mt105b_vle[1]

Mathematics 2M. AnthonyMT105b, 279005b

2011

Undergraduate study in Economics, Management, Finance and the Social Sciences

This subject guide is for a 100 course offered as part of the University of London International Programmes in Economics, Management, Finance and the Social Sciences. This is equivalent to Level 4 within the Framework for Higher Education Qualifi cations in England, Wales and Northern Ireland (FHEQ).

For more information about the University of London International Programmes undergraduate study in Economics, Management, Finance and the Social Sciences, see: www.londoninternational.ac.uk

This guide was prepared for the University of London International Programmes by:

Martin Anthony, Department of Mathematics, London School of Economics and Political Science.

This is one of a series of subject guides published by the University. We regret that due to pressure of work the author is unable to enter into any correspondence relating to, or arising from, the guide. If you have any comments on this subject guide, favourable or unfavourable, please use the form at the back of this guide.

University of London International ProgrammesPublications OfficeStewart House32 Russell SquareLondon WC1B 5DNUnited KingdomWebsite: www.londoninternational.ac.uk

Published by: University of London

University of London 2010

Reprinted with minor revisions 2011

The University of London asserts copyright over all material in this subject guide except where otherwise indicated. All rights reserved. No part of this work may be reproduced in any form, or by any means, without permission in writing from the publisher.

We make every effort to contact copyright holders. If you think we have inadvertently used your copyright material, please let us know.

Contents

Contents

1 General introduction 1

1.1 Mathematics 1 and Mathematics 2 . . . . . . . . . . . . . . . . . . . . . 1

1.2 Studying mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.3 Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.4 Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.5 How to use the subject guide . . . . . . . . . . . . . . . . . . . . . . . . 2

1.6 Recommended books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.6.1 Main text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.6.2 Other recommended texts . . . . . . . . . . . . . . . . . . . . . . 4

1.7 Online study resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.7.1 The VLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.7.2 Making use of the Online Library . . . . . . . . . . . . . . . . . . 6

1.8 Examination advice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.9 The use of calculators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 Further differentiation and integration, with applications 9

Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Revision of Mathematics 1 differentiation . . . . . . . . . . . . . . . . . . 9

2.3 Using derivatives for approximations . . . . . . . . . . . . . . . . . . . . 10

2.4 Taylors theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.5 Elasticities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.6 The effects of taxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.7 Revision of Mathematics 1 integration . . . . . . . . . . . . . . . . . . . 18

2.8 Definite integrals and areas . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.9 Consumer and producer surplus . . . . . . . . . . . . . . . . . . . . . . . 20

Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Sample examination/practice questions . . . . . . . . . . . . . . . . . . . . . . 23

Answers to activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

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Contents

Answers to Sample examination/practice questions . . . . . . . . . . . . . . . 25

3 Functions of several variables 33


Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Partial derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.3 Homogeneous functions and Eulers theorem . . . . . . . . . . . . . . . . 34

3.4 Optimisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.4.2 Unconstrained optimisation . . . . . . . . . . . . . . . . . . . . . 38

3.4.3 Applications of unconstrained optimisation . . . . . . . . . . . . . 38

3.4.4 Constrained optimisation . . . . . . . . . . . . . . . . . . . . . . . 40





4 Linear algebra and applications 51


Recommended reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2 Revision of Mathematics 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2.1 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2.2 Linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.3 Matrix inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.4 Inverse matrices and linear equations . . . . . . . . . . . . . . . . . . . . 53

4.5 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.5.1 The determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.5.2 Calculating determinants using row operations . . . . . . . . . . . 55

4.6 Cramers rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.7 The square linear system when A is not invertible . . . . . . . . . . . . . 58

4.8 Row operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

4.9 Calculating inverses using determinants . . . . . . . . . . . . . . . . . . . 62

4.10 Calculating inverses using row operations . . . . . . . . . . . . . . . . . . 64

4.11 Application: input-output analysis . . . . . . . . . . . . . . . . . . . . . . 66

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Contents

4.12 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.13 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.14 Finding eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . 67

4.15 Diagonalisation of a square matrix . . . . . . . . . . . . . . . . . . . . . 69





5 Differential equations 83


Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.2 Separable equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5.3 Linear equations and integrating factors . . . . . . . . . . . . . . . . . . 88

5.4 Second-order equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.5 Behaviour of solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

5.6 Coupled differential equations . . . . . . . . . . . . . . . . . . . . . . . . 93

5.6.1 Reducing to a second-order equation . . . . . . . . . . . . . . . . 94

5.6.2 Using diagonalisation . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.7 Applications of differential equations . . . . . . . . . . . . . . . . . . . . 99

5.8 Macroeconomics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

5.9 Continuous cash flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

5.10 Continuous price adjustment . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.11 Determining demand from elasticity . . . . . . . . . . . . . . . . . . . . . 102

5.12 Market trends . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103




Solutions to Sample examination/practice questions . . . . . . . . . . . . . . . 109

6 Difference equations 117


Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.2 Revision of Mathematics 1 . . . . . . . . . . . . . . . . . . . . . . . . . . 117

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Contents

6.2.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.2.2 Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

6.3 First-order difference equations . . . . . . . . . . . . . . . . . . . . . . . 118

6.4 Solving first-order difference equations . . . . . . . . . . . . . . . . . . . 120

6.5 Long-term behaviour of solution . . . . . . . . . . . . . . . . . . . . . . . 121

6.6 The cobweb model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

6.7 Financial applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

6.8 Homogeneous second-order difference equations . . . . . . . . . . . . . . 124

6.9 Non-homogeneous second-order equations . . . . . . . . . . . . . . . . . . 126

6.10 Behaviour of the solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

6.11 Coupled difference equations . . . . . . . . . . . . . . . . . . . . . . . . . 127

6.11.1 Solving by reducing to a second-order equation . . . . . . . . . . 127

6.11.2 Using diagonalisation . . . . . . . . . . . . . . . . . . . . . . . . . 129

6.12 Economic applications of second-order difference equations . . . . . . . . 130





A Sample examination paper 139

B Comments on the Sample examination paper 143

iv

1Chapter 1

General introduction

1.1 Mathematics 1 and Mathematics 2

If you are studying this subject, you will already have studied (or be concurrentlystudying) 05A Mathematics 1 or you will have a qualification which allowed youexemption from that course due to its equivalence. This subject builds uponMathematics 1. Everything in Mathematics 1 is essential to Mathematics 2. So,although Mathematics 2 is formally a separate subject, it is best thought of as anextension of Mathematics 1. Given this, it is essential that you have a goodunderstanding of Mathematics 1. In this subject guide, we will briefly review some ofthe important ideas and techniques from Mathematics 1 that we shall need forMathematics 2, but you should refer to the Mathematics 1 guide and the textbooks ifyou feel the need to refresh yourself on some of the more basic Mathematics 1 topics.

In Mathematics 2, we explore further some topics introduced in Mathematics 1 and westudy some more applications to the social sciences, particularly economics. Inparticular, we investigate further the applications of differentiation and integration, andfunctions of several variables: we shall see some new applications and also some newtechniques.

This subject also introduces some important new topics: for example, we shall meet newtechniques for solving linear equations, and we introduce differential and differenceequations.

This half course may not be taken with 76 Management mathematics, 173Algebra or 174 Calculus.

1.2 Studying mathematics

I make a number of points in the introductory chapter of the 05A Mathematics 1guide about the nature of studying mathematics, and these are worth repeating here.

The study of mathematics can be very rewarding. It is particularly satisfying to solve aproblem and know that it is solved. Unlike many of the other subjects you will study, inthe mathematics subjects, there is always a right answer. Although there may be onlyone right (final) answer, there could be a number of different ways of obtaining thatanswer, some more complex than others. Thus, a given problem will have only oneanswer, but many solutions (by which we mean routes to finding the answer).Generally, a mathematician likes to find the simplest solution possible to a givenproblem, but that does not mean that any other solution is wrong. (There may bedifferent, equally simple, solutions.)

1

1 1. General introduction

With mathematical questions, you first have to work out precisely what it is that thequestion is asking, and then try to find a method (hopefully a nice, simple one) whichwill solve the problem. This second step involves some degree of creativity, especially atan advanced level. You must realise that you can hardly be expected to look at everymathematics problem and write down a beautiful and concise solution, leading to thecorrect answer, straight away. For obvious reasons, teachers, lecturers, and textbooksrarely give that impression: they present the solution right there on the page or theblackboard, with no indication of the time a student might be expected to spendthinking or of the dead-end paths he or she might understandably follow before asolution can be found. It is a good idea to have scrap paper to work with so that youcan try out various methods of solution. You must not get frustrated if you cant solve aproblem immediately. As you proceed through the subject, gathering more experience,you will develop a feel for which techniques are likely to be useful for particularproblems. You should not be afraid to try different techniques, some of which may notwork, if you cannot immediately recognise which technique to use.

1.3 Aims and objectives

This half unit is designed to:

enable students to acquire further skills in the methods of calculus and linearalgebra (in addition to those in 05A Mathematics 1), as required for their use ineconomics-based subjectsprepare students for further courses in mathematics and/or related disciplines.

1.4 Learning outcomes

At the end of this half course and having completed the Essential reading and activities,you should have:

used the concepts, terminology, methods and conventions covered in the half courseto solve mathematical problems in this subjectthe ability to solve unseen mathematical problems involving understanding of theseconcepts and application of these methodsseen how mathematical techniques can be used to solve problems in economics andrelated subjects.

1.5 How to use the subject guide

This subject guide is absolutely not a substitute for the textbooks. It is only what itsname suggests: a guide to the study and reading you should undertake. In each of thesubsequent chapters, brief discussions of the syllabus topics are presented, together withpointers to recommended readings from the textbooks. It is essential that you usetextbooks. Generally, it is a good idea to read the texts as you work through a chapter

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11.6. Recommended books

of the guide. It is most useful to read what the guide says about a particular topic, thendo the necessary reading, then come back and re-read what the guide says to make sureyou fully understand the topic. Textbooks are also an invaluable source of examples foryou to attempt.

You should not necessarily spend the same amount of time on each chapter of the guide:some chapters cover much more material than others. The chapters are divided in theway they are in order to group together topics on particular central themes.

The discussion of some topics in this guide is rather more extensive than others. Often,this is not because those topics are more significant, but because the textbooktreatments are not as extensive as they might be.

Within each chapter of the guide you will encounter activities. You should carry outthese activities as you encounter them: they are designed to help you understand thetopic under discussion. Solutions to the activities are given near the ends of thechapters, but do make a serious attempt at them before consulting the solutions.

To help your time management, the chapters and topics of the subject are convertedbelow into approximate percentages of total time. However, this is purely forindicative purposes. Some of you will know the basics quite well and need to spend lesstime on the earlier material, while others might have to work hard to comprehend thevery basic topics before proceeding onto the more advanced.

Chapter Title % Time2 Further differentiation and integration 153 Functions of several variables 104 Linear algebra 255 Differential equations 256 Difference equations 25

At the end of each chapter, you will find a list of Learning outcomes. This indicateswhat you should be able to do having studied the topics of that chapter. At the end ofeach chapter, there are sample examination questions, which are based largely on pastexam questions. Some of the sample examination questions are really only samples ofparts of exam questions. Solutions are given at the ends of the chapters.

1.6 Recommended books

None of the books in the following reading list covers absolutely everything in thesyllabus. You should therefore ensure that you have access to a sufficient number of thebooks. At the beginning of each chapter, appropriate references are made to the booksdealing with the material of that chapter.

The main recommended text is the book by Anthony and Biggs. This covers most ofthe required material, and uses the same notations as this guide. This is therecommended book for 05A Mathematics 1, so most of you will already have a copy.

3


1.6.1 Main text

R Anthony, M. and N. Biggs, Mathematics for economics and finance. (Cambridge,UK: Cambridge University Press, 1996.) [ISBN 9780521559133].1

1.6.2 Other recommended texts

Please note that as long as you read the Essential reading you are then free to readaround the subject area in any text, paper or online resource. To help you readextensively, you have free access to the VLE and University of London Online Library(see below). Other useful texts for this course include:

R Binmore, K. and J. Davies, Calculus. (Cambridge, UK: Cambridge UniversityPress, 2001) [ISBN 9780521775410].

R Bradley, T. Essential mathematics for economics and business. (Chichester:Wiley, 2008) Third Edition. [ISBN 9780470018569].

R Dowling, Edward T. Introduction to mathematical economics. Schaums OutlineSeries. (New York; London: McGraw-Hill, 2000) Third Edition. [ISBN9780071358965].

R Ostaszewski, A. Mathematics in economics: models and methods. (Oxford, UK:Blackwell, 1993) [ISBN 9780631180562].

R Simon, C.P. and L. Blume, Mathematics for economists. (New York; London:W.W. Norton and Company Ltd, 1994) [ISBN 9780393957334].

The book by Anthony and Biggs covers most, but not all, of the required material. (Youwill need to refer elsewhere, as indicated in the guide, for the topics of Taylor series,and diagonalisation and its applications in differential and difference equations.) Eachchapter of Anthony and Biggs has a large section of fully worked examples, and aselection of exercises for the reader to attempt. Since this is the key text for 05AMathematics 1, it also will be useful for revision of that subject.

Binmore and Davies cover all the calculus you will need, and a lot more.

The book by Bradley covers most of the more basic material, and has plenty of workedexamples.

Dowlings book contains lots of worked examples. It is, however, less concerned withexplaining the techniques. It would not be suitable as your main text, but it is a goodsource of additional examples.

The book by Ostaszewski is very suitable for a number of the topics, and provides manyexamples.

The book by Simon and Blume is a large book covering everything in this subject andalso many topics outside the coverage of this subject.

There are many other books which cover the material of this subject, but those listedabove are the ones I shall refer to explicitly.

Detailed reading references in this subject guide refer to the editions of the settextbooks listed above. New editions of one or more of these textbooks may have been

1Recommended for purchase.

4

11.7. Online study resources

published by the time you study this course. You can use a more recent edition of anyof the books; use the detailed chapter and section headings and the index to identifyrelevant readings. Also check the virtual learning environment (VLE) regularly forupdated guidance on readings.

The single most important point to be made about learning mathematics is that tolearn it properly, you have to do it. Do work through the worked examples in atextbook and do attempt the exercises. This is the real way to learn mathematics. Inthe examination, you are hardly likely to encounter a question you have seen before, soyou must have practised enough examples to ensure that you know your techniques wellenough to be able to cope with new problems.

1.7 Online study resources

In addition to the subject guide and the Essential reading, it is crucial that you takeadvantage of the study resources that are available online for this course, including theVLE and the Online Library.

You can access the VLE, the Online Library and your University of London emailaccount via the Student Portal at:

http://my.londoninternational.ac.uk

You should receive your login details in your study pack. If you have not, or you haveforgotten your login details, please email [email protected] quoting yourstudent number.

1.7.1 The VLE

The VLE, which complements this subject guide, has been designed to enhance yourlearning experience, providing additional support and a sense of community. It forms animportant part of your study experience with the University of London and you shouldaccess it regularly.

The VLE provides a range of resources for EMFSS courses:

Self-testing activities: Doing these allows you to test your own understanding ofsubject material.

Electronic study materials: The printed materials that you receive from theUniversity of London are available to download, including updated reading listsand references.

Past examination papers and Examiners commentaries : These provide advice onhow each examination question might best be answered.

A student discussion forum: This is an open space for you to discuss interests andexperiences, seek support from your peers, work collaboratively to solve problemsand discuss subject material.

Videos: There are recorded academic introductions to the subject, interviews anddebates and, for some courses, audio-visual tutorials and conclusions.

5


Recorded lectures: For some courses, where appropriate, the sessions from previousyears Study Weekends have been recorded and made available.

Study skills: Expert advice on preparing for examinations and developing yourdigital literacy skills.

Feedback forms.

Some of these resources are available for certain courses only, but we are expanding ourprovision all the time and you should check the VLE regularly for updates.

1.7.2 Making use of the Online Library

The Online Library contains a huge array of journal articles and other resources to helpyou read widely and extensively.

To access the majority of resources via the Online Library you will either need to useyour University of London Student Portal login details, or you will be required toregister and use an Athens login:

http://tinyurl.com/ollathens

The easiest way to locate relevant content and journal articles in the Online Library isto use the Summon search engine.

If you are having trouble finding an article listed in a reading list, try removing anypunctuation from the title, such as single quotation marks, question marks and colons.

For further advice, please see the online help pages:

www.external.shl.lon.ac.uk/summon/about.php

1.8 Examination advice

Important: the information and advice given here are based on the examinationstructure used at the time this guide was written. Please note that subject guides maybe used for several years. Because of this we strongly advise you to always check boththe current Regulations for relevant information about the examination, and the VLEwhere you should be advised of any forthcoming changes. You should also carefullycheck the rubric/instructions on the paper you actually sit and follow thoseinstructions. Remember, it is important to check the VLE for:

up-to-date information on examination and assessment arrangements for this course

where available, past examination papers and Examiners commentaries for thecourse which give advice on how each question might best be answered.

A Sample examination paper may be found at the end of this subject guide. You willsee that from 200910, all of the questions on the paper are compulsory. Any furtherchanges to exam format will be announced on the VLE.

It is worth making a few comments about exam technique. Perhaps the most important,though obvious, point is that you do not have to answer the questions in any particular

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11.9. The use of calculators

order; choose the order that suits you best. Some students will want to do easyquestions first to boost their confidence, while others will prefer to get the difficult onesout of the way. It is entirely up to you.

Another point, often overlooked by students, is that you should always include yourworking. This means two things.

First, do not simply write down the answer in the exam script, but explain yourmethod of obtaining it (that is, what I called the solution earlier).

Secondly, include your rough working. You should do this for two reasons:

If you have just written down the answer without explaining how you obtainedit, then you have not convinced the Examiners that you know the techniques,and it is the techniques that are important in this subject. (The Examinerswant you to get the right answers, of course, but it is more important that youprove you know what you are doing: that is what is really being examined.)

If you have not completely solved a problem, you may still be awarded marksfor a partial, incomplete, or slightly wrong, solution: if you have written downa wrong answer and nothing else, no marks can be awarded. (You may havecarried out a lengthy calculation somewhere on scrap paper where you made asilly arithmetical error. Had you included this calculation in the exam answerbook, you would probably not have been heavily penalised for the arithmeticalerror.) It is useful, also, to let the Examiners know what you are thinking. Forexample, if you know you have obtained the wrong answer to a problem, butyou cant see how to correct it, say so!

As mentioned above, you will find that, wherever appropriate, there are sample examquestions at the end of the chapters. These are based in large part on the questionsappearing in past examination papers. As such, they are an indication of the types ofquestion that might appear in future exams. But they are just an indication. TheExaminers want to test that you know and understand a number of mathematicalmethods and, in setting an exam paper, they are trying to test whether you do indeedknow the methods, understand them, and are able to use them, and not merely whetheryou vaguely remember them. Because of this, you will probably encounter somequestions in your exam which seem unfamiliar. Of course, you will only be examined onmaterial in the syllabus. Furthermore, you should not assume that your exam will bealmost identical to the previous years: for instance, just because there was a question,or a part of a question, on a certain topic last year, you should not assume there will beone on the same topic this year. For this reason, you cannot guarantee passing if youhave concentrated only on a very small fraction of the topics in the subject. This mayall sound a bit harsh, but it has to be emphasised.

1.9 The use of calculators

You will not be permitted to use calculators of any type in the examination. This is notsomething that you should panic about: the Examiners are interested in assessing thatyou understand the key methods and techniques, and will set questions which do notrequire the use of a calculator.

7


In this guide, I will perform some calculations for which a calculator would be needed,but you will not have to do this in the exam questions. Look carefully at the answers tothe sample exam questions to see how to deal with calculations. For example, if theanswer to a problem is

2, then leave the answer like that: there is no need to express

this number as a decimal (for which one would need a calculator).

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2Chapter 2

Further differentiation and integration,with applications

Essential reading

(For full publication details, see Chapter 1.)

R Anthony and Biggs (1996) Chapters 6, 7, 8, 9, 25 and 26.

Further readingR Binmore and Davies (2001) Section 2.13 (on Taylors theorem) and Section 10.5

(on producer and consumer surplus)

R Bradley (2008) Chapters 6 and 8.

R Dowling (2000) Chapters 3, 4, 16 and 17.

R Ostaszewski (1993) Chapters 10 and 14.

R Simon and Blume (1994) Sections 3.5, 3.6 and 20.1.

2.1 Introduction

You will have studied 05A Mathematics 1 or its equivalent, and be familiar withdifferentiation and integration. In this chapter of the subject guide we will look at somemore applications of both differentiation and integration.

2.2 Revision of Mathematics 1 differentiation

From Mathematics 1, you should remember the meaning of the derivative, and knowhow to calculate derivatives. You should also know about some of the applications ofthe derivative.

Just to remind you, here are some of the important things we learned in Mathematics 1.

The derivative of a function f(x) at a point a is the instantaneous rate of change ofthe function at a. Formally,

f (a) = limh0

f(a+ h) f(a)h

.

9

22. Further differentiation and integration, with applications

We have the following standard derivatives:

f(x) f (x)xk kxk1

ex ex

lnx 1/xsinx cosxcosx sinx

We have the following rules for calculating derivatives

The sum rule: If h(x) = f(x) + g(x), then

h(x) = f (x) + g(x).

The product rule: If h(x) = f(x)g(x), then

h(x) = f (x)g(x) + f(x)g(x).

The quotient rule: If h(x) = f(x)/g(x) and g(x) 6= 0, then

h(x) =f (x)g(x) f(x)g(x)

g(x)2.

The composite function rule: if f(x) = s(r(x)), then

f (x) = s(r(x))r(x).

The derivative and second derivative can be used in optimisation and incurve-sketching. Particular applications in economics include profit-maximisation.

2.3 Using derivatives for approximations

Another way of looking at the definition of f is to think of it as an approximation,1

which tells us how a small change in the input x affects the output f(x). If we denote asmall change in x by x, then the resulting change in f(x) is

f = f(x+ x) f(x).Since f (x) is the limit of f/x as x approaches zero, for small values of x we have

f (x) ' fx

, or f ' f (x)x,

where the symbol ' means is approximately equal to. In d notation (in which thederivative is denoted by df/dx rather than f ), we have

f ' dfdx

x.

1See Anthony and Biggs (1996) Section 6.1.

10

22.3. Using derivatives for approximations

We made mention of this idea when we discussed marginal cost in Mathematics 1, andalso when we looked at the meaning of the Lagrange multiplier.

A couple of examples will serve to illustrate this simple use of the derivative forapproximations.

Example 2.1 Let us use the derivative to find the approximate change in thefunction f(x) = x4 when x changes from 3 to 3.005.

The derivative of f is f (x) = 4x3 and we therefore have the approximation

f ' f (3)x = 4(3)2 0.005 = 0.540.

The actual value of the change is

f = f(3.005) f(3) = (3.005)4 34 = 0.54135,

so our approximation is correct to two decimal places.

Example 2.2 Suppose the demand equation for a good is p3q = 8000 where q isthe number of units demanded (in thousands per week) and p is the price per unit,in dollars. If p is increased from $20 to $21, what will be the approximate fall insales? If, on the other hand, production were to be increased from 1000 units perweek to 1100, what would be the approximate fall in price?

The demand function and its derivative are

qD(p) = 8000p3, (qD)(p) = 8000 (3)p4 = 24000p4

.

Therefore when p = 20 and p = 1 we have

q ' (24000/p4)p = (24000/204)(1) = 0.15.

Remembering that q is measured in thousands of units, it follows that 150 fewerunits will be sold per week.

For the second question we have to consider p as a function of q, and so we need theinverse demand function and its derivative:

pD(q) = 20q1/3, (pD)(q) = 20 (1/3)q4/3.

So when q = 1 and q = 0.1 we have

p ' (20/3)q4/3 0.1 = 2/3.

The conclusion is that the price falls by about 67 cents.

Activity 2.1 Suppose that the demand for a good is given by the equation

p2q = 6000,

where q is the quantity (in thousands) and p is the price in dollars. If the price isincreased from $20 to $21, what is the approximate fall in expected sales?

11


2.4 Taylors theorem

We have just seen that the derivative may be used to give an approximation to a smallchange in value of a function. A more precise result along these lines is Taylorstheorem. This relates the function value not just to the derivative, but higher-orderderivatives (second-order, third-order, and so on).

Under certain circumstances, functions can be approximated well by certainpolynomials, referred to as power series.2 The appropriate way to do this is by usingTaylor series. To explain this, we first introduce a piece of notation. The derivative of afunction f is denoted f and the second derivative is denoted by f

. If we differentiate

the second derivative, we obtain the third derivative f (3), and if we differentiate this weobtain the fourth derivative f (4), and so on. In general, we have the nth derivative f (n).Taylors theorem states that if f is a function which can be differentiated n times, then,for certain ranges of x, the approximation

f(x) ' f(0) + f (0)x+ f (0)x2

2!+ f (0)

x3

3!+ + f (n)(0)x

n

n!,

is valid, where n! = n(n 1)(n 2) . . . 2 (called n factorial) is the product of all thepositive integers up to and including n. (Generally, the larger n, the better theapproximation.) The right-hand side of this approximation is known as the Taylor seriesfor f . Strictly speaking, the stated approximation is Taylors theorem about 0, oftencalled Maclaurins theorem, and the right-hand side is the Maclaurin series. Theright-hand side is known as a power series expansion of the function f .

A more general form of Taylors theorem is often useful. This states that (again, withcertain qualifications on the set of x for which the approximation is true), we have

f(x) ' f(a) + f (a)(x a) + f (a)(x a)2

2!+ + f (n) (x a)

n

n!.

This is called the Taylor expansion of f about a. Taking n = 1 we obtain

f(x) ' f(a) + f (a)(x a).Writing x = a+ h, and f = f(x) f(a), this becomes the simple rule we met earlier:

f ' f (a)x.Thus, Taylors theorem, for n > 1, is a generalisation of our simple approximation ruleand will, in general, be more accurate. When we refer simply to Taylors theorem orTaylor series we shall mean, for the sake of simplicity, those about 0 (that is, Maclaurinstheorem and Taylors theorem): so, unless it is otherwise specified, we take a = 0.

Example 2.3 The exponential function f(x) = ex has all its derivatives equal toex. Since e0 = 1, we therefore have that, for all n, f (n)(0)/n! = 1/n! and hence

ex ' 1 + x+ x2

2!+x3

3!+ + x

n

n!.

In fact, this approximation is valid for all x (though it requires mathematics outsidethe content of this subject to explain why).

2See, for example, Ostaszewski (1993) Sections 14.114.2, and Binmore and Davies (2001) Section2.13, for discussion of Taylors theorem.

12

22.4. Taylors theorem

Other important Taylor series are as follows:

ln(1 + x) ' x x2

2+x3

3 x

4

4+ . . . ,

valid for 1 < x 1,

sinx ' x x3

3!+x5

5! x

7

7!+ . . . ,

valid for all x, and

cosx ' 1 x2

2!+x4

4! x

6

6!+ . . . ,

valid for all x.

You should convince yourself that the Taylor series for these functions are as stated, bycalculating the derivatives. These standard Taylor series should be remembered.

The power series expansions for more complicated functions can often be determined byusing the standard power series given above, as the following example illustrates.

Example 2.4 We expand, as a power series up to x4, the function

f(x) = cos (ln(1 + x)) .

This means that we find the Taylor approximation with n = 4. One approach is tocalculate the first, second, third, and fourth derivatives of f and use Taylorstheorem directly. However, well take a different approach. (It is a good exercise tocheck that the other approach gives the same answer.) Well use the facts that

ln(1 + x) ' x x2

2+x3

3 x

4

4,

and

cos y ' 1 y2

2!+y4

4!.

It follows from these, by taking

y = ln(1 + x) ' x x2

2+x3

3 x

4

4,

that

cos (ln(1 + x)) = cos y

' 1 12y2 +

1

24y4

' 1 12

(x x

2

2+x3

3 x

4

4

)2+

1

24

(x x

2

2+x3

3 x

4

4

)4.

13


Because we only need the terms involving x4 or lower powers of x, the only relevantterms we get from the y2/2 term are given by

1

2

(x x

2

2+x3

3 x

4

4

)2=

1

2

(x x

2

2+x3

3 x

4

4

)(x x

2

2+x3

3 x

4

4

)=

1

2

((x)(x) 2

(x2

2

)(x) +

(x2

2

)2+ 2

(x3

3

)(x) +

)

=x2

2 x

3

2+

11

24x4 + ,

where indicates terms which involve powers of x greater than 4 (and termswhich we can therefore ignore).

Similarly, the only relevant term we get from the y4/24 term is x4/24. To explainwhy this is so, note that (

x x2

2+x3

3 x

4

4

)4,

is the bracketed expression (x x

2

2+x3

3 x

4

4

),

multiplied by itself four times. The terms which arise from this product are obtainedby multiplying together four objects, one from each occurrence of the bracketedexpression. Since the term with lowest power of x in each bracket is x, it is only bytaking the x from each bracket that we obtain a term with a degree of no more than4; and since all terms of degree greater than 4 are to be ignored, this is therefore theonly term we consider from y4/24.

Therefore, putting this all together, we have

cos (ln(1 + x)) ' 1(x2

2 x

3

2+

11

24x4)

+x4

24

= 1 x2

2+x3

2 5

12x4,

and this is the required expansion.

Activity 2.2 By taking x = 0.1 in the expansion just obtained, find anapproximate value for cos(ln(1.1)).

2.5 Elasticities

When the price of a commodity increases from p to p+ p, there is a change in thequantity demanded (usually a decrease). Suppose that the quantity demanded changes

14

22.5. Elasticities

from q to q + q (where q can be negative). Then, q = qD(p) andq + q = qD(p+ p). It follows that

q = qD(p+ p) qD(p) ' (qD)(p)p.The relative change in quantity is q/q and the relative change in price is p/p. Theratio of these is

(q/q)

(p/p)=p

q

q

p' pq

(qD)(p).

Since, generally, q will be negative, and since p > 0,

pq

q

p,

will be positive. We define the point elasticity of demand to be

= pq

dq

dp,

where q denotes the demand quantity qD(p).

For a typical demand function, the elasticity of demand is positive. It should be notedthat many texts omit the negative sign in the definition of elasticity of demand. Thedemand is said to be elastic if the elasticity is greater than 1, and inelastic if theelasticity is less than 1.

Example 2.5 Suppose that the demand curve has equation q = 2 p. Then thepoint elasticity of demand is

= pq

dq

dp= p

q(1) = p

q.

This can be written solely in terms of q as (2 q)/q or in terms of p as p/(2 p).

Example 2.6 Suppose that the demand function is given by q = 5/p2. Thendq/dp = 10/p3 and so the elasticity of demand is

= pq

dq

dp= p

q

(10p3

)= p

(5/p2)

(10p3

)= 2.

This is a constant, not depending on p or q.

Suppose a firm is producing a particular good, and let us think about how its totalrevenue changes if the selling price increases. If the price p rises, the quantity sold qfalls; but the revenue R = qp is the product of these two things, and it may rise or fall.We shall assume that a price rise applies uniformly to the entire supply of the goodunder consideration, which would certainly be the case if the firm is a monopoly. Usingthe product rule to differentiate total revenue TR = qp with respect to p, rememberingthat q is a function of p, we get

TR = (qp) = qp+ q.

15


Thus the condition that total revenue increases, TR > 0, is equivalent to qp+ q > 0,which is equivalent to

qpq< 1,

or < 1. Thus the elasticity determines whether revenue increases or decreases as priceincreases:

If < 1 (that is, demand is inelastic), then a small increase in price results in anincrease in revenue.

If > 1 (that is, demand is elastic), then a small increase in price results in andecrease in revenue.

The point elasticity of supply is defined in a way similar to that used to define pointelasticity of demand, but using the supply function qS rather than qD, and omitting thenegative sign. That is, the point elasticity of supply is given by

p

q(qS)(p) =

p

q

dqS

dp.

2.6 The effects of taxation

We now look at what happens in a market when a good is taxed. (This is not entirely adifferentiation topic, but this is as appropriate a chapter as any in which to place thistopic.)

When a fixed amount of tax is imposed on each unit of a good, we refer to this as anexcise tax or per-unit tax. The following example illustrates how we can determine thenew equilibrium price and quantity in the presence of such a tax.

Example 2.7 Suppose that the demand and supply functions for a good are givenby

qD(p) = 40 5p and qS(p) = 152p 10.

Then we can easily determine (by solving the equation q = qD(p) = qS(p)) that theequilibrium price is p = 4. Suppose that the government imposes an excise tax of Tper unit. How does this affect the equilibrium price?

The answer is found by noting that, if the new selling price is p, then, from thesuppliers viewpoint, it is as if the price were p T , because the suppliers revenueper unit is not p, but p T . In other words the supply function has changed: whenthe tax is T per unit, the new supply function qST is given by

qST (p) = qS(p T ) = 152

(p T ) 10.

Of course the demand function remains the same. Let us use qT and pT to denotethe new equilibrium values in the presence of the tax T . Then qT and pT satisfy theequations

qT = 40 5pT and qT = qST (pT ) = 152

(pT T ) 10.

16

22.6. The effects of taxation

Eliminating qT we get

40 5pT = 152

(pT T ) 10.Rearranging this equation, we obtain(

5 +15

2

)pT = 50 +

15

2T,

and so we have a new equilibrium price of

pT = 4 +3

5T.

For example, if T = 1, the equilibrium price rises from 4 to 4.6. Unsurprisingly, theselling price has risen (and, as you will discover from the Activity below, thequantity sold has decreased). But note that, although the tax is T per unit, theselling price has risen not by the full amount T , but by the fraction 3/5 of T . Inother words, not all of the tax is passed on to the consumer.

Activity 2.3 Show that the new equilibrium quantity is qT = 20 3T .

If we encountered a percentage tax rather than a per-unit tax, a similar sort of analysiswould apply. Again, the demand equation would remain unaltered, but the supplyequation would change: we would replace p by p(1 t) if the tax is 100t% of the sellingprice. (This is because, from the suppliers point of view, the revenue obtained fromeach item in other words, the effective price is not p, but p minus 100t% of p,which is p(1 t).)Suppose that a government wishes to raise revenue by imposing an excise tax on agood. Clearly, a small tax will bring in little revenue but, on the other hand, if the taxis too large consumption will fall dramatically and the revenue will also be hit. We canoften use differentiation to determine which level of tax will maximise the tax revenueto the government. Suppose that, as in the example above, we have demand and supplyfunctions given by

qD(p) = 40 5p and qS(p) = 152p 10.

We have seen that the equilibrium price and quantity in the presence of an excise tax Tare

pT = 4 +3

5T and qT = 20 3T.

(The determination of qT was Activity 2.3.) The revenue R(T ) is the product of thequantity sold qT and the excise tax T . That is,

R(T ) = qT T = (20 3T )T = 20T 3T 2.To find the value of Tm where this is a maximum, we first set R

(T ) = 0. We haveR(T ) = 20 6T , so that 10/3 is the only critical point. The second derivative isR(T ) = 6, which is negative, so Tm = 10/3 is the maximum point. The maximumrevenue is R(Tm) = 100/3.

17


2.7 Revision of Mathematics 1 integration

We now very briefly review the key topics in integration that were covered inMathematics 1.

Suppose the function f is given, and the function F is such that F (x) = f(x). Then wesay that F is an anti-derivative of f . Any two anti-derivatives of a given function fdiffer only by a constant. The general form of the anti-derivative of f is called theindefinite integral of f(x), and denoted by

f(x) dx.

Often we call it simply the integral of f . It is of the form F (x) + c, where F is anyparticular anti-derivative of f and c is an arbitrary constant, known as a constant ofintegration. The process of finding the indefinite integral of f is usually known asintegrating f , and f is known as the integrand.

Just as for differentiation, we shall have a list of standard integrals and some rules forcombining these.

f(x)f(x) dx

xn (n 6= 1) xn+1

n+ 1+ c

1/x ln |x|+ cex ex + csinx cosx+ ccosx sinx+ c

Note that the integral of 1/x is ln |x|+ c rather than lnx+ c because one cannot takethe logarithm of a negative number.

We also have the simple rules(f(x) + g(x)) dx =

f(x) dx+

g(x) dx,

for any functions f and g, andkf(x) dx = k

f(x) dx,

for any constant k.

If f is a function with an anti-derivative F , then the definite integral 3 of the function fover the interval [a, b] is defined to be b

a

f(x) dx = F (b) F (a).

In Mathematics 1, we met three important techniques for integration. First, we have thesubstitution or change of variable method. Formally, this uses the fact that when we


18

22.8. Definite integrals and areas

change the variable by putting x = x(u), we havef(x) dx =

f(x(u))x(u)du.

The rule for integration by parts is:u(x)v(x) dx = u(x)v(x)

u(x)v(x) dx.

The third technique is partial fractions. This involves rewriting integrands of the formp(x)/q(x), where p and q are polynomials, in a simpler form which makes them easier tointegrate. If p(x) is linear (that is, of the form ax+ b) and q(x) is a quadratic with twodifferent roots, then the method of partial fractions applies. Suppose thatq(x) = (x a1)(x a2), where a1 6= a2 and C is some number. Then it is possible towrite

p(x)

q(x)=

p(x)

(x a1)(x a2) =A1

x a1 +A2

x a2 , (?)

for some numbers A1 and A2. Cross-multiplying equation (?), we get

p(x) = A1(x a2) + A2(x a1).

The numbers A1 and A2 may be found by substituting x = a1, x = a2 in turn into thisidentity. When p(x)/q(x) is expressed in this way, it is easy to evaluate the integral. Wehave

p(x)

q(x)dx =

(A1

x a1 +A2

x a2

)dx

= A1 ln |x a1|+ A2 ln |x a2|+ c.

2.8 Definite integrals and areas

There is a useful relationship between the definite integral and the area under a curve.Suppose that f(x) 0 for x in the interval [a, b]. Then the area enclosed by the curvey = f(x), the x-axis and the vertical lines x = a and x = b is equal to

baf(x) dx.

The same result is essentially true even when the function is negative on part of theinterval, but in this case the area enclosed by that part of the graph of the function andthe x-axis is assigned a negative sign.

Example 2.8 What is the area under the curve y = x2 between x = 1 and x = 2?

To answer this, we first note that an anti-derivative of x2 is x3/3. So the area is 21

x2 dx =

[x3

3

]21

=23

3 1

3

3=

8

3 1

3=

7

3.

19


Activity 2.4 What is the area enclosed by the curve with equation y = sinx, thex-axis, the y-axis and the line x = pi?

2.9 Consumer and producer surplus

Figure 2.1 illustrates typical demand and supply curves for a good. Here, theequilibrium price, p, and quantity, q, are also indicated.

q

p

q

p

Demand

Supply

Figure 2.1: Typical supply and demand curves

At equilibrium, the consumers buy q units of the good at price p per unit, and hencethe total amount they pay is pq. This is the area denoted by R in Figure 2.2

q

p

q

p

DemandR

Figure 2.2: For q items at price p, consumers pay only pq, the area R

However, it can be argued4 that the total value the consumers place on q items of thegood is the area A of Figure 2.3, the area of the region that is bounded by the demandcurve, the q-axis, and the vertical lines q = 0 and q = q.

The difference between this area and the amount they actually pay is called theconsumer surplus. Since pq is the area of the rectangle R of length q and height p,the consumer surplus is the area of the region denoted CS in Figure 2.4, which is AR.


20

22.9. Consumer and producer surplus

q

p

q

p

Demand

A

Figure 2.3: Value of q items at price p is (approximately) the area A

q

p

q

p

DemandR

CS

Figure 2.4: The consumer surplus is the area CS (i.e. it is just AR)

Now, CS = AR. Area A is the area under the demand curve from q = 0 to q =, soA =

q0pD(q) dq, and area R is pq. Hence the consumer surplus is given by the

formula:

CS =

( q0

pD(q) dq

) pq.

An alternative way of viewing the consumer surplus is as the region enclosed by thedemand curve, the p-axis, p = p, and p = pD(0). Thus, we also have

CS =

pD(0)p

qD(p) dp.

Note that here we need to express q as a function of p on the demand curve, whereasthe first formula expresses p as a function of q. Whichever formula you use, try notsimply to memorise it, but rather remember what the relevant region is, and computeits area in the easiest way for a particular problem. (For example, if the demand curveis a straight line, then the region is question is triangular. Then, to compute its areaand hence the consumer surplus, we dont need a complicated formula involvingintegrals: we would simply use the well-known fact that the area of a triangle is one halfof the base times the height.)

There is a counterpart to the consumer surplus, known as the producer surplus. It canbe argued that the total cost to the manufacturers of producing q items is representedby the area under the supply curve between q = 0 and q = q. Yet they receive total

21


receipts, in equilibrium, of pq for producing and selling q units. The differencebetween these two values is known as the producer surplus. It is given by the formula

PS = pq q0

pS(q) dq.

Activity 2.5 The demand for a commodity is given by p(q + 1) = 231, and thesupply is given by p q = 11.Calculate the equilibrium price and quantity and the consumer surplus.

Learning outcomes

At the end of this chapter and the relevant reading, you should be able to:

as in Mathematics 1, know what is meant by the derivativeas in Mathematics 1, state the standard derivativesas in Mathematics 1, calculate derivatives using sum, product, quotient, andcomposite function (chain) rulesas in Mathematics 1, calculate derivatives by taking logarithmsas in Mathematics 1, establish the nature of the critical/stationary points of afunction.as in Mathematics 1, use the derivative to help sketch functionsas in Mathematics 1, know the terminology surrounding marginals in economics,and be able to find fixed costs and marginal costs, given a total cost functionas in Mathematics 1, know what is meant by the breakeven point and be able todetermine thisas in Mathematics 1, make use of the derivative in order to minimise or maximisefunctions, including profit functionsuse the derivative as a means of approximating small changes in a functionquote Taylors theorem, quote the standard Taylor (Maclaurin) series (for ex,ln(1 + x), sin x, and cosx, where the series are about x = 0), and know the rangesof x for which they are validdetermine power series approximations for functions using Taylors theorem and bymanipulating the standard Taylor seriesquote the definitions of point elasticity of demand and point elasticity of supplydetermine these elasticities given (respectively) the demand and supply equationsor setsknow what is meant by the terms elastic and inelastic, and be able to determinewhen demand is elastic or inelasticdetermine the new resulting equilibrium price and quantity in the presence of anexcise (per-unit) tax or a percentage taxas in Mathematics 1, understand the meaning of an (indefinite) integral and adefinite integralas in Mathematics 1, state the standard integralsas in Mathematics 1, use integration by substitution

22

22.9. Sample examination/practice questions

as in Mathematics 1, use integration by partsas in Mathematics 1, integrate using partial fractionsunderstand the connection between areas and definite integralscompute areas using definite integrationknow what is meant by consumer and producer surplus, and be able to calculateconsumer and producer surpluses

Sample examination/practice questions

Question 2.1

Use differentiation to find the approximate change inx as x increases from 100 to 101.

Show, more generally, that when n is large, the change inx as x increases from n to

n+ 1 is approximately 1/(2n).

Question 2.2

Express ln

1 x1 + x

as a series of terms in ascending powers of x up to and including x5.

Use your series to obtain an approximate value for ln(3/

11), giving your answercorrect to five decimal places.

Question 2.3

Expand as a power series, in terms up to x4, the function f(x) = cos(sin x).

Question 2.4

The demand quantity q for a good is given by

q(1 + p2) = 100,

where p is the price. Determine the point elasticity of demand as a function of p. Forwhat values of p is the demand elastic?

Question 2.5

The supply and demand functions for a good are

qS(p) = bp a, qD(p) = c dp,

where a, b, c, d are all positive, and bc > ad. Suppose the government wishes to raise asmuch money as possible by imposing an excise tax on the good. What should be thevalue of the excise tax? What is the resulting government revenue?

Question 2.6

Find the area enclosed by the curves y = 1/t2, y = t3, the t-axis and the lines t = 1/2and t = 2.

23


Question 2.7

The demand relationship for a product is

p =50

q + 5,

and the supply relationship is

p =q

10+

9

2.

On the same diagram, produce graphs of the supply and demand curves. Determine theconsumer and producer surpluses.

Question 2.8

Suppose that the demand and supply functions for a good are

qS(p) = bp a and qD(p) = c dp,

where a, b, c, d are positive constants. Find an expression for the consumer surplus.

Answers to activitiesFeedback to activity 2.1The demand function and its derivative are

qD(p) = 6000p2 and (qD)(p) = 6000 (2)p3 = 12000p3

.

Therefore when p = 20 and p = 1 we have

q '(12000

p3

)p = 12000

203= 1.5.

So the price rise will result in approximately 1500 fewer items being sold.

Feedback to activity 2.2You should obtain 0.995458. This compares well with the true value of 0.995461. (Ofcourse, this needs a calculator, but I stress again that such calculations will not berequired in the examinations, in which the use of calculators is not permitted.)

Feedback to activity 2.3The corresponding new equilibrium quantity is

qT = 40 5pT = 40 5(4 + (3/5)T ) = 20 3T.Feedback to activity 2.4The required area is the definite integral

pi0

sinx dx. This is calculated as follows: pi0

sinx dx = [ cosx]pi0 = cos(pi) ( cos(0)) = (1) + 1 = 2.

24

22.9. Answers to Sample examination/practice questions

Feedback to activity 2.5The inverse demand function is pD(q) = 231/(q + 1). The equilibrium quantity q is thesolution to the equation

231

q + 1= q + 11,

obtained by equating pD(q) and pS(q). So, q satisfies the equation

(q + 11)(q + 1) = 231

= q2 + 12q 220 = 0= (q + 22)(q 10) = 0.

Since q cannot be negative, q = 10. The equilibrium price is p = q + 11 = 21. Theconsumer surplus is then

CS =

q0

pD(q) dq pq

=

100

231

q + 1dq (21)(10)

=[231 ln(q + 1)

]100 210

= 231 ln(11) 231 ln(1) 210= 231 ln(11) 210,

which is approximately 343.9.

Answers to Sample examination/practice questions

Answer to question 2.1

The derivative of f(x) =x is f (x) = 1/(2

x). Therefore, by the approximation

f ' dfdx

x,

we have

f(101) f(100) ' f (100)(1) = 12

100=

1

20.

The approximate change is therefore 1/20 = 0.05. (Incidentally, the exact change is101100, which is 0.04987562, so the approximation is good.)

Generally, if n is large and x increases from n to n+ 1, then the change is givenapproximately by

f(n+ 1) f(n) ' f (n)((n+ 1) n) = 12n,

as required.

25



We first note that

ln

1 x1 + x

=1

2ln(1 x) 1

2ln(1 + x).

Now we use the following expansions:

ln(1 x) = x x2

2 x

3

3 x

4

4 x

5

5,

ln(1 + x) = x x2

2+x3

3 x

4

4+x5

5,

where we have omitted powers of x higher than x6 (noting that the question requires usto work with powers up to x5 only). It follows that we have, approximately,

1

2

(x x

2

2 x

3

3 x

4

4 x

5

5

) 1

2

(x x

2

2+x3

3 x

4

4+x5

5

),

and simplifying this, we see that the Taylor (Maclaurin) series is

ln

1 x1 + x

' x x3

3 x

5

5.

For the next part of the question, it would appear that we have to express 3/

11 in theform

(1 x)/(1 + x). For this to be so, we need

(1 x)/(1 + x) = (3/

11)2 = 9/11,

so 11(1 x) = 9(1 + x), or 20x = 2, so we take x = 0.1. Then,

ln

(311

)' (0.1) (0.1)

3

3 (0.1)

5

5

= 0.1 0.0013 0.00001

5= 0.1 0.000333 0.000002= 0.100335.

(Incidentally, the true value is 0.1003353480, so the approximation is correct to fivedecimal places.)


The series for sinx is

sinx ' x x3

3!+x5

5! x

7

7!+ ,

but since we are only interested in powers of x up to x4, we shall use the approximation

sinx ' x x3

3!= x x

3

6.

The series for cos y is

cos y ' 1 y2

2!+y4

4! y

6

6!+ .

26


To get an approximation to cos(sinx) up to terms in x4, we substitute

y = x x3

6

in the expansion for cos y and ignore any powers of x higher than x4. (Ignored termswill be denoted by .) We get

cos(sinx) ' 1 12

(x x

3

6

)2+

1

24

(x x

3

6

)4+

= 1 x2

2

(1 x

2

6

)2+x4

24

(1 x

2

6

)4+

= 1 x2

2

(1 x

2

3+x4

36

)+x4

24(1 + ) +

= 1 x2

2+x4

6+x4

24+

= 1 12x2 +

5

24x4 + .

Be careful about what can be ignored and what cannot. Consider, for example, the term

x4

2

(1 x

2

3+x4

36

).

In expanding this, since we are ignoring powers of x higher than 4, we can ignore all theterms inside the brackets apart from the first, since they lead to a power of x at least ashigh as x6. But you also have to be careful to include every term that needs to beincluded.


We haveq = 100/(1 + p2) = 100(1 + p2)1,

sodq

dp= 100(2p)(1 + p2)2 = 200p(1 + p2)2,

and the elasticity of demand is

= pq

dq

dp

= pq

( 200p(1 + p2)2

)=

200p2

q(1 + p2)2

=200p2

(100/(1 + p2)) (1 + p2)2

=2p2

1 + p2.

27


Note that in order to obtain the elasticity as a function of p, we needed to express q interms of p in the second-last of these equations. Now, the demand is elastic when > 1,which means 2p2/(1 + p2) > 1, or 2p2 > 1 + p2. So we have elastic demand when p2 > 1which, since p 0, means when p > 1.


The tax revenue is R(T ) = TqT . In order to calculate qT , we first note that when theexcise tax is imposed, the selling price at equilibrium, pT , is such that

qT = b(pT T ) a = c dpT .Solving for pT , we obtain

pT =c+ a

b+ d+

bT

b+ d.

Then

qT = c dpT = bc adb+ d

bdTb+ d

,

so that

R(T ) =

(bc adb+ d

)T

(bd

b+ d

)T 2.

Setting R(T ) = 0, we discover that there is only one critical point,

Tm =bc ad

2bd.

The second derivative R(T ) = (2bd)/(b+ d) is constant, and negative (since b and dare positive). Hence R(Tm) < 0 and Tm is a maximum point. Therefore, Tm is the levelof excise tax the government should impose. The resulting government revenue is

R(Tm) =(bc ad)24bd(b+ d)

.


It is a good idea to sketch the region described. Heres what it looks like.

Note that the curves y = 1/t2 and y = t3 intersect when t5 = 1 which, in the positivequadrant, means t = 1. The region were interested in then divides naturally into twoparts. The first is for t from 1/2 to 1, where the curve 1/t2 lies above t3 and the regionis therefore bounded by t = 1/2, t = 1 and y = t3. The second part is for t ranging from1/2 to 1: here, the curve y = t3 lies above y = 1/t2, and so the region is bounded byt = 1, t = 2 and y = 1/t2. The easiest way to compute the area, A, of the region is tocalculate each of the areas of the two parts, which well call A1 and A2, separately. ThenA = A1 + A2. We have

A1 =

11/2

t3 dt =

[t4

4

]11/2

=1

4(1)4 1

4

(1

2

)4=

15

64,

and

A2 =

21

1

t2dt =

[1t

]21

= 12(1

1

)=

1

2.

28


It follows that

A = A1 + A2 =47

64.


The supply and demand curves are as follows. Note that the supply curve is anupward-sloping straight line.

We need to find the equilibrium price and quantity. To find the equilibrium quantity, wesolve

50

q + 5=

q

10+

9

2.

Multiplying both sides by 10(q + 5), we obtain

500 = q(q + 5) + 45(q + 5) = q2 + 50q + 225,

29


soq2 + 50q 275 = 0.

The solutions are 5 and 55 and clearly it is the positive solution were interested in. Sothe equilibrium quantity is 5 and the equilibrium price is 50/(5 + 5) = 5.

Now, the consumer surplus is

CS =

50

50

q + 5dq (5)(5)

= [50 ln(q + 5)]50 25= (50 ln 10 50 ln 5) 25= 50 ln 2 25.

The producer surplus is easy to calculate, since it is the area of the triangular regionbounded by the p-axis, the line p = 5 and the portion of the supply curve between q = 0and q = 5. Since the supply curve intersects the p-axis at 4.5, this triangle has height 0.5and base 5, so its area is (1/2)(0.5)(5), which is 1.25. So the producer surplus is 1.25.


We first calculate the consumer surplus by elementary methods, using only the fact thatthe area of a triangle is half its base times its height. As is easily seen, the equilibriumprice is

p =c+ a

b+ d,

and the equilibrium quantity is

q = c dp = bc adb+ d

.

The inverse demand function is

pD(q) =c qd

.

The consumer surplus is the area of the triangular region bounded by the lines p = p

and q = 0, and by the demand curve. (Sketch the curves to see that this is so! The factthat the supply demand curve is a straight line means the region in question istriangular.) Since the demand curve crosses the p-axis at (0, pD(0)), this area is

CS =1

2(pD(0) p)q = 1

2

(c

d c+ ab+ d

)(bc adb+ d

).

That is,

CS =1

2

(cb+ cd cd ad

d(b+ d)

)(bc adb+ d

)=

(bc ad)22d(b+ d)2

.

We can also calculate the consumer surplus by definite integration. (This turns out tobe more difficult in this particular case, but definite integration really is needed fordemand functions more complex than the simple linear one discussed here.) We have

CS =

q0

pD(q) dq pq = q0

c qd

dq pq.

30


Now, q0

c qd

dq =

q0

( cd qd

)dq =

[c

dq q

2

2d

]q0

=c

dq (q

)2

2d,

so

CS =

(c

dq (q

)2

2d

) pq

=q

2d(2c q 2pd)

=q

2d

(2c bc ad

b+ d 2c+ a

b+ dd

)=

q

2d(b+ d)(2cb+ 2cd bc+ ad 2cd 2ad)

=q

2d(b+ d)(bc ad)

=(bc ad)22d(b+ d)2

,

as above.

31


32

3Chapter 3

Functions of several variables

Essential reading

(For full publication details, see Chapter 1.)

R Anthony and Biggs (1996) Chapters 11, 12, 13, 21 and 22.

Further readingR Binmore and Davies (2001) Sections 6.6 and 6.8.

R Bradley (2008) Chapter 7.

R Dowling (2000) Chapters 5 and 6.

R Ostaszewski (1993) Chapters 12 and 15.

3.1 Introduction

In this chapter of the guide, we briefly review the material from 05A Mathematics 1concerning functions of several variables, and we introduce some new topics andapplications related to functions of several variables.

3.2 Partial derivatives

We briefly review some key ideas from Mathematics 1 (though in the slightly moregeneral context of n-variable functions rather than 2-variable functions). A function ofn variables, for n 2, takes inputs (x1, x2, . . . , xn) and returns an output valuef(x1, x2, . . . , xn). In Mathematics 1, we concentrated almost entirely on the case inwhich n = 2, and we would often use x, y to denote the variables, rather than x1, x2. Inthis chapter we shall sometimes consider specific examples where n > 2. In particular,when n = 3, we shall often use x, y, z to represent the three variables rather thanx1, x2, x3.

For a function f(x1, x2, . . . , xn), the rate of change of f with respect to x1, when theother variables are fixed, is the partial derivative of f with respect to x1, denotedf/x1. The partial derivatives with respect to the other variables are similarly defined.

As we saw in Mathematics 1, calculating partial derivatives is only slightly moredifficult than calculating standard derivatives, and you should be proficient in this fromyour study of Mathematics 1. For example, to calculate the partial derivative of a

33

33. Functions of several variables

function f(x, y, z) with respect to x, you just treat y and z both as if they were fixednumbers, and differentiate with respect to x. Here is an example, to refresh yourmemory, and to indicate how to deal with 3-variable functions.

Example 3.1 Suppose f(x, y, z) = x2yy2 + z2. Writing this in the form

x2y(y2 + z2)1/2 makes it slightly easier to work with. We regard this as a product oftwo functions, where the second one is a composition. Applying the product andchain rules appropriately, we have

f

x= 2xy(y2 + z2)1/2 = 2xy

y2 + z2,

f

y= x2(y2 + z2)1/2 + x2y

(1

2

)(y2 + z2)1/2(2y)

= x2y2 + z2 +

x2y2y2 + z2

,

f

z= x2y

(1

2

)(y2 + z2)1/2(2z) =

x2yzy2 + z2

.

3.3 Homogeneous functions and Eulers theorem

Suppose we have a function of two variables, such as f(x, y) = 3x2y + 7xy2, and wemultiply the inputs x and y by a constant c. In this case we get output

f(cx, cy) = 3(cx)2(cy) + 7(cx)(cy)2 = c3(3x2y + 7xy2) = c3f(x, y).

Thus, for this particular f , multiplying the inputs by c results in the output beingmultiplied by c3. In general, if a function h is such that

h(cx, cy) = cDh(x, y),

then we say that h is homogeneous of degree D. The number D is called the degree ofhomogeneity of h. The function f given by the formula above is homogeneous of degree3. Note that many (indeed most) functions are not homogeneous. The notion of ahomogeneous function is related to the idea of returns to scale in economics. In thecase of a production function, we say that there are constant returns to scale if aproportional increase in k and l results in the same proportional increase in q(k, l); thatis, if

q(ck, cl) = cq(k, l).

This means, for example, that doubling both capital and labour doubles the production.This is the same as saying that q is homogeneous of degree 1. If q is homogeneous ofdegree D > 1 then the proportional increase in q(k, l) will be larger than that in k andl, and we say that there are increasing returns to scale. On the other hand, if D < 1 wesay that there are decreasing returns to scale.

34

33.3. Homogeneous functions and Eulers theorem

Example 3.2 Consider the function f(x, y) = x2y2 + y2x4 + y4. Then f is

homogeneous of degree 4 because

f(cx, cy) = (cx)2(cy)2 + (cy)2

(cx)4 + (cy)4

= c2x2c2y2 + c2y2c4(x4 + y4)

= c4x2y2 + c2y2c2x4 + y4

= c4x2y2 + c4y2x4 + y4

= c4f(x, y).

Informally speaking, a function such as the one just considered, which involves onlypowers of the variables and powers of combinations of the variables, is homogeneous ifthe total degree or power of each constituent term of the function is the same. In theexample we just saw, the total degree of the first part of the function, x2y2 is 2 + 2 = 4,the sum of the degree of x and the degree of y. The degree of the second term is2 + (1/2)4 = 4 (where the factor 1/2 comes from the square root). Since these are thesame, and both equal to 4, the function is homogeneous of degree 4. To see why thisinformal approach holds, just think about what happens when we substitute cx for xand cy for y. Perhaps it would also be useful to see why a function with parts ofdiffering degrees cannot be homogeneous. Consider the function f(x, y) = x+ y2. (Thishas terms of differing degree 1 and 2.) Then f(cx, cy) = cx+ c2y2. But there is no wayin which cx+ c2y2 can be written as cD(x+ y2), for all values of c, for some D.

Activity 3.1 Which of the following functions are homogeneous? If homogeneous,what are their degrees?

x2y +x4x2 + y2

,x

y+ 5 + ex

2/y2 , x2 + x sin y.

Example 3.3 Suppose we are told that the function

f(x, y) =xy + x2y

xy + y2,

is homogeneous of degree 2. Then we can use this fact to determine the numbers, and . There are several ways to do this. Think about what happens when x, yare replaced by cx, cy. We have

f(cx, cy) =c+1xy + c2+x2y

c1+xy + c2y2.

Now, if the function is homogeneous of degree 2, this should simply equal c2f(x, y).But for this to be true, three things must hold:

35


1. The degree of each of the two terms on the numerator (top line) should equaleach other; lets say this common degree is d1.

2. The degree of each of the two terms on the denominator (bottom line) shouldequal each other; lets say this degree is d2.

3. The degrees d1 and d2 should cancel to give degree 2; that is, the cd1 factor from

the top and the cd2 factor on the bottom should cancel to give c2. This meansthat d1 d2 = 2.

These conditions are equivalent to saying that

The numerator must be homogeneous.

The denominator must be homogeneous.

The degree of the numerator must be 2 more than the degree of thedenominator.

Respectively, these three conditions mean:

1. + 1 = 2 + .

2. 1 + = 2.

3. We can write four possible equations here. For example, since the degree of thefirst term on the numerator is + 1 and the degree of the first term on thedenominator is 1 + , we have + 1 (1 + ) = 2. By considering other terms(such as the second term on the numerator, and the first on the denominator,and so on), we obtain three other equations: + 1 2 = 2, 2 + (1 + ) = 2,and 2 + 2 = 2. But by the first and second observations above, these fourequations are all equivalent and we need only use one of them.

Choosing the second of the four equations in 3. (because it is simple) we obtain = 3. Then the equation in 1. tells us that = 2, and the equation in 2. tells usthat = 1. So the function is

f(x, y) =x3y + x2y2

xy + y2.

There is a useful result about homogeneous functions known as Eulers theorem. Thisstates that a function f(x, y) is homogeneous of degree D if and only if

xf

x+ y

f

y= Df.

Example 3.4 Consider the function f(x, y) = (x2 + y2)3/2x1/2y1/2. We can showthat this is homogeneous of degree 4, and verify that

xf

x+ y

f

y= 4f.

36

33.4. Optimisation

To show that the function is homogeneous of degree 4, we observe that

f(cx, cy) = ((cx)2 + (cy)2)3/2(cx)1/2(cy)1/2

= (c2)3/2(x2 + y2))3/2c1/2x1/2c1/2y1/2

= c4f(x, y).

Now,f

x=

3

22x(x2 + y2)1/2(x1/2y1/2) + (x2 + y2)3/2

(1

2x1/2y1/2

).

f

y=

3

22y(x2 + y2)1/2(x1/2y1/2) + (x2 + y2)3/2

(1

2x1/2y1/2

).

It follows that

xf

x+ y

f

y= (3x2 + 3y2)(x2 + y2)1/2(x1/2y1/2) + (x1/2y1/2)(x2 + y2)3/2

= 3(x2 + y2)3/2(x1/2y1/2) + (x2 + y2)3/2(x1/2y1/2)

= 4(x2 + y2)3/2(x1/2y1/2) = 4f(x, y),

as predicted by Eulers theorem.

Activity 3.2 That was a difficult example. Try an easier one for yourself, byverifying Eulers theorem for the function f(x, y) = x2y3 + x4y.

More generally, a function f(x1, x2, . . . , xn) of n variables is said to be homogeneous ofdegree D if for all c,

f(cx1, cx2, . . . , cxn) = cDf(x1, x2, . . . , xn).

Eulers theorem in this case is: f(x1, x2, . . . , xn) is homogeneous of degree D if and onlyif

x1f

x1+ x2

f

x2+ + xn f

xn= Df(x1, x2, . . . , xn).

3.4 Optimisation

3.4.1 Introduction

We now briefly review the optimisation techniques we met in Mathematics 1, and againwe shall describe them a little more generally, in the context of n-variable functions.Here, we will look at constrained optimisation of three-variable rather than simplytwo-variable functions. For unconstrained optimisation of such functions, there aretechniques for testing whether a critical point is a maximum, a minimum, or a saddlepoint. However, for functions of more than two variables, these techniques are outsidethe scope of this subject. Also not covered in this subject are tests for ensuring that thesolution to a constrained optimisation problem is a maximum or a minimum.

37


3.4.2 Unconstrained optimisation

The local maxima and minima of a function f(x1, x2, . . . , xn) occur at points where all

the partial derivativesf

xi(for i = 1, 2, . . . , n) are equal to 0. Such points are called

critical points or stationary points. A critical point which is neither a local maximumnor a local minimum is a saddle point. If n = 2, then there is a fairly simple test todetermine the nature of a critical point: Suppose that (a, b) is a critical point of f(x, y).Then the following statements are true, where the partial derivatives are evaluated at(a, b):

If2f

x22f

y2(2f

xy

)2> 0 and

f

x< 0, it is a maximum.

If2f

x22f

y2(2f

xy

)2> 0 and

f

x> 0, it is a minimum.

If2f

x22f

y2(2f

xy

)2< 0, it is a saddle point.

3.4.3 Applications of unconstrained optimisation

In Mathematics 1, we described the problem of maximising profit for a firm making twoproducts, X and Y . Generally, if pX and pY are the selling prices of one unit of X andone unit of Y , then the total revenue obtained by producing amounts x and y is

TR(x, y) = xpX + ypY .

The joint total cost function TC(x, y) will tell us how much it costs the manufacturer toproduce x units of X and y of Y . Then, the profit function is

(x, y) = TR(x, y) TC(x, y) = xpX + ypY TC(x, y),and we maximise this function of x and y using the techniques described above.1

There are many other interesting applications of unconstrained optimisation ineconomics. We now consider the case of a firm producing just one good, but selling thatgood in two markets, such as the domestic market and the export market.

Let us suppose that the firm is a monopoly in each of the two markets, and that thedemand curves for its product are different in the two markets. The firm may decide toset different prices in the domestic market and in the export market, or it may set thesame price in both. In the former case, we say that we have price discrimination. Thefollowing example demonstrates how to deal with such problems.

Example 3.5 Suppose the demand functions for a firms domestic and foreignmarkets are given by

P1 = 30 4Q1P2 = 50 5Q2.

1See Anthony and Biggs (1996) Chapter 13, for many examples.

38

33.4. Optimisation

and the total cost function isTC = 10 + 10Q,

where Q = Q1 +Q2. We shall determine the prices which maximise profit when wehave price discrimination, and when we have no price discrimination. First, we notethat the total revenue is P1Q1 + P2Q2, and hence the profit function is

(Q1, Q2) = (P1Q1 + P2Q2) (10 + 10Q)= (30 4Q1)Q1 + (50 5Q2)Q2 (10 + 10Q1 + 10Q2)= 20Q1 + 40Q2 4Q21 5Q22 10.

To find the maximum, we solve

Q1= 20 8Q1 = 0 and

Q2= 40 10Q2 = 0,

obtaining Q1 = 5/2 and Q2 = 4, and hence P1 = 20, P2 = 30. We can see that this isindeed a maximum of profit by using the second derivative test (check this!). Themaximum profit with price discrimination is therefore (5/2, 4) = 95. (Here, we havefound an expression for profit in terms of Q1 and Q2. An alternative approach wouldbe to determine as a function of P1 and P2 and maximise the resulting function ofP1 and P2.)

Now, when there is no price discrimination, we have P1 = P2 = P , say. From thedemand equations, we have

P = 30 4Q1 = 50 5Q2,so

Q1 =15

2 P

4and Q2 = 10 P

5,

and

Q = Q1 +Q2 =35

2 9P

20.

It follows that the profit, as a function of P , is given by

(P ) = P (Q1 +Q2) (10Q+ 10)= PQ 10Q 10

= P

(35

2 9P

20

) 10

(35

2 9P

20

) 10

= 22P 920P 2 185.

This is a function of the single variable P . (We could just as easily have expressed in terms of the single variable Q: both approaches are fine.) To maximise, we setd/dP = 0. This yields

22 910P = 0,

so P = 220/9. This gives a maximum of the profit function, sinced2/dP 2 = (9/10) < 0. The corresponding profit is (220/9) = 755/9, which is

39


83.88 to two decimal places. This is, as we would expect, less than the maximumprofit when price discrimination is allowed.

It is certainly possible to use Lagrangean methods, with the constraint P1 P2 = 0expressed in terms of Q1 and Q2, to deal with the case of no price discrimination.However, it is probably easier to take the approach indicated in the example justgiven, where the equation P1 = P2 is used to reduce the optimisation problem to aone-variable problem.

Activity 3.3 Follow up the comment just made by using the Lagrangean methodfor the last part of the example, in which there is no price discrimination.

3.4.4 Constrained optimisation

Suppose that f(x1, x2, . . . , xn) has to be minimised or maximised subject to theconstraint g(x1, x2, . . . , xn) = 0. This means we want to find the maximum (orminimum) value of the function f at points (x1, x2, . . . , xn) which satisfy the conditiong(x1, x2, . . . , xn) = 0. Then we may use the method of Lagrange multipliers.

2 To find theoptimal points, we first find the critical points of the (n+ 1)-variable function

L(x1, x2, . . . , xn, ) = f(x1, x2, . . . , xn) g(x1, x2, . . . , xn).

The function L is known as the Lagrangean (sometimes spelt Lagrangian) and isknown as the Lagrange multiplier. In other words, we find the points at which thefirst-order conditions

L

xi= 0 (for i = 1, 2, . . . , n) and

L

= 0.

Then the theory of Lagrange multipliers asserts that the required optimal points of f ,subject to the constraint, are to be found among these critical points.

In Mathematics 1, you studied the Lagrangean method quite extensively fortwo-variable functions. Here, we dont introduce much new material on this subject, butfor Mathematics 2, you should be able to deal with more challenging 2-variableconstrained optimisation problems, and with Lagrangean problems involving more thantwo variables. The following example serves as an illustration of the technique for suchfunctions.

Example 3.6 Use Lagrange multipliers to find the minimum value of

1

2x2+

1

3y2+

1

6z2,

for x, y, z > 0 subject to1

2x+

1

3y +

1

6z = c,

where c > 0 is a constant.

2See Anthony and Biggs (1996) Chapters 21 and 22.

40

33.4. Learning outcomes

To solve this, we form the Lagrangean

L =1

2x2+

1

3y2+

1

6z2

(1

2x+

1

3y +

1

6z c

),

and we solve the equations

L

x= 1

x3 1

2 = 0

L

y= 2

3y3 1

3 = 0

L

z= 1

3z3 1

6 = 0

L

= 1

2x 1

3y 1

6z + c = 0

The first three equations give three expressions for :

= 2x3

= 2y3

= 2z3,

from which it follows that x3 = y3 = z3 and hence x = y = z. Then, settingy = z = x in the final equation (which is simply the constraint) gives

1

2x+

1

3x+

1

6x = c,

which is x = c. Therefore the optimum (in this case minimum) is whenx = y = z = c, and the optimum value is

1

2c2+

1

3c2+

1

6c2=

1

c2.

Learning outcomes

At the end of this chapter and the relevant reading, you should be able to:

as in Mathematics 1, understand the concept of a function of many variablesas in Mathematics 1, calculate partial derivatives, and use implicit differentiationstate precisely what is meant by a homogeneous function of n variables, and beable to show that a gi

mt105b_vle[1]

Documents

revision of mathematics

london school of economics

department of mathematics

studying mathematics

copyright material

use of calculators

series of subject guides

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