Basic mathematics for economists

Basic Mathematics for Economists Economics students will welcome the new edition of this excellent textbook Given that many students come into economics courses without having studied mathematics for a number of years, this clearly written book will help to develop quantitative skills in even the least numerate student up to the required level for a general Economics or Business Studies course All explanations of mathematical concepts are set out in the context of applications in economics This new edition incorporates several new features, including new sections on: • • • financial mathematics continuous growth matrix algebra Improved pedagogical features, such as learning objectives and end of chapter questions, along with an overall example-led format and the use of Microsoft Excel for relevant applications mean that this textbook will continue to be a popular choice for both students and their lecturers Mike Rosser is Principal Lecturer in Economics in the Business School at Coventry University © 1993, 2003 Mike Rosser Basic Mathematics for Economists Second Edition Mike Rosser © 1993, 2003 Mike Rosser First edition published 1993 by Routledge This edition published 2003 by Routledge 11 New Fetter Lane, London EC4P 4EE Simultaneously published in the USA and Canada by Routledge 29 West 35th Street, New York, NY 10001 Routledge is an imprint of the Taylor & Francis Group This edition published in the Taylor & Francis e-Library, 2003 © 1993, 2003 Mike Rosser All rights reserved No part of this book may be reprinted or reproduced or utilised in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying and recording, or in any information storage or retrieval system, without permission in writing from the publishers British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Cataloging in Publication Data A catalog record for this book has been requested ISBN 0-203-42263-5 Master e-book ISBN ISBN 0-203-42439-5 (Adobe eReader Format) ISBN 0–415–26783–8 (hbk) ISBN 0–415–26784–6 (pbk) © 1993, 2003 Mike Rosser Contents Preface Preface to Second Edition Acknowledgements Introduction 1.1 Why study mathematics? 1.2 Calculators and computers 1.3 Using the book Arithmetic 2.1 Revision of basic concepts 2.2 Multiple operations 2.3 Brackets 2.4 Fractions 2.5 Elasticity of demand 2.6 Decimals 2.7 Negative numbers 2.8 Powers 2.9 Roots and fractional powers 2.10 Logarithms Introduction to algebra 3.1 Representation 3.2 Evaluation 3.3 Simplification: addition and subtraction 3.4 Simplification: multiplication 3.5 Simplification: factorizing 3.6 Simplification: division 3.7 Solving simple equations 3.8 The summation sign 3.9 Inequality signs © 1993, 2003 Mike Rosser Graphs and functions 4.1 Functions 4.2 Inverse functions 4.3 Graphs of linear functions 4.4 Fitting linear functions 4.5 Slope 4.6 Budget constraints 4.7 Non-linear functions 4.8 Composite functions 4.9 Using Excel to plot functions 4.10 Functions with two independent variables 4.11 Summing functions horizontally Linear equations 5.1 Simultaneous linear equation systems 5.2 Solving simultaneous linear equations 5.3 Graphical solution 5.4 Equating to same variable 5.5 Substitution 5.6 Row operations 5.7 More than two unknowns 5.8 Which method? 5.9 Comparative statics and the reduced form of an economic model 5.10 Price discrimination 5.11 Multiplant monopoly Appendix: linear programming Quadratic equations 6.1 Solving quadratic equations 6.2 Graphical solution 6.3 Factorization 6.4 The quadratic formula 6.5 Quadratic simultaneous equations 6.6 Polynomials Financial mathematics: series, time and investment 7.1 Discrete and continuous growth 7.2 Interest 7.3 Part year investment and the annual equivalent rate 7.4 Time periods, initial amounts and interest rates 7.5 Investment appraisal: net present value 7.6 The internal rate of return 7.7 Geometric series and annuities © 1993, 2003 Mike Rosser 7.8 7.9 7.10 Perpetual annuities Loan repayments Other applications of growth and decline Introduction to calculus 8.1 The differential calculus 8.2 Rules for differentiation 8.3 Marginal revenue and total revenue 8.4 Marginal cost and total cost 8.5 Profit maximization 8.6 Respecifying functions 8.7 Point elasticity of demand 8.8 Tax yield 8.9 The Keynesian multiplier Unconstrained optimization 9.1 First-order conditions for a maximum 9.2 Second-order condition for a maximum 9.3 Second-order condition for a minimum 9.4 Summary of second-order conditions 9.5 Profit maximization 9.6 Inventory control 9.7 Comparative static effects of taxes 10 Partial differentiation 10.1 Partial differentiation and the marginal product 10.2 Further applications of partial differentiation 10.3 Second-order partial derivatives 10.4 Unconstrained optimization: functions with two variables 10.5 Total differentials and total derivatives 11 Constrained optimization 11.1 Constrained optimization and resource allocation 11.2 Constrained optimization by substitution 11.3 The Lagrange multiplier: constrained maximization with two variables 11.4 The Lagrange multiplier: second-order conditions 11.5 Constrained minimization using the Lagrange multiplier 11.6 Constrained optimization with more than two variables 12 Further topics in calculus 12.1 Overview 12.2 The chain rule 12.3 The product rule 12.4 The quotient rule © 1993, 2003 Mike Rosser 12.5 12.6 12.7 Individual labour supply Integration Definite integrals 13 Dynamics and difference equations 13.1 Dynamic economic analysis 13.2 The cobweb: iterative solutions 13.3 The cobweb: difference equation solutions 13.4 The lagged Keynesian macroeconomic model 13.5 Duopoly price adjustment 14 Exponential functions, continuous growth and differential equations 14.1 Continuous growth and the exponential function 14.2 Accumulated final values after continuous growth 14.3 Continuous growth rates and initial amounts 14.4 Natural logarithms 14.5 Differentiation of logarithmic functions 14.6 Continuous time and differential equations 14.7 Solution of homogeneous differential equations 14.8 Solution of non-homogeneous differential equations 14.9 Continuous adjustment of market price 14.10 Continuous adjustment in a Keynesian macroeconomic model 15 Matrix algebra 15.1 Introduction to matrices and vectors 15.2 Basic principles of matrix multiplication 15.3 Matrix multiplication – the general case 15.4 The matrix inverse and the solution of simultaneous equations 15.5 Determinants 15.6 Minors, cofactors and the Laplace expansion 15.7 The transpose matrix, the cofactor matrix, the adjoint and the matrix inverse formula 15.8 Application of the matrix inverse to the solution of linear simultaneous equations 15.9 Cramer’s rule 15.10 Second-order conditions and the Hessian matrix 15.11 Constrained optimization and the bordered Hessian Answers Symbols and terminology © 1993, 2003 Mike Rosser Preface Over half of the students who enrol on economics degree courses have not studied mathematics beyond GCSE or an equivalent level These include many mature students whose last encounter with algebra, or any other mathematics beyond basic arithmetic, is now a dim and distant memory It is mainly for these students that this book is intended It aims to develop their mathematical ability up to the level required for a general economics degree course (i.e one not specializing in mathematical economics) or for a modular degree course in economics and related subjects, such as business studies To achieve this aim it has several objectives First, it provides a revision of arithmetical and algebraic methods that students probably studied at school but have now largely forgotten It is a misconception to assume that, just because a GCSE mathematics syllabus includes certain topics, students who passed examinations on that syllabus two or more years ago are all still familiar with the material They usually require some revision exercises to jog their memories and to get into the habit of using the different mathematical techniques again The first few chapters are mainly devoted to this revision, set out where possible in the context of applications in economics Second, this book introduces mathematical techniques that will be new to most students through examples of their application to economic concepts It also tries to get students tackling problems in economics using these techniques as soon as possible so that they can see how useful they are Students are not required to work through unnecessary proofs, or wrestle with complicated special cases that they are unlikely ever to encounter again For example, when covering the topic of calculus, some other textbooks require students to plough through abstract theoretical applications of the technique of differentiation to every conceivable type of function and special case before any mention of its uses in economics is made In this book, however, we introduce the basic concept of differentiation followed by examples of economic applications in Chapter Further developments of the topic, such as the second-order conditions for optimization, partial differentiation, and the rules for differentiation of composite functions, are then gradually brought in over the next few chapters, again in the context of economics application Third, this book tries to cover those mathematical techniques that will be relevant to students’ economics degree programmes Most applications are in the field of microeconomics, rather than macroeconomics, given the increased emphasis on business economics within many degree courses In particular, Chapter concentrates on a number of mathematical techniques that are relevant to finance and investment decision-making Given that most students now have access to computing facilities, ways of using a spreadsheet package to solve certain problems that are extremely difficult or time-consuming to solve manually are also explained © 1993, 2003 Mike Rosser Solution The Lagrange objective function will be G = Q(x, y, z) + λ(5000 − 8x − 12y − 6z) As there are three variables in the objective function and HB is × then the secondorder conditions for a maximum require that the determinant of the bordered Hessian of second-order partial derivatives |HB | < Therefore Qxx Qyx |HB | = Qzx −8 Qxy Qyy Qzy −12 Qxz Qyz Qzz −6 −8 −12 Thus, when the first row and column have been eliminated from HB , this problem also requires that Qyy Qzy −12 Qyz Qzz −6 −12 −6 > 0 Constrained optimization with any number of variables and constraints All the constrained optimization problems that you will encounter in this text have only one constraint and usually not have more than three variables in the objective function However, it is possible to set up more complex Lagrange functions with many variables and more than one constraint Second-order conditions requirements for optimization for the general case with m variables in the objective function and r constraints are that the naturally ordered border preserving principal minors of dimension m of HB must have the sign (−1)m−r for a maximum r for a minimum (−1) ‘Border preserving’ means not eliminating the borders added to the basic Hessian, i.e the last column and the bottom row, which typically show the prices of the variables These requirements only apply to the principal minors of order ≥ (1 + 2r) For example, if the problem was to maximize a utility function U = U(X1 , X2 , X3 ) subject to the budget constraint M = P1 X1 + P2 X2 + P3 X3 then, as there is only one constraint, r = Therefore we would just need to consider the principal minors of order greater than three since (1 + 2r) = (1 + 2) = As the full-bordered Hessian in this example with three variables is 4th order then only HB itself plus the first principal minor need be considered, as this is the only principal minor with order equal to or greater than © 1993, 2003 Mike Rosser The second-order conditions will therefore require that for a maximum: For the full bordered Hessian m = and so |HB | must have the sign (−1)m−r = (−1)4−1 = (−1)3 = −1 < and the determinant of the 3rd order naturally ordered principal minor of |HB | must have the sign (−1)m−r = (−1)3−1 = (−1)2 = +1 > These are the same as the basic rules for the three variable case stated earlier The last set of problems, below, just require you to use the bordered Hessian to check that second-order conditions for optimization are met in some numerical examples with a small number of variables to familiarize you with the method Those students who go on to study further mathematical economics will find that this method will be extremely useful in more complex constrained optimization problems Test Yourself, Exercise 15.11 A firm has the production function Q = K 0.5 L0.5 and buys input K at £12 a unit and input L at £3 a unit and has a budget of £600 Use the Lagrange method to find the input combination that will maximize output, checking that second-order conditions are satisfied by using the bordered Hessian A firm operates with the production function Q = 25K 0.5 L0.4 and buys input K at £20 a unit and input L at £8 a unit Use the Lagrange method to find the input combination that will minimize the cost of producing 400 units of Q, using the bordered Hessian to check that second-order conditions are satisfied A consumer has the utility function U = 20X 0.5 Y 0.4 and buys good X at £10 a unit and good Y at £2 a unit If their budget constraint is £450, what combination of X and Y will maximize utility? Check that second-order conditions are satisfied by using the bordered Hessian A consumer has the utility function U = 4X 0.75 Y 0.25 and can buy good X at £12 a unit and good Y at £2 a unit Find the combination of X and Y that they should purchase to minimize the cost of achieving a utility level of 580 and check that second-order conditions are met using the bordered Hessian matrix © 1993, 2003 Mike Rosser Answers Chapter 2.1 3,555 865 92,920 23 2.2 919 440 225 101 164 627 2.3 840 £13,800 17 598 172 £176 122 2.4 73 168 13 21 39 18 40 2.5 2.6 2.7 2.8 (a) 101 252 37 60 10 (b) 4 19 30 12 (d) 14 (c) 12 13 (e) 11 (a) (b) 1 15, , , , , , , 11 13 15 36.914 751.4 435.1096 0.09675 610 140 (a) 0.1 (b) 0.001 (c) 0.000001 (a) 0.452 (b) 2.431 (c) 0.075 (d) 0.002 −2 −24 −33 0.45 −117 −157 140 −330 17 10 16 3600 0.25 123 64 11.641754 531,441 0.0015328 36 −618.47021 10 25.000655 © 1993, 2003 Mike Rosser 36,082 0.35 2.9 2.10 ±25 2 96 10 5.2780316 0.03423 10 87.977857 270,818.98 1.2683 5.1331868 220.9478 16,552,877 2.8563 × 109 93.696376 0.2 2.4494897 1.5728 × 108 4.38228 Chapter 3.1 3.2 3.3 (a) 0.01x (b) 0.5x (c) 0.5x 0.01rx + 0.5wx + 0.5mx xp x (b) (a) 0.1x kg (b) 0.3x kg (c) x(0.1m + 0.3p) (a) 12 12 0.5w + 0.25 Own example 10.5x + 6y 3q − 6000 456 77.312 r + z, 9% Own example 1.094 £465.58 £2,100 (a) 99 + 0.78M 30x + 24x − 18y + 7xy − 12 (b) £2,166 6x + 5y − 650 9H − 120 3.4 6x − 24x x + 4x + 42x − 16y − 34xy + 6y 2x + 6x + xy + 3y 33x + 2y − 20y + 62xy − 21 120 + 2x + 54y + 40z − x + 6y + xy + 4xz + 8yz 200q − 2q 13x + 11y 8x + 60x + 76 10 4,000 + 150x 3.5 (x − 3y)2 (x + 4)2 Does not factorize Does not factorize Own example 3.6 3x + − 20x −1 3.8 11 40p £3,062.50 14 10 m 11 26 82 Hi , 173.7 cm 35 20 p 33% 60 i=1 n 6,000(0.9)i−1 ,16,260 tonnes (a) i=1 3.9 Own example n n (b) 200x −1 + 21 4y + x + 12 2(x + 3) + − x − − x + = 179x 3.7 x + n n Ri , £4,425 i=1 n−1 Ri , £4,933 13.25%, 8.2 n−3 (a) ≤ (b) < (a) Q1 < Q2 © 1993, 2003 Mike Rosser (c) ≥ (d) > (b) Q1 = Q2 (a) > (b) ≥ (c) Q1 > Q2 (c) > (d) > P2 > P1 Chapter 4.1 (a) Quantity demanded depends on the price of tea, average exp., etc (b) Qt dependent, all others independent (c) Qt = 99 − 6Pt − 0.5Yt + 0.8A + 1.2N + 1.4P c (suggested number assumes tea is an inferior good) (a) 202 (b) (c) 6, x ≥ 4.2 ◦ F = 32 + 1.8◦ C Yes; no P = 2,400 − 2Q It is not monotonic, e.g TR = 200 when q = or 10 T = (0.0625X − 25)2 ; no 4.3 Own example (Answers to to give intercepts on axes) x = 13 , y = −40 x = −12, y = P = 150, Q = 750 P = 60, Q = 300 K = 24, L = 40 Goes through origin only Goes through (Q = 0, TC = 200) and (Q = 10, TC = 250) Horizontal line at TFC = 75 Own example 10 (a) and (d); both slope upwards and have positive intercepts on P axis 4.4 Q = 90 − 5P ; 50; Q ≥ 0, P ≥ 4.5 3.75, 0.75, 0.375, −0.75 (c) (i) (a) (ii) (b) £6,440 P = 12, Q = 40; £4.50; 10 (a), (d) (a) 0.263 (b) 0.714 (c)1.667 4.6 C = 30 + 0.75Y P = 12 − 0.015Q By £20 to £100 (a) 2/3 (b) APC = 400Y −1 + 0.5 > 0.5 = MPC Own example −1.5 (a) becomes −1 (b) becomes −1 (c) no change (d) no change K = 100, L = 160, PK = £8, PL = £5 Cost £520 > budget; PL reduced by £10 to £30 (a) −10 (b) −1 (c) −0.1 (d) −0.025 (e) No change Height £120, base 12, slope −10 = −(wage) 4.7 Sketch graphs Like y = x 4.8 −1 Sketch graphs ; £260 Sketch graphs (a) 40 = 3250q Own example Steeper Own example π = 50x − 100 − 0.4x ; inverted U Own example −1 (b) Original firms’ π per unit = £27.50 but new firms’ AC = £170 > price 4.9 Plot Excel graphs 4.10 (a) 16L−1 (b) 0.16 (a) 57,243.34L−15 (a) 322.54L −1 (c) constant (b) 57.243 (b) 3.2254 (a) 3,125L−1.25 (b) 9.882 © 1993, 2003 Mike Rosser (c) constant (c) increasing (c) increasing (a) 23,415,916L−1.6667 −1.7714 (a) 4,093.062L 4.11 (b) 10,868.71 (c) decreasing (b) 1.173 (c) decreasing MR = 33.33 − 0.00667Q for Q ≥ 500 MR = 76 − 0.222Q for Q ≥ 22.5 MR = 80 − 0.555Q for Q ≥ 562.5 MC = 30 + 0.0714Q for Q ≥ 56 MC = 56 + 0.1333Q for Q ≥ 30 MC = + 0.0714Q for Q ≥ 59 Chapter 5.1 q = 40, p = 5.2 q = 118, p = 256 Own example x = 67, y = 17 (approximately) No solution exists (a) q = 80, p = 370 (b) q falls to 78, p rises to 376 x = 2.102, y = 62.25 (a) 40 (b) rises to 50 5.3 A = 24, B = 12 x = 190, y = 60 5.4 x = 30, y = 60 5.5 x = 24, y = 14.4, z = 19.2 200 A = 6, B = 36 A = 6, B = 22, C = x = 25, y = 20 x = 4, y = 6, z = 4 x = 17, y = 4, z = A = 82.5, B = 35, C = 6, D = 5.6 q = 500, p = 275 K = 17.5, L = 16, R = 10 (a) p rises from £8 to £10 (b) p rises to £9 Y = £3,750 m; government deficit £150 m Y = £1,625 m; balance of payments deficit £15 m L = 80, w = 52 5.7 p = 184 + 0.2a, q = 43.2 + 0.06a, p = 216, q = 52.8 p = 84 + 0.2t, q = 32 − 0.4t, p = 85, q = 30 p = 122.4 + 0.2t, q = 13.8 − 0.1t, p = 123.4, q = 13.3 (a) Y = 100/(0.25 + 0.75t), Y = 250 (b) Y = 110/(0.25 + 0.75t), Y = 275 p = (4200 + 3800v)/(9 + 5v), q = (750 − 50v)/(9 + 5v) p = 494.30, q = 76.94 5.8 q1 = 60, q2 = 80, p1 = £10, p2 = £8 q1 = 40, q2 = 50, p1 = £6, p2 = £4 p1 = £8.75, q1 = 60, p2 = £6.10, q2 = 550 © 1993, 2003 Mike Rosser £81 for extra 65 units £7.50 for extra 25 units q1 = 48, q2 = 39, p1 = £12, p2 = £8.87 (a) 190 units (b) £175 for extra 75 units 5.9 q1 = 180, q2 = 200, p = £39 q1 = 1,728, q2 = 780, p = £190.70 q1 = 1,510, q2 = 1,540, qA = 800, qB = 2,250, PA = £500, PB = £625 q1 = 160, q2 = 600, qA = 293 13 , qB = 266 23 , qC = 200, PA = £95, PB = £80, Pc = £60 q1 = 15.47, q2 = 27.34, q3 = 26.17, p = £14.20 5A.1 8.4 of A, 4.64 of B (tonnes); (a) no change (b) no B, 12.16 of A A = 13, B = 27 12 of A, of B of A, 32 of B Own example 22.5 of A, 7.5 of B 13.64 of A, 21.82 of B; £7092; surplus 2.72 of R, 22.72 of mix additive Produce 15 of A, 21 of B 30 of A, B = 10 Objective function parallel to first constraint 11 24,000 shares in X, 18,000 shares in Y, return £8,640 12 Own example 5A.2 C = 70 when A = 1, B = 1.5, slack in x = 30 A = 3, B = Q = 2.5, R = 1.5; excesses 62.5 mg of B, 27.5 mg of C 10 of A, of B; space for 50 extra loads of X Zero R, 15 tonnes of T; G exceeds by 45 kg 100 of A, 40 of B 5A.3 of A, of B Own example 7.5 of X and 15 of Y (tonnes) Own example Chapter 6.1 or 10 or 60 6.2 10 or −12.5 When x = 0.5 £16.353 (a) 1.01 or 98.99 (b) 11.27 or 88.73 (c) no solution exists Own example 6.3 x = 15, y = 15 or x = −3, y = 249 x = 1.75, y = 3.15 or x = −1.53, y = 20.97 q1 = 3.2, q2 = 4.8, p1 = £136, p2 = £96 p1 = £15, q1 = 80, p2 = £8.50, q2 = 70 6.4 52 1069 © 1993, 2003 Mike Rosser 10 16.4 Chapter 7.1 7.2 7.3 £4,630.50 £314.70 £17,623.16 £744.71 £40,441.40 £5,030.03 £43,747.41; 12.68% £501,159.74; 7.44% (a) APR 11.35% £2,083.61; 19.25% £625; 5.09% 19.28% 0.01467% £494,531.25; 4.5% £6,301.69 £355.89 No, A = £9,106.27 £6,851.65 (a) £9,638.58 (b) £11,579.83 (c) £13,318.15 5.27 years 11 10.7% 12 9.5% 12.1 years 5.45 years 13 7.5% 14 0.8% 10 3.42 years 15 10.3% 16 8.4% 17 (b) as PV = £5,269.85 7.4 (b) −£100.07 (a) £90.75 (f) £877.33 (g) £791.25 B, PV = £6,569.10 (c) −£474.01 (d) £622.86 (e) £1,936.87 (h) £992.16 (a) All viable (b) A best, NPV = £6,824.68 Yes, NPV = £7,433.56 Yes, NPV = £4,363.45 (a) Yes, NPV = £610.02 (b) no, NPV = −£522.30 B, NPV = £856.48 7.5 rA = 20%, rB = 41.6%, rC = 20%, rD = 20%; B consistently best, but others have same IRR with different NPV ranking (a) A, rA = 21.25%, rB = 20.42% NPVA = £2,291.34 7.6 (b) B, NPVB = £2,698.94, IRR = 16.93% (a) 2.5, 781.25, 50,857.3 (b) 3, 121.5, 14,762 (d) 0.8, 19.66, 267.8 (e) 0.75, 0.57, 9.06 (c) 1.4, 10.756, 139.6 5,741 (to nearest whole unit) A, £1,149.32; B, £2,980.91; C, £45,216.47 Yes, NPV = £3,774.71 7.7 7.8 £4,149.20 (a) k = 1.5, not convergent (b) k = 0.8, converges on 600 (c) k = −1.5, not convergent (d) k = (e) converges on 961.54 (f) not convergent converges on 54 £3,076.92 Yes, NPV = £50,000 (a) £240,000 (b) £120,000 (c) £80,000 (d) £60,000 £152.59 £197.38 £794.66 (a) 14.02% (b) 26.08% (c) 23.86% (d) 14.71% £3,500 £191.46 Loan is marginally better deal (PV of payments = £6,348.33 + £1,734 deposit = £8,082.33, less than cash price by £12.67) © 1993, 2003 Mike Rosser 7.9 6.82 years After 15.21 years 4% Yes, sum of infinite GP = 1,300 million tonnes 4.85% Chapter 8.1 36x 8.2 3x + 60 0.2x 192 −3 250 + 0.6x 8.3 120 − 6q, 20 8.4 7.5 260x 21.6 Own example −4x −2 − −0.7 Own example 25 14,400 £200 Own example 12q − 40q + 60 (a) 1.5q − 6q + 25 MC constant at 0.8 (a) 80 (b) 0.5q − 3q + 25 + 20q −1 (c) q − − 20q −2 Own example 8.5 (b) 158.33 8.6 50 − 23 q 8.7 0.8 8.8 £77.50 Own example Rise, maximum TY when t = £39 8.9 (a) 0.8 (b) 4,400 (d) 120 900 Proof (c) 40 or 120 24 − 1.2q 0.16667 (c) (e) Yes, both 940 Chapter 9.1 62.5 150 9.2 1,200, max 9.3 6, (a) 500 (b) 600 (c) 300 25, max 14.4956 50 4,096, max 0, 4, not max 3, not No stationary point exists 9.4 (a) (b) MC = 2q − 28q + 222, when q = 7, MC = 124 AVC = 23 q − 14q + 222, when q = 10.5, AVC = 148.5 (c) AFC = 50q −1 , when q → ∞ =, AFC → (d) TR = 200q − 2q , max when q = 50, TR = 5,000 (e) MR = 200 − 4q, no turning point, end-point max when q = (f) π = − 23 q + 12q − 22q − 50, max when q = 11, π = 272.67 π when q = 1, π = −60 23 Own example (a) 16 (b) (c) 12 No turning point but end–point when q = No turning point but end–point when q = Max when x = 63.33, no minimum © 1993, 2003 Mike Rosser 9.5 π max when q = (theoretical when q = −1.67 not realistic) (a) Max when q = 10 (b) no exists π max when q = 12.67, gives π = −48.8 5,075 when q = 10 27.6 when q = 37 9.6 15 orders of 400 560 480 every 4.5 months 140 9.7 (i) (a) q = 90 − 0.2t, p = 270 + 0.4t (b)&(c) q = 90, p = 270 (ii) (a) q = 250 − 1.25t, p = 125 + 0.375t (b)&(c) q = 250, p = 125 (iii) (a) q = 25 − 0.9615t, p = 160 + 0.385t (b)&(c) q = 25, p = 160 (there is no tax impact for (b) and (c) in all cases) q = 100, p = 380 (no tax impact) Chapter 10 10.1 (a) + 8x, 16 + 4z MPL = 4.8K MPK = 12K MPR = 16K 0.4 L −0.7 0.3 −0.6 (b) 42x z2 , 28x z (c) 4z + 6x −3 z3 , 4x − 9x −2 z2 , falls as L increases L R 0.4 , MPL = 12K 0.3 L−0.7 R 0.4 , 0.3 L0.3 R −0.6 MPL = 0.7, does not decline as L increases 10.2 No 1.2xj−0.7 (a) 0.228 (b) falls to 0.224 (c) inferior as ∂q/∂m < (d) elasticity with respect to ps = 0.379 and so a 1% increase in both prices would cause a percentage rise in q of 0.379 − 0.228 = 0.151% (a) Yes, MUA and MUB will rise at first but then fall; (b) no, MUA falls but MUB continually rises, therefore law not obeyed; (c) yes, both MUA and MUB continually fall No, MU will never reach zero for finite values of A or B 3,738.46; balance of payments changes from 4.23 deficit to 68.85 surplus 25 + 0.6q12 + 2.4q1 q2 10.3 0.45; 1.81818; 55 QLL = 6.4, MPL function has constant slope; QLK = −2K 0.6 L−1.5 , 2.4K −0.4 L−0.5 35 + 2.8K, position of MPL will rise as K rises; QKK = 2.8L, MPK has constant slope, actual value varies with L; QKL = 35 + 2.8K, increase in L will increase MPK , effect depends on level of K TC11 = 0.008q32 , TC22 = 0, TC33 = 0.008q12 TC12 = 1.2q3 = TC21 , TC23 = + 1.2q1 = TC32 TC31 = 0.016q1 q3 + 1.2q2 = TC13 10.4 q1 = 12.46; q2 = 36.55 p1 = 97.60, p2 = 101.81 q1 = 0, q2 = 501.55 (mathematical answer gives q1 = −1,292.24, q2 = 1,701.77 so rework without market 1) £575.81 when q1 = 47.86 and q2 = 39.01 q1 = 1,580.2, q2 = 1,791.8 © 1993, 2003 Mike Rosser q1 = 266.67, q2 = 333.33 K = 2,644.2, L = 3,718.5 £29,869.47 when K = 1,493.47 and L = 2,489.12 Because max π = £18,137.95 when K = 2,176.5 and L = 2,015.22 10 K = 10,149.1, L = 9,743.1 10.5 (a) 12K −0.4 L0.4 dK + 8K 0.6 L−0.6 dL (b) 14.4K −0.7 L0.2 R 0.4 dK + 9.60.3 L−0.8 R 0.4 dL + 19.2K 0.3 L0.2 R −0.6 dR (c) (4.8K −0.2 + 1.6KL2 )dK + (3.5L−0.3 + 1.6K L)dL (a) Yes (b) no, surplus (c) no, surplus 40x −0.6 z−0.45 + 12x 0.4 z−0.7 ∂QA ∂QA dM + ∂PA ∂M dPA Chapter 11 11.1 K = 12.6, L = 21 K = 500, L = 2,500 141.42 when K = 25, L = 50 A = 6, B = Own example (a) K = 1,000, L = 50 (b) K = 400, L = 20 1,950 when K = 60, L = 120 11.2 See answers to 11.1 11.3 See answers to 11.1 L = 241, K = 201, TC = £3,617 L = 38.8, K = 20.7, TC = £3,104.50 C1 = £480,621, C2 = £213,609 L = 19.04, K = 8.18, TC = £1,145.30 11.4 x = 30, y = 30, z = 90 877.8 when K = 15, L = 45, R = 13 x = 50, y = 100, z = 150 79,602.1 when x = 300, y = 300, z = 1,875 K = 26.7, L = 33.3, R = 8.9, M = 55.6 Own example L = 60, K = 45, R = 40 Chapter 12 12.1 Answer given −0.5 −0.5 0.5(6 + x) £8 12.2 3M(1 + i)2 MRPL = 60L 0.6x(3 + 0.6x )−0.5 − 8, L = 16 169 units 0.000868 (6x + 7)−0.5 (39x + 36.4x − 5.7) −0.5 76.5L (0.5K Own example 0.8 0.5 −0.4 + 3L ) 12 312.5 £190 (a) −0.05(60 − 0.1q)−0.5 (b) rate of change of slope = −0.0025(60 − 0.1q)−0.5 < when q < 600 (c) 400 12.3 (24 + 6.4x − 4.5x 1.5 − 3x 2.5 )(8 − 6x1.5 )−1.5 (18,000 + 360q)(25 + q)−1.5 © 1993, 2003 Mike Rosser −0.113 q = 1,333 13 , d2 TR/dq = −0.00367 L = 4.8, H = 7.2 Adapt proof in text for MC and AC to AVC = TVC(q)−1 12.4 (a) 12.5x + C (d) 42x + 18x −1 (b) 5x + 0.6x + 0.05x + C +C (e) 60x + 220x (b) 42q − 9q + 2q (a) 4q + 0.05q (d) 62q − 8q + 0.5q 12.5 1.5 −0.2 (c) 24x − 15x + C +C (c) 35q + 0.3q (e) 185q − 12q + 0.3q (a) £750,000 (b) £81,750 (c) £250,000 (d) £67,750 £49,600 Own example Chapter 13 13.1 20 No production in period 13.2 Pt = + 0.25(−2)t 13.3 2,790.625; yes 13.4 2,460.79 (a) Unstable (b) stable Stable, 118.54 39,946.789 404.64 492.57 1,848.259 Ptx = 562 − 63(0.83)t , 555.27 No, 1,976.67 < 1,980 Chapter 14 14.1 64.44 million 61,062 units 16.8 million tonnes Usage in million units: (a) 94.6, yes (d) 291.31, no 14.2 14.3 0.48% (c) 200.2, no e31,308.07 56,609 units 2%; 9.84 million; no £122,197.54 (b) 137.6, yes 9%, 401,767,300 barrels 587 2.05%; 3.49% e6,446.39 million 0.83%, 621.43 million tones 8.8% 5.83% 6.18% 9% discrete (equivalent to 8.62% continuous) 14.4 (a) 200e0.2t , 1477.81 (d) 40e1.32t , 21614597.49 (e) 128e−0.03t , 99.69 14.5 −20e0.4t + 200, 52.22, unstable −20e 14.6 −0.75t + 120, 119.53, stable 7e−0.325t + 30, stable (c) 14e−0.4t ,0.26 (b) 45e1.2t , 7323965.61 10 %, 6.77 −19.2e−1.5t + 32, 31.99, stable 75e0.08t − 300, −188.11, unstable −7.25e−0.96t + 26.25, difference 0.01 Yes, as predicted spot price is $27.56 32.54e−0.347t + 17.46, £23.20, periods © 1993, 2003 Mike Rosser $44.01 14.7 25e−0.2t + 180, 183.38 −63.33e−0.195t + 1583.33, 1574.32 12.22e−0.036t + 2027.78, 2036.3 −9.49e−0.226t + 141.49, 140.495 −18.154e−0.176t + 346.15, 343.015 Chapter 15 15.1 23 26 27 a yes,  1.4 1.2 0.8 15.2 27 Yes  14 3    c yes,   14  −3  70 23 0.6 0.6 0.24 39 b no 0.2 1.6 0.4 a  0.8 0.5 98 163 c not possible PR = 17.5 a 61 19 84 84 134 b 52 62 30 31.5  15.3  460 291  c  907 114 93  15.4 45 130 579 1077 4505 20 133 a 2  c 9 4 299 240 400 466 165 136 b 130 59 400 418 1030 560 213 56 71 67.5 47.5 10 2110 2155 9932 151 135     95 x 4 y  = 32 61 z     95 x 0 y  = 32 61 z 71 68 24 13 17 43 55 90 51   40 44 36 3181 1166   7386   318  −215  b 3 −8     56 x 4 y  = 28 34 z d not possible (b) and (c) (b) not square, (c) rows linearly dependent 15.5 A B.0 C.56 15.6 a −39 b.15 c −4 A.636 B −101 © 1993, 2003 Mike Rosser D.137 E.119 d 28 e 50 C −4462 f  −5 −5 C =  −8 −19 20    −8 0.6 0.4 −19 A−1 = 0.8 1.2 AdjA =  −5 −5 20 −1 −1   −0.5 0.5 −0.5 −0.5  0.1075 −0.0215 0.1505 −0.0538    1.7742 −1.3548 0.4839 1.6129  −0.3978 0.2796 0.043 −0.3011  15.7 15.8 0.16 −0.2 −0.3 0.5 x = 5, y =  −1.6 −3.8 x1 = 10, x2 = 6, x3 = β1 = −0.5, β2 = 1, β3 = 0.4, Q = 9.5 β1 = −300, β2 = 75, β3 = 400, β4 = −100, β5 = 0.2, β6 = 10, Q = 8370 15.9 x = 3, y = 15.10 q1 = 389.6, q2 = 62.3, max SOC met as |H1 | = −0.6, |H2 | = 0.448 See 15.8 answers q1 = 216.8, q2 = 435.8, max SOC met as |H1 | = −0.5, |H2 | = 0.19 q1 = 6.485, q2 = 2.376, q3 = 5.4, max SOC met as |H1 | = −16.8, |H2 | = 21.2, |H3 | = −16.8 q1 = 5.2, q2 = 35.4, q3 = 20.8, max SOC met as |H1 | = −4.08, |H2 | = 0.4832, |H3 | = −0.0225 15.11 K = 25, L = 100, max SOC met as |HB | = 0.72 K = 16, L = 32, SOC met as |HB | = −14.618 X = 25, Y = 100, max SOC met as |HB | = 4.543 K = 121.93, L = 243.86, SOC met as |HB | = −0.945 © 1993, 2003 Mike Rosser Symbols and terminology |x| [ ]m dy dx ex f1 > ≥ absolute value 60 definite integral 388 derivative 247 exponential function 432 first-order derivative 295 greater than inequality 59 greater than or equal to weak inequality 59 ≡ identity 52 ∞ infinity 69 integral 390 © 1993, 2003 Mike Rosser λ < ≤ log ln ∂y ∂x f11 x √ √ x y Lagrange multiplier 342 less than inequality 59 less than or equal to weak inequality 59 logarithm (base 10) 29–32 natural logarithm (base e) 440–1 partial derivative 291 second-order partial derivative 297 small change in x 253–4, 257 square root 26 xth root of y 27 summation 56 ... Basic Mathematics for Economists Economics students will welcome the new edition of this excellent textbook Given that many students come into economics courses without having studied mathematics. .. Solution 1p = £0.01 Therefore 0.01p = £0.0001 In mathematics a decimal format is often required for a value that is usually specified as a percentage in everyday usage For example, interest rates... expressions for economic concepts involving unknown values Simplify and reformulate basic algebraic expressions Solve single linear equations with one unknown variable Use the summation sign Perform basic