Elements of Mathematics for Economics and Finance (PDF)

Vassilis C. Mavron and Timothy N. Phillips Elements of Mathematics for Economics and Finance With 77 Figures Vassilis C. Mavron, MA, MSc, PhD Timothy N. Phillips, MA, MSc, DPhil, Institute of Mathematical and Physical DSc Sciences...

Vassilis C. Mavron and Timothy N. Phillips Elements of Mathematics for Economics and Finance With 77 Figures Vassilis C. Mavron, MA, MSc, PhD Timothy N. Phillips, MA, MSc, DPhil, Institute of Mathematical and Physical DSc Sciences Cardiff School of Mathematics University of Wales Aberystwyth Cardiff University Aberystwyth SY23 3BZ Senghennydd Road Wales, UK Cardiff CF24 4AG Wales, UK Mathematics Subject Classification (2000): 91-01; 91B02 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library Library of Congress Control Number: 2006928729 ISBN-10: 1-84628-560-7 e-ISBN 1-84628-561-5 Printed on acid-free paper ISBN-13: 978-1-84628-560-8 © Springer-Verlag London Limited 2007 Apart from any fair dealing for the purposes of research or private study, or criticism or review, as permitted under the Copyright, Designs and Patents Act 1988, this publication may only be reproduced, stored or transmitted, in any form or by any means, with the prior permission in writing of the publishers, or in the case of reprographic reproduction in accordance with the terms of licences issued by the Copyright Licensing Agency. Enquiries concerning reproduction outside those terms should be sent to the publishers. The use of registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant laws and regulations and therefore free for general use. The publisher makes no representation, express or implied, with regard to the accuracy of the information contained in this book and cannot accept any legal responsibility or liability for any errors or omissions that may be made. Printed in the United States of America (HAM) 9 8 7 6 5 4 3 2 1 Springer Science + Business Media, LLC springer.com  Preface The mathematics contained in this book for students of economics and finance has, for many years, been given by the authors in two single-semester courses at the University of Wales Aberystwyth. These were mathematics courses in an economics setting, given by mathematicians based in the Department of Mathematics for students in the Faculty of Social Sciences or School of Man- agement. The choice of subject matter and arrangement of material reflect this collaboration and are a result of the experience thus obtained. The majority of students to whom these courses were given were study- ing for degrees in economics or business administration and had not acquired any mathematical knowledge beyond pre-calculus mathematics, i.e., elementary algebra. Therefore, the first-semester course assumed little more than basic pre- calculus mathematics and was based on Chapters 1–7. This course led on to the more advanced second-semester course, which was also suitable for students who had already covered basic calculus. The second course contained at most one of the three Chapters 10, 12, and 13. In any particular year, their inclusion or exclusion would depend on the requirements of the economics or business studies degree syllabuses. An appendix on differentials has been included as an optional addition to an advanced course. The students taking these courses were chiefly interested in learning the mathematics that had applications to economics and were not primarily in- terested in theoretical aspects of the subject per se. The authors have not at- tempted to write an undergraduate text in economics but instead have written a text in mathematics to complement those in economics. The simplicity of a mathematical theory is sometimes lost or obfuscated by a dense covering of applications at too early a stage. For this reason, the aim of the authors has been to present the mathematics in its simplest form, highlighting threads of common mathematical theory in the various topics of v vi Elements of Mathematics for Economics and Finance economics. Some knowledge of theory is necessary if correct use is to be made of the techniques; therefore, the authors have endeavoured to introduce some basic theory in the expectation and hope that this will improve understanding and incite a desire for a more thorough knowledge. Students who master the simpler cases of a theory will find it easier to go on to the more difficult cases when required. They will also be in a better position to understand and be in control of calculations done by hand or calculator and also to be able to visualise problems graphically or geometrically. It is still true that the best way to understand a technique thoroughly is through practice. Mathematical techniques are no exception, and for this reason the book illustrates theory through many examples and exercises. We are grateful to Noreen Davies and Joe Hill for invaluable help in prepar- ing the manuscript of this book for publication. Above all, we are grateful to our wives, Nesta and Gill, and to our chil- dren, Nicholas and Christiana, and Rebecca, Christopher, and Emily, for their patience, support, and understanding: this book is dedicated to them. Vassilis C. Mavron Timothy N. Phillips Aberystwyth Cardiff United Kingdom United Kingdom March 2006  Contents 1. Essential Skills.............................................. 1 1.1 Introduction............................................. 1 1.2 Numbers................................................ 2 1.2.1 Addition and Subtraction........................... 3 1.2.2 Multiplication and Division.......................... 3 1.2.3 Evaluation of Arithmetical Expressions................ 4 1.3 Fractions................................................ 5 1.3.1 Multiplication and Division.......................... 7 1.4 Decimal Representation of Numbers........................ 8 1.4.1 Standard Form..................................... 10 1.5 Percentages.............................................. 10 1.6 Powers and Indices....................................... 12 1.7 Simplifying Algebraic Expressions.......................... 16 1.7.1 Multiplying Brackets................................ 16 1.7.2 Factorization...................................... 18 2. Linear Equations........................................... 23 2.1 Introduction............................................. 23 2.2 Solution of Linear Equations............................... 24 2.3 Solution of Simultaneous Linear Equations................... 27 2.4 Graphs of Linear Equations................................ 30 2.4.1 Slope of a Straight Line............................. 34 2.5 Budget Lines............................................ 37 2.6 Supply and Demand Analysis.............................. 40 2.6.1 Multicommodity Markets............................ 44 vii viii Contents 3. Quadratic Equations........................................ 49 3.1 Introduction............................................. 49 3.2 Graphs of Quadratic Functions............................. 50 3.3 Quadratic Equations...................................... 56 3.4 Applications to Economics................................. 61 4. Functions of a Single Variable.............................. 69 4.1 Introduction............................................. 69 4.2 Limits.................................................. 72 4.3 Polynomial Functions..................................... 72 4.4 Reciprocal Functions...................................... 75 4.5 Inverse Functions......................................... 81 5. The Exponential and Logarithmic Functions................ 87 5.1 Introduction............................................. 87 5.2 Exponential Functions.................................... 88 5.3 Logarithmic Functions.................................... 90 5.4 Returns to Scale of Production Functions.................... 95 5.4.1 Cobb-Douglas Production Functions.................. 97 5.5 Compounding of Interest.................................. 98 5.6 Applications of the Exponential Function in Economic Modelling...................................... 102 6. Differentiation.............................................. 109 6.1 Introduction............................................. 109 6.2 Rules of Differentiation.................................... 113 6.2.1 Constant Functions................................. 113 6.2.2 Linear Functions................................... 114 6.2.3 Power Functions................................... 114 6.2.4 Sums and Differences of Functions.................... 114 6.2.5 Product of Functions............................... 116 6.2.6 Quotient of Functions............................... 117 6.2.7 The Chain Rule.................................... 117 6.3 Exponential and Logarithmic Functions..................... 119 6.4 Marginal Functions in Economics........................... 121 6.4.1 Marginal Revenue and Marginal Cost................. 121 6.4.2 Marginal Propensities............................... 123 6.5 Approximation to Marginal Functions....................... 125 6.6 Higher Order Derivatives.................................. 127 6.7 Production Functions..................................... 129 Contents ix 7. Maxima and Minima....................................... 137 7.1 Introduction............................................. 137 7.2 Local Properties of Functions.............................. 138 7.2.1 Increasing and Decreasing Functions.................. 138 7.2.2 Concave and Convex Functions...................... 138 7.3 Local or Relative Extrema................................. 139 7.4 Global or Absolute Extrema............................... 144 7.5 Points of Inflection....................................... 145 7.6 Optimization of Production Functions....................... 146 7.7 Optimization of Profit Functions........................... 151 7.8 Other Examples.......................................... 154 8. Partial Differentiation...................................... 159 8.1 Introduction............................................. 159 8.2 Functions of Two or More Variables........................ 160 8.3 Partial Derivatives........................................ 160 8.4 Higher Order Partial Derivatives........................... 163 8.5 Partial Rate of Change.................................... 165 8.6 The Chain Rule and Total Derivatives...................... 168 8.7 Some Applications of Partial Derivatives.................... 171 8.7.1 Implicit Differentiation.............................. 171 8.7.2 Elasticity of Demand............................... 173 8.7.3 Utility............................................ 176 8.7.4 Production........................................ 179 8.7.5 Graphical Representations........................... 181 9. Optimization............................................... 185 9.1 Introduction............................................. 185 9.2 Unconstrained Optimization............................... 186 9.3 Constrained Optimization................................. 193 9.3.1 Substitution Method................................ 193 9.3.2 Lagrange Multipliers................................ 197 9.3.3 The Lagrange Multiplier λ: An Interpretation.......... 201 9.4 Iso Curves............................................... 204 10. Matrices and Determinants................................. 209 10.1 Introduction............................................. 209 10.2 Matrix Operations........................................ 209 10.2.1 Scalar Multiplication................................ 211 10.2.2 Matrix Addition.................................... 212 10.2.3 Matrix Multiplication............................... 212 10.3 Solutions of Linear Systems of Equations.................... 220 x Contents 10.4 Cramer’s Rule........................................... 222 10.5 More Determinants....................................... 223 10.6 Special Cases............................................ 230 11. Integration................................................. 233 11.1 Introduction............................................. 233 11.2 Rules of Integration....................................... 236 11.3 Definite Integrals......................................... 241 11.4 Definite Integration: Area and Summation................... 243 11.5 Producer’s Surplus....................................... 250 11.6 Consumer’s Surplus....................................... 251 12. Linear Difference Equations................................ 261 12.1 Introduction............................................. 261 12.2 Difference Equations...................................... 261 12.3 First Order Linear Difference Equations..................... 264 12.4 Stability................................................. 267 12.5 The Cobweb Model....................................... 270 12.6 Second Order Linear Difference Equations................... 273 12.6.1 Complementary Solutions........................... 274 12.6.2 Particular Solutions................................ 277 12.6.3 Stability.......................................... 282 13. Differential Equations...................................... 287 13.1 Introduction............................................. 287 13.2 First Order Linear Differential Equations.................... 288 13.2.1 Stability.......................................... 292 13.3 Nonlinear First Order Differential Equations................. 292 13.3.1 Separation of Variables.............................. 294 13.4 Second Order Linear Differential Equations.................. 296 13.4.1 The Homogeneous Case............................. 297 13.4.2 The General Case.................................. 300 13.4.3 Stability.......................................... 302 A. Differentials................................................ 305 Index........................................................... 309  1 Essential Skills 1.1 Introduction Many models and problems in modern economics and finance can be expressed using the language of mathematics and analysed using mathematical tech- niques. This book introduces, explains, and applies the basic quantitative meth- ods that form an essential foundation for many undergraduate courses in eco- nomics and finance. The aim throughout this book is to show how a range of important mathematical techniques work and how they can be used to explore and understand the structure of economic models. In this introductory chapter, the reader is reacquainted with some of the basic principles of arithmetic and algebra that formed part of their previous mathematical education. Since economics and finance are quantitative subjects it is vitally important that students gain a familiarity with these principles and are confident in applying them. Mathematics is a subject that can only be learnt by doing examples, and therefore students are urged to work through the examples in this chapter to ensure that these key skills are understood and mastered. 1 2 Elements of Mathematics for Economics and Finance 1.2 Numbers For most, if not all, of us, our earliest encounter with numbers was when we were taught to count as children using the so-called counting numbers 1, 2, 3, 4,.... The counting numbers are collectively known as the natural numbers. The natural numbers can be represented by equally spaced points on a line as shown in Fig. 1.1. The direction in which the arrow is pointing in Fig. 1.1 indicates the direction in which the numbers are getting larger, i.e., the natural numbers are ordered in the sense that if you move along the line to the right, the numbers progressively increase in magnitude. X 1 2 3 4 5 6 7 Figure 1.1 The natural numbers. It is sometimes useful and necessary to talk in terms of numbers less than zero. For example, a person with an overdraft on their bank account essentially has a negative balance or debt, which needs to be cancelled before the account is in credit again. In the physical world, negative numbers are used to report temperatures below 00 Centigrade, which is the temperature at which water freezes. So, for example, −50 C is 50 C below freezing. If the line in Fig. 1.1 is extended to the left, we can mark equally spaced points that represent zero and the negatives of the natural numbers. The nat- ural numbers, their negatives, and the number zero are collectively known as the integers. All these numbers can be represented by equally spaced points on a number line as shown in Fig. 1.2. If we move along the line to the right, the numbers become progressively larger, while if we move along the line to the left, the numbers become smaller. So, for example, −4 is smaller than −1 and we write −4 < −1 where the symbol ‘ −4 where the symbol ‘>’ means ‘is greater than’. Note that these symbols should not be confused with the symbols ‘≤’ and ‘≥’, which mean ‘less than or equal to’ and ‘greater than or equal to’, respectively. 1. Essential Skills 3 X -4 -3 -2 -1 0 1 2 3 4 5 Figure 1.2 Integers on the number line. 1.2.1 Addition and Subtraction Initially, numerical operations involving negative numbers may seem rather confusing. We give the rules for adding and subtracting numbers and then appeal to the number line for some justification. If a and b are any two numbers, then we have the following rules a + (−b) = a − b, (1.1) a − (+b) = a − b, (1.2) a − (−b) = a + b. (1.3) Thus we can regard −(−b) as equal to +b. We consider a few examples: 4 + (−1) = 4 − 1 = 3, and 3 − (−2) = 3 + 2 = 5. The last example makes sense if we regard 3 − (−2) as the difference between 3 and −2 on the number line. Note that a − b will be negative if and only if a < b. For example, −2 − (−1) = −2 + 1 = −1 < 0. 1.2.2 Multiplication and Division If a and b are any two positive numbers, then we have the following rules for multiplying positive and negative numbers: a × (−b) = −(a × b), (1.4) (−a) × b = −(a × b), (1.5) (−a) × (−b) = a × b. (1.6) So multiplication of two numbers of the same sign gives a positive number, while multiplication of two numbers of different signs gives a negative number. 4 Elements of Mathematics for Economics and Finance For example, to calculate 2 × (−5), we multiply 2 by 5 and then place a minus sign before the answer. Thus, 2 × (−5) = −10. It is usual in mathematics to write ab rather than a × b to express the multi- plication of two numbers a and b. We say that ab is the product of a and b. Thus, we can write (1.6) in the form (−a)(−b) = ab. These multiplication rules give, for example, (−2) × (−3) = 6, (−4) × 5 = −20, 7 × (−5) = −35. The same rules hold for division because it is the same sort of operation as multiplication, since a 1 =a×. b b So the division of a number by another of the same sign gives a positive number, while division of a number by another of the opposite sign gives a negative number. For example, we have (−15) ÷ (−3) = 5, (−16) ÷ 2 = −8, 2 ÷ (−4) = −1/2. 1.2.3 Evaluation of Arithmetical Expressions The order in which operations in an arithmetical expression are performed is important. Consider the calculation 12 + 8 ÷ 4. Different answers are obtained depending on the order in which the operations are executed. If we first add together 12 and 8 and then divide by 4, the result is 5. However, if we first divide 8 by 4 to give 2 and then add this to 12, the result is 14. Therefore, the order in which the mathematical operations are performed is important and the convention is as follows: brackets, exponents, division, multiplication, addition, and subtraction. So that the evaluation of expressions within brackets takes precedence over addition and the evaluation of any number or expressions raised to a power (an exponential) takes prece- dence over division, for example. This convention has the acronym BEDMAS. However, the main point to remember is that if you want a calculation to be done in a particular order, you should use brackets to avoid any ambiguity. 1. Essential Skills 5 Example 1.1 Evaluate the expression 23 × 3 + (5 − 1). Solution. Following the BEDMAS convention, we evaluate the contents of the bracket first and then evaluate the exponential. Therefore, 23 × 3 + (5 − 1) = 23 × 3 + 4 = 8 × 3 + 4. Finally, since multiplication takes precedence over addition, we have 23 × 3 + (5 − 1) = 24 + 4 = 28. 1.3 Fractions A fraction is a number that expresses part of a whole. It takes the form a/b where a and b are any integers except that b = 0. The integers a and b are known as the numerator and denominator of the fraction, respectively. Note that a can be greater than b. The formal name for a fraction is a rational number since they are formed from the ratio of two numbers. Examples of statements that use fractions are 3/5 of students in a lecture may be female or 1/3 of a person’s income may be taxed by the government. Fractions may be simplified to obtain what is known as a reduced fraction or a fraction in its lowest terms. This is achieved by identifying any common factors in the numerator and denominator and then cancelling those factors by dividing both numerator and denominator by them. For example, consider the simplification of the fraction 27/45. Both the numerator and denominator have 9 as a common factor since 27 = 9 × 3 and 45 = 9 × 5 and therefore it can be cancelled: 27 3×9 3 = =. 45 5×9 5 We say that 27/45 and 3/5 are equivalent fractions and that 3/5 is a reduced fraction. To compare the relative sizes of two fractions and also to add or subtract two fractions, we express them in terms of a common denominator. The com- mon denominator is a number that each of the denominators of the respective fractions divides, i.e., each is a factor of the common denominator. Suppose we wish to determine which is the greater of the two fractions 4/9 and 5/11. The common denominator is 9 × 11 = 99. Each of the denominators (9 and 6 Elements of Mathematics for Economics and Finance 11) of the two fractions divides 99. The simplest way to compare the relative sizes is to multiply the numerator and denominator of each fraction by the denominator of the other, i.e., 4 4 × 11 44 5 5×9 45 = = , and = =. 9 9 × 11 99 11 11 × 9 99 So 5/11 > 4/9 since 45/99 > 44/99. We follow a similar procedure when we want to add two fractions. Consider the general case first of all in which we add the fractions a/b and c/d with b = 0 and d = 0: a c a×d c×b + = + b d b×d d×b a×d+b×c =. b×d Therefore, we have a c ad + bc + =. (1.7) b d bd For example, 2 3 2×5+3×7 10 + 21 31 + = = =. 7 5 7×5 35 35 The result for the subtraction of two fractions is similar, i.e., a c ad − bc − =. (1.8) b d bd Example 1.2 Simplify 13 5 −. 24 16 Solution. The idea is to express each of these fractions as equivalent fractions having a common denominator. Therefore, we have 13 5 13 × 16 5 × 24 − = − 24 16 24 × 16 16 × 24 208 − 120 = 384 88 = 384 11 × 8 = 48 × 8 11 =. 48 1. Essential Skills 7 Note that a smaller common denominator, namely 48, could have been used in this example since the two denominators, viz. 16 and 24, are both factors of 48. Thus 13 2 × 13 26 = = 24 2 × 24 48 and 5 3×5 13 = =. 16 3 × 16 48 Therefore, 13 5 26 − 15 11 − = =. 24 16 48 48 1.3.1 Multiplication and Division To multiply together two fractions, we simply multiply the numerators together and multiply the denominators together: a c a×c ac × = =. (1.9) b d b×d bd To divide one fraction by another, we multiply by the reciprocal of the divisor where the reciprocal of the fraction a/b is defined to be b/a provided a, b = 0. That is a c a d a×d ad ÷ = × = =. (1.10) b d b c b×c bc Example 1.3 Simplify the following fractions 5 16 1. × , 8 27 9 12 2. ÷. 13 25 Solution. 1. The product is the fraction 5 × 16. 8 × 27 To simplify this fraction, we note that 8 is a factor of the numerator and denominator (since 16 = 8 × 2) and can be cancelled. Therefore, we have 5 16 5 × 16 5×8×2 10 × = = =. 8 27 8 × 27 8 × 27 27 8 Elements of Mathematics for Economics and Finance 2. Using the rule (1.10) for the division of two fractions, we have 9 12 9 25 9 × 25 ÷ = × =. 13 25 13 12 13 × 12 Then noting that 3 is a common factor of the numerator and denominator, we have 5 16 3 × 3 × 25 3 × 25 75 × = = =. 8 27 13 × 4 × 3 13 × 4 52 1.4 Decimal Representation of Numbers A fraction or rational number may be converted to its equivalent decimal rep- resentation by dividing the numerator by the denominator. For example, the decimal representation of 3/4 is found by dividing 3 by 4 to give 0.75. This is an example of a terminating decimal since it ends after a finite number of digits. The following are examples of rational numbers that have a terminating decimal representation: 1 = 0.125, 8 and 3 = 0.12. 25 Some fractions do not possess a finite decimal representation – they go on forever. The fraction 1/3 is one such example. Its decimal representation is 0.3333... where the dots denote that the 3s are repeated and we write 1 = 0.3̇, 3 where the dot over the number indicates that it is repeated indefinitely. This is an example of a recurring decimal. All rational numbers have a decimal representation that either terminates or contains an infinitely repeated finite sequence of numbers. Another example of a recurring decimal is the decimal representation of 1/13: 1 = 0.0769230769230... = 0.07̇69230̇, 13 where the dots indicate the first and last digits in the repeated sequence. All numbers that do not have a terminating or recurring decimal represen- tation √ are known as irrational numbers. Examples of irrational numbers are 2 and π. All the irrational numbers together with all the rational numbers 1. Essential Skills 9 form the real numbers. Every point on the number line in Fig. 1.2 corresponds to a real number, and the line is known as the real line. To convert a decimal to a fraction, you simply have to remember that the first digit after the decimal point is a tenth, the second a hundredth, and so on. For example, 2 1 0.2 = = , 10 5 and 375 3 0.375 = =. 1,000 8 Sometimes we are asked to express a number correct to a certain number of decimal places or a certain number of significant figures. Suppose that we wish to write the number 23.541638 correct to two decimal places. To do this, we truncate the part of the number following the second digit after the decimal point: 23.54 | 1638. Then, since the first neglected digit, 1 in this case, lies between 0 and 4, then the truncated number, 23.54, is the required answer. If we wish to write the same number correct to three decimal places, the truncated number is 23.541 | 638, and since the first neglected digit, 6 in this case, lies between 5 and 9, then the last digit in the truncated number is rounded up by 1. Therefore, the number 23.541638 is 23.542 correct to three decimal places or, for short, ‘to three decimal places’. To express a number to a certain number of significant figures, we employ the same rounding strategy used to express numbers to a certain number of decimal places but we start counting from the first non-zero digit rather than from the first digit after the decimal point. For example, 72,648 = 70,000 (correct to 1 significant figure) = 73,000 (correct to 2 significant figures) = 72,600 (correct to 3 significant figures) = 72,650 (correct to 4 significant figures), and 0.004286 = 0.004 (correct to 1 significant figure) = 0.0043 (correct to 2 significant figures) = 0.00429 (correct to 3 significant figures). Note that 497 = 500 correct to 1 significant figure and also correct to 2 signif- icant figures. 10 Elements of Mathematics for Economics and Finance 1.4.1 Standard Form The distance of the Earth from the Sun is approximately 149,500,000 km. The mass of an electron is 0.000000000000000000000000000911 g. Numbers such as these are displayed on a calculator in standard or scientific form. This is a shorthand means of expressing very large or very small numbers. The standard form of a number expresses it in terms of a number lying between 1 and 10 multiplied by 10 raised to some power or exponent. More precisely, the standard form of a number is a × 10b , where 1 ≤ a < 10, and b is an integer. A practical reason for the use of the standard form is that it allows calculators and computers to display more significant figures than would otherwise be possible. For example, the standard form of 0.000713 is 7.13 × 10−4 and the standard form of 459.32 is 4.5932 × 102. The power gives the number of decimal places the decimal point needs to be moved to the right in the case of a positive power or the number of decimal places the decimal point needs to be moved to the left in the case of a negative power. For example, 5.914 × 103 = 5914 and 6.23 × 10−4 = 0.000623. Returning to the above examples, the Earth is about 1.495 × 108 km from the Sun and the mass of an electron is 9.11 × 10−28 g. Similarly, a budget deficit of 257,000,000,000 is 2.57 × 1011 in standard form. 1.5 Percentages To convert a fraction to a percentage, we multiply the fraction by 100%. For example, 3 3 = × 100% = 75%, 4 4 and 3 3 = × 100% = 23.077% (to three decimal places). 13 13 To perform the reverse operation and convert a percentage to a fraction, we divide the number by 100. The resulting fraction may then be simplified to obtain a reduced fraction. For example, 45 9 45% = = , 100 20 where the fraction has been simplified by dividing the numerator and denomi- nator by 5 since this is a common factor of 45 and 100. 1. Essential Skills 11 To find the percentage of a quantity, we multiply the quantity by the number and divide by 100. For example, 25 25% of 140 is × 140 = 35, 100 and 4 4% of 5, 200 is × 5, 200 = 208. 100 If a quantity is increased by a percentage, then that percentage of the quantity is added to the original. Suppose that an investment of £300 increases in value by 20%. In monetary terms, the investment increases by 20 × 300 = £60, 100 and the new value of the investment is £300 + £60 = £360. In general, if the percentage increase is r%, then the new value of the invest- ment comprises the original and the increase. The new value can be found by multiplying the original value by the factor r 1+. 100 It is easy to work in the reverse direction and determine the original value if the new value and percentage increase is known. In this case, one simply divides by the factor r 1+. 100 Example 1.4 The cost of a refrigerator is £350.15 including sales tax at 17.5%. What is the price of the refrigerator without sales tax? Solution. To determine the price of the refrigerator without sales tax, we divide £350.15 by the factor 17.5 1+ = 1.175. 100 So the price of the refrigerator without VAT is 350.15 = £298. 1.175 12 Elements of Mathematics for Economics and Finance Similarly, if a quantity decreases by a certain percentage, then that percent- age of the original quantity is subtracted from the original to obtain its new value. The new value may be determined by multiplying the original value by the quantity r 1−. 100 Example 1.5 A person’s income is e25,000 of which e20,000 is taxable. If the rate of income tax is 22%, calculate the person’s net income. Solution. The person’s net income comprises the part of his salary that is not taxable (e5,000) together with the portion of his taxable income that remains after the tax has been taken. The person’s net income is therefore
 22 78 5,000 + 1 − × 20,000 = 5,000 + × 20,000 100 100 = 5,000 + 78 × 200 = 5,000 + 15,600 = e20,600. 1.6 Powers and Indices Let x be a number and n be a positive integer, then xn denotes x multiplied by itself n times. Here x is known as the base and n is the power or index or exponent. For example, x5 = x × x × x × x × x. There are rules for multiplying and dividing two algebraic expressions or numerical values involving the same base raised to a power. In the case of multiplication, we add the indices and raise the expression or value to that new power to obtain the product rule xa × xb = xa xb = xa+b. For example, x2 × x3 = (x × x) × (x × x × x) = x5. 1. Essential Skills 13 In the case of division, we subtract the indices and raise the expression or value to that new power to obtain the quotient rule xa xa ÷ xb = = xa−b. xb For example, x×x 1 x2 ÷ x4 = = 2, x×x×x×x x and using the quotient rule we have x2 = x2−4 = x−2. x4 More generally, we have 1 = x−n. xn Suppose now that we multiply an expression with a fractional power as many times as the denominator of the fraction. For example, multiply x1/3 by itself three times. We have x1/3 × x1/3 × x1/3 = x1/3+1/3+1/3 = x1 = x. However, the number that when multiplied by itself three times gives x is known √ as the cube root of x, and an alternative notation for x1/3 is 3 x. The symbol √n x, which sometimes appears on a calculator as x1/n , is known as the nth root √ of x. In the case n = 2, the n is omitted in the former symbol. So we write x √ rather than 2 x for the square root x1/2 of x. Suppose we wish to raise an expression with a power to a power, for example (x2 )4. We may rewrite this as (x2 )(x2 )(x2 )(x2 ) = x2+2+2+2 = x8 , using the product rule. More generally, we have (xm )n = xmn. These rules for simplifying expressions involving powers may be used to evaluate arithmetic expressions without using a calculator. For example, 23 = 2 × 2 × 2 = 8, 4 3 = 3 × 3 × 3 × 3, √ 81 = 9, √ 3 27 = 3, 1 1 2−3 = =. 23 8 Note the following two conventions related to the use of powers: 14 Elements of Mathematics for Economics and Finance 1. x1 = x (An exponent of 1 is not expressed.) 2. x0 = 1 for x = 0 (Any nonzero number raised to the zero power is equal to 1.) To summarise, we have the following rules governing indices or powers: Rules of Indices xa xb = xa+b (1.11) xa = xa−b (1.12) xb a b (x ) = xab (1.13) 1 = x−a (1.14) xa √a 1 x = x a (1.15) √a b xb = x a (1.16) Finally, consider the product of two numbers raised to some power. For example, consider (xy)3. Now (xy)3 = (x × y) × (x × y) × (x × y) = (x × x × x) × (y × y × y) = x3 y 3 , since it does not matter in which order numbers are multiplied. More generally, we have (xy)a = xa y a. Similarly, we have
a x xa = a. y y Example 1.6 Simplify the following using the rules of indices: x2 1. , x3/2 x2 y 3 2.. x4 y 1. Essential Skills 15 Solution. 1. Using the quotient rule (1.12), we have x2 √ 3/2 = x2−3/2 = x1/2 = x x 2. Using the quotient and reciprocal rules, we have x2 y 3
2
3 x y = x4 y x4 y = (x2−4 )(y 3−1 ) (using the quotient rule (1.12)) = x−2 y 2 y2 = (using the reciprocal rule (1.14)) x2  y 2 =. x Example 1.7 Write down the values of the following without using a calculator: 1. 3−3 2. 163/4 3. 16−3/4 −1/3 4. 27 5. 43/2 6. 190. Solution. 1 1 1. 3−3 = 3 =. 3 27 √ 2. 163/4 = (161/4 )3 = ( 4 16)3 = 23 = 8. 1 1 3. 16−3/4 = =. 163/4 8 1 1 1 4. 27−1/3 = = √ =. 271/3 3 27 3 √ 5. 43/2 = (41/2 )3 = ( 4)3 = 23 = 8. 6. 190 = 1. Note that we could also evaluate 43/2 as follows: √ 43/2 = (43 )1/2 = 641/2 = 64 = 8. 16 Elements of Mathematics for Economics and Finance 1.7 Simplifying Algebraic Expressions In the algebraic expression 7x3 , x is called the variable, and 7 is known as the coefficient of x3. Expressions consisting simply of a coefficient multiplying one or more variables raised to the power of a positive integer are called monomials. Monomials can be added or subtracted to form polynomials. Each of the monomials comprising a poly- nomial is called a term. For example, the terms in the polynomial 3x2 + 2x + 1 are 3x2 , 2x, and 1. The coefficient of x2 is 3, the coefficient of x is 2, and the constant term is 1. To add or subtract two polynomials, we collect like terms and add or sub- tract their coefficients. For example, if we wish to add 7x + 2 and 5 − 2x, then we collect the terms in x and the constant terms: (7x + 2) + (5 − 2x) = (7 + (−2))x + (2 + 5) = 5x + 7. Example 1.8 Simplify the following: 1. (3x2 + 2x + 1) + (5x2 − x − 7), 2. (9x4 + 12x3 + 6x + 1) − (x4 + 2x2 − 4), 3. (x3 + 4x − 5) + (2x2 − x + 8). Solution. 1. (3 + 5)x2 + (2 − 1)x + (1 − 7) = 8x2 + x − 6. 2. (9 − 1)x4 + 12x3 − 2x2 + 6x + (1 + 4) = 8x4 + 12x3 − 2x2 + 6x + 5. 3. x3 + 2x2 + (4 − 1)x + (−5 + 8) = x3 + 2x2 + 3x + 3. 1.7.1 Multiplying Brackets There are occasions when mathematical expressions may be simplified by re- moving any brackets present. This process, which is also known as expanding the brackets or multiplying out the brackets, culminates in an equivalent ex- pression without brackets. The removal of brackets is based on the following basic rule: a(b + c) = ab + ac, (1.17) 1. Essential Skills 17 where a, b, and c are any three numbers. Since the order in which multiplication is performed is not important, we also have (b + c)a = ba + ca, (1.18) The rules (1.17) and (1.18), which are examples of what is known as the dis- tributive law, may be generalized to include expressions involving polynomials. For example, 3(x + 2y) = 3x + 6y, and −2(3x2 − 5y) = −6x2 + 10y. It is important to take care multiplying out brackets when there is a negative sign outside the brackets. In this case, the sign of each term inside the brackets is changed when the brackets are removed. For example, −(2x2 − 3x − 2y + 5) = −2x2 + 3x + 2y − 5. We also have the following rule for multiplying two brackets: (a + b)(c + d) = ac + bc + ad + bd, (1.19) where a, b, c, and d are any three numbers. So to multiply out two brackets we simply multiply each term in the second bracket by each term in the first bracket and add together all contributions. For example, (x + 2)(2x − 3) = (x)(2x) + (2)(2x) + (x)(−3) + (2)(−3) = 2x2 + 4x − 3x − 6 = 2x2 + x − 6. The rule (1.19) extends to brackets containing more than two terms. The impor- tant thing to remember is that each term in the second bracket is multiplied by each term in the first before all contributions are added together. For example, (2x − y + 5)(x − 3) = (2x)(x) + (−y)(x) + (5)(x) +(2x)(−3) + (−y)(−3) + (5)(−3) = 2x2 − xy + 5x − 6x + 3y − 15 = 2x2 − xy − x + 3y − 15. Example 1.9 Multiply out the brackets and simplify the following: 1. (2x + 3)(7 − 5x), (120 − 24x) 2. , 4.8 3. (x + 3y)(2x − 5y − 1). 18 Elements of Mathematics for Economics and Finance Solution. 1. Using the rule (1.19), we have (2x + 3)(7 − 5x) = (2x)(7) + (3)(7) + (2x)(−5x) + (3)(−5x) = 14x + 21 − 10x2 − 15x = 21 − x − 10x2. 2. In this example, we just note that division of 120 − 24x by 0.48 is the same as multiplication of 120 − 24x by 1/(4.8), and therefore we can use the rule (1.17): (120 − 24x) 1 = (120 − 24x) 4.8 4.8 120 −24x + 4.8 4.8 = 25 − 5x. 3. Using the generalization of rule (1.19), we have (x + 3y)(2x − 5y − 1) = (x)(2x) + (3y)(2x) + (x)(−5y) +(3y)(−5y) + (x)(−1) + (3y)(−1) = 2x2 + 6xy − 5xy − 15y 2 − x − 3y = 2x2 + xy − 15y 2 − x − 3y. 1.7.2 Factorization Factorization is the reverse process to multiplying out the brackets. It involves taking a mathematical expression and rewriting it by expressing it in terms of a product of factors. There are a number of techniques that can be used to factorize an expression: 1. The simplest technique is to identify a common factor in two or more terms. The equivalent factorized expression can then be written in terms of the common factor multiplying a bracketed expression. For example, a) ab − ac = a(b − c), b) 4x2 + 6x = 2x(2x + 3), c) ax2 − a2 x = ax(x − a), d) −36x2 − 9x = −9x(4x + 1), 1. Essential Skills 19 5x + 10y 5(x + 2y) x + 2y e) = =. 10x − 5y 5(2x − y) 2x − y 2. The second technique is based on the following identity involving the difference of two squares: a2 − b2 = (a − b)(a + b). An identity is a formula valid for all values of the variables; in this case, a and b. The following are examples of the application of this identity: a) x2 − 36 = (x − 6)(x + 6); b) 9a2 − 16x2 = (3a)2 − (4x)2 = (3a − 4x)(3a + 4x); c) 9 − 36x2 = 9(1 − 4x2 ) = 9(12 − (2x)2 ) = 9(1 − 2x)(1 + 2x). An additional technique that can be used for factorizing quadratic expressions of the form ax2 + bx + c or ax2 + bxy + cy 2 will be discussed in Chapter 3. EXERCISES 1.1. Evaluate 35 − 8 ÷ 22 + 5 + 23 × 4. 1.2. Express the following fractions using decimal notation: 3 a) , 10 5 b) , 16 3 c) , 4 3 d) , 13 2 e) , 7 1 f). 19 1.3. Simplify the following fractions: 2 3 a) + , 5 8 5 3 b) − , 16 32 20 Elements of Mathematics for Economics and Finance 15 18 c) × , 54 35 32 56 d) ÷. 49 21 1.4. Find which is the larger of the two fractions: 11/32, 7/24 by express- ing the numbers as: a) fractions with the same denominator; b) decimals. 1.5. Write each of the following numbers correct to two decimal places: a) 51.2361 b) 7.896 c) 362.275 1.6. Write each of the following numbers correct to three significant fig- ures: a) 5,889 b) 0.0002817 c) 72,961 d) 0.09274 1.7. Write each of the following numbers in standard form: a) 495,200 b) 0.000000837 1.8. The computing equipment belonging to a company is valued at $45,000. Each year, 12% of the value is written off for depreciation. Find the value of the equipment at the end of two years. 1.9. Death duties of 20% are paid on a legacy to three children of £180,000. The eldest child is bequeathed 50%, the middle child 30%, and the youngest child the remainder. How much does each child re- ceive? What percentage of the original legacy does the youngest child receive? 1.10. Simplify the following: a) x2/3 x7/3 , x5 b) , x2 1. Essential Skills 21 c) (x2/3 )6 , x3 y 2 d). x2 y 5 1.11. Write down the values of the following without using a calculator: a) 165/4 , b) 811/4 ,
2/3 27 c) , 125 d) 81−3/4. 1.12. Multiply out the brackets and simplify the following: a) (2x + 9)(3x − 8), b) (x + 4)(6x + 3), c) (3x − 2)(11 − 4x), (15 − 24x + 18y) d) , 0.75 e) (x − 4y + 7)(5x − 2y − 3). 1.13. Factorize the following expressions: a) 96x − 32, b) −21x + 49x2 , c) 4x2 − 49.  2 Linear Equations 2.1 Introduction In this book, we will be concerned primarily with the analysis of the relation- ship between two or more variables. For example, we will be interested in the relationship between economic entities or variables such as – total cost and output, – price and quantity in an analysis of demand and supply, – production and factors of production such as labour and capital. If one variable, say y, changes in an entirely predictable way in terms of an- other variable, say x, then, under certain conditions (to be defined precisely in Chapter 4), we say that y is a function of x. A function provides a rule for providing values of y given values of x. The simplest function that relates two or more variables is a linear function. In the case of two variables, the linear function takes the form of the linear equation y = ax + b for a = 0. For example, y = 3x + 5 is an example of a linear function. Given a value of x, one can determine the corresponding value of y using this functional rela- tionship. For instance, when x = 2, y = 3 × 2 + 5 = 11 and when x = −3, y = 3 × (−3) + 5 = −4. We will say more about functions in Chapter 4. Lin- ear equations or functions may be portrayed by a straight line on a graph. In this chapter, we introduce graphs and give a number of examples showing how linear equations can be used to model situations in economics and how to interpret properties of their graphs. 23 24 Elements of Mathematics for Economics and Finance 2.2 Solution of Linear Equations A mathematical statement setting two algebraic expressions equal to each other is called an equation. The ability to solve equations is one of the most impor- tant algebraic techniques to master. Equipped with this skill, you will be able to solve a range of economic problems. The simplest type of equation is the linear equation in a single variable or unknown, which we will denote by x for the moment. In a linear equation, the unknown x only occurs raised to the power 1. The following are examples of linear equations: 1. 5x + 3 = 11, 2. 1 − 4x = 3x + 7, 2 + 3x 2x − 1 3. =. 5 6 A linear equation may be solved by rearranging it so that all terms involving x appear on one side of the equation and all the constant terms appear on the other side. This is achieved by performing a series of algebraic operations. The key is to remember that you must perform the same operations to both sides of the equation. You must be completely impartial so that each stage of the rearrangement process yields an equivalent equation. Two equations are said to be equivalent if and only if when one holds then so does the other. Equivalent equations, therefore, have precisely the same solutions if they have any at all. However, it is important that you never multiply or divide through an equation by 0. For example, take the equation 1 = 2, which is not valid, and multiply both sides by 0. Then we obtain the equation 0 = 0, which is true. So the two equations are not equivalent. If an equation contains a fraction, then the equation may be simplified by multiplying through by the denominator. Remember that the value of a fraction a/b is the same if the numerator and denominator are multiplied (or divided) by the same nonzero number. That is, a ta = , b tb for any number t = 0. It is instructive to look at an example. Example 2.1 Solve the equation 7x − 4 = 2x + 4. 2 2. Linear Equations 25 Solution. To determine the value of x that satisfies this equation, we rearrange the equation so that all terms involving the unknown x appear on the one side of the equation and all the constant terms appear on the other. 1. Multiply both sides by 2, which is the denominator of the fraction on the left-hand side of this equation: 7x − 4 = 2 × (2x + 4) = (2 × 2)x + (2 × 4) = 4x + 8. 2. Subtract 4x from both sides so that all terms involving x are on the left- hand side: 7x − 4 − 4x = 4x + 8 − 4x, 3x − 4 = 8. 3. Add 4 to both sides so that all the constant terms are on the right-hand side: 3x − 4 + 4 = 8 + 4, 3x = 12. 4. Finally divide both sides by 3: 3x 12 = , 3 3 x = 4. So the solution to this equation is x = 4. We can check to see if this answer is correct by replacing x by 4 in the original equation. If x = 4 is the correct solution, then the left- and right-hand sides of the equation should give the same numerical value. (7 × 4) − 4 LHS = 2 28 − 4 = 2 24 = 2 = 12 RHS = 2×4+4 = 12. 26 Elements of Mathematics for Economics and Finance Example 2.2 Solve the equation x x − 3 = + 1. (2.1) 4 5 Solution. Again, we go through the solution step-by-step. The idea is to rearrange the equation so that all terms involving x appear on the left-hand side and all the constant terms appear on the right-hand side. Once this is done, the terms involving fractions are simplified. 1. Subtract x/5 from both sides: x x − −3=1 4 5 2. Add 3 to both sides x x − =1+3=4 4 5 3. Simplify the left-hand side by expressing it as a single fraction. This is achieved by expressing each of the fractions in terms of their lowest common denominator, 20. In the case of the first fraction, both the numerator and denominator are multiplied by 5, and in the case of the second fraction they are both multiplied by 4, i.e., x 5x 5x x 4x 4x = = and = =. 4 5×4 20 5 4×5 20 Therefore 5x 4x − = 4 20 20 5x − 4x = 4 20 x = 4. 20 4. Finally multiply both sides by 20: x = 80. The solution to this equation is x = 80. Again we can check that this is the correct solution by substituting x = 80 into the left- and right-hand sides of (2.1). 2. Linear Equations 27 2.3 Solution of Simultaneous Linear Equations A number of economic models are built on linear relationships between vari- ables. For example, the economic concept of equilibrium requires the solution of a system of equations. The next degree of difficulty is to solve two linear equations in two un- knowns. Suppose the two unknowns are denoted by x and y. The most general form of system of simultaneous linear equations in the unknowns x and y is a1 x + b1 y = c1 , (2.2) a2 x + b2 y = c2. (2.3) where a1 , b1 , c1 , a2 , b2 , and c2 are constants. In the first equation (2.2), the coefficient of x is a1 and that of y is b1. We are going to describe the elim- ination method for solving this system of equations. As its name suggests, the method involves eliminating one of the variables from the system. This allows us to determine the value of the unknown that remains by solving a single linear equation in one unknown. The value of the eliminated unknown is then determined by substituting the known value into either of the original equations and solving another linear equation. Suppose we wish to eliminate the variable y from (2.2)–(2.3). To do this, we multiply (2.2) by b2 and (2.3) by b1 so that the coefficients of y in the equivalent equations are the same: b2 a1 x + b2 b1 y = b2 c1 , (2.4) b1 a2 x + b1 b2 y = b1 c2. (2.5) Next we eliminate the variable y by subtracting (2.5) from (2.4): (b2 a1 − b1 a2 )x = b2 c1 − b1 c2 , (2.6) from which we deduce b2 c1 − b1 c2 x=. (2.7) b2 a1 − b1 a2 Note that we can only perform this last step provided that (b2 a1 − b1 a2 ) = 0. The quantity (b2 a1 − b1 a2 ) is known as the determinant (see Chapter 10) of the system of equations (2.2)–(2.3). The condition for this system to possess a unique solution is that the determinant is nonzero. Similarly, we can eliminate x from equations (2.2)–(2.3) to obtain c2 a1 − c1 a2 y= ; (2.8) b2 a1 − b1 a2 or we can obtain y by substituting the value of x we have obtained (2.7) in either (2.2) or (2.3) and solving the resulting linear equation. 28 Elements of Mathematics for Economics and Finance There is no guarantee that a system of two or more simultaneous equations will possess a unique solution. Consider the system of equations 2x + y = 10, 2x + y = 5. This system of equations does not have a solution. In fact, the equations are inconsistent. They cannot hold simultaneously since 10 = 5! We shall see later in this chapter that the solution of a system of simultaneous linear equations may be interpreted as the point of intersection of two straight lines. For the example under consideration, the two lines are parallel and therefore never intersect. Next consider the system of equations 2x + y = 10, −6x − 3y = −30. At first sight this might seem to be an innocuous system of equations. However, the second equation is just a multiple of the first; obtained by multiplying the first equation by −3. In this case, the equations are not independent. The second equation does not provide any additional information over the first equation. Since there are two unknowns to be determined, there is no unique solution – in fact there are infinitely many solutions. For the above system one can verify that x = s and y = 10 − 2s is a solution for any number s. To obtain a unique solution to a system of simultaneous linear equations, the equations must be consistent and independent and there must be as many equations as unknowns (variables). Example 2.3 Solve the system of equations 3x + 2y = 1 −2x + y = 2. Solution. We solve this system of equations using the elimination method in which we eliminate the variable x. To do this, we arrange for the coefficients of x in both equations to differ only in sign by multiplying the two equations by appropriate factors. The variable can then be eliminated by adding or sub- tracting the two equations. For example, suppose we multiply the first equation by 2 and the second by 3: 6x + 4y = 2 −6x + 3y = 6. 2. Linear Equations 29 The variable x is eliminated by adding the two equations: 7y = 8, which, after division by 7, gives 8 y=. 7 This value can now be substituted in either of the original two equations to obtain the corresponding value of x. Let us use the first equation, then
 8 3x + 2 = 1 7 16 3x + = 1 7 16 3x = 1 − 7 7 − 16 3x = (since 1 = 7/7) 7 9 3x = − 7
 1 9 x = × − 3 7 3 x = − 7 Therefore, the solution is x = −3/7, y = 8/7. Of course, we can check that we have the correct solution by substituting it back into the original set of equations and checking that the equations are satisfied. An alternative but equivalent method for solving simultaneous linear equa- tions is known as the substitution method. The idea is to rearrange one of the equations in order to isolate one of the variables on the left-hand side. The expression for this variable is then substituted into the second equation to yield a linear equation for the other variable. We demonstrate this by means of an example. Example 2.4 At the beginning of the year, an investor had £50,000 in two bank accounts, each of which paid interest annually. The interest rates were 4% and 6% per annum, respectively. If the investor has made no withdrawals during the year and has earned a total of £2,750 interest, what was the initial balance in each of the two accounts? 30 Elements of Mathematics for Economics and Finance Solution. Let x and y denote the initial balances in the accounts with interest rates 4% and 6%, respectively. Since the total amount invested at the start of the year was £50,000, we have x + y = 50,000. The amount of interest earned on the two bank accounts during the year is given by 0.04x and 0.06y, respectively. Since the total amount of interest earned during the year is £2,750, 0.04x + 0.06y = 2,750, or, after multiplying through by 100 4x + 6y = 275,000. Therefore, we have two equations with which to determine initial balances in the two bank accounts: x+y = 50,000 (2.9) 4x + 6y = 275,000. (2.10) Multiplying (2.9) by 4, we obtain 4x + 4y = 200,000. (2.11) Then subtracting (2.11) from (2.10) yields 2y = 75,000, so that y = 37,500. Finally, it follows from (2.9) that x = 12,500. Therefore, the initial balance in each of the two accounts was £12,500 and £37,500, re- spectively. 2.4 Graphs of Linear Equations Consider the linear equation y = 3x − 2. Given a value of x, one can use this equation to determine the corresponding value of y. For example, when x = 0, y = 3 × 0 − 2 = −2, and when x = 2, y = 3 × 2 − 2 = 6 − 2 = 4. The collection of all such pairs of values of x and y that satisfy this linear equation can be represented on a graph. 2. Linear Equations 31 Consider the two perpendicular lines shown in Fig. 2.1. The horizontal line is referred to as the x-axis and the vertical line as the y-axis. The point where these lines intersect is known as the origin and is denoted by the letter O. At this point, both variables take the value zero. Each axis is assigned a numerical scale that is chosen appropriately for the situation being considered. On the x-axis, the scale takes positive values to the right of the origin and negative values to the left. Moreover, the further we move away from the origin, the larger these values become. On the y-axis, the scale takes positive values above the origin and negative values below. Again, the further we move away from the origin in the vertical direction, the larger these values become. These axes enable us to define uniquely any point, P , in terms of its coordinates, (x, y). We write the coordinates (x, y) alongside the point P as in Fig. 2.1. The first number, x, denotes the horizontal distance along the x-axis and the second number y denotes the vertical distance along the y-axis. The arrows on the axis denote the positive direction. The collection of all points (x, y) satisfying a linear equation lie on a straight line. That is, any equation of the form y = ax + b, (2.12) where a and b are constants is a linear equation and can be represented by a straight line graph. We sometimes say that y is a linear function of x since in the equation defining y, the variable x only occurs linearly. Note also that the equation x = k, where k is any constant, is also repre- sented by a straight line graph: the ‘vertical’ line, parallel to the y-axis, through the point (k, 0). Example 2.5 Plot the following points A : (−2, 3), B : (−3, −4), C : (3, 5), D : (1, −4). Solution. The position of A is determined by the pair of values x = −2 and y = 3, and therefore it is located 2 units in the negative x-direction and 3 units in the positive y-direction as shown in Fig. 2.2. The other points are plotted in a similar way. The general form of a linear equation is cx + dy = e, (2.13) where c, d, and e are constants. We assume that c and d are not both zero. This equation contains multiples of x and y and a constant. These are the only terms involving x that are present in a linear equation; otherwise the equation 32 Elements of Mathematics for Economics and Finance y P:(x,y) y 0 x x Figure 2.1 The coordinate axes and the position of a general point P. is said to be nonlinear. The values c and d are referred to as the coefficients of x and y, respectively,. For example, the coefficients of the linear equation 2x − y = −3 are 2 and −1. More specifically, the coefficient of x is 2 and the coefficient of y is −1. Any equation of the form (2.13) can be rearranged into the form (2.12) provided d = 0. First subtract cx from both sides of (2.13): dy = −cx + e. Then divide both sides by d provided d = 0: c e y =− x+. (2.14) d d If we now compare this equation with (2.12) by comparing the coefficients of x and the constant terms in both equations, we see that (2.14) is just (2.12) with c e a=− , b=. d d 2. Linear Equations 33 y 6 5 C 4 A 3 2 1 0 -4 -3 -2 -1 1 2 3 4 -1 x -2 -3 B D -4 -5 Figure 2.2 The location of the points specified in Example 2.5. Note that when d = 0, the linear equation (2.13) reduces to e cx = e or x =. c This is represented by a straight line parallel to the y-axis passing through the point (e/c, 0) on the x-axis. To sketch the graph of a straight line, it is sufficient to draw a line through any two points lying on it. Example 2.6 Sketch the graph of the straight line y = 2x + 3, for values of x lying between 0 and 4. Solution. We determine the coordinates of two points on the line. When x = 0, we have that y = 3 and when x = 4, we have y = 11. Therefore, the points (0, 3) and (4, 11) lie on the line. The graph is formed by drawing a straight line through these points as shown in Fig. 2.3. 34 Elements of Mathematics for Economics and Finance y 12 (4,11) 10 8 6 4 (0,3) 2 0 1 2 3 4 x Figure 2.3 The graph of the equation y = 2x + 3. Example 2.7 Sketch the straight line 2x + y = 5. Solution. Setting x = 0 gives y = 5. Hence (0, 5) lies on the line. Setting y = 0 gives 2x = 5 or x = 5/2. Hence (5/2, 0) lies on the line. 2.4.1 Slope of a Straight Line The coefficients a and b in the linear equation y = ax + b of (2.12) have special significance and can be related to features of its graph. When x = 0, y = b and therefore the constant b represents the intercept on the y-axis, i.e., it is the value of y corresponding to the point of intersection of the straight line with the y-axis. The value of x for which y = 0 is the solution of the linear equation ax + b = 0. This equation has solution x = −b/a, provided a = 0. 2. Linear Equations 35 y 5 4 3 2 1 -1 0 1 2 3 x Figure 2.4 The graph of the equation 2x + y = 5. The coefficient a in the equation y = ax + b defines the slope or gradient of the straight line with that equation. The slope of a straight line provides important information about the behaviour of the relationship between the variables x and y. Let A : (x1 , y1 ) and B : (x2 , y2 ) be any two distinct points lying on a straight line as shown in Fig. 2.5. The slope or gradient of the line measures the ratio of the change in the vertical direction with respect to the change in the horizontal direction as one moves from A to B. We illustrate this with reference to Fig. 2.5. Since y1 = ax1 + b and y2 = ax2 + b, then y2 − y1 = ax2 − ax1 = a(x2 − x1 ). Therefore, BC y2 − y1 a(x2 − x1 ) = = = a, AC x2 − x1 x2 − x1 i.e. y2 − y1 BC a= =. (2.15) x2 − x1 AC The value of a is independent of the choice of points A, B on the line. Positive values of a correspond to straight lines where y increases as x increases, while negative values of a correspond to straight lines where y decreases as x increases. Larger values of a correspond to straight lines with steeper slopes. For example, 36 Elements of Mathematics for Economics and Finance y B:(x2 ,y ) 2 A:(x 1,y1 ) C 0 x Figure 2.5 The graph of a linear equation and its slope. the slope of the straight line y = 6x−3 is steeper than that of y = x+3. Another way of viewing the slope a is that it is the change in y when x increases by one unit, as then x2 − x1 = 1 and therefore a = y2 − y1. Example 2.8 Determine the slope and intercept of the straight line 9x + 3y = 4. Solution. We need to write this equation in the form y = ax + b. 9x + 3y = 4 3y = −9x + 4 4 y = −3x + 3 One can say immediately that the slope of this straight line is −3 and the intercept is 4/3. 2. Linear Equations 37 Example 2.9 Find the slope of the straight line that passes through the points (2, −1) and (−2, −11). Solution. The slope of a straight line passing through the points (x1 , y1 ), (x2 , y2 ) is y2 − y1 a =. x2 − x1 Therefore the required slope is −11 − (−1) −10 5 a = = =. −2 − 2 −4 2 2.5 Budget Lines Suppose that a company or an individual has a given budget, B, that can be used to purchase two goods. If the cost or price of each of these goods is known, then it is possible to determine the different combinations of the two goods that can be bought with the given budget. Suppose that the two goods are denoted by X and Y , and their respective prices are PX and PY. The quantities purchased of these goods is also denoted by X and Y. Then the equation of the budget line is PX X + PY Y = B. (2.16) Example 2.10 An electrical company has a budget of £6,000 a week to spend on the manu- facture of toasters and kettles. It costs £5 to manufacture a toaster and £12 to manufacture a kettle. Write down the equation of the budget line and sketch its graph. Solution. Let T and K denote the number of toasters and kettles that are manufactured each week. Then the cost of manufacture and the available bud- get means that the budget line has the equation 5T + 12K = 6,000. 38 Elements of Mathematics for Economics and Finance T 1200 1000 800 600 400 200 0 0 100 200 300 400 500 K Figure 2.6 The graph of the budget line 5T + 12K = 6,000. To sketch the graph of this budget line, it is sufficient to determine the coor- dinates of two points on the line. When T = 0, 12K = 6,000 and therefore K = 500. Similarly, when K = 0, 5T = 6,000 and therefore T = 1,200. The graph of the budget line is given by the straight line joining the points T = 0, K = 500 and T = 1,200, K = 0. The graph of the budget line is sketched in Fig. 2.6. Example 2.11 A person has £120 to spend on two goods (X, Y ) whose respective prices are £3 and £5. 1. Draw a budget line showing all the different combinations of the two goods that can be bought with the given budget (B). 2. What happens to the original budget line if the budget falls by 25%? 3. What happens to the original budget line if the price of X doubles? 4. What happens to the original budget line if the price of Y falls to £4? Draw the new budget lines in each case. 2. Linear Equations 39 Solution. 1. The general equation of a budget line is PX X + PY Y = B where PX is the price of X and PY is the price of Y. Now if PX = 3, PY = 5, B = 120, then the equation of the budget line is 3X + 5Y = 120. We can rearrange this equation to give 3 Y = − X + 24. 5 The graph of this budget line is represented by the solid line in Fig. 2.7. 2. If the budget falls by 25% it is reduced by 25% of £120, i.e., £30. The new budget B = £120 − £30 = £90. The equation for the new budget line is 3X + 5Y = 90, which, after rearrangement, can be written in the form 3 Y = − X + 18. 5 This line has the same slope as the original budget line but lies to the left of it. This is the dashed line in Fig. 2.7. 3. If PX = 6 the budget equation becomes 6X + 5Y = 120 or 6 Y = − X + 24. 5 This time the intercept remains the same as the original budget line but the slope is steeper – the slope is −6/5 compared with the slope of −3/5 of the original budget line. The graph of this budget line is represented by the long dashed line in Fig. 2.7. 4. If PY = 4 , then the budget equation is 3X + 4Y = 120, or 3 Y = − X + 30. 4 This time both the slope and the intercept change. See the dash-dot line in Fig. 2.7. 40 Elements of Mathematics for Economics and Finance Y 40 (1) (2) (3) 30 (4) 20 10 0 10 20 30 40 X Figure 2.7 The graph of the budget lines in Example 2.11. 2.6 Supply and Demand Analysis Microeconomics is concerned with the analysis of the economic theory and policy of individual firms and markets. The mathematics we have introduced so far can be used to calculate the market equilibrium in which the demand and supply of a particular good balance. The quantity demanded, Q, of a particular good depends on the market price, P. We shall refer to the way Q depends on P as the demand equation or demand function. Functions will be defined in more detail later in the book (Chapter 4). Economists normally plot the relationship between price and quantity with Q on the horizontal axis and P on the vertical axis. We assume that this relationship is linear, i.e., P = aQ + b, for some appropriate constants (parameters) a and b. A graph of a typical linear demand function is the dashed line in Fig. 2.8. Elementary theory shows that demand usually falls as the price of the good rises so the slope of the line is negative, i.e., a < 0. We say that P is a decreasing function of Q. Similarly, the supply equation or supply function is the relation between the quantity, Q, of a good that producers plan to bring to the market and 2. Linear Equations 41 P supply equation demand equation Po Qo Q Figure 2.8 The graph of typical linear demand and supply equations. The point of intersection provides the point of equilibrium for the model. the price, P , of the good. A typical linear supply curve is the solid line in Fig. 2.8. Economic theory indicates that as the price rises, so does the supply. Mathematically, P is then said to be an increasing function of Q. Note that the supply Q is zero when P = b. It is only when the price exceeds this threshold level that the producers decide that it is worth supplying any good whatsoever. We are interested in the interplay between supply and demand. Of particular significance is the point of intersection of the demand and supply curves (see Fig. 2.8). At this point, the market is said to be in equilibrium because the quantity demanded is equal to the quantity supplied. The corresponding price, P0 , and quantity, Q0 , are called the equilibrium price and quantity. It is also of interest to observe the effect of a shift of the market price away from its equilibrium price. Example 2.12 The demand and supply equations of a good are given by 4P = −Qd + 240, 5P = Qs + 30. 42 Elements of Mathematics for Economics and Finance Determine the equilibrium price and quantity. Solution. At market equilibrium, we have Qd = Qs = Q, say , where Q is the equilibrium quantity. In this case, the demand and supply equations become 4P = −Q + 240, 5P = Q + 30. This is a system of two simultaneous equations in the unknowns P and Q. We can eliminate Q from the system by adding the two equations. This gives 9P = 270. Then, dividing both sides by 9 gives the equilibrium price P = 30. Finally, the equilibrium quantity Q is determined by substituting this value into either of the demand or supply equations. The supply equation gives 5 × 30 = Q + 30, which, after rearrangement yields the equilibrium quantity Q = 120. Example 2.13 The demand and supply functions of a good are given by P = −Qd + 125, 2P = 3Qs + 30. Determine the equilibrium price and quantity. Determine also the effect on the market equilibrium if the government decides to impose a fixed tax of £5 on each good. Who pays the tax? 2. Linear Equations 43 Solution. At market equilibrium, we have Qd = Qs = Q, say , where Q is the equilibrium quantity. In this case, the demand and supply equations become P = −Q + 125, (2.17) 2P = 3Q + 30. (2.18) This is a system of two simultaneous equations in the unknowns P and Q. We can eliminate Q from the system by multiplying the demand equation (Eq. (2.17)) by 3: 3P = −3Q + 375, (2.19) and adding the resulting equation (2.17) to the supply equation (2.18). This gives 5P = 405, which, after dividing both sides by 5 gives the equilibrium price P = 81. Finally, the equilibrium quantity Q is determined by substituting this value into either of the demand or supply equations. The demand equation gives 81 = −Q + 125, which, after rearrangement yields the equilibrium quantity Q = 125 − 81 = 44. If the government imposes a fixed tax of £5 on each good, then the original supply equation needs to be modified. This is because the amount the supplier receives as a result of each sale is the amount that the consumer pays (P ) less the tax (£5), i.e., P −5. Thus, the new supply equation is obtained by replacing P by P − 5 in the original supply equation: 2(P − 5) = 3Qs + 30. (2.20) This equation can be simplified by multiplying out the bracket on the left- hand side and taking the constant term to the right-hand side. The new supply equation becomes 2P − 10 = 3Qs + 30, or 2P = 3Qs + 40. (2.21) 44 Elements of Mathematics for Economics and Finance We then proceed as before to determine the equilibrium price and quantity for the new situation. At market equilibrium, we have Qd = Qs = Q, say , where Q is the equilibrium quantity. In this case, the demand and supply equations become P = −Q + 125, (2.22) 2P = 3Q + 40. (2.23) We can eliminate Q from the system by multiplying the demand equation (Eq. (2.22)) by 3: 3P = −3Q + 375, (2.24) and adding the resulting equation (2.24) to the supply equation (2.23). This gives 5P = 415, which, after dividing both sides by 5 gives the equilibrium price P = 83. Finally, the equilibrium quantity Q is determined by substituting this value into either of the demand or supply equations. The demand equation gives 83 = −Q + 125, which, after rearrangement yields the equilibrium quantity Q = 125 − 83 = 42. The influence of government taxation on the equilibrium price is to increase it from £81 to £83. Therefore, not of all of the tax is passed on to the consumer. The consumer pays an extra £2 per good after tax has been imposed. The remaining part of the tax is borne by the supplier. 2.6.1 Multicommodity Markets At the beginning of this section, we looked at supply and demand analysis for a single good. We extend these ideas now to a multicommodity market. Suppose that there are two goods in related markets, which we call good 1 and good 2. 2. Linear Equations 45 The demand for either good depends on the prices of both good 1 and good 2. If the corresponding demand functions are linear, then Qd1 = a1 + b1 P1 + c1 P2 Qd2 = a2 + b2 P1 + c2 P2 where Pi and Qdi denote the price and demand for the ith good, and ai , bi , and ci are constants depending on the model. For the first equation a1 > 0, because there is a positive demand when the prices of both goods are zero. Also b1 < 0, because the demand of a good falls as its price rises. The sign of c1 depends on the nature of the two goods. If the goods are substitutable, then an increase in the price of good 2 would mean that consumers would switch from good 2 to good 1, causing Qd1 to increase. Substitutable goods are therefore characterized by a positive value of c1. On the other hand, if the goods are complementary, then a rise in the price of either good would see the demand fall so c1 is negative. Similar results apply to the signs of a2 , b2 and c2. Example 2.14 The demand and supply functions for two interdependent commodities are given by Qd1 = 145 − 2P1 + P2 Qs1 = −45 + P1 Qd2 = 30 + P1 − 2P2 Qs2 = −40 + 5P2 where Qdi , Qsi , and Pi denote the quantity demanded, quantity supplied, and price of good i, respectively. Determine the equilibrium price and quantity for this two-commodity model. Are these goods substitutable or complementary? Give reasons for your answer. Solution. At equilibrium, the quantity supplied is equal to the quantity de- manded for each good, so that Qd1 = Qs1 and Qd2 = Qs2. Let us write these respective common values as Q1 and Q2. Then for good 1 we have