .4596 · 4945.25 ≈ 12163.3. 1 4. Offer (a) is better. The second offer has present value 4600 1



.

t

.

.



.

t

.

.



.



10.6 2. Using (2) we get a

= 1 −(0(.107+/120.)07· /80000 ≈ 928.87. 12)−120 ·

12 000 1.115 [1 (1.115)−8 ] 67 644.42. 0.115 7000 Schedule (c) has present value 22 000 [1 (1.115)−12 ] 66 384.08. 0.115 Thus schedule (c) is cheapest. When the interest rate becomes 12.5 %, schedules (b) and (c) have present values equal to 65907.61 and 64374.33, respectively.

4. Schedule (b) has present value

+

− −

≈ ≈

10.7 2. Equation (1) is here

a

a + + ·· · = −a0 , which yields a/r = −a0 , so r = −a/a0 . (1 + r)2 1+r





+ (1/1.175)2 + · · · + (1/1.175)7 ≈ 1 546 522.94.) 6. Applying (10.5.2) with a = 1000 and n = 5 gives the equation P 5 = ( 1000/r ) 1 − 1/(1 + r)5 = 4340 to be solved for r . For r = 0 .05%, the present value is $4329 .48; for r = 0.045%, the present value is 4. $ 1.55 million. (400 000 1/1.175

$4389.98. Because d P 5 /dr < 0, it follows that p is a little less than 5%.






42

C H A P T ER 1 0

I N T E R ES T R A T E S A N D P R E S EN T V A L U E S

Review Problems for Chapter 10 2. (a) 8000 1.053

·

= 9261

(b) 8000 1.0513

·

∗ (c) ( 1.05)t

≈ 15085.19

= 4, so t ∗ = ln 4/1.05 ≈ 28.5

4. 15 000e0.07·12

≈ 34745.50 6. (a) 44/(1 − 0.56) = 100 (b) 20/[1 − (1.2)−1 ] = 120 (c) 3/(1 − 2/5) = 5 (d) 400/(1 − 20−1 ) = 8000/19 5000 8. (a) 5000(1.04)4 = 5849.29. (b) [(1.04)4 − 1] = 21 232.32 0.04 (c) The last payment will be on 1st January 2006, when the initial balance of 10 000 will have earned interest for 10 years. So K must solve 10 000 (1.04)10 K [(1.04)8 1]/0.04 70 000. We find that K 5990.49.

≈

·

3200 [1 0.08

+

−

=

− (1.08)−10] = 21 472.26. 3000 (b) Present value: 7000 + [1 − 1.08−5 ] = 18 978.13. 0.08 4000 (c) Four years ahead the present value is [1 − ( 1.08)−10 ] = 26 840.33. The present value when 0.08 − Lucy makes her choice is 26 840.33 · 1.08 4 = 19 728.44. She should choose option (a) .

10. (a) Present value:


C H A P T ER 1 1

F U N C T I ON S O F M A N Y V A R I A B LE S

43

Chapter 11 Functions of Many Variables In principle, this chapter represents the first occasion on which the student will ever have seen functions of many variables. For this reason, the rather slow start provided by Section 11.1 does seem required. In fact, the first three sections deal exclusively with functions of two variables. Though definition (1) is abstract, it is immediately followed by concrete examples. As usual, first comes the mechanical Example 1, which prepares the student for problems 1–4. Then come, in rapid succession, a demand function with both price and income as variables, a Cobb–Douglas production function, and an example that prepares for the later discussion of homogeneous functions. Problem 5 is especially relevant at this stage. The final discussion of domains may seem to deserve lower priority, but difficulties may arise later if this topic is neglected. Section 11.2 introduces the key idea of partial differentiation. We try to make it as easy as possible by borrowing from earlier work for functions of one variable. In addition to mastering the technique for partially differentiating simple functions, the student should be told how important it is to understand the basic definitions of the partial derivatives given in (2) and (3), as well as the approximations in (4) to (6). In particular, economists often interpret the partial derivative f 1 (x,y) as approximately equal to the change in f ( x , y ) that results from increasing x by one unit while holding y constant. (Of course, it is better in principle to interpret f 1 (x,y)h as approximately equal to f (x h) f ( x , y ), for h small.) After the mechanics of partial derivatives, Section 11.3 turns to geometric representations of functions. Graphs are useful for functions of one variable. They are less useful for functions of two variables (or more) because of the difficulty of drawing in three (or more) dimensions. Nevertheless, for many functions of two variables, level curves are a useful device that often arises in economic contexts—especially as indifference curves, isoquants, etc. The relationship between partial derivatives, graphs, and level curves is important. Section 11.4 explains how an equation in three variables can be represented, in principle, by a surface in three dimensional space. An easy example is the budget plane if there are three commodities. The distance formula in three dimensional space is derived and the equation for a sphere in three dimensions follows easily. Following all the work in Sections 11.1–11.3 with functions of two variables, we hope that the extension to n variables in Sections 11.5 and 11.6 will seem routine. The general Cobb–Douglas function and its loglinear transformation are important in economics. Example 11.5.2 concerns the arithmetic, geometric, and harmonic means. Our discussion of continuity and the Euclidean n-dimensional space n (for n > 3) is brief. These concepts will tax the powers of most undergraduate students if taken beyond this very intuitive level. The only really new concept in Section 11.6 is the Hessian matrix. Young’s Theorem 11.6.1 is stated carefully, as is the formal definition of the partial derivative, and the concept of a C k function. The latter is used occasionally in later chapters. Section 11.7 brings in economic applications, starting with marginal products. The notation F K , F L , F T  is preferred for these (when K is capital, L is labour, and T is land) because it makes clear what factor is being varied in each case. Brief mention is made of complementary factors and, more importantly, of diminishing marginal products. These are applications of second partial derivatives, of course. This chapter ends with a brief discussion of partial elasticities in Section 11.8. These, of course, arise in many economic contexts.

+ −

Answers to Even-Numbered Problems 11.1 2. f (0, 1) 0, f ( 1, 2) 4, f (104 , 10−2 ) and f (a , b k) f(a,b) 2 abk ak 2 .

=

− = − + − =


+

= 1, f (a,a) = a 3, f (a + h,b) = (a + h)b2 = a b2 + hb2 ,

44

C H A P T ER 1 1


= 1, f (a,a) = 4a2 , f (a + h,b) − f(a,b) = 2(a + b)h + h2 = (tx)2 + 2(tx)(ty) + (ty)2 = t 2 (x2 + 2xy + y 2 ) = t 2 f(x,y) for all t , including t = 2. 6. (a) y  = x − 2 (b) x 2 + y 2 ≤ 2 (c) 1 ≤ x 2 + y 2 ≤ 4. The domains in (b) and (c) are the shaded sets 4. (a) f ( 1, 2) (b) f (tx,ty)

−

shown in Figs. M11.1.6b and c.

z y

y x2

+ y2

≤ √  2

2

y

x

1

Figure M11.1.6b

x

x

2

Figure M11.1.6c

Figure M11.3.6

11.2 2. (a) ∂z/∂x (c) ∂z/∂x (e) ∂z/∂x (g) ∂z/∂x

= 2x , ∂z/∂y = 6y (b) ∂z/∂x = y , ∂z/∂y = x = 20x 3 y 2 − 2y 5 , ∂z/∂y = 10 x 4 y − 10xy 4 (d) ∂z/∂x = ∂z/∂y = e + = y e , ∂z/∂y = x e (f) ∂z/∂x = e /y, ∂z/∂y = −e /y 2 = ∂z/∂y = 1/(x + y) (h) ∂z/∂x = 1/x , ∂z/∂y = 1/y 4. (a) z  = 3, z  = 4, and z  = z  = z  = z  = 0 (b) z  = 3 x 2 y 2 , z  = 2 x 3 y , z  = 6 xy 2 , z  = 2 x 3 , and z  = z  = 6 x 2 y (c) z  = 5 x 4 − 6xy , z  = −3x 2 + 6y 5 , z  = 20 x 3 − 6y , z  = 30 y 4 , and z  = z  = −6x (d) z  = 1 /y , z  = −x/y 2 , z  = 0, z  = 2 x/y 3 , and z  = z  = −1/y 2 (e) z = 2y(x + y )−2 , z = −2x(x + y )−2 , z = −4y(x + y )−3 , z = 4x(x + y )−3 , and z = z = 2(x − y)(x + y) −3 (f) z = x(x 2 + y 2 )−1 2 , z = y(x 2 + y 2 )−1 2 , z = y 2 (x 2 + y 2 )−3 2 , z = x 2 (x 2 + y 2 )−3 2 , and z  = z  = −xy(x 2 + y 2 )−3 2 6. (a) F  = 2 .26· 0.44S −0 56 E 0 48 = 0 .9944S −0 56 E 0 48 , F  = 2 .26· 0.48S 0 44 E −0 52 = 1 .0848S 0 44 E −0 52 (b) S F  + EF  = S · 2.26 · 0.44S −0 56 E 0 48 + E · 2.26 · 0.48S 0 44 E −0 52 = 0 .44 F + 0.48 F = 0 .92 F , so k = 0 .92. 8. Here ∂z/∂x = x/(x 2 + y 2 ), ∂z/∂y = y/(x 2 + y 2 ), ∂ 2 z/∂x 2 = (y 2 − x 2 )/(x 2 + y 2 )2 , and ∂ 2 z/∂y 2 = (x 2 − y 2 )/(x 2 + y 2 )2 . Thus, ∂ 2 z/∂x 2 + ∂ 2 z/∂y 2 = 0. x

x y

xy

xy

y

xx

x

x

xy

y

yx

yy

y

x

yy

xx

x

xy

xx

y

xx

x

yy

xy

xy

S

y

yx

.

.

.

E

.

.

yx

yx

yy

/

x

.

S

xy

xx

/

yy

yx

yy

y

yx

x

E

xy

/

/

.

.

/

xx

.

.

.

.

11.3 2. (a) A straight line through ( 0, 2, 3) parallel to the x -axis. (b) A plane parallel to the z -axis whose intersection with the xy -plane is the line y 2

4. f(x,y) e x −y of f at height e c

=

2

= x .

+ (x2 − y 2 )2 = e + c2 when x 2 − y 2 = c , so the last equation represents a level curve + c2 .


c

C H A P T ER 1 1


45

6. Generally, the graph of g(x,y) f (x) in 3-space consists of a surface traced out by moving the graph of z f (x) parallel to the y -axis in both directions. The graph of g(x,y) x is the plane through the y -axis at a 45◦ angle with the xy -plane. The graph of g(x,y) x 3 is shown in Fig. M11.3.6. (Only a

=

=

portion of the the unbounded graph is indicated, of course.)

=

= −

8. (a) f (2, 3)

= 8. f(x, 3) = 8 has the solutions x = 2 and x = 5 (b) As y varies with x = 2 fixed, the minimum of f (2, y) is 8 when y = 3. (c) A: f 1 (x,y) > 0, f 2 (x,y) > 0. B : f 1 (x,y) < 0, f 2 (x,y) < 0. C : f 1 (x,y) = 0, f 2 (x,y) = 0. At A, f 1 ≈ 2 /1 = 2 and f 2 ≈ 2 /0.6 = 10 /3. 1 2 10. F (1, 0) = F (0, 0) + 0 F 1 (x, 0) dx ≥ 2, F (2, 0) = F (1, 0) + 1 F 1 (x, 0) dx ≥ F (1, 0) + 2, 1 1 F (0, 1) = F (0, 0) + 0 F 2 (0, y ) d y ≤ 1, F (1, 1) = F (0, 1) + 0 F 1 (x, 1) dx ≥ F (0, 1) + 2, 1 F (1, 1) = F (1, 0) + 0 F 2 (1, y ) d y ≤ F (1, 0) + 1. 11.4

  

 



√ = (4 − (−1))2 + (−2 − 2)2 + (0 − 3)2 = √ 25 + 16 + 9 = 50 = 5√ 2. 4. (x − 2)2 + (y − 1)2 + (z − 1)2 = 25 6. (x − 4)2 + (y − 4)2 + (z − 12 )2 measures the square of the distance from the point (4, 4, 12 ) to the point 2. d

(x,y,z) on the paraboloid.

11.5 2. (a) F (K 1, L , T ) F ( K , L , T ) is the increase in output from increasing capital input by one unit. (b) F (K , L , T ) AK a Lb T c , where A , a , b , and c are constants, A > 0. (c) F ( tK , t L, t T ) t a +b+c F ( K , L , T )

+

−

= = 4. You drive (5/60) · 0 + (10/60) · 30 + (20/60) · 60 + (15/60) · 80 = 45 kilometres in 5 + 10 + 20 + 15 = 50 minutes, so the average speed is 45 × 60/50 = 54 kph. 6. (a) Each machine would produce 60 units per day, so each unit produced would require 480 /60 = 8 minutes. (b) Total output is =1 (T/t ) = T =1 (1/t ). If all n machines were equally efficient, the time needed for each unit would be nT T =1 (1/t ) = n =1 (1/t ), the harmonic mean of



n i

t 1 , . . . , tn .

i

 

n i n i

i

i



n i

i

11.6 2. (a) f 1 (c) f 1 (e) f 1 (f) f 1

= 2x , f 2 = 3y 2 , and f 3 = 4z3 (b) f 1 = 10x , f 2 = −9y 2 , and f 3 = 12z3 = y z, f 2 = x z, and f 3 = xy (d) f 1 = 4x 3 /yz , f 2 = −x 4 /y2 z, and f 3 = −x 4 /yz 2 = 12x(x 2 + y 3 + z4)5 , f 2 = 18y 2 (x 2 + y 3 + z4 )5 , and f 3 = 24z3 (x 2 + y 3 + z4 )5 = y ze , f 2 = x ze , and f 3 = xye 4. First-order partials: w1 = 3yz + 2xy − z3 , w 2 = 3xz + x 2 , w 3 = 3 xy − 3xz 2 . Second-order partials:  = 2 y , w  = w  = 3z + 2x , w = w  = 3 y − 3z2 , w  = 0, w  = w  = 3x , w  = −6xz . w11 12 21 13 31 22 23 32 33 2a 0 0 a(a − 1)g/x 2 abg/xy acg/xz 2 6. (a) (b) abg/xy b(b − 1)g/y bcg/yz , in concise form. 0 2b 0 acg/xz bcg/yz c(c − 1)g/z2 0 0 2c 8. f  = y x −1 , f  = zy −1 (ln x)x , f  = y (ln x)( ln y)x



x

xyz

z yz

xyz

 

y

z


xyz

yz

z

z

yz



46

C H A P T ER 1 1


11.7 2. Simplified answers: (a) K Y K

+ LY  = a Y

(b) K Y K

+ LY  = (a + b)Y

L

L

(c) K Y K

+ LY  = Y L

4. ∂D/∂p and ∂E/∂q are normally negative, because the demand for a commodity goes down when the

price of that commodity increases. If the commodities are substitutes, this means that demand increases when the price of the other good increases. So the usual signs are ∂D/∂q > 0 and ∂E/∂p > 0. 6. KY K Y L KY K

+ LY  = mY . (Y  = −(m/ρ)a(−ρ)Ae K − −1 [aK − + (1 − a)L− ]− = −(m/ρ)(1 − a)(−ρ)Ae L− −1 [aK − + (1 − a)L− ]− −1, so + LY  = −(m/ρ)(−ρ) [aK − + (1 − a)L− ]− = mY. L

λt

K

λt

ρ

ρ

ρ

ρ

ρ

ρ

L

ρ

ρ

(m/ρ) 1

−

and

(m/ρ)

m/ρ

11.8 2. Let z

g

=u

ax1d bx2d cx3d . Then El 1 z Elu ug El1 u bdgx2d /u and El3 z cdgx3d /u,soEl1 z El2 z El3 z

with u

=

+

+ =

= +

= g(x1 /u)adx1 −1 = adgx1 /u. = dg(ax1 + bx2 + cx3 )/u = d g. d

d

d d d Similarly, El2 z (This result follows easily from the fact that the function is homogeneous of degree dg (see Problem 12.7.2(b) and the elasticity form (12.7.3) of the Euler equation.)

4.



∂

=

pD(p, m)

=

 (p,m) mDm

+

− D(p, m) = pD(p, m) [El

m D(p, m) 1] > 0iffElm D(p, m) > 1, ∂m m m2 m2 so pD/m increases with m if El m D > 1. (Using the formulas in Problem 7.7.10, the result also follows from the fact that El m (pD(p, m)/m) El m D(p,m) Elm m El m D(p, m) 1.)

p

=

−

−

=

−


= −10, f (2a, 2a) = −4a2 , f ( a , b + k) − f(a,b) = −6bk − 3k2 , f(tx,ty) − t 2 f(x,y) = 0. 4. (a) ∂Y/∂K ≈ 0 .083K 0 356 S 0 562 and ∂Y/∂S ≈ 0 .035K 1 356 S −0 438 . (b) The catch becomes 2 1 356+0 562 = 2 1 918 ≈ 3 .779 times higher. 6. F  = aF/K , F  = bF/L, and F  = cF/M , so K F  + LF  + MF  = (a + b + c)F . 8. (a) g(2, 1, 1) = −2, g(3, −4, 2) = 352, and g(1, 1, a + h) − g( 1, 1, a) = 2 ah + h2 − h. (b) ∂g/∂x = 4x − 4 y − 4, ∂g/∂y = −4x + 20 y − 28, ∂g/∂z = 2z − 1. The Hessian matrix of the 4 −4 0 second-order partials is: −4 20 0 . 0 0 2 √ √ √ √ √ √ √ 10. (a) ∂z/∂x = 10 xy 4 (x 2 y 4 + 2)4 (b) K(∂F/∂K) = 2 K( K + L)(1/2 K) = K + L (c) K F  + LF  = K (1/a)aK −1 (K + L )1 −1 + L(1/a)aL −1 (K + L )1 −1 = (K + L )(K + L )1 −1 = F (d) ∂g/∂t = 3 /w + 2wt , so ∂ 2 g/∂w∂t = −3/w2 + 2t  = 4 xy + 4xz (e) g 3 = t 3 (t 12 + t 22 + t 32 )−1 2 (f) f 1 = 4 xyz + 2xz 2 , f 13 12. If x − y = c , then F(x,y) = ln (x − y) 2 + e2 − = ln c2 + e2 , a constant. 2x 2 2y 2 14. (a) El z = 3, El z = −4 (b) El z = 2 , El z = 2 (x + y 2 ) ln(x 2 + y 2 ) (x + y 2 ) ln(x 2 + y 2 ) 2 2 2 (c) El z = El (e e ) = El e = x , El z = y (d) El z = x /(x + y ), El z = y 2 /(x 2 + y 2 ) −

2. f ( 1, 2)

.

.

.

.

K

.

L

.

M

K



K

a

a

a

a

L

M



a

L

.

a

a

/a

a

a

a

/a

/a

/

(x y)

x x

y

x

x y

c

x

x

x


y

y

x

y

CHAPTER 12

T O O L S F O R C O M P A RA T I V E S T A T I C S

47

Chapter 12 Tools for Comparative Statics Chapter 12 gathers together a number of tools and techniques from the calculus of many variables that are much used by economists. It also introduces important concepts such as elasticity of substitution, homogeneous and homothetic functions, linear approximations, and differentials. First, however, Section 12.1 deals with the chain rule for differentiating functions of many variables, when these variables all depend on a single variable such as time. Of course, it begins with the two variable case, and Examples 1 and 2 where the validity of the rule is checked by direct differentiation. Examples 3 and 4 are just two of the many obvious economic applications that spring to mind immediately. Finally, the section closes with an attempt (in small print) to justify the chain rule. Next, Section 12.2 considers chain rules for functions of many variables that are themselves functions of many variables. It is necessary to learn how to keep notation straight, and under control! Section 12.3 deals with implicit differentiation, generalizing the results from Section 7.1. It introduces formula (1), giving the slope of a level curve for the two-variable case. This formula is then applied mechanically in Examples 1 to 3, and to the economic Example 4 on the incidence of taxation in a simple model of supply and demand. Next comes a formula for the second derivative. For those students who have already learned about determinants, the second derivative can also be expressed as a bordered Hessian (though we do not use this term). Section 12.4 develops more general results on implicit differentiation. Two interesting examples are a standard profit maximization model, and then a simple search model. Section 12.5 introduces the elasticity of substitution, with standard applications to the Cobb–Douglas and CES (constant elasticity of substitution) functions. Homogeneous functions receive much attention in economics. Section 12.6 introduces them for the case of two variables, along with Euler’s theorem and other properties. Some geometric properties that may be helpful are discussed. Section 12.7 considers homogeneous functions of many variables. After proving that a function is homogeneous if and only if it satisfies the Euler equation, we show that the other relevant properties of functions of two variables generalize appropriately. Economic applications are then considered. In particular, the relationship between homogeneity and returns to scale is considered, as well as homogeneity of demand functions. The section concludes with an introduction to homothetic functions, and how they generalize homogeneous functions. Next, Section 12.8 considers linear approximations and tangent planes. Section 12.9 turns to differentials of functions of several variables. Rules for differentials are presented, as well as an explanation of how they are natural counterparts of rules for derivatives. The invariance of the differential, however, deserves more careful consideration, though most economists have neglected it. Section 12.10, on systems of equations, begins with an elementary discussion of degrees of freedom, and the associated counting rule. Some examples illustrate difficulties with the counting rule for nonlinear functions.1 1

We do not discuss the concept of functional dependence. Most texts in mathematics for economists state erroneously that functional dependence of n functions is equivalent to the vanishing of the associated Jacobian determinant. For a correct account, see J. E. Marsden and M. J. Hoffman: Elementary Classical Analysis, 2. ed. W. H. Freeman and Co. (1993).


48

C H A P T ER 1 2

T O O L S F O R C O M P A RA T I V E S T A T I CS

The main part of Section 12.11 consists of Examples 1–3 showing how to find partial derivatives from differentials. Whereas Examples 1 and 2 are purely mechanical, with no economic distractions, Example 3 is based on a rather standard macroeconomic model. The general case of a system of structural equations is briefly discussed, as is the distinction between endogenous and exogenous variables. Finally, the concept of reduced form is defined in this setting.

Answers to Even-Numbered Problems 12.1 dz

2. (a)

dt dz

(b)

dt

4. dY/dt K L

= +  · +  +  = + + + + + + = + = +  +  −  = − √ =  − y

ln y

1

x

x

1

Aae at

x

1

y

y

Bbebt

1 1/2 2K

−

10L 5.

1

ln x

a

t

ln (ln t)

ln t t 1

1 1 /2 2L

0.5e0.1t

1

t

ln(t

t ln t

1)

t

b

0.2

10K

−

35

7 5/100 when t

= 0 and so

= = 6. Let U(x) = u(x, h(x)) = ln[ x + (ax 4 + b) 3 ] − 3 ln(ax 4 + b). αx −1 (3b − ax 4 )   (x ∗ ) = 0 at x ∗ = √ 3b/a , whereas U  (x) > 0 for U Then U (x) = . So 3[x + (ax 4 + b) 3 ](ax 4 + b) α

α

α/

α

4

α

α/

x < x ∗ and U  (x) < 0 for x > x ∗ . Hence x ∗ maximizes U .

12.2

= y 2 + 2xy 2ts = 5t 4 s2 + 4t 3 s 4 , ∂z/∂s = y 2 2s + 2xyt 2 = 2t 5s + 4t 4 s 3 ∂z 2(1 − s)e + + ∂z 2(1 − t)e + + = (e + + e )2 and ∂s = (e + + e )2 (b) ∂t 4. ∂z/∂t 1 = F  (x)f 1 (t 1 , t 2 ), ∂z/∂t 2 = F  (x)f 2 (t 1 , t 2 ) ∂C ∂Q1 ∂Q2 ∂Q1 − − −1 p + α bBp −1 p− 6. = + + = − a b 2cQ1 α1 A(a + 2cAp1 p2 )p1 2 2 1 2 ∂p1 ∂p1 ∂p1 ∂p1 ∂C −1 − β bBp p− −1 − − = β 1 A(a + 2cAp1 p2 )p1 p2 2 1 2 ∂p 2. (a) ∂z/∂t

t s t s

t s

t s t s

ts

t s

ts

α1

α1

β1

α1

β1

β1

β1

α1

α2

β2

β2

α2

2

8. (a)

∂u ∂r

∂f ∂x ∂f ∂y ∂f ∂z ∂f ∂w = ∂x + + + ∂r ∂y ∂r ∂z ∂r ∂w ∂r

(b)

∂u ∂r

= y zw + xzw + xyws + xyz(1/s) = 28

12.3  2. See the answers to Problem 7.1.1. (For (a): Put F(x,y) x 2 y . Then F 1 x 2 , F 11 2xy , F 2   0, so y  F 1 /F 2 2y , F 12 2x , F 22 2xy/x 2 2y/x . Moreover, according to (3),          3 2 2 6 y (1/(F 2 ) ) F 11 (F 2 ) 2F 12 F 1 F 2 F 22 (F 1 ) (1/x )[2yx 4 2(2x)( 2xy)x 2 ] 6 y/x 2 .)

= −

=



=

−

=−

=

=−

=−

+

=

=−

−

=

=

=

− 3y 2 4. −6xy + 3y 2 + 6y = −1 at (x, y) = (1, 1). 6. Differentiating the equation w.r.t. x gives (i) 1 − az  = f  (y − bz)(−bz ). Differentiating w.r.t. y = 0, solving (i) for f  and inserting it into (ii) yields gives (ii) −az  = f  (y − bz)(1 − bz ). If bz  = 0, so b = 0 and then again az  + bz = 1. If bz = 0, then (i) implies az  = 1. But then z    az + bz = 1 . 6x

h (x) = −

x

y

x

y

x

y

y

x


x

x

x

x

CHAPTER 12

49


12.4 2. Differentiating partially w.r.t. x yields ( ) 3 x 2 3z2 zx 3zx 0, so z x x 2 /(1 z2 ). By symmetry,  , differentiate ( ) w.r.t. y to obtain 6zz z 3 z2 z 3 z 0, so zy y 2 /(1 z 2 ). To find zxy y x xy xy   2 2 2 3 2 2 zxy 2 zx y /(1 z ) . (Alternatively, differentiate z x x /(1 z ) w.r.t. y .)

= =

∗

−

+ ∗

−

=

=

−

+

−

−

=

+ z ln z x ln x + zy −1  4. and z = − y ln y + xz −1 y ln y + xz −1 F  (1, 3) 6. (a) F (1, 3) = 4. The equation for the tangent is y − 3 = −  (x − 1) with F  (1, 3) = 10 and F (1, 3) yx z = − x

−

− =

y 1

x

z

y

x

y

z

z

x

x

x

y

F y (1, 3)

= 5, so y = −2x + 5.

∂y

αy = , ∂K K(1 + 2c ln y)

(b)

∂y

βy = ∂L L(1 + 2c ln y)

12.5 2. (a) Ryx

= (x/y) −1 = (y/x)1− a

a

(b) σ yx

= El

Ryx (y/x)

= El

Ryx (Ryx )

−

1/(1 a)

= 1/(1 − a)

12.6 2. x(tp,tr)

= A(tp)−1 5 (tr)2 08 = At −1 5 p−1 5 t 2 08 r 2 08 = t −1 5 t 2 08 Ap−1 5r 2 08 = t 0 58 x(p,r), so the .

.

.

.

.

.

.

.

.

.

.

function is homogeneous of degree 0 .58 . (Alternatively, use the result in Example 11.1.4.)

4. Checking definition (1) shows that f is homogeneous of degree 0. Then using the partial derivatives found in Example 11.2.1(b), we confirm that xf 1 (x,y) yf 2 (x,y) 0.

+

= 6. Definition (1) requires that for some number k one has, t 3 x 3 + t 2 xy = t (x 3 + xy) for all t > 0 and all (x, y). In particular, for x = y = 1, we must have t 3 + t 2 = 2 t . For t = 2, we get 12 = 2 · 2 , or 2 = 6. For t = 4, we get 80 = 2 · 4 , or 4 = 40. But 2 = 6 implies 4 = 36. So the two values of k k

k

k

k

k

k

k

k

must actually be different, implying that f is not homogeneous of any degree.

8. Let C and D denote the the numerator and the denominator in the expression for σ yx in Problem 12.5.3. Because F is homogeneous of degree one, Euler’s theorem implies that C F 1 F 2 F , and ( ) implies   and yF    . Hence, D xy (F  )2 F  2F  F  F  (F  )2 F  that xF 11 yF 12 xF 21 xF 12 22 2 11 1 2 12 1 22          2 2 2 2 2 2 F 12 y (F 2 ) 2xyF 1 F 2 x (F 1 ) F 12 (xF 1 yF 2 ) F 12 F , using Euler’s theorem again. 2       Hence σ xy C /D ( F 1 F 2 F)/( F 12 F ) F 1 F 2 /F F 12 .

−

= −



=

+

= −

= − +

−



= − =−

= =−

+

=

= − −



+

12.7 2. (a) The function is homogeneous of degree 1 because (tx1 tx2 tx3 )2

= (tx )4 + (tx )4 + (tx )4

F(tx1 , tx2 , tx3 )

1 6

2

3

t (x1 x2 x3 )2

= t 4 (x 4 + x 4 + x 4 ) 1t 1

(b) G(tx1 , tx2 , tx3 ) dg

= t

=



a(tx1 )d

2

d

3



1

x1

tx1

1

1

+ tx + tx 2

2

3

t d (ax1d

G(x1 , x2 , x3 ), so G is homogeneous of degree dg .


1

+ x1 + x1

 

d g

+ b(tx2 ) + c(tx3 ) =



= d

3



tF (x1 , x2 , x3 ) d

+ bx2 + cx3 )



g

∗

=

50

C H A P T ER 1 2

4. vi a


 

n i



n i



n i

= u − a/(x1 + · · · + x ), so =1 x v = =1 x u − =1 ax /(x1 + · · · + x ) = − [a/(x1 + · · · + x )] =1 x = a − a = 0. By Euler’s theorem, v is homogeneous of degree 0. 6. Let  = ln C(t w, y) − ln C(w, t). It suffices to prove that  = ln t , because then C(t w,y)/C(w, y) = ln e = t . We find that i

n

n n i

i i

i

i i

i

n

t

n

 =



ai [ln(twi )

i 1

=

−

− ln w ] + i

n

n



1 2

aij [ln(twi ) ln(twj )

i,j 1

=

= +

−

Because ln (twi ) ln wi ln t ln wi ln wi ln t ln wi ln t ln wj , this can be reduced to

+

n

= ln t





ai

i 1

=

+

1 (ln t)2 2

n



+

aij

i,j 1

Finally, this simplifies to  parameters a i , a ij , and b i .

1 (ln t) 2

=

− ln w ln w ] + ln y i

j



bi [ln(twi )

i 1

=

− ln w ] i

= ln t and also ln(tw ) ln(tw ) − ln w ln w = ( ln t)2 + i

n

n

  ln wi

j 1

i 1

=

=

1 (ln t) 2

+

aij

j

i

n

j

n

  ln wj

i 1

n

+ ln y ln t

aij

j 1

=

=



bi

i 1

=

= ln t + 0 + 0 + 0 + 0 = ln t because of the specified restrictions on the

12.8





≈ Ax0 y0 + aAx0 −1 y0 (x − x0 ) + bAx0 y0 −1 (y − y0 ) = Ax0 y0 1 + a x −x x0 + b y −y y0 0 0 4. f (0.98, −1.01) ≈ −5 − 6(0.98 − 1) + 9(−1.01 + 1) = −4.97. The exact value is −4.970614, so the 2. f(x,y)

a b

a

a b

b

a b

error is 0.000614.

≈ v(1, 0) + v1 (1, 0) · 0.01 + v2 (1, 0) · 0.02 = −1 − 1/150 8. g(0) = f (x0 ), g(1) = f (x). Using formula (12.2.3), it follows that g  (t) = f 1 (x0 + t (x − x0 ))(x1 − x10 ) + ·· · + f  (x0 + t (x − x0 ))(x − x 0 ). Putting t = 0 gives g  (0) = f 1 (x0 )(x1 − x10 ) + · · · + f  (x0 )(x − x 0 ), and the conclusion follows. 6. v( 1.01, 0.02)

n

n

n

n

n

n

12.9

= 3x 2 dx + 3y 2 dy (b) d z = e (dx + 2xy dy) (c) d z = x 2 −2 y 2 (x dx − y dy) 4. Let T (x , y , z ) = [x 2 + y 2 + z2 ]1 2 = u1 2 , where u = x 2 + y 2 + z2 . Then dT = 12 u−1 2 du = 1 −1 2 (2xd x + 2 yd y + 2 zdz). For x = 2, y = 3, and z = 6, we have u = 49, T = 7 and dT = 2u 1 (xdx + ydy + zdz) = 17 (2dx + 3dy + 6dz). Thus, T (2 + 0.01, 3 − 0.01, 6 + 0.02) ≈ T (2, 3, 6) + 7 1 · 0.01 + 3(−√ 0.01) + 6 · 0.02] = 7 + 17 · 0.11 ≈ 7.015714. (A calculator or computer gives a better 7 [2 ≈ . . y2

2. (a) d z

/

/

/

/

approximation:

49 2206

7 015739.)

6. (a) d X AβN β −1 et dN AN β e t dt (b) d X1 B EX E −1 N 1−E dX B( 1 E)X E N −E dN

= + = + − 8. d(ln z) = a 1 d(ln x1 ) + · · · + a d(ln x ), so dz/z = a 1 dx1 /x1 + a2 dx2 /x2 + · · · + a dx /x . n

n

n

n

n

12.10 2. There are 6 variables Y , C , I , G , T , and r , and 3 equations. So there are 6 © Knut Sydsæter and Peter Hammond 2006

− 3 = 3 degrees of freedom.

CHAPTER 12

51


12.11 2. (a) Differentiating yields: u3 dx x 3u2 du dv 2 y dy and 3v du Solving for d u and d v yields, with D 9 xu 3 3v ,

+

+ = = −

= D −1 (−3u4 − 1) dx + D−1 6yudy

du

and

+ 3u dv − dx = 0.

= D −1 (3xu 2 + 3u3 v ) d x + D−1 (−6yv)dy

dv

= D −1 (−3u4 − 1), v = D −1 (3xu2 + 3u3 v) (c) u = 283/81 and v  = −64/27 4. With a fixed, I  (r)dr = S  ( Y ) d Y and a dY + L (r)dr = d M . Solving for d Y and d r in terms of d M (b) u x

x

x

gives

∂Y ∂M

I  (r)

= aI  (r) + L (r)S  (Y )

∂r

and

∂M

x

S  (Y )

= aI  (r) + L (r)S  (Y )

6. (a) Differentiation yields the equations: dY dC dI dG, dC F Y  dY F T  dT F r d r , and dI f Y  dY f r d r . Hence, d Y F T  dT dG (F r f r ) dr /(1 F Y  f Y  ). (b) ∂Y/∂T F T  /(1 F Y  f Y  ) < 0, so Y decreases as T increases. But if d T dG with d r 0,     then d Y 1 F T dT/(1 F Y f Y ), which is positive provided that F T > 1.

=

= + + + + +

= 

+ = = +

 −

− − −

 =−

+ − −

+

=

=

8. (a) There are 3 variables and 2 equations, so there is (in general) one degree of freedom. (b) Differentiation gives 0 lPdy L (r)dr and S y dy S r dr S g dg I y dy I r dr . We find

=

+

+

− L (r)S  = L (r)(S  − I  ) − lP(S  − I  ) , dg dy

g

y

y

r

r

dr dg

+

=

+

lP S g

= L (r)(S  − I  ) − lP(S  − I  ) y

y

r

r

 dx1 U  d x2 p 1 dλ λdp1 ; 10. Differentiating while putting dp2 d m 0, gives: (i) U 11 12   (ii) U 21 dx1 U 22 d x2 p 2 dλ; (iii) p1 dx1 dp1 x1 p2 d x2 0. Solving for d x1 , we obtain

+

=

= =

+

+

+ =

=

+

 − p1 U  ) + x1 (p2 U 12 22 = 2 2    ∂p1 p1 U 22 − 2p1 p2 U 12 + p2 U 11 ∂x1

λp22


= G1 (u, v)φ1 (t,s) and ∂z/∂s = G1(u, v)φ2 (t,s) + G2 (u, v)ψ (s) 4. dX/dN = g(u) + g  (u) ϕ  (N) − u , where u = ϕ(N)/N , 2 d 2 X/dN 2 = ( 1/N)g  (u) ϕ  (N) − u + g  (u)ϕ  (N) 2. ∂z/∂t

 

6. Differentiating each side w.r.t. x while holding y constant gives 3 x 2 ln x x 2 (6z2 ln z 2 z2 )z1 . When x y z e, this gives z1 1/2. Differentiating a second time gives 6 x ln x 5x    2 2 2  (12z ln z 10z)(z1 ) (6z ln z 2z )z11 . When x y z e and z1 1 /2, this gives z11 11 /16e.

+

=

= = = = + + + = = = = 8. (a) MRS = R = 2 y/ 3x (b) MRS = R = y/(x + 1) (c) MRS = R = (y/x)3 10. Since y /x = (R )1 3 , σ = El (y/x) = 1 /3. yx

yx

yx

12. (a) 1

(b) k

/

yx

(c) 0


Ryx

yx

+ + =

=

52

C H A P T ER 1 2


14. Differentiate f (tx1 , . . . , t xn )

= g(t)f (x1 , . . . , x ) w.r.t. t and put t = 1, as in the proof of Euler’s   theorem (Theorem 12.7.1). This yields =1 x f (x1 , . . . , x ) = g (1)f(x1 , . . . , x ). Thus, by Euler’s theorem, f must be homogeneous of degree g  (1). In fact, g(t) = t where k = g  (1).



n

n i

i i

n

n

k

16. (a) Differentiating, then gathering all terms in dp and d L on the left-hand side, yields

(i) F  (L) dp

+ pF (L)dL = d w

(ii) F(L) dp

+ (pF  (L) − w)dL = L dw + dB

Since we know that pF  (L) w, (ii) implies that dp (Ldw dB)/F(L). Substituting this into (i) and solving for d L, we obtain d L [ (F(L) LF  (L))dw F  (L)dB ]/pF(L)F  (L). It follows that

=

∂p ∂w

L

= F(L) ,

∂p ∂B

=

−

1

∂L

= F(L) ,

∂w

=

=

F(L)

−

+

− LF (L) ,

pF(L)F  (L)

(b) We know that p > 0, F  (L) > 0, and F  (L) < 0. Also, F(L) clear that ∂p/∂w > 0, ∂p/∂B > 0, and ∂L/∂B > 0.

∂L ∂B

F  (L)

= − pF(L)F  (L)

= (wL + B)/p > 0. Hence, it is

To find the sign of ∂ L/∂w, we need the sign of F(L) LF  (L). From the equations in the model, we get F  (L) w/p and F(L) (wL B)/p , so F(L) LF  (L) B/p > 0. Therefore ∂L/∂w < 0.

=

=


+

− −

=

C H A P T ER 1 3

M U L T I V AR I A B L E O P T I M I ZA T I O N

53

Chapter 13 Multivariable Optimization Chapter 13 is devoted to multivariable optimization. The first five sections treat the case of only two variables. Section 13.1 begins with a graphical explanation of the first-order conditions stated in Theorem 13.1.1. These are first applied to a simple problem without any particular economic interpretation, and then to simple profit maximization in Examples 2–4. We believe the economic interpretations of conditions ( ) in Example 3 are worth the student’s serious attention. Example 5 looks at a monopolist who sells his product in two separate markets. Again, the economic interpretations of the first-order conditions are important. In identifying global maxima or minima, the theory of concave or convex functions provides sufficient conditions that are enormously important in economic analysis. Theorem 13.2.1 gives the precise result for functions of two variables. It is applied in Example 5 to a cost minimization problem where one of three variables has to be eliminated. Stationary points can be local or global maxima, local or global minima, or saddle points. Section 13.3 discusses ways of distinguishing these different possibilities using second-order conditions that are formally stated in Theorem 13.3.1. (The proof is indicated at the end of the section.) Several examples and many problems are supplied. Probably the most important application of multivariable calculus to economic theory is as a tool for solving optimization problems. So interesting economic examples abound. Of these, Section 13.4 presents a small selection, including some simple models of discriminating monopolists and monopsonists. Section 13.5 begins with a brief discussion of some essential topological concepts in the plane—namely, interior points, open sets, boundary points, closed sets, bounded sets, and compact sets. It is pointed out that closed sets are typically determined by weak inequalities applied to continuous functions, and open sets by strict inequalities applied to such functions. The extreme-value theorem claims that a continuous function defined on a closed and bounded set in the plane has both a maximum and minimum. This result was stated for functions of one variable in Theorem 8.4.1. The corresponding statement for functions of two variables is given in Theorem 13.5.1. Frame (1) outlines a typical procedure for finding maximum and minimum values. It pays careful attention to the possibility that these may occur at boundary points, by insisting on finding maxima and minima on the boundary as well as in the interior. Examples 1 and 2 illustrate the procedure. Section 13.6 discusses the extension of results in earlier sections to functions of n variables. It begins with formal definitions of maxima and minima that must be clearly understood. Topology in n is briefly mentioned; the main difference from 2 is that the word “circle” (or “disk”) must be replaced by “ball” in any definition. Only the most essential concepts receive attention, however. Theorem 13.6.3 states the invariance of any maximum point to (strictly) increasing transformations of the maximand. This result often gets overlooked despite its enormous importance and usefulness in economics. Indeed, the different role of strictly increasing and merely increasing transformations, illustrated in Problem 2, should also be noted carefully. Finally, Section 13.7 discusses the envelope theorem which appears so often in economics. Problem 4 is typical of the way in which the envelope theorem can give interesting economic conclusions which are far from evident to the uninitiated, and impossible even to understand without some facility in multivariable calculus.

∗∗


54

C H A P T ER 1 3


Answers to Even-Numbered Problems 13.1 2. (a) (x,y) (3, 4). (f 1 (x,y) 2x 6 and f 2 (x,y) 2y 8, so the only stationary point is 2 2 2 2 (x,y) ( 3, 4).) (b) f(x,y) x 6x 3 y 8y 4 35 32 42 (x 3)2 (y 4)2 10 10 for all (x, y), whereas f (3, 4) 10, so ( 3, 4) minimizes f .

= − = −

= − = + = − + + + + + − − = − + + + ≥ − = − 4. (a) P (10, 8) = P (12, 10) = 98 (b) (x, y) = ( 11, 9) maximizes P , and P (11, 9) = 100. 13.2

2. (a) Profit: π(x, y) 24x 12y C(x,y) 2x 2 4y 2 4xy 64x 32y 514. Maximum at    π  x 40, y 24. (Then π(40, 24) 1150.) Since π11 4 0, π22 8 0, and π11 22  2 (π12 ) 16 0, we have found the maximum. (b) x 34, y 20. (With y 54 x , profits are 2 2 2 π 2x 4(54 x) 4x( 54 x) 64x 32(54 x) 514 10x 680x 10450, which has a maximum at x 34. Then y 54 34 20. The maximum value is 1110.)

=

=

+

−

= −

−

+ + + − = − ≤ = − ≤ = = = − − − = − + −

= = − = ≥ ˆ = − − − + − + + = = − = 4. (a) π(x,y) = p x + qy − C(x,y) = (25 − x)x + (24 − 2y)y − (3x 2 + 3xy + y 2 ) = −4x 2 − 3xy − 3y 2 + 25x + 24y . (b) π1 = −8x − 3y + 25 = 0 and π2 = −3x − 6y + 24 = 0 when (x, y) = ( 2, 3).  = −8 ≤ 0, π  = −6 ≤ 0, and π  π  − (π  )2 = (−8)(−6) − (−3)2 = 39 ≥ 0. Moreover, then π 11 22 11 22 12 So (x, y) = ( 2, 3) maximizes profits. 6. (a) The first-order conditions are: π 1 = p − 2αx = 0, π2 = q − 2βy = 0. Thus the only stationary point is  = −2α ≤ 0, π  = −2β ≤ 0, and π  π  − (π  )2 = 4 αβ ≥ 0, x ∗ = p/2α , y ∗ = q /2β . Moreover, π11 22 11 22 12 ∗ ∗ so (x , y ) = (p/2α,q/2β) maximizes profits. (b) π ∗ (p,q) = px ∗ + qy ∗ − α(x ∗ )2 − β(y ∗ )2 = p 2 /4α + q 2 /2β . Hence ∂ π ∗ (p, q)/∂p = p/2α = x ∗ : Increasing the price p by one unit increases the optimal profit by x ∗ , the optimal production level of the first good. ∂π ∗ (p, q)/∂q y ∗ has a similar interpretation.

=

   abAx a−1 y b−1 , and 8. The second-order partial derivatives are f 11 a(a 1 )Ax a −2 y b , f 12 f 21  b(b 1)Ax a y b−2 . Thus, f  f  (f  )2 a bA2 x 2a −2 y 2b−2 1 (a b) . Suppose that a b 1. f 22 11 22 12  0 and f  0, and f  f  (f  )2 0. Then a 1 and b 1 as well. If x > 0 and y > 0, then f 11 22 11 22 12 We conclude from Note 13.2.2 that f is concave for x > 0, y > 0 when a b 1.

=

≤

−

= − =

−

≤

≤

 − =+  = ≤ + ≤

+ ≤ − ≥

13.3  2, f  4 y , f  4 x 4 2. (a) f 1 2 x 2y 2 , f 2 4 xy 4y , f 11 12 22 (b) f 2 0 4y(x 1) 0 x 1 or y 0. If x 1, then f 1 0 for y y 0, then f 1 0 for x 0. Thus we get the three stationary points classified in the table:

= + = ⇐⇒ = =

= + = = + = ⇐⇒ = − =

Type of stationary point:

4

8

Local minimum point

4

0

−16

Saddle point

−4

0

−16

Saddle point

A

B

C

(0, 0)

2

0

( 1, 1)

−

2

− −1)

2


=

− B 2

(x, y)

( 1,

= + = = −

AC

= ±1. If

C H A P T ER 1 3


55

= 24x 2 − 12y = 12(2x 2 − y ), f 2 (x,y) = −12x + 3y 2 = 3(y2 − 4x). The stationary points are thus the solutions of the system (i) 2 x 2 − y = 0, (ii) −4x + y 2 = 0. From (i) we get y = 2 x 2 which inserted into (ii) yields 4 x(x 3 − 1) = 0. The only solutions of this equation are x = 0 and x = 1.  (x,y) = 48x , f  (x,y) = −12, and The stationary points are thus (0, 0) and (1, 2). Moreover, f 11 12  f 22 (x,y) = 6 y . The stationary points are classified in the table:

4. (a) f 1 (x,y)

− B 2

(x, y)

A

B

C

(0, 0)

0

−12

0

−144

(1, 2)

48

−12

12

432


AC

Saddle point Local minimum point

≥

=

(b) Since g(x,y) is a square, g(x,y) 0 for all x and y . On the other hand, g (0, 0) 0, so ( 0, 0) is a global minimum point for g . (There are many global minimum points for g , namely all points (x, y) satisfying 8x 3 12xy y 3 0.)

−

+ =

=

= = = ≥

6. In all three cases we easily find that ( 0, 0) is a stationary point where z 0 and A B C 0, so 2 AC B 0. In case (a), z 0 for all (x,y), so the origin is a maximum point. In case (b), z 0 for all (x,y), so the origin is a minimum point. In (c), z takes positive and negative values at points arbitrarily

− =

≤

close to the origin, so it is a saddle point. 2

= 0 and for y > −1/x . f 1 = f 2 = 0 at all points ( 0, b) with b ∈ .

8. (a) f is defined for x

(b) f  (x,y) =

2xy

and f  (x,y) =

x2

. Here 2 + x 22y 1 + x2y (c) Because AC − B = 0 when (x, y) = (0, b), the 1

1

second-derivative test fails. But by studying the function directly, we see that (0, b) is a local maximum point if b < 0; a saddle point if b 0; and a local minimum point if b > 0. See Fig. M13.3.8.

=

z

= ln(1 + x2 y)

z

y

x

Figure M13.3.8

13.4

= −bp2 − dq 2 + (a + βb)p + (c + βd)q − α − β(a + c), p∗ = a +2bβb , q ∗ = c +2d βd .  = −2b, π  = 0, and π  = −2d . The second-order conditions are obviously satisfied because π 11 12 22 2 2 π = −bp − dp + (a + βb)p + (c + βd)p − α − β(a + c) and the price (b) The new profit function is ˆ a + c + β(b + d) which maximizes profits is pˆ = . 2(b + d)

2. (a) π


56

C H A P T ER 1 3


a2

c2

a+c + − α . Moreover, pˆ = with 2b 2d 4b 4d 2(b + d) 2 − bc)2 ≥ 0. Note that the difference is 0 when ˆπ (p)ˆ = 4(a(b++c)d) − α, and π (p∗, q ∗) − πˆ (p)ˆ = 4(ad bd(b + d) ad = bc, in which case p ∗ = q ∗ , so the firm wants to charge the same price in each market anyway. 4. (a) Let (x0 , y0 ) = (0, 11.29), (x1 , y1 ) = (1, 11.40), (x2 , y2 ) = (2, 11.49), and (x3 , y3 ) = (3, 11.61),

(c) If β = 0, then p ∗ =

a

c

, q ∗ =

, and π (p∗ , q ∗ ) =

so that x 0 corresponds to 1970, etc. (The numbers y t are approximate, as are most subsequent results.) 1 1 We find that µx 1 2 3 ) 1.5, µy 11 .40 11 .49 11 .61) 11.45, 4 (0 4 (11.29 1 and σ xx [(0 1 .5)2 ( 1 1 .5)2 ( 2 1 .5)2 ( 3 1 .5)2 ] 1.25. Moreover, we find σ xy 4 1 ( 1.5)(11.29 11.45) ( 0.5)(11.40 11.45) (0.5)(11.49 11.45) (1.5)(11.61 11.45)], which 4[ is equal to 0.13125, and so a σ xy /σ xx 0 .105 and b µ y aµ x 11 .45 0.105 1.5 11 .29.

= − −

+ + + = = + + + = = + − + − + − = = − +− − + − + − ˆ = − ˆ ≈ ˆ = = − · = (b) With z0 = ln 274, z1 = ln 307, z2 = ln 436, and z3 = ln 524, we have (x0 , z0 ) = (0, 5.61), (x1 , z1 ) = ( 1, 5.73), (x2 , z2 ) = ( 2, 6.08), and (x3 , z3 ) = ( 3, 6.26). As before, µ = 1 .5 and σ = 1 .25. Moreover, µ = 14 (5.61 + 5.73 + 6.08 + 6.26) = 5 .92 and x

xx

z

≈ 41 [(−1.5)(5.61 − 5.92) + (−0.5)(5.73 − 5.92) + (0.5)(6.08 − 5.92) + (1.5)(6.26 − 5.92)] = 0.2875 Hence ˆc = σ /σ = 0 .23, d ˆ = µ − cˆµ = 5 .92 − 0.23 · 1.5 = 5 .575. (c) With ln (GNP) = 0 .105x + 11.25, GNP = e 11 25 e0 105 = 80017e0 105 . Likewise, FA = 256e0 23 . The requirement that FA = 0.01 GNP implies that e0 23 −0 105 = 80017/25600, and so 0.125x = ln(80017/25600). Thus x = ln(80017/25600)/0.125 = 9.12. Since x = 0 corresponds to 1970, the σ xz

xz

xx

z

x

.

.

x

.

.

x

.

x

.

x

x

goal would have been reached in 1979.

13.5 2. (a) Stationary points occur where both (i) f 1 (x,y) 3 x 2 9y 0 and (ii) f 2 (x,y) 3 y 2 9x 0. 1 2 4 From (ii) we get x 27 y , whose only solutions are y 0 and 3 y , which inserted into (i) yields y y 3. So the only interior stationary point is ( 3, 3). Along the edges x 0 and y 0 of the boundary, the function has no extreme point except at a corner. Along the edges x 4 and y 4, we find two candidates for extreme points at (4, 2 3) and (2 3, 4). There are also the four corners (0, 0), (4, 0), (0, 4), and (4, 4). Comparing the values of f at all these points, we find that f has maximum 91 at (0, 4) and at (4, 0), and minimum 0 at (3, 3). (Extreme points exist by the extreme-value theorem.)

=

=

=

√

− = =

√

=

= =

=

− = =

=

(b) At any stationary point, f 1 2x 1 0 and f 2 4 y 0, so f is stationary at the interior point ( 21 , 0) of the domain. Along the boundary x 2 y 2 1, the behaviour of f is described by the function g(x) x2 2(1 x 2 ) x 2 x x 2 , with x in [ 1, 1]. We see that g  (x) 1 2x 0 at 1 x , so the points ( 12 , 12 3) are optimality candidates. Moreover, g( 1) 2 and g(1) 0, 2 so ( 1, 0) and ( 1, 0) are also optimality candidates. Comparing the values of f at all these points, we conclude that f has maximum 9/4 at ( 1/2, 3/2) and at ( 1/2, 3/2). The minimum is 1/4 at (1/2, 0).

= − =

= + = − −

−

− = √ − − − ± −

+ = √

= = −

= − − = − = =

−

−√

−

(c) At any stationary point, f 1 3 x 2 2x 0 and f 2 2y 0, so f is stationary at ( 23 , 0) (and at the boundary point ( 0, 0)). Along the edge x 0, y [ 1, 1], f (0, y) 3 y 2 , which is largest at y 0 and smallest at y 1. Along the edge x 2 y 2 1, x [0 , 1], one has f ( x , y ) 2 x 3 , which is strictly increasing, smallest at x 0, largest at x 1. Comparing the values of f at all these points, we conclude that the maximum is 3 at (0, 0) and at ( 1, 0). The minimum is 2 at (0, 1) and at (0, 1).

=

=

= ±


− = =

=

= − = ∈ − + = ∈ =

= −

= +

−

C H A P T ER 1 3 2


2

2

57 2

(d) f 1 (x,y) ( 2x 2 5x 3)ex −x (2y 1)e(y −2) and f 2 (x,y) 2 (2y 2 5y 3)(x 2)ex −x e(y −2) , so the function f has five stationary points, with (x, y) (1, 1), ( 32 , 1), ( 1, 32 ), ( 32 , 32 ), or ( 2, 12 ). None of these are in the interior of the domain, however. So any maximum or minimum must be on one of the 1 four edges, or at the four corners. Obviously f 0 along the edges x 2 and y 2 , as well as at the three corners which make up the ends of these edges. At the other corner, f (0, 0) 2 e4 . Along the edge x 0 one has f 2 < 0 for all y in [0, 12 ], so only the corners are relevant. It remains to consider the edge y 0, along which f 1 < 0 for 0 x < 1 and for 32 < x 2, but f 1 > 0 for 1 < x < 32 . Note that 1 19/4 f (1, 0) e 4 > 0 and f ( 32 , 0) e < 2 e4 . We conclude that the maximum is 2 e4 at ( 0, 0) and the 2 minimum is 0 at (x, 1/2) for all x [0 , 2], and at (2, y) for all y [0 , 1/2].

=

− +

−

=

=

=

= =

=

=

− +

=

≤

−

= =

≤

∈ ∈ 4. (a) The first-order conditions 2 axy + by + 2 y 2 = 0 and ax 2 + bx + 4 xy = 0 must have (x,y) = (2/3, 1/3) as a solution. So a = 1 and b = −2. Also c = 1 /27, so that f (2/3, 1/3) = −1/9. Because  (2/3, 1/3) = 2 /3, B = f  (2/3, 1/3) = 2/3, and C = f  (2/3, 1/3) = 8 /3, Theorem 13.3.1 A = f 11 12 22 shows that this is a local minimum.

(b) Maximum 193/27 at (2/3, 8/3). Minimum 1/9 at (2/3, 1/3). (The extreme points must be stationary points in the interior of the admissible set or else points on the boundary. Note that we must check if there are stationary points in the interior other than (2/3, 1/3). The first-order conditions reduce to 2y(x 1 y) 0 and x (x 2 4y) 0. The stationary points satisfying these equations are ( 0, 0), (0, 1), ( 2, 0), and ( 2/3, 1/3). Three of these four are on the boundary. The maximum is a point on the boundary line 2x y 4.)

−

− + =

− + =

+ =

6. (a) Closed and bounded, so compact.

(d) Closed and unbounded.

(b) Open and unbounded. (c) Closed and bounded, so compact. (e) Closed and unbounded. (f) Open and unbounded.

8. (a) The domain D is shown in Fig. M13.5.8. The first order partials are 1 1 2 2 f 1 (x,y) e − 4 (x +y ) 1 14 (x y) , f 2 (x,y) e − 4 (x +y ) 1 12 y(x y) . (b) The unique stationary point (7/2, 1/2) gives the maximum value 4 e−15/16 .

−

=



+

−

=

+



y

D y

= 1 − x

x

Figure M13.5.8

13.6 x2

x2

= e − and g(x) = F (f (x)) = ln (e− ) = −x 2 both have a unique maximum at x = 0. (b) Only x = 0 maximizes f (x). But g(x) = 5 is maximized at every point x because it is a constant. 4. f  = −6x 2 + 30x − 36, f  = 2 − e , f  = −3 + e . The 8 stationary points are (x,y,z) = √ , ±√ ) (x,y,z) = ( , ±√ , ±√ ) ( ,± 2. (a) f (x)

x

3

y2

y

ln 2

ln 3 ,and

z2

z

2

ln 2

ln 3 , where allcombinations of signs areallowed.

13.7

= 278 p3 r −3 , L∗ = 41 p2 w−2 , T ∗ = 3√ 1 3 p 3 2 q −3 2 1 (b) Q ∗ = 49 p2 r −2 + 21 pw −1 + √ p1 2 q −1 2 , so ∂Q∗/∂r = − 89 p2 r −3 = −∂K ∗/∂p 3

2. (a) K ∗

/

/


/

/

58

C H A P T ER 1 3


4. (a) First-order conditions: (i) R 1  R  C  < 0 and D (b) π11 11 11

=

− C1 + s = 0, (ii) R2 − C2 − t = 0. = π11 π22 − (π12 )2 = (R11 − C11 )(R22 − C22 ) − (R12 − C12 )2 > 0.

−

(c) Taking the total differentials of (i) and (ii) yields

 (R11

− C11 )dx 1∗ + (R12 − C12 )dx2∗ = −ds,

 (R21

− C21 )dx 1∗ + (R22 − C22 )dx2∗ = d t

Solving for d x1∗ and d x2∗ yields, after rearranging,

 − C  )ds − (R − C  )dt − (R22 ∗ 22 12 12 dx = , 1

D

dx ∗ = 2

 (R21

− C21 )ds + (R11 − C11 )dt D

From this we find that the partial derivatives are ∂x1∗ ∂s

 + C  − R22 22 = > 0 ,

∂x1∗

D

∂t

 + C  − R12 12 = > 0 , D

∂x2∗ ∂s

=

 R21

− C21 < 0, D

∂x2∗ ∂t

=

 R11

− C11 < 0 D

where the signs follow from the assumptions in the problem and the fact that D > 0 from (b). Note that these signs accord with economic intuition. For example, if the tax on good 2 increases, then the production of good 1 increases, while the production of good 2 decreases.  R  and C  C  . (d) Follows from the expressions in (c) because R 12 21 12 21

=

=

Review Problems for Chapter 13 2. (a) The profit is π (Q1 , Q2 ) 120 Q1 90Q2 0.1Q21 0.1Q1 Q2 0.1Q22 . The first-order conditions give Q1 500, Q2 200. We easily see that the second-order conditions are satisfied. (b) The profit is now π (Q1 , Q2 ) P 1 Q1 90 Q2 0 .1Q21 0 .1Q1 Q2 0 .1Q22 . The first-order conditions give π1 P 1 0 .2Q1 0 .1Q2 0 and π2 90 0 .1Q1 0 .2Q2 0. If we require Q1 400, then P 1 80 0.1Q2 0 and 90 40 0.2Q2 0. It follows that Q2 250 and so P 1 105. Second-order conditions are easily checked.

=

=

=

+

ˆ ˆ = − − −

= − =

−

+ =

−

−

− − ˆ = − − − =

−

−

=

= = =  = 12 x 2 + 4y − 6, 4. (a) f 1 = 2 (x + y − 2) + 4x(x 2 + y − 2), f 2 = 2 (x + y − 2) + 2(x 2 + y − 2), f 11  = f  = 4 x +2, f  = 4. (b) f  = f  = 0 implies that x +y −2 = −2x(x 2 +y −2) = −(x 2 +y −2). f 12 21 22 1 2 1 The last equality gives x = 2 or y = 2 − x 2 . Inserting each of these into f 2 = 0 yields three stationary points, which are classified in the following table.

− B 2


4

4

Local minimum point

6

4

4

Local minimum point

4

4

−2

(x, y)

A

B

C

(0, 2)

2

2

(1, 1)

10

(1/2, 13/8)

7/2

(c) f (0, 2)

AC

Saddle point

= f (1, 1) = 0, and f(x,y) ≥ 0 for all (x, y), so these are (global) minimum points.


C H A P T ER 1 3


59

6. (a) f is stationary when 3 x 2 2xy x 2 2y 0. Substituting 2 y x 2 in the first expression yields 3x 2 x 3 , implying that x 0 or x 3. There are two stationary points at (0, 0 and ( 3, 9/2). (b) ( 0, 0), ( 12 2, 2), ( 12 2, 2). (The stationary points must satisfy y (8x 2 5xy 1) 0 and x( 2y 2 5xy 1) 0. Notice that at any stationary point, x 0 implies y 0, and conversely. So apart from (0, 0), any other stationary point must satisfy 5 xy 1 8x 2 2y 2 . Since no solution of these last two equations can satisfy y 2x , the only possibility is that y 2 x and so 2x 2 1.)

=

−

8.



− = − + = √ √ − =√ −√ = + =

=

= −

= − =



= =

=

−

− − − e −x 1 e −x −y 2 1  f  −− −x −y − e −y . f 21 e 22 2  −  x 1 −y It follows that f 11 = e ( 2 e − 1) < 0 because e −y < 1 < 2. In the same way, f 22 f  f  − (f  )2 = 1 e−x −y (2 − e−x − e−y ) > 0, so f is concave.  f 11

 f 12

11 22

12

1 e x y 2 1 e x y 2

=

+ = =

< 0. Moreover,

2

= f 2 = 0 implies that 2 x − y − 3x 2 = −2y − x = 0. So y = − 12 x and 52 x − 3x 2 = 0, implying = 0 or 56 . There are two stationary points, which are classified in the following table.

10. (a) f 1 that x

(x, y)

A

B

C

(0, 0)

2

−1

−2

(5/6,

− B 2


−5

Saddle point

AC

−5/12) −3 −1 −2 5 Local maximum point  = 2 − 6x ≤ 0 ⇐⇒ x ≥ 1 /3, while f  = −2 ≤ 0, and f  f  − (f  )2 = 12 x − 5 ≥ 0 ⇐⇒ (b) f 11 22 11 22 12 x ≥ 5 /12. We conclude that f is concave if and only if x ≥ 5 /12. The largest value of f in S is 125/432, obtained at ( 5/6, −5/12).


60

C H A P T ER 1 4

C O N S T RA I N E D O P T I M I ZA T I O N

Chapter 14 Constrained Optimization Constrained optimization is the central tool of mathematical economics. Economics is about making the best use of scarce resources, and scarcity is naturally represented by constraints. The usual technique for solving constrained optimization problems is the Lagrange multiplier method. Most of this chapter is concerned with equality constraints; the last two sections offer a brief introduction to inequality constraints. For the case of two variables, the Lagrange multiplier method is the subject of Section 14.1. The method is set out as a procedure in a large frame that should be memorized by the students. Partly because it easily leads to errors when there are inequality constraints, we prefer not to set to zero the partial derivative of the Lagrangian w.r.t. the multiplier (as is done in many other textbooks). 2 Example 2 is important because it explains what needs to be done when the equation system has multiple solutions. The consumer demand problem with a Cobb–Douglas utility function is considered in Example 3, and the appropriate demand functions are derived. Section 14.2 presents the standard economic interpretation of Lagrange multipliers as shadow prices. For the case when the (optimal) value function is differentiable, we present an easy proof that its derivative is equal to the Lagrange multiplier. Example 1 illustrates its use in economics. The section ends with a rich selection of problems. Section 14.3 is devoted to explaining first with a geometric argument why the Lagrangian method works— when it does. Then an analytic argument for the first-order conditions is given, and a precise result is formulated in Theorem 14.3.1. Example 1 considers a utility maximization problem in which a mechanical application of the Lagrangian method fails to give the right answer. Problem 1 shows that the solution to a constrained maximization problem need not maximize the Lagrangian, even though (under usual conditions) it must make the Lagrangian stationary. This is another point that is rather commonly misunderstood. Let the teacher beware of misleading attempts to motivate the Lagrangian technique! Fortunately, the error found in the quotation in Problem 2 is far less common. In Section 14.4 we turn from necessary to sufficient conditions for optimality. Of these, the simplest to state is that a global maximum (or minimum) of the Lagrangian which satisfies the constraint is optimal. This could have been stated as a theorem in its own right. In practice, however, there are difficulties in verifying that a global maximum (or minimum) has been found, except in the important case when the Lagrangian happens to be concave (or convex). This is the case mentioned in Theorem 14.4.1. (The appropriate generalization to quasi-concave functions is not, in our view, an “essential” topic for an undergraduate course.) The concept of a local constrained extremum is introduced. Theorem 14.4.2 then gives sufficient conditions for the Lagrangian method to find a local constrained extremum. In smaller print the sufficient conditions in Theorem 14.4.2 are expressed in terms of the sign of the bordered Hessian determinant. Sections 14.1–14.4 consider only functions of two variables. The extension to n variables in Section 14.5 is mostly routine, though actually solving problems with many variables is often very complicated. Example 3 discusses how to derive individual demand functions in a general n good setting. Then the case of m equality constraints is considered. The Lagrangian needs as many Lagrange multipliers as there are constraints, of course. This gives the right number of equations when the first-order conditions are combined with the constraints themselves. In Example 5 we study a welfare maximization problem in a simple exchange economy. Problem 3 is an interesting economic example, involving the optimal choice of leisure as well as 2

The equality constraint is an obvious necessary condition for optimality. There is no reason to differentiate anything to demonstrate this. © Knut Sydsæter and Peter Hammond 2006

C H A P T ER 1 4


61

consumption of two goods. It also shows the possibility of a corner solution with zero labour supply in case unearned income is sufficiently high. Section 14.6 deals with sensitivity analysis. First, the standard interpretation of Lagrange multipliers as shadow prices is mentioned. With m constraints, the value function depends on m variables. Where it is differentiable, the m Lagrange multipliers are partial derivatives of the value function. Example 1 verifies the properties of Lagrange multipliers numerically for a case with two constraints. Adjusting the right-hand side of an equality constraint is just one of many perturbations to which a constrained optimization problem can be subjected. Other perturbations are considered in this section; in economics they feature constantly in comparative static propositions, of course. For such perturbations, frame (6) states one version of the “envelope theorem”, which often arises in economic analysis. Examples 2, 3, and 4 involve interesting economic applications. The final Sections 14.7 and 14.8 give a brief introduction to nonlinear programming—i.e., optimization problems where the contraints are expressed as inequalities, rather than equalities. For the case of two variables and one constraint, a frame sets out a recipe for solving the problem based on the Lagrange multiplier method. Part of the recipe involves a complementary slackness condition. Theorem 14.7.1 emphasizes how this method leads to a sufficient condition when the Lagrangian is concave. Section 14.7 concludes with a graphical explanation of the complementary slackness condition. In Section 14.8, the recipe of Section 14.7 is extended to general problems with n variables and m constraints. Particular attention is paid to nonnegativity restrictions on the variables. The Lagrange multipliers are related to partial derivatives of the (optimal) value function.


= (4/5, 8/5) with λ = 8/5. (With L(x,y) = x 2 + y 2 − λ(x + 2y − 4), the first-order conditions are L1 = 2 x − λ = 0 and L2 = 2 y − 2λ = 0. From these equations we get 2 x = y , which inserted into the constraint gives x + 4x = 4. So x = 4 /5 and y = 2 x = 8 /5, with λ = 2 x = 8 /5.) (b) The same method as in (a) gives 2 x − λ = 0 and 4y − λ = 0, so x = 2 y . From the constraint we get x = 8 and y = 4, with λ = 16. 4. (a) With L = x 2 + y 2 − 2x + 1 − λ(x 2 + 4y 2 − 16), the first-order conditions are 2 x − 2 − 2λx = 0 and √ 2y − 8λy = 0. One possibility is λ = 1 /4, in which case x = 4 /3andso y = ±4 2/3. Alternatively, y = , 3/4), (ii) (x,y,λ) = 0, in which case x = ±4. So there are four solution candidates: (i) (x,y,λ) = ( 4, 0√ √ (−4, 0, −5/4), (iii) (x, y, λ) = ( 4/3, 4 2/3, 1/4), and (iv) (x, y, λ) = ( 4/3, −4 2/3, 1/4). Of these, 2. (a) (x,y)

the second is the maximum point (while (iii) and (iv) are the minimum points). (b) The first-order conditions imply that 2 x 3y λ 3 x 2y . So (x, y) ( 50, 50) with λ

+ = = + = = 250. 6. (a) With L(x,y) = 100 − e− − e− − λ(px + qy − m), L = L  = 0 when e − = λp and e − = λq . Hence, x = − ln(λp) = − ln λ − ln p, y = − ln λ − ln q . Inserting these expressions for x and y into the constraint, then solving for ln λ, yields ln λ = −(m + p ln p + q ln q)/(p + q) . Therefore x(p,q,m) = [ m + q ln (q/p)]/(p + q) , y(p, q, m) = [ m + p ln(p/q)]/(p + q). (b) x(tp,tq,tm) = [ tm + tq ln (tq/tp)]/(tp + tq) = x(p, q, m), so x is homogeneous of degree 0. In x

y

x

y

x

y

the same way we see that y(p, q, m) is homogeneous of degree 0.

14.2

= x + y − λ(x2 + y − 1). The equations L1 = 1 − 2λx = 0, L2 = 1 − λ = 0, and x 2 + y = 1

2. (a) L(x,y)


62

C H A P T ER 1 4


have the solution x 1/2, y 3/4, and λ 1. (b) The solution is illustrated in Fig. M14.2.2. The minimization problem has no solution because f (x, 1 x 2 ) x 1 x 2 as x . (c) The new solution is x 0 .5 and y 0 .85. The change in the value function is f ∗ (1.1) f ∗ (1) (0.5 0.85) (0.5 0.75) 0 .1. Because λ 1, one has λ dc 1 0.1 0 .1. So, in this case, (3) is satisfied with equality. (This is because of the special form of the functions f and g .)

=

+

−

=

=

+

=

=

=

− = + − → −∞

=

→ ∞ −

· = · =

=

y

6 5 4 3

y

2

y

= 1 − x

d 1

( 12 , 43 )

2

−1

−3−2 −−11 −2 −3 −4 −5 −6

x

1

+ y = 45

x

+ y = 21

x

Figure M14.2.2

4. (a) x ∗

y2

= x (x + 1)2

(x,y)

1 2 3 4

x

Figure M14.3.4

= 4, y ∗ = 24, λ = 1 /4.

ˆ =

ˆ =

=

−

=

(b) y 97/4, x 4. U 105/4 104/4 1/4, the value of the Lagrange multiplier from (a). (There is exact equality here because U is linear in one of the variables.) (c) x ∗ q 2 /4p, y ∗ m/q q/ 4p. (Note that y ∗ > 0 iff m > q 2 /4p.)

=

=

−

14.3 2. The problem with systems of three equations and two unknowns is not that they are merely difficult to solve but that they are usually inconsistent—i.e., it is impossible to solve them. The equations f x (x,y) f y (x,y) 0 are NOT valid at the optimal point.

=

=

= − = + + − = − + + =

4. The minimum is 1 at (x, y) ( 1, 0). Actually, this problem is quite tricky. The Lagrangian can be written in the form L (x 2)2 (1 λ)y 2 λx(x 1)2 . The only stationary point which satisfies the constraint is (0, 0), with λ 4, and with f (0, 0) 4. (In fact, L2 0 only if λ 1 or y 0. For λ 1, L1 3 (x 1)2 2 > 0 for all x . For y 0, the constraint gives x 0 or x 1. But x 1    gives L1 2, so x 0 is necessary for a stationary point.) Yet at ( 1, 0) both g 1 ( 1, 0) and g2 ( 1, 0)

=

= =

+

=

+ =

=

−

=

= = − −

=

= − −

are 0, and the Lagrange multiplier method fails. The given problem is to minimize (the square of) the distance from ( 2, 0) to a point on the graph of g(x, y) 0. But the graph consists of the isolated point ( 1, 0) and a smooth curve, as illustrated in Fig. M14.3.4.

−

−

=

14.4 2. With L(x,y) x 2 y 2 λ(x 2 y a) the first-order conditions are 2 x λ 0 2y 2 λ. So (x,y) (a/5, 2a/ 5) with λ 2a/ 5. Applying Theorem 14.4.2 with f(x,y) x2 y 2 and g(x,y) x 2y , we find that D(x, y, λ) 10, so the only stationary point is a local minimum. (By

=

= = +

+ −

+ − =

− = = − = +

=

reducing the problem to a one-variable problem, it is easy to see that it is actually a global minimum.) Geometrically, the problem is to find that point on the straight line x 2y a that is closest to the origin.

+ =


C H A P T ER 1 4

63


14.5

= x + 4y + 3z − λ(x2 + 2y 2 + 13 z2 − b). So necessary conditions are: = 0, (i) L1 = 1 − 2 λx = 0; (ii) L2 = 4 − 4 λy = 0; (iii) L3 = 3 − 23 λz = 0. It follows that λ  2 and so x = 1/2λ, y = 1/λ, z = 9/2λ. Inserting these values into the constraint yields λ = 9/b, so √ √ λ = ±3/ b. The value of the objective function is x + 4y + 3z = 18 /λ, so λ = −3/ b determines the √ minimum point. This is (x, y, z) = (a, 2a, 9a) , where a = − b/6. 4. The constraints reduce to h + 2 k +  = 0 and 2h − k − 3  = 0, so k = −h and  = h. But then f(x,y,z) = x 2 + y 2 + z2 = 200 + 3h2 ≥ 200 for all h , so f is maximized for h = 0. Then k =  = 0 also, and we conclude that (x, y, z) = ( 10, 10, 0) solves the minimization problem. 6. (a) c1 = α1 r /p1 , c2 = α2 r /p2 . (Write the utility function in the equivalent form U (c1 , c2 ) = ln (c1 ) (c2 ) . Using Theorem 13.6.3, the result follows immediately from (∗∗) in Example 14.1.3, recalling that α 1 + α2 = 1.) (b) In Example 5, λ1 = α1 /e1 , λ2 = α2 /e2 . So p 1 = λ1 and p2 = λ2 implies c1 = α1 r /p1 = r e1 = β e1 provided that r = β , which also gives c2 = β e2 . Because p1 e1 + p2 e2 = λ 1 e1 + λ2 e2 = α 1 + α2 = 1, it follows that r = β (p1 e1 + p2 e2 ) for i = 1 , 2. 8. (a) With a Cobb–Douglas utility function, U  (x) = α U (x)/x , so from (6) (with j = 1), we have p /p1 = U  (x)/U 1 (x) = α x1 /α1 x . Thus p x = (a /a1 )p1 x1 . Inserted into the budget constraint, we have p 1 x1 + (a2 /a1 )p1 x1 + · · · + (a /a1 )p1 x1 = m , which implies that p 1 x1 = a 1 m/(a1 + · · · + a ). Similarly, p x = a m/(a1 + · · · + a ) for k = 1, . . . , n . (b) From (6) (with j = 1), we get x −1 /x1 −1 = p /p1 and so x /x1 = (p /p1 )−1 1− or p x /p1 x1 = (p /p1 )1−1 1− = (p /p1 )− 1− . The budget constraint gives − 1− . = m. So p1 x1 = mp1− 1− p1 x1 1 + (p2 /p1 )− 1− + · · · + (p /p1 )− 1− p 2. Here L



i

i

i α1

i α2

i



i

i

i

i

i

i

i

i

i

i

k

k

k

k

k k

i

i

i

k

k

i

k

k

n

k k

/(

k

k

a)

n

a/(

k



n

a/(

a k a)

a

a)

a/(

n

Arguing similarly for each k , one has x k

k

= mp−1 k

k

a)



−

/(1 a)

/(

k

a)

k k

n

a/(



a)

=

a/(

a)

i

i 1

=

n

i 1



−a/( 1−a) for k = 1, . . . , n.

pi

10. Differentiating the constraint w.r.t. x 1 (keeping x2 fixed) yields g1 g3 (∂x3 /∂x 1 ) 0, which implies (i) ∂ x3 /∂x 1 g1 /g3 . Similarly, (ii) ∂ x3 /∂x 2 g2 /g3 . The first-order conditions for the maximization of z f (x1 , x2 , x3 ), where x3 is a function of (x1 , x2 ), are (iii) ∂z/∂x1 f 1 f 3 (∂x 3 /∂x1 ) 0 and f 2 f 3 (∂x 3 /∂x2 ) 0. Substitute from (i) and (ii) into (iii) and (iv), letting λ f 3 /g3 . (iv) ∂z/∂x2 This yields the equations f 1 λg1 0 and f 2 λg2 0. Also, by definition of λ, f 3 λg3 0. These are the conditions (3) for n 3.

=

= −

= −

= +

= − = =

+

=

= +

− =

= = − =

14.6



= aM/α, y = bM/β, z = cM/γ , λ = 1/(2M), where M = √ L/ a2 /α + b2 /β + c2 /γ . (The first-order conditions give x = a /2λα, y = b/2λβ√ , z = c/ 2λγ . Substituting in the constraint and solving for λ gives the solution.) (b) We find that M = L/5, and the given values of x , y , and z follow. For L = 100 one has M = 2 and λ = 1/4. The increase in the maximal value as L increases from 100 to √ √ 101, is approximately λ · 1 = 0 .25. The actual increase is 5 ( 101 − 100 ) ≈ 0 .249378. 4. K ∗ = 2 1 3 r −1 3 w1 3 Q4 3 , L ∗ = 2 −2 3 r 2 3 w−2 3 Q4 3 , C ∗ = 3 · 2−2 3 r 2 3 w1 3 Q4 3 , λ = 2 4 3 r 2 3 w 1 3 Q1 3 . The equalities ( ∗) are easily verified. 2. (a) x

/

/

/

/

/

/

/

/


/

/

/

/

/

/

/

/

64

C H A P T ER 1 4


6. (a) Given L ex y z λ(x y z 1) µ(x 2 y 2 z2 1), the first-order conditions are (i) e x λ 2µx 0; (ii) 1 λ 2µy 0; (iii) 1 λ 2µz 0. From (ii) and (iii) it follows that µ(y z) 0. If µ 0, then λ 1, x 0, and y z 1, y 2 z2 1. The stationary points occur 1 where (x, y, z) ( 0, 1, 0) or (0, 0, 1), in which case e x y z 2. If µ 0, then y z x) , 2 (1 1 1 implying that x 2 2 (1 x) 2 1, so x 1 or 3 . The maximum is at (1, 0, 0) where e x y z e . Another stationary point is at ( 13 , 23 , 23 ) where ex y z e −1/3 43 < 73 < e. The relevant Lagrange 1 (e 1). (b) f ∗ λ (0.02) µ ( 0.02) 0 .01(3 e) 0 .0028. multipliers are λ 1, µ 2

= + + − + + − − − = − − = − = = = = = + − = = − = = −

−

−

+ + − − − = + = + = + + = =

+ + = + ≈ · + ·−

=

= = − + + = − ≈

14.7 2. (a) The Lagrangian is L(x,y) x 2 2 y 2 x λ(x 2 y 2 1 ). Conditions (2)–(4) take the form: 2x 1 2 λx 0; 4y 2 λy 0; λ 0 (λ 0 if x 2 y 2 < 1). (b) Candidates: (1/2, 0) with λ 0; ( 1, 0) with λ 1/2; ( 1, 0) with λ 3/2; and ( 1/2, 3/2) with λ 2. Maximum 9/4 at ( 1/2, 3/2) and at ( 1/2, 3/2). The extreme-value theorem confirms that this is the solution.

= + − − + − − − = − = ≥ = + √ = √ = −√ = − ± = − − − 4. (a) f 1 (x,y) ≥ 0 (f 1 (x,y) = 0 if x < b) and f 2 (x,y) = 0 (b) f 1 (x,y) = 0 and f 2 (x,y) ≤ 0 (f 2 (x,y) = 0 if y > a ) 14.8 2. (a) The admissible set is the shaded region in Fig. M14.8.2. x y x y (b) With the constraints g1 (x,y) 4, g2 (x,y) 1, g3 (x,y) 1, the x x +y Lagrangian is L x y e e λ1 ( x y 4) λ2 ( x 1) λ3 ( y 1). The first-order conditions are that there exist nonnegative numbers λ 1 , λ 2 , and λ 3 such that: (i) Lx 1 ex ex +y λ1 λ2 0; (ii) Ly 1 ex +y λ1 λ3 0; (iii) λ 1 ( x y 4) 0; (iv) λ 2 ( x 1) 0; (v) λ 3 ( y 1) 0. (We formulate the complementary slackness conditions as ex λ 3 e x > 0. in (14.7.5).) From (ii), ex +y 1 λ 1 λ 3 . Inserting this into (i) yields λ2 Because λ 2 > 0, (iv) implies that x 1. So any solution must lie on the line (II) in the figure, which shows that the third constraint must be slack. (Algebraically, because x y 4 and x 1, we have 4 x +y y 4 x 5 > 1.) So from (v) we get λ 3 0, and then (ii) gives λ1 e 1 e 1 > 0. Thus from (iii), the first constraint is active, so y 4 x 5. Hence the only possible solution is ∗ ∗ (x , y ) ( 1, 5). Because L(x,y) is concave, we have found the optimal point.

= − − ≤ − = − ≤ = − ≤ − = + − − − − − + − − − − − +

= − − − − =

+ + = − + = = + + = −

= −

+ + =

− − + =

= +

≥ − =

=

= −

+ ≥ =

= − =

≥

= − − ≥ −

y

7 II ( 1, 5)

−

6 5 4 3 I

2

(3, 1)

1

−2 −1

III 1

2

3

4

5

6

7

x

Figure M14.8.2

= − x 2 + λy + µ(y − x + 2) − ν(y 2 − x), which is stationary when + λ + µ − 2νy = 0. Complementary slackness requires in addition,

4. (a) The Lagrangian is L y (i) 2x µ ν 0; (ii) 1

− − + =


C H A P T ER 1 4


65

(iii) λ 0 (λ 0 if y > 0), (iv) µ 0 (µ 0 if y x 2 > 0); (v) ν 0 (ν 0 if y 2 < x ). From (ii), 2νy 1 λ µ > 0, so y > 0. Then (iii) implies λ 0, and 2νy 1 µ. From (i), 1 x µ). But x y 2 > 0, so ν > µ 0, and from (v), y 2 x . 2 (ν Suppose µ > 0. Then y x 2 y y 2 2 0 with roots y 1 and y 2. Only y 2 is feasible. Then x y 2 4. Because λ 0, conditions (i) and (ii) become µ ν 8 and µ 4ν 1, so ν 7/3, which contradicts ν 0, so (x, y) (4, 2) is not a candidate. Therefore µ 0 after 1 ν y 2 and, by (ii), 1 all. Thus x 2νy 4y 3 . Hence y 4−1/3 , x 4−2/3 . This is the only 2 remaining candidate. It is the solution with λ 0, µ 0, and ν 1 /2y 4 −1/6 .

≥

=

≥

=

− +

≥ =

=

= + + = + = − ≥ ≥ = − + = − + = = − = = = = = − + = − = − =− ≥ = = = = = = = = = = = = (b) We write the problem as max xe − − 2ey subject to y ≤ 1 + 12 x , x ≥ 0, y ≥ 0. The Lagrangian is − − 2ey − λ(y − 1 − x/ 2), so the first-order conditions (4) and (5) are: L = x e (i) L = e − − xe − + 21 λ ≤ 0 (= 0 if x > 0 ); (ii) L = x e − − 2e − λ ≤ 0 ( = 0 if y > 0 ); (iii) λ ≥ 0 (λ = 0 if y < 1 + 12 x). From (i) we have x ≥ 1 + 12 λe − ≥ 1, so (iv) (x − 1 )e − = 12 λ. Suppose λ > 0. Then (iii) implies (v) y = 1 + 21 x > 0. From (ii) and (iv) we then have xe − = 2e + λ = e − + 12 λ and so λ = 2 e − − 4e = 2 e(e− − 2), by (v). But then λ > 0 implies that e− > 2, which contradicts x ≥ 0. When λ = 0, (iv) gives x = 1. If y > 0, then (ii) yields e −1 = 2 e, and so y − 1 = ln (2e) > 12 x when x = 1. Thus feasibility requires that y = 0, so we see that (x, y) = ( 1, 0) is the only point satisfying all the conditions, with λ = 0. y x

y x

y x

x

y x

y x

y

x y

y x

y x

y x

1 2x

y x

1 2x

y

Review Problems for Chapter 14 2. (a) x

= 25mp , y = 35mq

(b) x

= 3mp , y = 23mq

(c) x

= 35mp , y = 25mq

4. (a) Interpretation: Maximize the volume of a rectangular box with a given surface area A. (b) With the Lagrangian L xyz λ(2xy 2xz 2yz A), the first-order conditions yield yz λ(2y 2z) 0, x z λ(2x 2z) 0, xy λ(2x 2y) 0. (The optimal solution obviously yz xz xy requires all three variables to be positive.) It follows that λ . The other y z x z x y

= −

− + + + = −

− − + = + = = + = + = + required equality is just the constraint divided by 2. (c) From yz/(y + z) = xz/(x + z) we get y (x + z) = x(y + z), or y z = xz , so x = y because z > 0. Similarly, from xz/(x + z) = xy/(x + y) we get y = z. Inserting y = z = x into the constraint and √ √ solving for x yields x = A/6, and so the optimal solution is x = y = z = A/6. The optimal box is a cube whose six faces all have area A/ 6.



6. With the Lagrangian L (1/x) 4y 4z λ(x y z 2) µ x 2 conditions yield (i) 1 /x 2 λ 2µx 0; (ii) 4 λ 2µy 0; (iii) 4 and (iii) give µy µz , i.e. (iv) µ(y z) 0.



= − + + − + + − − + y 2 + z2 − 3/2 , the first-order − − = − − = − λ − 2µz = 0. Equations (ii) = − = = 0. Then (iv) implies y = z, which inserted into the constraints gives x = 2 − 2y and A. Suppose µ  x 2 + 2y 2 = 3/2, so ( 2 − 2y) 2 + 2y 2 = 3/2, or 6y 2 − 8y + 5/2 = 0. The solutions of this equation are y = 5/6 and y = 1/2. This gives the following solution candidates: P 1 = (1/3, 5/6, 5/6) and P 2 = ( 1, 1/2, 1/2). The function values are are 11/3 and 3 respectively. B. Suppose µ = 0. From (ii) we have λ = 4, and then(i) gives1 /x 2 = 4,so x = ±1/2. Inserting x = 1 /2 into the first constraint gives z = 3 /2 − y , and then the second constraint gives 2 y 2 − 3y + 1 = 0, with the roots y = 1 and y = 1 /2. The new solution candidates are P 3 = ( 1/2, 1, 1/2) and P 4 = ( 1/2, 1/2, 1). The function values are both 4.


66

C H A P T ER 1 4


It remains to consider the case x 1/2. But then the first constraint gives z 5 /2 y , and the 2 second constraint gives the equation 2 y 5y 5 0, which has no real solutions. By evaluating the objective function f at all the four candidate points, we find that f has its largest value at P 3 and P 4 , with f max f (1/2, 1, 1/2) f (1/2, 1/2, 1) 4.

= − − + =

=

−

=

= = 8. (a) The maximum value is 20 attained at (x, y, z) = (4, 0, 0). The minimum value is 7.5 attained at all points (x, y, z) satisfying x = −1 and y 2 + z2 = 7 .5. (With λ as the Lagrange multiplier, the first-order conditions imply (i) 2 x + 1 = 2 λx ; (ii) 2y = 4 λy; (iii) 2 z = 4 λz. From (ii), y = 0 or λ = 1 /2. From (iii), z = 0 or λ = 1/2. Suppose first that λ = 1/2. Then (i) implies x = −1 and 2(y 2 + z 2 ) = 15, = 1/2, then (ii) and (iii) give y = z = 0. Then which gives the minimum value. On the other hand, if λ  x = ±4, and x = 4 gives the maximum. By the extreme value theorem, maximum and minimum values exist in (a) as well as in (b).) (b) The function f has a unique stationary point in the interior of S at (x,y,z) ( 1/2, 0, 0), with f ( 1/2, 0, 0) 1/4. We conclude that the maximum point of f in S is at ( 4, 0, 0) and the minimum is at ( 1/2, 0, 0).

−

−

= −

= −

=

−

− − = = ∗∗ ∗ ∗∗ = − − + = + + = = − − + 12. (a) See Fig. M14.R.12. Note that x ≥ ln (3/2). (b) With λ and µ as the Lagrange multipliers, the firstorder conditions are: (i) −(2x + 1) + λe− = 0; (ii) −y + λ − µ = 0; (iii) λ ≥ 0 (λ = 0 if e − < y ); (iv) µ ≥ 0 (µ = 0 if y < 2 /3). From (i), λ = (2x + 1 )e ≥ 3/2, because of (a). From (ii), µ = λ − y ≥ 3/2 − 2 /3 > 0, so y = 2/3 because of (iii). Solution: (x ∗ , y ∗ ) = (ln(3/2), 2/3), with λ = 3[ln(3/2) + 1/2], µ = 3 ln(3/2) + 5/6. (The Lagrangian is concave so this is the solution.) 10. (a) The Lagrangian is L(x,y) U(x,y) λ[py w(24 x) ]. The first-order conditions imply   pU 1 wU 2 λwp, which immediately yields ( ). (b) Differentiating ( ) and ( ) w.r.t. w gives  24 x wx  and p(U  x  U  y  ) U  w(U  x  U  y  ). Solving for x  ∂x/∂w pyw w w 11 w 12 w 2 21 w 22 w (4ln18 5)/(ln18 1). yields the given formula. (c) ∂x/∂w x

x

x

y y

= e −x

2

1 y

= 2/3

(x ∗ , y ∗ )

1 Figure M14.R.12


2

x

C H A P T ER 1 5

M A T R IX A N D V E C T O R A L G E B RA

67

Chapter 15 Matrix and Vector Algebra Chapter 15 begins our treatment of matrix algebra by presenting matrices, vectors, and simple algebraic rules for working with them. Section 15.1 introduces a general notation for linear simultaneous equations, and motivates them through a simple case of the Leontief input–output system. Matrices are defined in Section 15.2 as rectangular arrays of numbers. (The idea that matrices represent linear transformations is, in our view, better postponed to a more advanced and abstract book on linear algebra.) Particular examples are the coefficient matrix for a system of linear simultaneous equations, and a “spreadsheet” recording the value of retail sales for different commodities at different sales outlets. Then equality, addition, and multiplication by a scalar are defined in obvious ways. The rules for matrix multiplication are far from intuitively obvious to a student seeing them for the first time. Section 15.3 therefore takes some pains to explain how C AB is defined for a 2 3 matrix A and a 3 2 matrix B in order that the 2 2 matrix C should satisfy Cx A(Bx) for all 2-vectors x . This, of course, builds on the rules for multiplying a coefficient matrix by a vector, which are also explained in Section 15.3. Thereafter, the rule for multiplying general m n and n p matrices is set out and illustrated. Attention is drawn to the important fact that matrix multiplication is generally noncommutative—i.e., generally AB BA. The section concludes by reviewing how a general system of equations can be expressed concisely in matrix form. Next, Section 15.4 turns to the associative law and (since multiplication is non-commutative) the two versions of the distributive law for matrix multiplication. Then powers of square matrices are defined, as is the identity matrix of any finite dimension. The section concludes by warning against algebraic errors such as replacing AB by BA, or asserting that if AB 0 , then either A 0 or B 0 (or both). Section 15.5 gives a brief discussion of transpose matrices, including general rules for transposition, and of symmetric matrices. A powerful general method for solving linear systems of equations is Gaussian elimination, which is the subject of Section 15.6. The “staircase” method is explained and a convenient notation for elementary operations is introduced. Next, Section 15.7 introduces n -dimensional vectors as lists of n numbers. The rules of addition, subtraction, and multiplication by a scalar are described. Linear combinations are defined. Particularly for two-dimensional vectors, the fact that there is a simple geometric interpretation can help students enormously. This is the topic of Section 15.8, which also discusses the parallelogram law of vector addition, and how to represent multiplication by a scalar. The section concludes with some remarks about n-dimensional geometry. Calculating the inner (or scalar) product of two n-vectors is an important step on the way to finding the product of two matrices. In economics, it is common to consider corresponding vectors of prices and of quantities. Then the inner product of a price and quantity vector represents the value of a commodity bundle. This is used in Section 15.8 to motivate the concept. Rules for manipulation are presented. The length of a vector is defined. The Cauchy–Schwarz inequality is stated. The section concludes with a brief discussion of orthogonality, a concept that is useful in econometrics. Geometric ideas receive more attention in Section 15.9. A diagram is used to motivate the algebraic definition of a line in a space of three dimensions. This algebraic definition is then extended to n dimensions. Similarly, a diagram of a plane in three dimensions is used to motivate the algebraic definition, which is then extended to n dimensions.

×

= =

×

×

=


×

×

=

=

=

68

C H A PT E R 1 5

M A TR I X A N D V E C TO R A L G EB R A

Answers to Even-Numbered Problems 15.1 2. Yes, the system is linear in a , b , c , and d . 4. 2x1 3x1 4x1

+ 3x2 + 4x3 = 1 + 4x2 + 5x3 = 2 + 5x2 + 6x3 = 3

6. (a)The commodity bundle owned by individual j . (b) ai 1 ai 2 pm amj The first case is when i 1. (c) p 1 a1j p2 a2j

=

+

+ · · · +

+ +···+ a

in

is the total stock of commodity i .

15.2

= 3 and v = −2. (Equating the elements in row 1 and column 3 gives u = 3. Then, equating those in row 2 and column 3 gives u − v = 5 and so v = −2. The other elements then need to be checked, but

2. u

this is obvious.) 4. A

 + =

15.3 2. (i)

4. (a)

(c)

B

1 2

0 4



4 , A 16

 − = −

1 2

B

2 2

−6 −2



, and 5A

 = −

− 3B

3 10

8 12

−20

8



−      = = = − −    =    −    =   −    −   =   −   1 6

1 3

15 13

1 5

(ii) and (iii): AB

x1 x2

2 1

3 1

=

0.2875 0.2250 0.4875

6. T(Ts)

1 1

3 5

(b)

x1 x2 x3

0 0

C(AB)

1 1 2

2 1 3

1 1 1

0 , since AB 0

x1 x2 x3

0 0

0 . 0

4 5 1

0 0

15.4 2. (ax 2

+ by2 + cz2 + 2dxy + 2exz + 2fyz) (a 1 × 1 matrix) 4. Equality in (a) aswell asin (b) ifand onlyif AB = BA. ((A + B)(A − B) = A 2 − AB + BA− B2  = A2 − B2 unless AB = BA. The other case is similar.) 6. (a) Direct verification by matrix multiplication. (b) AA = (AB)A = A(BA) = AB = A, so A is idempotent. Then just interchange A and B to show that B is idempotent. A, which is true for k (c) As the induction hypothesis, suppose that A k AA A , which completes the proof by induction.

=


=

= 1. Then A +1 = A A = k

k

C H A P T ER 1 5

15.5 2. A =


69

 −  =   + =   =  −  − −      = = = = − 3 2

1 , B  5

0 2

2 , (A 2

3 4

B)

1 , (αA) 7

6 4

4 10 4 10 B  A , and A B , (AB) 10 8 10 8 Verifying the rules in (2) is now very easy.

2 10

AB

4. Symmetry requires a 2

2 , 10

4 14

− 1 = a + 1 and a 2 + 4 = 4 a. The second equation has the unique root a = 2,

which also satisfies the first equation.

6. By the associative law and rule (15.5.2)(d) for transposition, (A1 A2 A3 ) (A1 (A2 A3 )) (A2 A3 ) A1 (A3 A2 )A1 A 3 A2 A1 . In general, for any natural number n > 3, one has An ) ((A1 A2 An−1 )An ) An (A1 A2 An−1 ) . As the induction hypothesis, suppose the (A1 A2 result is true for n 1. Then the last expression becomes A n An−1 , A2 A1 , so the result is true for n.

·· ·

=

=

−

· ··

=

···

=

=

· ··

15.6 2. Gaussian elimination yields:

 

1 1 1

1 1 2

−

−1

2

1 2

 − − ← ← −  − − ← 1

1

 ∼ 

1 0 0

1 2 1

−

−1

3 a 1

 − −  −  − 1 1

1 2

+ b 1 −1 1 −1 1 1 1 1 1 ∼ 0 1 −3/2 1/2 −3/2 1 ∼ 0 1 1/2 0 1 a+1 b 1 0 0 a + 5/2 b 1/2 For any z , the first two equations imply that y = − 12 + 23 z and x = 1 − y + z = 32 − 12 z. From the last = − 52 , there is a unique solution with z = (b − 12 ) (a + 52 ). For a = − 25 , equation we see that for a  = 12 , but z is arbitrary (one degree of freedom) if b = 12 . there are no solutions if b  a

b



4. (a) After moving the first row down to row number three, Gaussian elimination yields the matrix



1 2 1 b2 3 1 0 1 2 b b 2 2 2 3 3 0 0 3 4a b1 (2a 2 )b2 ( 12 Obviously, there is a unique solution iff a

− −

+

−

−

+ − a)b3 = 3/4.



(b) Put a 3 /4 in part (a). Then the last row in the matrix in (a) becomes (0, 0, 0, b1 14 b3 ). It follows 1 1 that if b1 4 b3 there is no solution. If b1 4 b3 there are an infinite number of solutions. In fact, 3 1 x 2b2 b3 5t , y 2t , z t , with t  . 2 b2 2 b3

= = = − + −

=

− +

= =

−

∈

15.7

+ b + c = (−1, 6, −4), a − 2b + 2c = (−3, 10, 2), 3 a + 2b − 3c = ( 9, −6, 9), −a − b − c = −(a + b + c) = ( 1, −6, 4). 4. (a) x = 0 for all i . (b) Nothing, because 0 · x = 0 for all x. 6. 4x − 2x = 7 a + 8b − a, so 2x = 6 a + 8b, and x = 3 a + 4b. 8. The scalar product of the two vectors is x 2 + (x − 1)x + 3 · 3x = x 2 + x 2 − x + 9x = 2 x 2 + 8x = 2 x(x + 4), which is 0 for x = 0 and x = −4. 2. a

i


70

C H A PT E R 1 5

M A TR I X A N D V E C TO R A L G EB R A

10. (a) The firm’s revenue is p z. Its costs are p x. (b) Profit = revenue– costs = p z p x If p y < 0, the firm makes a loss equal to p y.

·

·

·

· − · = p ·(z−x) = p ·y.

− ·

15.8

=

=

=

=

=

=

2. (a) λ 0 gives x (3, 1), λ 1/4 gives x (2, 5/4), λ 1/2 gives x (1, 3/2), λ 3 /4 gives x ( 0, 7/4), and λ 1 gives x ( 1, 2). See Fig. M15.8.2. (b) As λ runs through [0, 1], the vector x will run through all points on the line segment S between a and b. In fact, according to the point–point formula, the line L through ( 3, 1) and ( 1, 2) has the equation 1 7 x2 x x 4x2 7. The line segment S is traced out by having x1 run through [3, 1] as x2 4 1 4 or 1 runs through [1, 2]. Now, (1 λ)a λb ( 3 4λ, 1 λ). Any point (x1 , x2 ) on L satisfies x1 4x2 7 1 and equals (3 4λ, 1 λ) for λ x1 ) x 2 1. Any point on the segment of this line between 4 (3 a ( 3, 1) and b ( 1, 2) equals (3 4λ, 1 λ) for some λ [0 , 1].

=

=

= − +

=

=

= −

−

+ = − + = − + − + = − = − = − − +

− + =

∈

y λ

= 1

= 3/4 λ = 1/2

λ

λ

1

b

= 1/4

λ

= 0

a

−1

1

2

3

x

Figure M15.8.2

+ −

− = − + −

− = − = − = =

= = − − = − =

4. (a) x 1 (1, 2, 1) x2 ( 3, 0, 2) (x 1 3x2 , 2x1 , x1 2x2 ) ( 5, 4, 4) when x 1 2 and x 2 1. (b) x 1 and x2 must satisfy x1 (1, 2, 1) x2 ( 3, 0, 2) ( 3, 6, 1). Then x1 3x2 3, 2 x1 6, and x1 2x2 1. The first two equations yield x 1 3 and x 2 2; then the last equation is not satisfied.

− =

6. The vectors are orthogonal iff their scalar product is 0, that is iff x 2 which is the case for x 2 and x 4.

− x − 8 − 2x + x = x 2 − 2x − 8 = 0,

= − = 8. (||a||+||b||)2 − ||a + b||2 = ||a||2 + 2||a||·||b||+||b||2 − (a + b)(a + b) = 2 (||a||·||b|| − a · b) ≥ 2(||a||·||b||−|a · b|) ≥ 0 by the Cauchy–Schwarz inequality (2). 15.9

=

2. (a) Direct verification. (To show that a lies on L, put t 0.) (b) The direction of L is given by the vector ( 1, 2, 1), and the equation of the orthogonal plane is ( 1)(x1 2) 2(x2 ( 1)) 1 (x3 3) 0, or x1 2x2 x3 1. (c) We must have 3( t 2) 5(2t 1) (t 3) 6, and so t 4 /3. The point of intersection is then (2/3, 5/3, 13/3).

− − +

− −+ +

+ = −

4. (a) Direct verification.

= ( −2, 1, −1) + t (−1, 2, 3) = ( −2 − t, 1 + 2t, −1 + 3t).

(b) (x 1 , x2 , x3 )

Review Problems for Chapter 15 2. (a) A

  − − = −   − = B

(d) C(AB)

3 2 2 6

−

2 2 1 8

− + − − + · − = − − + = =

(b) A (e) AD


  − − + − = − −   = 3 2

2C

B

2 0

2 2

2 3

4 8

(c) AB

  − =

(f) DC is not defined.

2 2

−

4 3

C H A P T ER 1 5

4. (a)

       −   =       =    +      − =  +  + + = + + + + + + = + + + 2 5

a

(c)

1 3

5 8

1 2 4

a

1

x y z

p3

p2 q

1 7

1 3 p 2

6. (a) TS

3 5

x1 x2

p

(b)

1 1 1 2

1 3 4 0

1 2 8 1

1 4 0 1

71


x y z t

a b c d

b1 b2 b3

2p 2 q

1 2 1 2 p p q 2 2 3 2

p q

2pq 2

p2 q pq pq 2 2p2 q 2pq 2

pq 2

1 pq 2 2

pq

q3

1 2 1 3 q q 2 2 2 3

S because p

q

+ q = 1.

1 T A similar argument shows that T 2 2 1 individual elements: T3 TT 2 T( 2 T

+ 12 S. To prove the last equality, we do not have to consider the = = + 12 S) = 12 T2 + 12 TS = 12 ( 21 T + 21 S) + 21 S = 14 T + 34 S. (b) The appropriate formula is (∗) T = 21− T + ( 1 − 2 1− )S. This formula is correct for n = 1 (and for n = 2, 3). Suppose (∗) is true for n = k . Then using the two first equalities in (a), T +1 = TT = T(21− T + ( 1 − 2 1− )S) = 21− T2 + ( 1 − 2 1− )TS = 21− ( 12 T + 12 S) + ( 1 − 2 1− )S = 2− T + 2− S + S − 2 · 2− S = 2 − T + (1 − 2− )S, which is formula (∗) for n = k + 1. −2 ∼ 1 4 1 1 4 1 1 4 1 ← 1 0 5 8. (a) ∼ ∼ −4 0 1 −1 2 2 8 ← 0 −6 6 −1/6 0 1 −1 The solution is x 1 = 5, x 2 = −1. − 2 −3 2 2 −1 2 ← 1 −3 1 0 (b) 1 −3 1 0 ← ∼ 2 2 −1 2 ← 3 4 −1 1 3 4 −1 1 ← − − 1 3 1 0 1 3 1 0 ∼ 0 8 −3 2 1/8 ∼ 0 1 −3/8 1/4 −13 −4 1 ← 0 13 −4 1 0 13 ← 1 −3 1 0 1 −3 1 0 ∼ 0 1 −3/8 1/4 ∼ 0 1 −3/8 1/4 3 −18/7 0 0 7/8 −9/4 8/7 0 0 1 ← 1 0 −1/8 3/4 1 0 0 3/7 ∼ 0 1 −3/8 1/4 ← ∼ 0 1 0 −5/7 −18/7 3/8 1/8 0 0 1 −18/7 0 0 1 The solution is x 1 = 3 /7, x 2 = −5/7, x 3 = −18/7. −5 ∼ 1 3 4 0 1 3 4 0 (c) 5 1 1 0 ← 0 −14 −19 0 −1/14 14 0 ∼ 10 31 19/414 00 ←−3 ∼ 01 01 −191//14 0 The solution is x 1 = ( 1/14)x3 , x 2 = −(19/14)x3 , where x 3 is arbitrary. (One degree of freedom.) a √ = √ 35, √ b = √ 11, and c = √ 69. Also, |a · b| = |(−1) · 1 + 5 · 1 + 3 · (−3)| = | − 5| = 5, and 10. √ = =

n

n

n

k

k

k

k

k

k

k

k

k

k

k

k

k

                              

35 11



385 is obviously greater than 5, so the Cauchy–Schwarz inequality is satisfied.

12. If PQ QP P, then PQ QP P , and so P2 Q P(PQ) P(QP P) (QP P)P P2 QP 2 2P2 . Thus, P2 Q QP2 2 P2 . Moreover, P3 Q P(P2 Q) (PQ)P2 2P3 (QP P)P2 2P3 QP 3 3P3 Hence, P 3 Q QP3 3 P3 .

− = + + = + =

+ +

=

+

+

=


−

+

=

=

=

−

= =

+

= (PQ)P + P 2 = = P(QP2 + 2P2 ) =

72

C H A P T ER 1 6

D E T E RM I N A N T S A N D I N V E R SE M A T R I CE S

Chapter 16 Determinants and Inverse Matrices In this chapter, Section 16.1 begins with determinants of order 2, whose definition is motivated by the form of the typical solution to two linear simultaneous equations in two unknowns. In fact, determinants are defined so that Cramer’s rule for solving such equations is valid. Then it is shown that the determinant of a 2 2 matrix is equal to the area of the parallelogram generated by its two columns, regarded as a pair of 2-vectors. Next, Section 16.2 moves on to three equations in three unknowns. This gives rise to a complicated formula, which is then broken down so that it appears as the expansion of the determinant by the cofactors of its first row. It is claimed that the determinant of a 3 3 matrix equals the volume of the (3-dimensional) parallelepiped generated by its columns, regarded as three 3-vectors. Sarrus’s rule for evaluating 3 3 determinants is then introduced. Section 16.3 proposes a formula for n n determinants that is essentially the sum of n! terms, each term being the product of n elements, of which one is chosen from each row and one from each column of the matrix. Then a sign rule is given to determine whether this product should have a or attached in forming the overall sum. Of course, this formula is virtually useless except in a few special cases of “sparse” matrices where virtually all the n ! terms collapse to 0. In fact, Section 16.3 really helps clear the ground for the fundamental “Rules for Determinants” that are enunciated in Theorem 16.4.1. Particularly important is the rule saying that determinants are unchanged by the elementary row (or column) operations of adding a multiple of one row (or column) to any other row (or column). These rules are illustrated by showing how they apply to particular 2 2 matrices. There are also two pages of discussion concerning how to prove most parts of Theorem 16.4.1. We envisage, however, that most students will prefer to take on trust the key results set out in the theorem. The topic of Section 16.5 is expansion by cofactors in general—not just the expansion of 3 3 determinants by the first row that was considered in Section 16.2. Also important is the fact that expansion by “alien” cofactors always gives zero, because it is like finding the determinant of a matrix in which one row or column has been repeated. The section concludes with a proof that the rule for expanding cofactors is valid, though once again, this is a result that we expect most students will prefer to take on trust. By the end of this section students should have learned how to find the determinant of a matrix in any one of several equivalent ways— expansion by cofactors, the full expansion of n ! terms, or by elementary row or column operations designed to produce a “sparse” matrix—even one that is upper or lower triangular, and so with determinant equal to the product of its diagonal terms. Having completed the discussion of determinants per se, Section 16.6 moves on to the inverse of a square matrix. After the definition, it is claimed that an inverse exists iff the matrix is non-singular in the sense of having a non-zero determinant. It is shown that, if an inverse exists, it must be unique. The general formula for the inverse of a 2 2 matrix is given. Then Theorem 16.6.1 sets out some key properties of inverse matrices. These include the results that the inverse of the inverse is the original matrix, that the inverse of a matrix product is the reverse product of the inverses, and that the inverse of the transpose is equal to the transpose of the inverse. Finally, Theorem 16.6.2 states how to solve linear equations by matrix inversion, in case the relevant matrix is non-singular. An explicit formula for the inverse of a matrix is the main topic of Section 16.7. Indeed, based on the rules for expansion by cofactors (including alien cofactors), it is shown that one can find the adjoint matrix, defined as the transpose of the matrix of cofactors, then divide the adjoint by the (non-zero) determinant of the (non-singular) matrix. This is the “general formula for the inverse” set out in Theorem 16.7.1. The section concludes by outlining how to find the inverse matrix by the generally more practical procedure of elementary row operations.

×

±

×

±

×

×

+ −

×

×

×


C H A P T ER 16

D E T E RM I N A N T S A N D I N V E R S E M A T R I CE S

73

Section 16.8 explains how a general linear system of n equations in n unknowns with a non-singular coefficient matrix can be solved by means of Cramer’s rule (Theorem 16.8.1). This result is of theoretical interest, but for systems with more than 4 unknowns, the calculation of the appropriate determinants usually becomes computationally impractical. Finally, whereas a linear system of n equations in n unknowns is usually soluble only when its coefficient matrix is non-singular, a system of homogeneous equations (with zero right-hand sides) has a non-trivial solution iff the coefficient matrix is singular. This is Theorem 16.8.2. To conclude the chapter, Section 16.9 gives a brief introduction to the Leontief model.

Answers to Even-Numbered Problems 16.1 2. See Fig. M16.1.2. The shaded area is 18

 = 



3 2

0 . 6

( 2, 6)

(3, 0)

Figure M16.1.2

 =

+ +



+ + +

a11 b11 a12 b21 a11 b12 a12 b22 4. The matrix product is AB , implying that a21 b11 a22 b21 a21 b12 a22 b22 AB (a11 b11 a12 b21 )(a21 b12 a22 b22 ) (a11 b12 a12 b22 )(a21 b11 a22 b21 ). On the other hand, A B (a11 a22 a 12 a21 )(b11 b22 b 12 b21 ). A tedious process of expanding each expression, then cancelling four terms in the expression of A B , reveals that the two expressions are equal.

| |= | || | =

+ −

+ −

−

+

| || |

6. We write the system as

− C = I 0 + G0 −bY + C = a Y

Then Cramer’s rule yields

 + =  − I 0

G0

1

a

Y

 + +  = − −  −1

1

b

1 1

a

I 0

1

b

 = − − 1

G0

,

C

I 0

b

 −  =

+ G0 a

1 b

1 1

a

+ b(I 0 + G0 ) 1−b

The expression for Y is most easily found by substituting the second equation into the first, and then solving for Y . Then use C a bY to find C .

16.2

 − =

2. AB

1 7 5

= +

−1 −1 13 9

13 10

 | | = − | | = | , A


2, B

3, AB

| = |A| · |B| = −6

74

C H A P T ER 1 6


4. By Sarrus’ rule the determinant is (1

reduces to the given expression.

+ a)(1 + b)(1 + c) + 1 + 1 − (1 + b) − (1 + a) − (1 + c), which

6. (a) Substituting T d tY into the expression for C gives C C in the expression for Y then yields Y a b(Y d tY )

= +

= + − −

to derive the answers given in (b) below. 1 1 0 (b) We write the system as b 1 b t 0 1

 −  =   − −  − − = −−  − − − A0 a d

Y

1

b t

1

b t

T

1

b t

1 1 0 1 1 0 1 1 0 1 1 0

0 b

1 0 b

1

= a − bd + b(1 − t)Y . Substituting for + A0 . Then solve for Y , T , and C in turn

 −      − = −    = − + =  −−  − −  − − −   = + + −  − − Y C T

A0 a d

1

a

bd

A0

1

b(1

t)

,

b t

C

1

b t

A0 a d

t (a

0

(1

A0 )

1

b(1

. Then Cramer’s rule yields

A0 a d

0 b

1 1 0 1 b 0 1

  = − +  − a

bd

1

A0 b(1

b(1

− t)

− t)

b)d

t)

b

1

(This problem is meant to train you in using Cramer’s rule. Note how systematic elimination is much more efficient.)

16.3 2.

+a12 a23 a35a41 a54 (Four lines between pairs of boxed elements rise as one goes to the right.)

4. With A

 = 

a11

0 .. .

0

a12 . . . a22 . . . .. .. . . 0 ...

a1n a2n .. . ann

 

and B

 = 

b11

0 .. .

b12 . . . b22 . . . .. .. . . 0 ...

b1n b2n .. . b nn

 

,

0 the product AB is easily seen to be upper triangular, with the elements a 11 b11 , a 22 b22 , . . . , ann bnn on the main diagonal. The determinant AB is, according to (4), the product of these n numbers. On the other hand, A a 11 a22 ann , and B b 11 b22 bnn , so the required equality follows immediately.

| |=

16.4

 =

2. A

2 1 3

1 0 1

| | | |=

· ··

1 2 5

· ··

 | | =| | =− , A

A

2

| | = |A||B| = −12, 3|A| = 9, | − 2 B| = (−2)3 (−4) = 32, |4A| = 43 |A| = 43 · 3 = 192, 1 0 0 |A|+|B| = −1, whereas |A + B| is not determined. (As an example showing this, let A1 = 0 1 0

4. AB




0

0

3



C H A P T ER 16

and B1

A2

 =

 = 3 0 0

 −

1 0 0 0 1 0 . Then A1 0 0 4 0 0 2 0 1 0 and B 2 0 2 0 1 0 0



 =


75

| | = 3, |B1 | = −4, and |A1 + B1 | = −4. On the other hand, if −

0 0 1



, then A2

| | = 3, |B2 | = −4, and |A2 + B2 | = −10.)

6. (a) The first and the second columns are proportional, so the determinant is 0 by part E of Theorem 16.4.1.

(b) Add the second column to the third. This makes the first and third columns proportional. (c) The term x y is a common factor for each entry in the first row. If x y , the first two rows are proportional. If x y , the first row has all elements 0. In either case the determinant is 0.

− =

=

| | = a (a2 + 1) + 4 + 4 − 4(a2 + 1) − a − 4 = a 2 (a − 4), so |A1 | = −3

8. By Sarrus’s rule, for example, Aa A1 6 ( 3)6 729. and A61

| | =| | = − = 10. (a) Because A 2 = I it follows from part G of Theorem 16.4.1 that |A|2 = |I | = 1, and so |A| = ±1. n

n

(b) Direct verification by matrix multiplication. (c) (In A)(In A) I n2 A A A2 I n

+ = + − − = − A2 = 0 ⇐⇒ A2 = I . 12. Add each of the last n − 1 rows to the first row. Each element in the first row then becomes na + b. So −

= (na + b)

Dn

 

n

1

1

+ b

a .. .

a

a

.. . a

...

  + 1

... ... .. .

a .. .

a

b

Next, add the first row multiplied by a to all the other n 1 rows. The result is an upper triangular matrix whose diagonal elements are 1 ,b,b,...,b, with product equal to b n−1 . The conclusion follows easily.

−

−

16.5 2. (a)

−abc

· · · · = 360

(b) abcd (c) 6 4 3 5 1

16.6 2. Multiply the two matrices to get I3 . 4. (a)

(b)

  =  −    =  −   =     =  − −   =  −    =   − −     | |= = = − =   x y

2 3

3 4

4 3

x y

3 2

−1

3 5

4 3

8 11

1 2

3 2

3 5

(c)

x y

3 1

0 0

0 1 1 1 1 2 3 6. (a) A 1, A 1 1 2 , A 2 2 3 . Direct verification yields A3 1 1 1 1 2 2 (b) The equality shown in (a) is equivalent to A(A I)2 I , so A−1 (A I)2 . 2

(c) Choose P = (A − I)−1 =


0 1 0

0 0 1

1 1 0

, so that A

− 2A2 + A − I = 0.

= −

= [(A − I)2 ]−1 = P 2 . The matrix −P also works.

76

C H A P T ER 1 6


8. (a) A 2

= (PDP−1 )(PDP−1 ) = PD(P−1 P)DP−1 = PDIDP−1 = P(D)2 P−1 . (b) Suppose the formula is valid for m = k . Then A +1 = AA = PDP −1 (PD P−1 ) = PD(P−1 P)D P−1 = PDID P−1 = PDD P−1 = PD +1 P−1. 10. (a) If C 2 + C = I , then C(C + I) = I , and so C−1 = C + I = I + C. (b) Because C 2 = I − C, it follows that C 3 = C 2 C = (I − C)C = C − C2 = C − (I − C) = −I + 2C. Moreover, C 4 = C 3 C = ( −I + 2C)C = −C + 2C2 = −C + 2(I − C) = 2 I − 3C. s + 17 4t − 16 0 1 = I3, and so T = A−1 , for s = −5, t = 4. 12. (a) AT = 12 2s + 10 5t − 8 0 3s + 15 4t − 16 12 BX X = 2 + C ⇐⇒ BX − 2X = C ⇐⇒ (B − 2I)X = C . But it is easy to see that B − 2I = A . (b) So BX = 2 X + C ⇐⇒ AX = C ⇐⇒ X = A −1 C = TC , when s = −5 and t = 4. Hence, −5 4 3 2 3 0 1 −1/2 0 0 1/6 1 X = 12 7 −8 3 1 0 3 1 = 1/2 3 −3 1/6 1 4 −3 0 5 −4 1 1/2 −1 2 1/6 k

k

k

k

k

k

k







 

 



16.7 2. The determinant is 72. The cofactors are C11 C21 C31

 =   = −   = 

so the inverse is

 = − = −  = −  −   −    = = =− − −   −  = = −  =    −  = − | | − 0 1

3 1

3 1

2 1

3 0

2 3

3,

5 ,

9 ,

C11 C12 C13

1

A

C21 C22 C23

6 4

C12

3 1

C22

2 4

C32

2 6

C31 C32 C33

18 ,

2 1

2 3

3 18 6

1 72

×

6,

18 ,

5 6 14

9 18 18

 =   = −  −  =  − 6 4

C13

0 1

 =  =  = − 6

C23

2 4

3 1

C33

2 6

3 0

14 18

.

4. Let B denote the n p matrix whose i th column has the elements b 1i , b2i , . . . , bni . The p systems of n equations in n unknowns can be expressed as AX B , where A is n n and X is n p . Following the method illustrated in Example 2, exactly the same row operations that transform the n 2n matrix (A : I) into (I : A−1 ) will also transform the n (n p) matrix (A : B) into (I : B∗ ), where ∗ . (In fact, because these row operations are together equivalent to B∗ is the matrix with elements bij A−1 B.) When k premultiplication by A −1 , it must be true that B ∗ r , the solution to the system is ∗ ∗ ∗ x1 b 1r , x 2 b 2r , . . . , x n b nr .

=

×

×

× +

=

=

=

=

×

=

16.8 2. The determinant of the system is equal to

D1

 = 

b1 b2 b3

−

1 1 3


  −

0 2 , 1

−10, so the solution is unique. The determinants in (2) are here D2

 = 

3 1 2

b1 b2 b3

  −

0 2 , 1

 = 

D3

3 1 2

−

1 1 3

b1 b2 b3

 

C H A P T ER 16

77


Expanding each of these determinants by the column (b 1 , b2 , b3 ), we find that D 1 1 1 1 D2 5 b1 3b2 6b3 , D3 5 b1 7b2 4b3 . Hence, x1 b b b x 2 1 10 2 5 3, 2 1 7 2 x3 2 b1 10 b2 5 b3 .

= − − = = − + +

− −

=

−

= −5b1 + b2 + 2b3, = − 12 b1 + 103 b2 + 35 b3,

−

16.9 2. (a) No sector delivers to itself. (b) The total amount of good i needed to produce one unit of each good.

(c) This column vector gives the number of units of each good which are needed to produce one unit of good j . (d) No meaningful economic interpretation. (The goods are usually measured in different units, so it is meaningless to add them together.) 4. The Leontief system for this three-sector model is as follows:

− −

− 0.2x2 − 0.1x3 = 85 + 0.8x2 − 0.2x3 = 95 − 0.2x2 + 0.9x3 = 20

0.9x1 0.3x1 0.2x1

which does have the claimed solution. 6. The quantity vector x0 must satisfy ( ) (In A)x0 b and the price vector p0 must satisfy ( ) p0 (In A) v  . Multiplying ( ) from the right by x0 yields v x0 (p0 (In A))x0 p 0 ((In A)x0 ) p0 b.

− =

∗

∗∗

−

=

=

−

=

∗∗ =

−

Review Problems for Chapter 16 2. (a)

−4

(b) 1. (Suggestion: Subtract row 1 from rows 2 and 3. Then subtract twice row 2 from row 3. The resulting determinant has only one non-zero term in its third row.) (c) 1. (Suggestion: Use exactly the same row operations as in (b).)

4. (a) At

| | = t + 1, so A has an inverse iff t = −1. (b) Multiplying the given equation from the right by 0 0 −1 A1 yields BA1 + X = I 3 . Hence X = I 3 − BA1 = 0 0 −1 . −2 −1 0 −2 4 −t 6. The determinant of the coefficient matrix is −3 1 t = 5t 2 − 45t + 40 = 5(t − 1)(t − 8). t − 2 −7 4 = 1 and t = 8. So by Cramer’s rule, there is a unique solution iff t  8. (a) (I + aU)(I + bU) = I 2 + bU + aU + abU2 = I + (a + b + nab)U, because U2 = nU, as is easily 4 3 3 verified. (b) Note first that 3 4 3 = I 3 + 3U. Moreover, (a) implies that n

t

n

n



3

3

4



 



n

 



+ 3U)(I3 + bU) = I3 + ( 3 + 10b)U = I3 for b = −3/10. Thus the inverse of 7 −3 −3 1 I3 − (3/10)U = −3 7 −3 . 10 −3 −3 7 11 −6 = A, so A2 + cA = 2I2 if c = −1. 10. (a) |A| = −2. A2 − 2I2 = 18 −10 (b) If B 2 = A , then |B|2 = |A| = −2, which is impossible. (I3



  




4 3 3

3 4 3

3 3 4



is

78

C H A P T ER 1 6

12. (a) For a


= 1 and a = 2, there is a unique solution. If a = 1, there is no solution. If a = 2, there are

infinitely many solutions. (These are a special case of the results in (b).) (b) Using Gaussian elimination (where we start by interchanging rows 1 and 3), we obtain the matrix





−3 b3 2 a +6 b2 − 2b3 2 −a + 3a − 2 b1 − b2 + (2 − a)b3 The last row corresponds to the equation ( −a 2 + 3a − 2)z = b 1 − b2 + (2 − a)b 3 . If −a 2 + 3a − 2 = −(a − 1)(a − 2) = 0, i.e. if a = 1 and a = 2, then the system has a unique solution. When a = 1 and b1 − b2 + b3 = 0, or when a = 2 and b 1 = b 2 , there are infinitely many solutions. Otherwise there is no 1 0 0

0 1 0

solution.

14. Using Gaussian elimination, one can show that there is a unique solution when a

solution is:

− 2a + 3 , = 2ba(b − 2)

x

= 2a b+−b 2− 9 ,

y


2b − 3 = 2ab −a(b2a − − 2) ,

z

= 0 and b = 2. The = 7b−−22a

u

C H A P T ER 1 7

L I N E A R P R O G RA M M I N G

79

Chapter 17 Linear Programming Chapter 17 presents some of the main ideas behind linear programming (LP). But it does not cover the simplex method. This we regard as an important computational technique that is suitable for a specialist course, but not really essential knowledge for most economists. Section 17.1 begins by considering problems with only two choice variables, which can therefore be solved graphically. Example 1 is a detailed discussion of a simple introductory “baker’s problem”; Example 2 moves rather faster. Then, after a brief discussion of when graphical techniques will work, the latter part of the section presents the general LP problem. It also gives an n -dimensional geometric interpretation of such problems, enabling the basic idea of the simplex method to be visualized. Duality theory is introduced in Section 17.2. Example 1 revisits the baker’s problem and explicitly calculates the marginal profit from relaxing each of the three main constraints. The resulting numbers (which will be interpreted as shadow prices later) are then used to verify algebraically the solution found by geometric techniques. Then it is shown that, in this same example, these shadow prices solve the dual problem. This is set up as a problem facing somebody who wants to pay as little as possible to take over the resources available to the baker, while offering just enough to ensure that the baker is compensated sufficiently for these resources. The resulting prices then enable the baker to earn just as much profit as by continuing in business. After the example, it is shown how to set up the dual of a general LP problem. Finally, the concise matrix formulations of the primal and dual problems are presented. Section 17.3 begins with a further discussion of the baker’s problem. This time, the point is made that any price vector that is feasible for the dual problem must give a value for the resources that is not less than the profit earned from any activity vector that is feasible for the primal. Indeed, this is stated as Theorem 17.3.1, and proved in general linear programs by a very easy argument (which does, however, involve recognizing how the same double sum can be written in two different ways). Moreover, Theorem 17.3.2 shows that if the value of primal and dual are made equal, then both problems have been solved. This has an even easier proof. The section concludes with the Duality Theorem 17.3.3, claiming: (i) that if the primal has an optimal solution, then so does the dual; (ii) the values of both primal and dual must be equal when both are solved; and (iii) unless there is an upper bound in the value of the primal, the dual is infeasible. Proving (i) and (ii) involves relatively advanced arguments (or at least an understanding of the simplex method) so the proofs are omitted. An easy proof of (iii) is provided, however. Section 17.4 provides a standard economic interpretation of any LP problem in terms of activity vectors and scarce resources. It also interprets shadow prices as those ensuring that the “shadow profit” on any activity is non-positive, and interprets the dual problem in rather the same way as was done in Section 17.2. Finally, it is shown how shadow prices can be used to estimate changes in the optimized value of the primal from changes in the resources. The last topic of the chapter is complementary slackness, the subject of Section 17.5. The baker’s problem is used to illustrate how positive activity levels are associated with binding constraints in the dual, and positive shadow prices with binding constraints in the primal. Then, for the case of a general primal with two variables and three constraints, the part of the duality theorem claiming equality of the values of primal and dual is invoked. It shows that the pattern observed in the baker’s problem always holds. This complementary slackness property is then stated and proved as Theorem 17.5.1. (The fairly long proof is in small print, especially as it uses vector and matrix notation at the beginning. But it is not hard.) The standard interpretations of the complementary slackness conditions are then presented. Example 1 is provided to show how, given the solution to the dual problem with two variables (which can be found graphically, of course), complementary slackness can be used to identify the set of binding constraints in the primal, and also the activity levels which © Knut Sydsæter and Peter Hammond 2006

80

C H A P T ER 1 7


must be zero. Then, of course, the primal can be solved from simultaneous equations. This is an important technique for solving (and setting) examination questions! Next, it is shown that the Kuhn–Tucker conditions discussed in Chapter 14 can be applied to LP problems, and imply complementary slackness. This is important in linking the results of Chapters 17 and 14. Finally, there is a brief discussion of what happens when some constraints in the primal must hold with equality. Then the signs of the corresponding dual variables are no longer constrained to be nonnegative.

17.1

− + =

+ = 5.

2. (a) A graph shows that the solution is at the intersection of the lines 2x1 3x2 6 and x 1 x2 Hence max = 98/5 for (x1 , x2 ) ( 9/5, 16/5). (b) The solution satisfies 2 x1 3x2 13 and x 1 x2 6. Hence max = 49 for (x1 , x2 ) ( 5, 1) (c) The solution satisfies x 1 3x2 0 and x 2 2. Hence max = 10/3 for (x 1 , x2 ) ( 2, 2/3)

= + = − =

+ = =

−

=

=

4. (a) No maximum exists. Consider Fig. M17.1.4. By increasing c , the dashed level curve x1

+ x2 = c

moves to the north-east and so this function can take arbitrarily large values. (b) Yes, the maximum is at P (1, 0). The level curves are the same as in (a), but the direction of increase is reversed. x2

=

−x1 + x2 = −1

4 x1

3

+ x2 = c

−x1 + 3x2 = 3

2 1

−1

1

2

3

4

6

5

x1

−1

Figure M17.1.4

 

3x

+ 5y ≤ 3900 6. The LP problem is: max 700 x + 1000y subject to x ≥ 0 , y ≥ 0 x + 3y ≤ 2100 2x + 2y ≤ 2200 A figure showing the admissible set and an appropriate level line for the criterion function will show that the solution is at the point where the two lines 3 x + 5y = 3900 and 2x + 2y = 2200 intersect. Solving these equations yields x = 800 and y = 300. The firm should produce 800 sets of type A and 300 of type B .

17.2 2. min 8u1

+ 13u2 + 6u3 subject to

17.3





u1 2u1

+ 2u2 + u3 ≥ 8 + 3u2 + u3 ≥ 9

≥ 0, u2 ≥ 0, u3 ≥ 0

u1

+ 25x2 ≤ 10 000 x1 ≥ 0 , x2 ≥ 0 + 25x2 ≤ 8000 The solution can be found graphically. It is x 1∗ = 0, x 2∗ = 320, and the value of the criterion function is + 500x2 subject to

2. max 300x1

10x1 20x1

160 000, the same value found in Example 17.1.2 for the optimal value of the primal criterion function.


C H A P T ER 1 7


81

17.5 2. (a) Figure M17.5.2 shows the relevant portion of the feasible set with the different constraints numbered, as well as two dashed level curves for the criterion function Z y 1 2y2 . We see that the minimum is attained at the point (y 1∗ , y2∗ ) ( 3, 2).

= +

=

y2

(4) 10 (3) 5

(2)

y1

+ 2y2 = Z0

(3, 2)

(1) 10

5

y1

15

Figure M17.5.2

+ x2 − x3 + x4 ≤ 1 , x ≥ 0, j = 1, . . . , 4. 6x1 + x2 + x3 − 2x4 ≤ 2 ∗ ∗ Because y1 and y2 are both positive, both constraints are satisfied with equality. Furthermore, the two last constraints in the primal problem are satisfied with strict inequality at the optimum. Hence x3∗ = x 4∗ = 0. The system therefore reduces to the system x 1∗ + x2∗ = 1, 6x1∗ + x2∗ = 2, so x 1∗ = 1/5 and x 2∗ = 4/5. The maximum is thus at (x1∗ , x2∗ , x3∗ , x4∗ ) = ( 1/5, 4/5, 0, 0). (c) If the constraint is changed to y 1 + 6y2 ≥ 15 .1, the solution to the primal is still at the intersection of (b) The dual: max 15x1

+ 5x2 − 5x3 − 20x4 s.t.

x1

j

the lines (1) and (2) in Fig. M17.5.2, but with (1) shifted up slightly. The solution to the dual is completely unchanged. The optimal values in both problems increase by (15.1 15) x1∗ 0 .02.

−

· = 4. (a) For x 3 = 0, the solution is x 1 = x 2 = 1 /3. For x 3 = 3, the solution is x 1 = 1 and x 2 = 2. (b) Let zmax denote the maximum value of the objective function. If 0 ≤ x3 ≤ 7/3, then zmax (x3 ) = 2x3 + 5/3 for x 1 = 1 /3 and x 2 = x 3 + 1/3. If 7/3 < x3 ≤ 5, then z max (x3 ) = x3 + 4 for x 1 = x 3 − 2 and x 2 = 5 − x3 . If x 3 > 5, then z max (x3 ) = 9 for x 1 = 3 and x 2 = 0. Because z max (x3 ) is increasing, the maximum is 9 for x 3 ≥ 5. (c) The solution to the original problem is x 1 = 3 and x 2 = 0, with x 3 as an arbitrary number ≥ 5. Review Problems for Chapter 17

 − + ≤ − ≤  −− − − ≤ ≤ −−

x1 2x2 16 x1 2x2 6 2. (a) max x1 x2 s.t. x1 0 , x2 0 2x1 x2 8 4x1 5x2 15 The maximum value is 8 at (x1 , x2 ) ( 0, 8). (b) (y 1 , y2 , y3 , y4 ) ( 12 (b 1), 0, b, 0) for any b satisfying 0 b 1 /5. (c) The maximand for the dual becomes kx1 x2 . The solution is unchanged provided that k

− +

=

+


=

+

≥

≥

≤ ≤

≤ −1/2.

82 TEST I ( Elementary Algebra)

A certain familiarity with elementary algebra is an essential prerequisite for reading the textbook (and for understanding most modern economics texts). This test is designed for students and instructors to discover whether the students have the proper background. (In a number of countries, many beginning economics students’ background in elementary algebra appears to have become much weaker during the last few years. In fact, lecturers using this test (or similar ones) have been shocked by the results, and have had to readjust their courses.) At the head of each problem, immediately after the number, the relevant sections of the introductory chapters in the book are given in parentheses, followed in square brackets by the number of points for a correct answer to each separate part of the problem. In a 20–30 minute test, any student who scores less than 50 (out of 100) has serious problems with elementary algebra. Such students definitely need to review the relevant section of Chapters 1 and 2, or consult other elementary material. The correct answers are given on a separate page following the test.

1

(1.2) [Points: (a) 2; (b) 2; (c) 3; (d) 3] Calculate/simplify: 73 72 (a) 74

·

2

(b) (5.5

3

− 3.5)

(c)

 −  − −  2 5

2 5

2 5

219 (d) 19 2

− 217 + 217

(1.2–1.4) [Points: (a) 2; (b) 2; (c) 2; (d) 4]

√ = 5, then 4x 4y 2 =? √ (b)√ 11 % of 3500 is? (c) 132 − 122 =? 3+ 2 (d) Rationalize the denominator: √ √ (i.e., find a new fraction that is equal but has no square root 3− 2 (a) If 2x 2 y

in the denominator).

3

(1.3) [Points: (a) 2; (b) 2; (c) 2] Expand:

+ 2y)2

(a) (x

4

− 3y)2

(b) (2x

(c) (a

+ b)(a − b)

(1.3) [Points: (a) 2; (b) 3; (c) 3; (d) 4] Expand and simplify: (a) 5a (3a 2b) 2(a (c) (1 x) 2 (1 x) 2

− −

+ − − 3b) +

+ 2)2 + (x − 2)2 − 2(x + 2)(x − 2) − a)3

(b) (x (d) (2

5

(1.2) [Points: 4] If the GNP of a certain country in 2000 was 8 billion dollars, write down an expression for the GNP 6 years later if it increases by 5% each year.

6

(1.3) [Points: (a) 3; (b) 3; (c) 4] Factorize: (a) 5a 2 b

+ 15ab 2

(b) 9

− z2


(c) p 3 q

− 4p2 q 2 + 4pq 3

83 7

(1.4) [Points: (a) 2; (b) 2; (c) 2] Expand and simplify to a single fraction:

(a)

8

1 2

− 13

1

1 − 4 6

(c)

√ 3

27a 6

(d) p 1/5 (p4/5

− p −1 5 ) /

√ − =

− = +

(1.6) [Points: (a) 2; (b) 3; (c) 3] Solve the following inequalities: x 1 (a) 3x 2 < 5 (b) 0 (c) x 3 < x x 3

− ≤ +

(2.3) [Points: (a) 3; (b) 3; (c) 3] Solve the following equations:

− 9x 2 = 0

(a) 3x

12

(c)

(2.1) [Points: (a) 2; (b) 2; (c) 2] Solve the following equations for the unknown x : 3 1 3 (a) x 6 (b) (c) 3 x 2 x 1 5 2x 3

− +

11

− 10a + 320a

(b) (x 1/2 y −1/4 )4

= −

10

6a 5

− 13

(1.5) [Points: (a) 2; (b) 2; (c) 2; (d) 2] Calculate/simplify: (a) 251/2

9

(b)

1 2

(b) x 2

− 2x − 15 = 0

(c) 2P 2

= 2 − 3P

(2.4) [Points: (a) 3; (b) 4; (c) 4] Solve the following systems of equations: 3 3 3 p q 2x y 5 1.5p 0.5q 14 (a) (b) (c) 3 1 x 2y 5 2.5p 1.5q 28 7

− = + =

− +


= =

+ =

p

−q =

84 TEST II (Elementary Mathematics)

Students who now enter university or college courses in economics tend to have a wide range of mathematical backgrounds and aptitudes. At the low end, they may have no more than a shaky command of elementary algebra. Or, at the high end, they may already have a ready facility with calculus, though often it is some years since economics students took their last formal mathematics course. Experience suggests therefore that right from the start of the course, it is very important that the instructor, as well as each individual student, should get some impression of what the student knows well, what is vaguely familiar, and what seems to be more or less forgotten or perhaps never learned at all. The present test is meant to test the students’ actual knowledge in some elementary mathematical topics of interest to economists. The level is nevertheless more advanced than for Test I and the topics covered are discussed in the earlier chapters of the main text. The maximum total score is 100. 1

[Points: (a) 2; (b) 2; (c) 2; (d) 2] (a) 25

+ 25 = 2 ⇒ x = 132 − 52 =



(c) 2

(b) 3−15

x

(d)

+ 3−15 + 3−15 = 3 ⇒ y = 226 − 223 z = ⇒ z = 226 + 223 9 y

[Points: (a) 1+1; (b) 2+2] (a) Find the slopes of the following straight lines: (i)

= − 32 x + 4

y

(ii) 6x

− 3y = 5

(b) Find the equation of the straight line that: (i) passes through ( 2, 3) and has slope 2. (ii) passes through both (a, 0) and (0, b).

−

3

−

[Points: 5] Fill in the following table and sketch the graph of y

−2

x

−1

0

= −x 2 + 2x + 4.

1

2

3

2

= −x + 2x + 4

y

4

[Points: 4+4] Determine maximum/minimum points for:

= x 2 − 4x + 8

(a) y 5

= −2x 2 + 16x − 14

(b) y

[Points: 4+4] Perform the following polynomial divisions: (a) (2x 3

6 7

− 3x + 10) ÷ (x + 2) (b) (x4 + x) ÷ (x2 − 1) [Points: 5] f(x) = 43 x 3 − 15 x 5 . For what values of x is f  (x) = 0? [Points: 3+3+3+3] Sketch the graph of a function f in each of the following cases: (a) f  (x) > 0 and f  (x) > 0, (b) f  (x) > 0 and f  (x) < 0 (c) f  (x) < 0 and f  (x) > 0, (d) f  (x) < 0 and f  (x) < 0 © Knut Sydsæter and Peter Hammond 2006

4

85 8

[Points: 3+3] (a) The cost in dollars of extracting T tons of a mineral ore is given by C interpretation of the statement that f  (1000) 50.

=

= f (T ) . Give an economic

(b) A consumer wants to buy a certain item at the lowest possible price. Let P (t) denote the lowest price found after searching the market for t hours. What are the likely signs of P  (t) and P  (t)? 9

[Points: 2+2+2+2] Write down the general rules for differentiating the following:

= f (x) + g(x) ⇒ y  = (c) y = f (x)/g(x) ⇒ y  = (a) y

10

[Points: 2+2+2+2+2+2+2+2] Differentiate the following functions:

= x 2 (e) y = e (a) y

x

11

= x 5 /5 (f) y = ln x (b) y

= x +x 1 (g) y = 2 (c) y

x

= (x 2 + 5)6 (h) y = x

(d) y

x

[Points: 2+2+2+2+2] Which of the following statements are correct? (a) The rule which converts a temperature measured in degrees Fahrenheit into the same temperature measured in degrees Celsius is an invertible function. (b) A concave function always has a maximum. (c) A differentiable function can only have an interior maximum at a stationary point for the function. (d) If f  (a) 0, then a is either a local maximum point or a local minimum point. (e) The conditions f  (a) 0 and f  (a) < 0 are necessary and sufficient for a to be a local maximum point for f .

=

12

= f (x)g(x) ⇒ y  = (d) y = f (g(x)) ⇒ y  = (b) y

=

[Points: 2+2+2+2] In each of the following cases, decide whether the given formula is correct or not: (a) (b) (c) (d)

   

= 13 x 3 + C [f(x) + g(x)] dx = f (x ) d x + g(x)dx f(x)g(x)dx = f (x ) d x g(x)dx x d x = b − a x 2 dx

b a

   


86 Answers to TEST I

73 72 1. (a) 74 2 (c) 5

· = 73+2 = 75 = 75−4 = 71 = 7 (b) (5.5 − 3.5)3 = 23 = 8 74 74 − −2 −2 = −8 (= −0.064) (d) 219 − 217 = 217(22 − 1) = 3 5 5 125 219 + 217 217 (22 + 1) 5 = 3500 · 0.11 =√ 385 2. (a) 4√ x 4 y 2 = ( 2x 2 y) 2 = 25 (b) 11 % of 3500 is 3500 · 11/100√ √ √ + 12)(13 − 12) = 25 = 5 (or 132 − 122 = 169 − 144, etc.) (c)√ 132 − 122 = (13√ √ 2 = 13 − 12 = 1” is a SERIOUS mistake.) (“ √ 132 − √ 122 = √ 132 −√ 12√ √ √ √ √ 3+ 2 ( 3 + 2)( 3 + 2) 3+2 3 2+2 = (d) √ √ = √ √ √ √ = 5 + 2 6 3−2 3− 2 ( 3 − 2)( 3 + 2) 3. (a) (x + 2y) 2 = x 2 + 4xy + 4y 2 (b) (2x − 3y) 2 = 4 x 2 − 12xy + 9y 2 (c) (a + b)(a − b) = a 2 − b2 4. (a) 5a − (3a + 2b) − 2(a − 3b) = 5 a − 3a − 2b − 2a + 6b = 4 b (b) (x + 2)2 + (x − 2)2 − 2(x + 2)(x − 2) = [ (x + 2) − (x − 2)]2 = 4 2 = 16 (c) (1 − x) 2 (1 + x) 2 = [ (1 − x)( 1 + x) ]2 = ( 1 − x 2 )2 = 1 − 2x 2 + x 4 (d) (2 − a) 3 = ( 2 − a) 2 (2 − a) = (4 − 4a + a 2 )(2 − a) = 8 − 4a − 8a + 4a 2 + 2a 2 − a 3 = −a 3 + 6a 2 − 12a + 8

   

5. 8(1.05)6 billion dollars. 6. (a) 5a 2 b 15ab 2 5 ab(a 3b) (b) 9 (c) p3 q 4p 2 q 2 4pq 3 pq(p2 4pq

+ = + − z2 = (3 − z)(3 + z) − + = − + 4q 2 ) = pq(p − 2q)2 . (One point for the first equality.) 1 1 3 2 3 2 1 − = − = 7. (a) − = (b) 20 is the common denominator, so 2 3 2·3 3·2 6 6 6 6a 4 · 6a 24a − 2a + 3a a 3a 2a 3a 25a 5a − + = − + = = = 5 10 20 20 20 20 20 20 4 1 1 6 4 2 − − 3 12 12 12 = = = (c) 21 2 (or 21 − 13 = 2 ( 14 − 16 ), so the ratio is 2). 1 3 2 1 − 6 12 − 12 12 4 √ √ √ 8. (a) 251 2 = 5 (b) (x 1 2 y −1 4 )4 = x 1 2 ·4 y −1 4 ·4 = x 2 y −1 (c) 27a 6 = 27 a 6 = 3 a 2 (d) p 1 5 (p4 5 − p −1 5 ) = p 1 5 p4 5 − p 1 5 p−1 5 = p 1 5+4 5 − p1 5−1 5 = p − 1 9. (a) 3x = −30, so x = −10 (b) 2x + 3 = 3 (x − 1), so x = 6. (Neither denominator is 0 when x = 6.) √ (c) If 3 − x = 2, then 3 − x = 4, so x = −1. This is indeed a solution, as is easily checked. /

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3

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3

3

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−3x + 2 < 5 or −3x < 3, so that x > −1. (Recall that an inequality is reversed if multiplied

10. (a)

by a negative number.) (b) 3 < x 1. (Use a sign diagram. Note that the fraction is undefined if x 3.) 3 3 2 (c) x < 1 or 0 < x < 1. (x < x , or x x < 0, and so x (x 1) < 0, or x (x 1)(x Then use a sign diagram.)

−

−

≤

−

−

= − −

+ 1) < 0.

− 3x) = 0, so x = 0 or x = 1/3 (b) x = −3, x = 5 (c) P = −2 or P = 1/2 12. (a) x = 3, y = 1 (b) p = 10, q = 2 (c) Put x = 1/p, y = 1/q . Then 3x + 3 y = 3 and 3x − y = 7. The solution to this system is x = 2 and y = −1. But then p = 1 /2 and q = −1. 11. (a) 3x( 1


87 Answers to TEST II 1. (a) 25

+ 25 = 2 · 25 = 26 , so x = 6. (b) 3−15 + 3−15 + 3−15 = 3 · 3−15 = 3 −14 , so y = −14. √ √ √ (c) 132 − 52 = 169 − 25 = 144 = 12. 226 − 223 223 (23 − 1) 23 − 1 7 = = = (d) 26 , so z = 7. 2 + 223 223 (23 + 1) 23 + 1 9 2. (a) (i) −3/2 (ii) 2 (b) (i) y = −2x − 1 (ii) y = ( −b/a)x + b 3.

−2 −4

x 2

= −x + 2x + 4

y

−1

0

1

2

3

4

1

4

5

4

1

−4

The graph of the function is shown in the figure below. y

5 4

= −x2 + 2x + 4

y

3 2 1

−4 −3 −2 −1−1 −2 −3 −4

1

2

3

4

x

= 2. (Follows from x 2 − 4x + 8 = x 2 − 4x + 22 + 8 − 22 = (x − 2)2 + 4, or by using calculus.) (b) Maximum 18 for x = 4. (Follows from −2x 2 + 16x − 14 = −2(x 2 − 8x + 7) = −2(x 2 − 8x + 42 + 7 − 42 ) = −2[(x − 4)2 − 9] = −2(x − 4)2 + 18, or by using calculus.) 2x 3 − 3x + 10 5. (a) = 2 x 2 − 4x + 5 x + 2 4 x +x x + 1 1 = = x 2 + 1 + 2 x 2 + 1 + (b) 2 x −1 x −1 x − 1 6. f  (x) = 4 x 2 − x 4 = x 2 (4 − x 2 ) = x 2 (2 − x)( 2 + x) = 0 for x = 0 and for x = ±2. 4. (a) Minimum 4 for x

7. y

y

y

y

= f (x)

= f (x)

y y

= f (x)

y

= f (x)

x

(a)

y

x

(b)


x

(c)

x

(d)

All Solutions 1078

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