BBMP1103 Mathematics for Management

BBMP1103 MATHEMATICS FOR MANAGEMENT Assoc Prof Dr Zurni Omar Noraziah Haji Man Hawa Ibrahim Fatinah Zainon Azizan Saaban

Copyright © Open University Malaysia (OUM)

Project Directors:

Prof Dato’ Dr Mansor Fadzil Prof Dr Wardah Mohamad Open University Malaysia

Module Writers:

Assoc Prof Dr Zurni Omar Noraziah Haji Man Hawa Ibrahim Fatinah Zainon Encik Azizan Saaban Universiti Utara Malaysia

Moderators:

Fatinah Zainon Noraziah Haji Man Assoc Prof Dr Zurni Omar Hawa Ibrahim Universiti Utara Malaysia Raziana Che Aziz Open University Malaysia

Developed by:

Centre for Instructional Design and Technology Open University Malaysia

Printed by:

Meteor Doc. Sdn. Bhd. Lot 47-48, Jalan SR 1/9, Seksyen 9, Jalan Serdang Raya, Taman Serdang Raya, 43300 Seri Kembangan, Selangor Darul Ehsan

First Edition, November 2007 Second Edition, October 2011 Third Edition, April 2013 (rs) Copyright © Open University Malaysia (OUM), April 2013, BBMP1103 All rights reserved. No part of this work may be reproduced in any form or by any means without the written permission of the President, Open University Malaysia (OUM).


Table of Contents Course Guide Topic 1

ix-xiii

Matrix 1.1 Classifications/Types of Matrices 1.1.1 Row Matrix (Row Vector) 1.1.2 Column Matrix (Column Vector) 1.1.3 Square Matrix 1.1.4 Diagonal Matrix 1.1.5 Special Matrix 1.2 Matrix Operations 1.2.1 Equality of Matrix 1.2.2 Transpose 1.2.3 Matrix Addition 1.2.4 Matrix Subtraction 1.2.5 Scalar Multiplication 1.2.6 Matrix Multiplication 1.3 Determinant 1.3.1 Minor of Element aij 1.3.2

Topic 2

Cofactor of Element aij

1 3 3 3 4 4 6 6 7 7 8 9 9 12 13 15

1.4 1.5

Inverse Matrix Solving Linear Equation System Using Matrices 1.5.1 Matrix Equation 1.5.2 Inverse Matrix Method 1.5.3 Cramer's Rule Summary Key Terms

17 18 19 20 23 28 28

Linear and Quadratic Functions 2.1 Linear Equations and Graphs Sketching 2.1.1 Linear Equations 2.1.2 Slope 2.1.3 Types of Straight Lines 2.1.4 Graphs Sketching 2.2 Parallel and Perpendicular Lines 2.3 Quadratic Equations and Graphs Sketching 2.4 Intersection Point

29 30 30 31 32 33 35 39 46


iv  TABLE OF CONTENTS

Summary Key Terms

51 51

Topic 3

Application of Linear and Quadratic Functions 3.1 The Demand and Supply Function 3.1.1 The Market Equilibrium Point 3.2 Cost and Revenue Functions 3.2.1 Break-even Point Analysis 3.3 Maximum and Minimum Summary Key Terms

52 52 53 57 59 60 63 64

Topic 4

Exponential and Logarithmic Functions 4.1 Properties of Exponentials 4.2 Equations and Graphs 4.3 Logarithmic Functions 4.4 Properties of Logarithms 4.5 Equations and Graphs 4.5.1 Application on Growth and Decay Processes 4.5.2 Investment with Compound Interest Summary Key Terms

65 65 68 71 72 75 77 79 82 83

Topic 5

Differentiation 5.1 Constant Rule 5.2 Power Rule 5.3 Constant Times a Function Rule 5.4 The Rules of Sums and Differences of Functions 5.5 Product Rule 5.6 Quotient Rule 5.7 Chain Rule 5.8 Power Rule (Special Case of Chain Rule) Summary Key Terms

84 85 85 86 87 90 91 93 95 99 99

Topic 6

Application of Differentiation 6.1 Second and Third Degree Differentiation 6.2 Total Cost Function (C )

6.3

6.2.1 Average Total Cost Function ( C ) 6.2.2 Marginal or Ultimate Total Cost Function (CÊ ) 6.2.3 Minimising Total Cost Total Revenue Function (R ) 6.3.1 Average Total Revenue Function ( R ) Copyright © Open University Malaysia (OUM)

100 101 102 103 104 105 108 108

TABLE OF CONTENTS  v

6.3.2 Marginal or Ultimate Total Revenue Function (RÊ ) 6.3.3 Maximising Revenue Functions 6.4 Total Profit Function () 6.4.1 Average Total Profit Function (  ) 6.4.2 Ultimate Total Profit Function (Ê) 6.4.3 Maximising Total Profit Summary Key Terms

108 109 112 113 114 115 119 119

Topic 7

Integration 7.1 Anti-derivatives 7.2 Definite Integrals 7.3 Integration by Substitution Summary Key Terms

120 120 125 128 132 132

Topic 8

Application of Integration 8.1 Finding Area Under a Graph 8.2 Application of Integration in Economics and Business 8.2.1 ConsumersÊ and ProducersÊ Surpluses 8.2.2 Finding Function from Its Marginal Function Summary Key Terms

133 134 138 138 142 147 147

Topic 9

Partial Differentiation 9.1 Functions of Multi-variable 9.2 Partial Derivatives 9.3 Higher order Partial Derivatives Summary Key Terms

148 149 153 156 159 159

Topic 10

Application of Partial Differentiation 10.1 Maximum and Minimum for Functions of Two Variables 10.2 Lagrange Multiplier Summary Key Terms

160 161 167 172 173

Answers

174


vi  TABLE OF CONTENTS


COURSE GUIDE



COURSE GUIDE

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ix

COURSE GUIDE DESCRIPTION You must read this Course Guide carefully from the beginning to the end. It tells you briefly what the course is about and how you can work your way through the course material. It also suggests the amount of time you are likely to spend in order to complete the course successfully. Please keep on referring to the Course Guide as you go through the course material as it will help you to clarify important study components or points that you might miss or overlook.

INTRODUCTION BBMP1103 Mathematics for Management is one of the courses offered by Faculty of Business and Management at Open University Malaysia (OUM). This course is worth three credit hours and should be covered over 15 weeks.

COURSE AUDIENCE This is a core course for all students undertaking Bachelor Degree Business Admin and Bachelor Degree Business Accounting programme. As an open and distance learner, you should be acquainted with learning independently and being able to optimise the learning modes and environment available to you. Before you begin this course, please ensure that you have the right course material, and understand the course requirements as well as how the course is conducted.

STUDY SCHEDULE It is a standard OUM practice that learners accumulate 40 study hours for every credit hour. As such, for a three-credit hour course, you are expected to spend 120 study hours. Table 1 gives an estimation of how the 120 study hours could be accumulated.


x



COURSE GUIDE

Table 1: Estimation of Time Accumulation of Study Hours Study Activities

Study Hours

Briefly go through the course content and participate in initial discussion

3

Study the module

60

Attend 3 to 5 tutorial sessions

10

Online participation

12

Revision

15

Assignment(s), Test(s) and Examination(s)

20

TOTAL STUDY HOURS ACCUMULATED

120

LEARNING OUTCOMES By the end of this course, you should be able to: 1.

Apply the basic mathematical concepts;

2.

Construct mathematical formulae;

3.

Assess mathematical concepts in their specific course; and

4.

Apply the usage of mathematics in daily life.

COURSE SYNOPSIS This course is divided into 10 topics. The synopsis for each topic can be listed as follows: Topic 1 discusses the classification of matrices, matrix operation and determinants. These will be followed by solutions to the simultaneous linear equations detail using the method of matrix inverse and CramerÊs rule. Topic 2 discusses linear and quadratic functions, solves equations and sketch graphs. Topic 3 discusses the applications of linear and quadratic functions mainly in Economics.


COURSE GUIDE

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xi

Topic 4 discusses both exponential and logarithm functions and how these two functions associate with one another. Topic 5 discusses the rules of differentiation which make the process of obtaining the derivatives for various functions simpler. Topic 6 discusses the process of deriving higher level differentiation and the function of total cost, total revenue and total profit in order to arrive at the minimum cost as well as maximum revenue and profit. Topic 7 will introduce integration as reversal process of differentiation. Further discussion will be on indefinite integrals, and definite integrals, followed by integration on algebra, and exponential and logarithm functions. The method which will be introduced is integration by substitution. Topic 8 discusses the application of integration i.e. finding area under a curve and its application in Economics and Business. Marginal functions, producersÊ surplus and consumersÊ surplus are the functions that will be discussed in detail. Topic 9 discusses multi-variable functions with its derivatives which is known as partial derivatives. This topic explains the first and second degree of partial derivatives. Topic 10 discusses the application of partial differentiation i.e. how to determine the maximum and minimum points of functions of two variables as well as applications of the business optimising functions with constraints.

TEXT ARRANGEMENT GUIDE Before you go through this module, it is important that you note the text arrangement. Understanding the text arrangement will help you to organise your study of this course in a more objective and effective way. Generally, the text arrangement for each topic is as follows: Learning Outcomes: This section refers to what you should achieve after you have completely covered a topic. As you go through each topic, you should frequently refer to these learning outcomes. By doing this, you can continuously gauge your understanding of the topic. Self-Check: This component of the module is inserted at strategic locations throughout the module. It may be inserted after one sub-section or a few subsections. It usually comes in the form of a question. When you come across this Copyright © Open University Malaysia (OUM)

xii 

COURSE GUIDE

component, try to reflect on what you have already learnt thus far. By attempting to answer the question, you should be able to gauge how well you have understood the sub-section(s). Most of the time, the answers to the questions can be found directly from the module itself. Activity: Like Self-Check, the Activity component is also placed at various locations or junctures throughout the module. This component may require you to solve questions, explore short case studies, or conduct an observation or research. It may even require you to evaluate a given scenario. When you come across an Activity, you should try to reflect on what you have gathered from the module and apply it to real situations. You should, at the same time, engage yourself in higher order thinking where you might be required to analyse, synthesise and evaluate instead of only having to recall and define. Summary: You will find this component at the end of each topic. This component helps you to recap the whole topic. By going through the summary, you should be able to gauge your knowledge retention level. Should you find points in the summary that you do not fully understand, it would be a good idea for you to revisit the details in the module. Key Terms: This component can be found at the end of each topic. You should go through this component to remind yourself of important terms or jargon used throughout the module. Should you find terms here that you are not able to explain, you should look for the terms in the module. References: The References section is where a list of relevant and useful textbooks, journals, articles, electronic contents or sources can be found. The list can appear in a few locations such as in the Course Guide (at the References section), at the end of every topic or at the back of the module. You are encouraged to read or refer to the suggested sources to obtain the additional information needed and to enhance your overall understanding of the course.

PRIOR KNOWLEDGE There is no prerequisite requirement for this course.

ASSESSMENT METHOD Please refer to myVLE.


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xiii

REFERENCES Abdul Razak Yaakub. (2002). Pengenalan kepada matematik untuk pengurusan. (Edisi Kedua). Sintok: Universiti Utara Malaysia. Bittinger, M. L. & Ellenbogen, D. J. (2007). Calculus and its applications (9th ed). Boston: Pearson International Edition Haeussler, E. F., Richards, S. P., & Wood, R. J. (2004). Introductory mathematical analysis for business, economics, and the life and social sciences (11th ed.). New Jersey: Prentice Hall. Kamran Sa'yan, & Ku Ruhana Ku Muhamud. (2002). Matematik untuk pengurusan, ekonomi & sains sosial. Sintok: Universiti Utara Malaysia. Lial/Miller (1980). Essential calculus with application (2nd ed.). England: Scott, Foresman and Company. Malim, M. Rozi, Maidinshah, H., Ishak, F., & Jamal, M. (1999). Business calculus. Shah Alam: ITM. Margaret, L. L., Charles, D. M., & Raymond, N. G. (1992). Finite mathematics and calculus with applications (4th ed.). Harper Collins College Publishers. Rahela, R., Fatinah, Z., & Hawa, I. (2005). Penyelesaian soalan-soalan dalam matematik untuk pengurusan. Sintok: Universiti Utara Malaysia.

TAN SRI DR ABDULLAH SANUSI (TSDAS) DIGITAL LIBRARY The TSDAS Digital Library has a wide range of print and online resources for the use of its learners. This comprehensive digital library, which is accessible through the OUM portal, provides access to more than 30 online databases comprising e-journals, e-theses, e-books and more. Examples of databases available are EBSCOhost, ProQuest, SpringerLink, Books24x7, InfoSci Books, Emerald Management Plus and Ebrary Electronic Books. As an OUM learner, you are encouraged to make full use of the resources available through this library.


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COURSE GUIDE


Topic  Matrix

1

LEARNING OUTCOMES By the end of this topic, you should be able to:



1.

Explain the concept of matrix and the classification of matrices;

2.

Define matrix addition, scalar multiplication and multiplication of matrices and the properties related to these operations;

3.

Calculate the determinant for square matrix using the cross multiplication method and the cofactor expansion method;

4.

Determine the inverse of an invertible matrix and use inverses to solve the linear equation system; and

5.

Apply Cramer's rule to find the solution of a two-linear equation system.

INTRODUCTION

Data is an important source of information. Therefore, it is necessary for the data to be arranged in the most comprehensible and straightforward form. Matrix is one method which is frequently used. In economics, matrix is used in formulating problems and displaying data. For example, a manufacturer who produces products D, E and F could represent the units of labour and material involved in one weekÊs production of these items as shown in the table below:

Labour Material

D 10 5

Product E 12 9

F 16 7

10 12 16  5 9 7

More simply, the data can be represented by the matrix A  


2

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TOPIC 1

MATRIX

Matrices can be categorised into several classes or types. Later on, this topic will discuss matrix operations each with its own properties which differ from the operations of real numbers. The application of matrix is to solve simultaneous equation systems. Two methods will be employed to solve the simultaneous equation system, which are the inverse matrix method and the Cramer's rule. The knowledge on finding the determinant of a given matrix is essential in order to apply these methods. Matrix is a rectangular array of numbers consisting of m horizontal rows and n vertical columns,

 a11 a  21  .   .  .   am1

a12 ... a1n  a22 ... a2 n  . ... .   . ... .  . ... .   am 2 ... amn 

This matrix is called an m x n (read as „m by n‰) matrix or a matrix of order m x n. For any matrices, the data has to be placed in a rectangular table form, as shown below:

2 0 1  0 1 3   The above matrix consists of two rows and three columns. Hence, the dimension, order, size or degree of the above matrix is 2  3. The dimension of a matrix is the number of rows first, followed by the number of columns. The elements or entries of the above matrix in the first row are 2, 0 and 1. While, the elements for the second rows are 0,  1 and 3. Generally, a matrix is denoted by a capital letter. On the other hand, every element of the matrix will be denoted by smaller capitals with subscripts as shown below:

 a11 A   a21  a31

a12 a22 a32

a13  a23  a33 


TOPIC 1

MATRIX



3

Based on the matrix A above, we can conclude that A is a matrix with the dimension of 3  3, where its elements or entries:



a11 lies in the first row and the first column



a21 lies in the second row and the first column



a32 lies in the third row and the second column

The dimension of the matrix A can be written at the lower right side of the letter, in a form of a subscript, i.e. A33. In general, a matrix A with the dimension of m  n (matrix A with m rows and n columns) is often written as Amxn. The elements of matrix A are denoted by aij , where i = 1, 2, ⁄, m and j = 1, 2, ⁄, n.

SELF-CHECK 1.1 If numbers are arranged in a non-rectangular form, can that still be called a matrix? Explain.

1.1

CLASSIFICATIONS/TYPES OF MATRICES

Matrices can be categorised into several types based on its dimensions and elements. Let us find out more.

1.1.1

Row Matrix (Row Vector)

Row matrix or row vector is a matrix with only one row, as illustrated by the following matrices: (a)

B13  1 0 1

1.1.2

(b)

B14  1 3 2 1

Column Matrix (Column Vector)

A matrix with only one column is called column matrix or column vector. The following are samples of column matrices:


4

(a)



TOPIC 1

MATRIX

1  L31  0   3

1.1.3

(b)

 3  2 L41    1    0

Square Matrix

A matrix with equal numbers of rows and columns is called square matrix. The m x n matrix is square, if and only if, m = n . The examples of square matrices are as follows: 3 2 1  1 2  (b) S33   3 1 0  (a) S 22    4 0    2 1 4  The elements on the main diagonal of any given square matrix are all the elements which lie from the upper left corner to the lower right corner. The main diagonal elements for matrix S22 above are 1 and 0. While, the main diagonal of matrix S33 are 3, 1 and 4 .

1.1.4

Diagonal Matrix

If a square matrix has at least one non-zero element on its main diagonal and all the other elements are zero, the matrix is known as diagonal matrix or aij = 0 for i ≠ j. Below are samples of such matrices: (a)

P22

1.1.5

1 0    0 1

(b)

P33

1 0 0   0 0 0  0 0 3

Special Matrix

Identity Matrix, denoted by I is the diagonal matrix whose main diagonal entries are 1Ês. Let us now look at examples of identity matrices.

(a)

I 22

1 0    0 1 

(b)

I 33

1 0 0   0 1 0  0 0 1 


TOPIC 1

MATRIX



Zero matrix or null matrix, denoted by 0 is the matrix when all the elements of a matrix are set to zero. Examples are as shown below: (a)

032

0 0   0 0  0 0 

(b)

0 0 0  023    0 0 0 

SELF-CHECK 1.2 If there exists only one element in a matrix, can the matrix be called square matrix? Explain.

EXERCISE 1.1 1.

Given:

(a)

 3 1 B   6 2   1 0 

(b)

C   1 5 8

(c)

6 D   9   4 

State the order of each matrix. 2.

3. 4.

 a 0 0 Find the value(s) of a, that make  0 0 0  a diagonal matrix.  0 0 0  1 0 0  Is the matrix   an identity matrix? Clarify your answer. 0 1 0 Determine the classes for each of the matrices below:

(a)

0 0  0  0

0 0 0 0

0 0  0  0

(b)

0 0    2 

(c)

 1

2 1 0


5

6

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TOPIC 1

1.2

MATRIX

MATRIX OPERATIONS

In this subtopic, you will learn about the following matrix operations:



Equality of matrix



Transpose



Matrix addition



Matrix subtraction



Scalar multiplication



Matrix multiplication

1.2.1

Equality of Matrix

Matrices A = [aij] and B = [bij] are equal if and only if they have the same order and

aij= bij for each i and j. Thus, 1  5 2  5 0.5   3  0 1 1  0  2  2  but 1 2  1 2 0 A matrix equation can be defined as a system of equation. For example, suppose that

q  3  q  1 2r 

q  1  3 q  p  1  r 2r

5  q  4 

By equating corresponding entries, we must have

a13  q  1, b13  5 . Therefore a13  b13 , then

q 1  5 q  5 1 q4 Copyright © Open University Malaysia (OUM)

TOPIC 1

MATRIX



7

a21  q  1, b21  r . Therefore b21  a21 , then r  q 1 r  4 1 r 3

a23  p  1, b23  q  4 . Therefore a23  b23 , then p 1  q  4 p  q  4 1 p  q 5 p  4  5  1

1.2.2

Transpose

Given a matrix Amn, with elements aij, where i = 1, 2, ⁄, m and j = 1, 2, ⁄, n. The transpose matrix for Amn, which is denoted by AT nm, is a matrix with elements aij, where i = 1, 2, ⁄, n and j = 1, 2, ⁄, m. In other words, we just reverse the order of the row and column elements of matrix Amn so that the rows turn into columns and columns into rows. Examples: (a)

2 3 1 2 1 0   T If A =  1 2 0  , then A   3 2 1   0 1 3   1 0 3 

(b)

 1 3  1 0 1   T If B =  3 2 0  , then B   0 2     1 0  T

The transpose operation has the property:  AT   A .

1.2.3

Matrix Addition

Adding two or more matrices is only feasible if all the matrices are of the same dimension. The sum is obtained by adding the corresponding elements (i.e. the elements that lie on the same position). The addition operation cannot be done if the matrices have different dimensions. Copyright © Open University Malaysia (OUM)

8



TOPIC 1

MATRIX

Examples: (a)

 1 0   2 4   1 4   2 2    1 3    1 5       

(b)

 2 2 0   1 2 0   3 4 0   4 1 3    3 2 0    7 1 3       

(c)

 2 3  7 1 3   1 5    3 1 3  , cannot be performed because the two matrices have     different dimension.

1.2.4

Matrix Subtraction

Two or more matrices can be deducted from another, as long as they have the same dimension. To perform the subtraction operation, the corresponding elements that lie on the same position of the respective matrices are subtracted from the other. This subraction operation also cannot be performed if the dimensions of the matrices are different. Examples: (a)

(b)

 3 0 1   0 2 5   3 2 4   2 1 3    1 3 2    1 4 1         1 4 1  2 1 2   1 3 3   2 3  7 1 3   1 5    3 1 3  , the subraction cannot be carried out because the two     matrices have different dimension.

Properties of matrix addition and matrix subtraction are:

A B B A A B  B A  A  B   C  A  B  C 

 A  B   C  A  B  C  A O O  A  A A O  O  A


TOPIC 1

1.2.5

MATRIX



9

Scalar Multiplication

Scalar multiplication is obtained by multiplying each entry of the matrix by the scalar. For example: (a)

(b)

If A =

1 0 2 3  

5A

 1 0   5(1) 5(0)   5 0  5     2 3   5(2) 5(3)  10 15 

=

3 0 1  If B   2 1 3   1 4 1  3 0 1   1(3) 1(0) 1(1)   3 0 1 B  (1)  2 1 5    1(2) 1(1) 1(5)    2 1 5   6 4 7   1(6) 1(4) 1(7)   6 4 7 

1.2.6

Matrix Multiplication

Multiplication of two matrices is possible only if the number of columns of the first matrix is the same as the number of the row of the second matrix. Suppose the first matrix is denoted by Amn and the second matrix is Bst. The multiplication of Amn  Bst can be done if and only if n = s. The product of this multiplication is another matrix (say C) with m rows and t columns. If AB exists, then

AB  Am n  B s t  C m t Generally, suppose a AB   11  a21

a12 a22

c 11 c 12  c 21 c 22

b b a13   11 12 b b a23   21 22 b31 b32

b13  b23  b33 

c 13  c 23 


10 

TOPIC 1

MATRIX

where;

c11  a11 b11  a12 b21  a13 b31 c12  a11 b12  a12 b22  a13 b32 c13  a11 b13  a12 b23  a13 b33 c21  a21 b11  a22 b21  a23 b31 c22  a21 b12  a22 b22  a23 b32 c12  a21 b13  a22 b23  a23 b33 For example: Given A32

(a)

 1 1   0 2  ,  1 0 

1

1

1 0 2 

B 2 2    , and C 23   0 1 1   1 0   

(1)(1)  (1)( 1) (1)(1)  (1)(0)  AB = A32  B22 = (0)(1)  (2)(1) (0)(1)  (2)(0)  (1)(1)  (0)(1) (1)(1)  (0)(0) 

 2 1 =  2 0   1 1  32 (b) BA= B22  A32 is not possible because the number of columns of matrix B22 is not the same as the number of the rows of matrix A32.

(c)

AC33

(1)(1)  ( 1)(0) (1)(0)  ( 1)(1) (1)(2)  (1)(1)  = A32  C23 = (0)(1)  (2)(0) (0)(0)  (2)(1) (0)(2)  (2)(1)  (1)(1)  (0)(0) (1)(0)  (0)(1) (1)(2)  (0)(1)  1 1 1 = 0 2 2  1 0 2 

Properties of matrix multiplication are:

A  BC    AB C A B  C   AB  AC and  A  B C  AC  BC

AB  BA AI  IA  A Copyright © Open University Malaysia (OUM)

TOPIC 1

MATRIX

 11

There are cases where the multiplication of two matrices is the matrix itself, i.e.

A  A = A. This matrix A is known as idempotent matrix. Example:  1  Given A = 0   0 

0 1 2 1 2

 0  1 , then AA = 2 1  2

 1  0   0 

0 1 2 1 2

 0  1  1 0 2 1  0 2

0 1 2 1 2

  0  1   1   0 2  1   0 2 

0 1 2 1 2

 0  1 A 2 1  2

Hence, A is an idempotent matrix.

SELF-CHECK 1.3 List several examples of idempotent matrices.

EXERCISE 1.2 1.

Suppose:  4 2   2 1  2 1 3  4 1 2      A=   , B   5 1 3  , C   0 6  and D   3 5  4 0 1      1 3  3 2  Find:

2.

(a) (d)

3A

(g)

(AT)A

(b) (e)

AB

A+B (2A)(5C)

(c) (f)

CD (2A  B)D

Determine matrix A that satisfies the following equation. 1

0

 1 3  6 

A    1 3   3


12 

TOPIC 1

1.3

MATRIX

DETERMINANT

Determinant is defined only for square matrices. The determinant of matrix A is denoted by A and has a scalar value. This section will only emphasise on obtaining the determinants for square matrices with the dimensions until 3  3. Determinants are used to find the invertible matrices which are then used to explicitly describe the solution to the linear equation system.

If A =  a11  is a square matrix of order 1, then A  a11 a12  a Given matrix A22 =  11  . The determinant of a matrix can be obtained by  a21 a22  taking the difference between the multiplication of the elements on main diagonal (a11 and a22) and that of opposite diagonal ( a12 and a21 ). The determinant of the matrix A22 is given by: A22   a11  a22    a12  a21 

Example:  0 1 If A =   and B = 2 4 

1 2 , determine 0 1 

(a)

|A|

(b)

|B|

(c)

|AB|

(d)

|BA|

(e)

|A||B|

(f)

|B||A|

Solutions: (a)

|A| = 0(4) ă (ă1)(2) = 2

(b)

|B| = 1(1) ă 2(0) = 1

(c)

AB = 

(d)

BA = 

 0 1  . Therefore, AB  = 2 8 

(0)(8) ă (ă1)(2) = 2

4 7  . Therefore, BA 2 4

=

(4)(4) ă (7)(2)

= =

16 ă14 2


TOPIC 1

(e)

| A|| B| = (2)(1) = 2

(f)

| B|| A| = (1) (2) = 2

The determinant of matrix A33

 a11   a21  a31

a12 a22 a32

MATRIX

 13

a13  a23  , is obtained by: a33 

For entry a11 , we delete the entries in row 1 and column 1, as shown below:  a11 a  21  a31

a12 a22 a32

a13  a23  a33 

a This leaves the matrix  22  a32 called the minor of a11 .

a23  of order 2. The determinant of this matrix is a33 

Similarly, the minor of a12 is

a21 a31

a23 a , and for a13 is 21 a33 a31

a22 . a32

So, to find the determinant of any square matrix A of order 3 is given by: A  a11

a22 a32

a23 a  a12 21 a33 a31

a23 a  a13 21 a33 a31

a22 a32

 a11  a22 a33  a23 a32   a12  a21a33  a23 a31   a13  a21a32  a22 a31 

1.3.1

Minor of Element aij

The minor of element aij is the determinant of the sub-matrix left after omitting the i th row and j th column. For instance, suppose 1 4 1  A =  0 2 0   2 3 3 Copyright © Open University Malaysia (OUM)

14 

TOPIC 1

MATRIX

Then, element of minor is: 1

4

1

4

1

4

1

4

1

2 0 2 0 = (2)(3) ă (0)( ă3) = 6 3 3 2 3 3

m11 (i.e. 1) = 0

1

0 0 2 0 = (0)(3) ă (0)(2) = 0 2 3 2 3 3

m12 (i.e. 4) = 0

1

0 2 2 0 = (0)(3) ă (2)(2) = ă 4 2 3 2 3 3

m13 (i.e. 1) = 0

1

4 1 2 0 = (4)(3) ă (1)( ă3) = 15 3 3 2 3 3

m21 (i.e. 0) = 0

1

4

1

4

1

1 1 2 0  = (1)(3) ă (1)(2) = 1 2 3 2 3 3

m22 (i.e. 2) = 0

1

1 4 2 0  = (1)( ă3) ă (4)(2) = ă11 2 3 2 3 3

m23 (i.e. 0) = 0

1 4 1

m31 (i.e. 2) = 0 2 0  2 3 3 1

4

1

1

4

4 1 2 0

= (4)(0) ă (1)(2) = ă2

1 1 2 0 = (1)(0) ă (1)(0) = 0 0 0 2 3 3

m32 (i.e. -3) = 0

1

1 4 2 0 = (1)(2) ă (4)(0) = 2 0 2 2 3 3

m33 (i.e. ) = 0


TOPIC 1

MATRIX

 15

 6 0 4  The minor matrix A is Minor A =  15 1 11  2 0 2 

1.3.2

Cofactor of Element aij

The cofactor of element cij is the minor of element mij multiplied by (-1)i+j. Hence, the cofactor of element:

c11 = (ă1)1+1  m11 = 1  6 = 6 c12 = (ă1)1+2  m12 = ă1  0 = 0 c13 = (ă1)1+3  m13 = 1  (ă 4) = ă 4 c21 = (ă1)2+1  m21 = ă1  15 = ă15 c22 = (ă1)2+2  m22 = 1  1 = 1 c23 = (ă1)2+3  m23 = ă1  (ă11) = 11 c31 = (ă1)3+1  m31 = 1  (ă2) = ă2 c32 = (ă1)3+2  m32 = ă1  0 = 0 c33 = (ă1)3+3  m33 = 1  2 = 2  6 0  4 The cofactor matrix is Cofactor A =  15 1 11   2 0 2 

The following steps are required to compute the determinant of a matrix using the cofactor expansion method : Step 1:

Select one row or column to perform cofactor expansion. In general, we choose the row or column with many zeroes. In matrix A, the second row has many zeroes. Therefore, choose the second row of matrix A to perform the cofactor expansion.

Step 2:

Perform cofactor expansion by multiplying each element in the selected row or column with its coresponding cofactor. Hence, the determinant of matrix A is.


16 

TOPIC 1

MATRIX

Determinant A = |A|

Note:

= a21c21 + a22 c22 + a23c23 = 0(15) + 2(1) + 0(11) = 0 + 2(1) + 0 =2

The same value of determinant will be obtained although cofactor expansion is performed on a different row or column.

The transpose of a cofactor matrix is an adjoint matrix. Adjoint A = [Cofactor A]T  6   15  2 6   0   4

0  4 1 11  0 2 

T

15 2  1 0  11 2 

EXERCISE 1.3 1.

2.

Find the determinant for the following matrices: (a)

 2 5 1 3   

(c)

1 2 3  2 3 0    3 0 0 

(a)

(b)

(b)

 a b  b a   

(d)

3 2 1  0 3 2    0 0 3 

 1 a Calculate the value for a, given the determinant for   is 6.  2 4  a b  Find the determinant for  1 0  ?  2 4 


TOPIC 1

MATRIX

 17

INVERSE MATRIX

1.4

Inverse matrix is defined for square matrix only. However, not all the square matrices have an inverse. If the determinant of a square matrix is equal to zero, then the matrix has no inverse. A matrix without an inverse is known as singular matrix.

Inverse matrix of A is denoted by A1. a

a 

1 a

Let A   11 12  and A  a11a22  a12 a21 . Therefore, A1   22 A  a21  a21 a22  If A is a square matrix of order 3, then A1 

a12  a11 

1  Adjoint A A

In the above example:

 3  6 15 2   1  1 1 0    0 A   0  2   4 11 2    2 

15 2 1 2 11 2

 1  0   1 

When a matrix A is multiplied by its inverse A-1, the following properties are then true: (a)

A  A1 = I

(b)

A1  A = I


18 

TOPIC 1

MATRIX

EXERCISE 1.4 1.

2.

Find the inverse (if there exist any) for the following matrices. Then, prove that your answers are correct. (a)

 3 4  2 2   

(c)

 4 2 2  1 3 4     3 1 6 

(b)

 2 3 4   0 0 1   1 2 1 

(d)

 1 4 1  2 3 2     1 2 3 

Given:

2 3 . Find A1 and show that (A1) 1 = A. A   4 5 3.

Suppose:

a b  B  c d 

1.5

(a)

Determine B 1 .

(b)

State the properties required for the existence of B 1 .

(c)

Verify BB 1 = B 1 B = I.

SOLVING LINEAR EQUATION SYSTEM USING MATRICES

In this section, we shall illustrate methods by which matrices can be used to solve a system of linear equations. The two methods are Matrix Inverse method and Cramer's rule.


TOPIC 1

1.5.1

MATRIX

 19

Matrix Equation

Systems of linear equations can be represented by using matrix multiplication. For example, consider the matrix equation:

 x 1 4 2     4   2 3 1   y    3  z      x  4 y  2z   4   2 x  3 y  z    3     By equality of matrices, corresponding entries must be equal, so we obtain the system:

x  4 y  2z  4 2 x  3 y  z  3 Hence, this system of linear equations can be expressed in the form of matrix equation system AX = B. Where A is the matrix obtained from the coefficients of the variables, X is a column matrix obtained from the variables, and B is a column matrix obtained from the constants. Example: Given a two-linear equation system:

x + 2y = 0 2x  y = 5 1 2   x  0 It can be expressed as:       2 1  y   5  1 2  where: A   ,  2 1

x 0 X    and B    as a matrix equation system.  y 5

Example: Given a three-linear equation system: 2x ă y + 3z = 3 x + 2y ă z = 4 2x ă 2z =0


20 

TOPIC 1

MATRIX

The matrix form is:

 2 1 3  A  1 2 1 ,  2 0 2 

1.5.2

 2 1 3   x   3  1 2 1  y    4  ,where       2 0 2   z   0  x X   y  and  z 

3 B   4  .  0 

Inverse Matrix Method

A system of linear equations can be written in matrix form, AX=B, where A is the coefficient matrix. If we can determine the values of the entries in the unknown matrix X, we have a solution of the system. An inverse of matrix A, A1 is use to solve an equation of AX = B. Multiply both sides of equation AX=B by A1, A1 ( AX )  A1 B ( A1 A) X  A1 B

IX  A1 B where I is an identity matrix. Then, X  A1 B , is called method of inverse to solve a system of linear equations.

Example: Solve

x + 2y = 3 2x  y = 5

using the inverse matrix method. Solution: Step 1:

Convert the equation into a matrix equation form, which is 1 2   x  0 1 2   2 1  y    5  where A   2 1 ,       

Step 2:

 x 0 X    and B     y 5

Determine the inverse of matrix A, i.e. A1 . A  1 1   2  2   5


TOPIC 1

1  1 2   A1  5  2 1 

Step 3:

1 5  2  5

MATRIX

 21

2 5  1  5 

Use the formula X = A1B to obtain the solution: 1  x 5  y   2     5

2  5  0  1   5   5 

2    1

Hence, solutions for the above simultaneous linear equations are x = 2, y = 1. Example: Solve 2x ă y + 3z = 3 x + 2y ă z = 4 =0 2x ă 2z using the method of matrix inverse. Solution: Step 1:

Firstly, we have to convert the equation into matrix equation form, which is  2 1 3   x   3   1 2 1  y    4  , where       2 0 2   z   0   2 1 3   x 3     A   1 2 1 , X   y  and B   4   2 0 2   z   0 


22 

TOPIC 1

Step 2:

Determine the inverse of matrix A, i.e. A1. To do this we need to compute its determinant, i.e.

MATRIX

A   1 1

1 2

1 1 2 2

  2  1

2 2

2

3

2 2

  0  1

3 2

2

3

1 1

 1 1 2    1 2    2  2  2    3 2    0  1 0   2  10   20

The minor elements are:

n11  – 4, n12 = 0, n13 = ă 4, n21 = 2 and n22 = – 10, n23  2 n31 = –5, n32 = –5, n33 = 5. Therefore,  4 0 4  Minor A =  2 10 2  , Cofactor A =  5 5 5 

 4 0 4   2 10 2     5 5 5 

 4 2 5 Adjoint A =  0 10 5   4 2 5 

Using the formula: 1 A 1   Adjoint A A  4 2 5  1    0 10 5  20  4 2 5 

1 1  5 10  1 = 0  2  1 1  5 10

1  4   1  4  1   4  Copyright © Open University Malaysia (OUM)

TOPIC 1

Step 3:

MATRIX

 23

Use the formula X = A1B to obtain the solution for: 1 1   x   5 10  y  0 1    2  z   1 1   5 10

1  4   3  1   1    4    2  4   0  1  1     4 

Thus, the solutions for the given simultaneous equations are x = 1, y = 2 and z = 1.

1.5.3

Cramer's Rule

Another method which can be applied to solve the simultaneous equation AX = B is Cramer's rule. The following steps have to be taken to solve a system of n linear equations in n unknowns. Step 1:

Determine the determinant for coefficient matrix A, that isA. If A= 0, CramerÊs rule is no longer applicable.

Step 2:

Find Aiwhere Ai is the matrix formed when i th column in matrix A is substituted by matrix B as shown below:  a11 a  21   Ai         a1n

 

 

 

 

 

 

b1 b2    bn

     

a1n  a2 n          a2 n 



ith column

Step 3:

To obtain xi, we use the following formula: xi 

Ai A


24 

TOPIC 1

MATRIX

Example: Solve x + 2y = 0 2x  y = 5 using the Cramer's rule. Solution: Step 1:

1 2   x  0 Determine A for        2 1  y  5  A = (1)(1) ă (2)(2) = 5

Step 2:

Find A1and A2. 0 2  A1    . Therefore, A1= (0)( 1) ă (2)(5) = 10 5 1 1 0  A2    . Thus, A2= (1)(5) ă (0)(2) = 5 2 5

Step 3:

Obtain the value for x and y. x

A 1 10  2 5 A

y

A2 5   1 A 5

Example: Solve 2x ă y + 3z = 3 x + 2y ă z = 4 2x ă 2z =0 using the CramerÊs rule method. Solution:

Step 1:

Determine A for

 2 1 3   x   3   1 2 1  y    4        2 0 2   z   0  Copyright © Open University Malaysia (OUM)

TOPIC 1

 25

MATRIX

by using the cofactor expansion on the third row: A  2  1

31

1

3

2

1

  2  1

3 3

2 1 1

2

 2  1 1   2  3   2  2  2   1 1   2  5   2  5   20

Step 2:

Find A1,A2 and A3.

1 2 0

3 A1   4  0 A1  2  1

3 3

3 1 4

A2  2  1

31

3  2   1  , A2   1  2  2  2

3 4 0

3  2   1  , A3   1  2  2 

1 2 0

 2  3 2    1 4    20

3 3 3 31 2   2  1 4 1 1 4

 2  3 1   3 4     2   2  4    31 

 2  15   2  5   40 A3  2  1

Step 3:

31

1 3 2

4

 2  1 4    3 2    20

Obtain the value for x, y and z. x

A1

y

A2

z

A3

A

A

A



20 1 20



40 2 20



20 1 20


3 4  0 

26 

TOPIC 1

MATRIX

ACTIVITY 1.1 What are the advantages of using matrices to solve linear equation systems compared to algebra techniques (substitution/elimination)? Compile your answer and share it with your coursemates during tutorial.

EXERCISE 1.5 1.

Express the following linear equation systems in the form of matrix equations. Subsequently, solve the equations by using the matrix inverse method. (a)

x + 2y = 14 2x ă y = 5

(b)

x + 2y + z = 7 x+y+z=4 3x + y + z = 2

2.

Solve (1) using the Cramer's rule.

3.

Solve the following equation system by using appropriate method:

3 x  2 y  z  b1 3 x  2 y  z  b2

x  y  z  b3 where: (a)

b 1  2,

b 2  2,

b 3  4.

(b)

b 1  8,

b 2  3,

b 3  6.


TOPIC 1

MATRIX

 27

MULTIPLE CHOICE QUESTIONS 1.

1 2  4 7 and B     . Find: 3 2.5  3 1

Given A   (a)

A  2B . 8 9   6 2 

A.  (b)

7  9   3 0.5

6 3   3 4

9 11   9 4.5

B. 

C. 

D. 

B. ă25

C. ă17

D. 17

B.

 4 3   7 1

A. 

T

2.

3 1 2   4   2 0 4   2  =    

8    A. 4    0  3.

B. Does not exist

C. 16

4 0

16    D. 4    0 

 3 1 1   Given A  6 and its cofactor is  3 1 1 . Find the inverse of A .  6 4 2 

 3 3 6    A. 1 1 4    1 1 2 

 3 1 1 1  B.  3 1 1  6  6 4 2 

 3 3 6  1  C.  1 1 4   6  1 1 2 

 3 1 1   D. 6 3 1 1    6 4 2 


28 

4.

TOPIC 1

MATRIX

1 1 1  x 1       Given A  3 2 1 , X  y , B  2 , and A  1        z   1  2 5 3 Calculate the value of z. A. ă18

B. ă10

C. ă7

D. 18



You should realise the significance of matrix in facilitating them to better understand the data.



You should now be able to perform matrix operations well, to obtain the determinant and inverse of a matrix.



Ultimately, you are expected to know how to solve simultaneous linear equation systems using the matrix methods.

Matrix

Scalar Multiplication

Types/Classifications of Matrices

Matrix Multiplication

Equality of Matrix

Determinant

Transpose

Inverse Matrix

Matrix Addition

CramerÊs Rule

Matrix Substraction


Topic

2



Linear and Quadratic Functions




1.

Identify linear and quadratic functions;

2.

Find the slope of a line;

3.

Determine whether two lines are parallel or perpendicular;

4.

Sketch the graphs of linear and quadratic functions; and

5.

Find intersection point.

INTRODUCTION

In the beginning of the 17th century, Gottfried Wilhelm Leibniz has introduced the term „function‰ to mathematical vocabulary. This concept has become one of the most fundamental mathematical concepts. In general, a function is a special type of input-output relation that expresses how one quantity (the output) depends on another quantity (the input). A function is a rule that assigns each value of x (input) to only one value of y (output) which is denoted by symbol f (other symbols such as g and h are also used). Usually, the symbol x is used to represent independent variable as it is free to take any value, while the symbol y is used to denote dependent variable as its value depends on the value taken by x.

y  f  x   or  read as "y is a function of x" y  g  x  


30 

TOPIC 2

LINEAR AND QUADRATIC FUNCTIONS

LINEAR EQUATIONS AND GRAPHS SKETCHING

2.1

A linear function graph is a straight line.

SELF-CHECK 2.1 Is a linear equation a function? Explain.

2.1.1

Linear Equations

Linear equation is an equation where the highest power of x is equal to 1. General form: y = mx + c, where m is the slope and c is the y-intercept. Examples: Obtain the slope and y-intercept for each of the linear equations below: (a)

y = 6 ă 3x

(b)

2y + 6x = 9

Solutions: Express the following equations in general form, i.e. y = mx + c. Then, calculate the value for m (scalar for x) and the y-intercept, i.e. the value of c. (a)

y = 6 ă 3x y = ă 3x + 6

(General form)

Therefore, m = ă3 and c = 6. (b)

2y + 6x = 9 2y = ă 6x + 9 y  3 x 

9 2

Therefore, m = ă3 and c 

(General form) 9 . 2


TOPIC 2

2.1.2


 31

Slope

If two points A(x1, y1) and B (x2, y2) are given, a slope can be derived by using the formula below: m

y2  y1 x2  x1

Examples: Find the slope for each line that connects the two given points: (a)

A (1, 4) and B (ă2, 5)

(b)

C (0, ă3) and D (7, ă1)

(c)

E (ă6, 6) and F (1, 6)

Solutions: 54 2  1 1  3

(a)

m

(b)

m

(c)

m

1   3

70 1  3  7 2  7

66 1   6 

0 7 0 


32 

TOPIC 2


ACTIVITY 2.1 Why is it necessary to discuss the slope of a line? What is the significance of a slope? Discuss during your tutorial.

2.1.3

Types of Straight Lines

There are various types of straight lines. Let us now look at each type. (a)

Horizontal Line  y=a  Parallel to x-axis  Its slope is zero

Graph 2.1

Vertical Line  x= b  Parallel to y-axis  Its slope is undefined

(b)

Graph 2.2

(c)

Ascending-Slant Line  y = mx + c  Ascending line from left to right  Its slope is positive

Graph 2.3


TOPIC 2

(d)


 33

Descending-Slant Line  y = mx + c  Descending line from left to right  Its slope is negative

Graph 2.4

2.1.4

Graphs Sketching

The followings are steps for sketching a linear function graph: (a)

Find two different points that are on the line and plot them. (In general, these are the y-intercept and x-intercept) The y-intercept can be obtained by substituting x = 0 into the equation and calculating the corresponding value for y. The x-intercept can be obtained by substituting y = 0 into the equation and calculating the corresponding value for x.

(b)

Connect the two points to form a straight line.

Examples: Sketch graph for each of the following linear functions: (a)

y = 2x 1

(b)

y =  4x

Solutions: (a)

y = 2x  1 (i)

The first point: Find the y-intercept Let x = 0, y = 2(0)  1 y = 1  Hence, the first point is (0, 1).


34 

TOPIC 2

(ii)


The second point: Find the x-intercept Let y = 0, 2x ă 1 = 0 2x = 1 1 x = 2 1  Hence, the second point is ( , 0). 2

Graph 2.5

(b)

y =  4x (i)

The first point: Find the y-intercept Let x = 0, y =  4(0) y=0  Hence, the first point is (0,0).

(b)

The second point: Find any other point besides the x-intercept (as it is the same as the y-intercept) Let x = 2, y =  4(2) y = 8  Hence, the second point is (2, 8).


TOPIC 2


 35

Graph 2.6

2.2

PARALLEL AND PERPENDICULAR LINES

Parallel lines are distinct lines lying in the same plane, never intersecting each other. Parallel lines have the same slope. Perpendicular lines, on the other hand, are lines that intersect each other at right angles. Two lines are said to be parallel if and only if they have the same slope.

. Graph 2.7

Example: Is line 2y ă 3x + 6 = 0 parallel to another line 4y = 6x + 3 ? Solution: Find the slope of each line:


36 

TOPIC 2


2 y  3x  6  0 2 y  3x  6 3x 3 y 2 3  m1  2

4 y  6x  3 6 3 x 4 4 3 3 y x 2 4 3  m2  2 y

Since they have the same slope, these two lines are parallel. Example: Find an equation of a straight line that passes through point (ă2, 10) and parallel to another straight line 5x ă y = 0. Solution: Determine the slope for line 5x ă y = 0. 5x  y  0  y  5 x y  5x  m5

The equation of a straight line is y = mx + c. (The slope for this equation is also 5 since the slope of the two lines are the same) Now, substitute m = 5 into equation y = mx + c, i.e. y = 5x + c. This line passes through point (ă2, 10). So, substitute x = ă2 and y = 10 into y = 5x + c. 10 = ă10 ă c 10 + 10 = c

c = 20  The equation of a straight line that we are looking for is y = 5x + 20. So, now we know when two lines are said to be parallel. How about perpendicular? Let us see.

Two lines are said to be perpendicular if and only if the product of the two slopes is ă1. Copyright © Open University Malaysia (OUM)

TOPIC 2


 37

Graph 2.8

Example: Determine whether a straight line, y ă 2x = 1 is perpendicular to 2y + x = 2 or not. Solution: Calculate the slope for each line and multiply them. 2y  x  2

y  2x  1 y  2x  1  m1  2

2 y  x  2 x y 1 2 1  m2  2

Since the multiplication of the two slopes is equal to ă1, therefore the two lines are perpendicular. Example: Find an equation of a straight line that passes through point (1, 2) and perpendicular to a straight line x + 5y = 2. Solution: Determine the slope for line x + 5y = 2.


38 

TOPIC 2


x  5y  2 5y  x  2 x 2  y 5 5 1  m1  5

Find the slope of the required line, which is m2 .  1     m2   1  5 m2  5

(The multiplication of the two slopes is ă1 as the two lines are perpendicular)

Now, substitute m = 5 into equation y = mx + c, i.e. y = 5x + c. This line passes through point (1, 2). So, we substitute x = 1 and y = 2 into y = 5x + c. 2 2 2ă5 ă3

= 5(1) + c =5+c =c =c

 The equation of the straight line that we are looking for is y = 5x ă 3.

ACTIVITY 2.2 If the product of two slopes is equal to 1, are the lines perpendicular? Explain.


TOPIC 2


 39

EXERCISE 2.1 1.

For each of the following equations, determine the slope and

y - intercept: x 1 2

(a)

y

(c)

y = ă3x

(b)

y = ă5 ă 5x

(d)

3y = 5 ă 2x

2.

Find the equation of a straight line with slope ă1 that passes through point (3, 2).

3.

Given two points A (2, 4) and B (5, 12). Determine the equation of a straight line that passes through them.

4.

Find an equation of a straight line that passes through point (2, 1) and is parallel to line 2y + x = 5.

5.

Obtain an equation of a straight line that passes through point (3, ă2) and is perpendicular to line 3x ă y + 3 = 0.

EXERCISE 2.2 Sketch a graph for each of the linear functions below:

2.3

(a)

y = 3x + 2

(c)

3y + 2x = 2

(b)

y

x 2

QUADRATIC EQUATIONS AND GRAPHS SKETCHING

The general form of a quadratic equation is y = ax 2 + bx + c , where a, b and c are real numbers and a  0. The highest degree for x in a quadratic equation is 2.


40 

TOPIC 2


The graph of a quadratic function is a parabola. The direction of which the parabola opens depends on the value of a. If a is positive, then the parabola opens upward and the function has a minimum value.

Graph 2.9

Meanwhile, if a is negative, then the parabola opens downward and the function has a maximum value.

Graph 2.10

The followings are steps for sketching graph of quadratic function f ( x ) = ax 2 + bx + c . (a)

Determine the direction of which the parabola opens by observing the value of a.

(b)

Find the turning point (x, y) using the formula below: x

b , 2a

y

4ac  b 2 4a


TOPIC 2

(c)


 41

Find the y-intercept, at which x = 0. Substitute x = 0 into the quadratic function. f ( x ) = ax 2 + bx + c f (0 ) = a (0)2 + b (0) + c

=c Hence, (0, c) is the y-intercept. (d)

Find the x-intercept (if it exists). The graph crosses x-axis when y = 0, that is when ax2 + bx + c = 0. This equation can be solved by factoring or using quadratic formula. Quadratic Formula: x 

b  b 2  4ac 2a

The graph crosses the x-axis or not, depends on the value of b 2  4ac. (i)

When b 2  4ac > 0, the graph crosses the x-axis at two points.

(ii)

When b 2  4ac = 0, the graph crosses the x-axis at only one point.

(iii) When b 2  4ac < 0, the graph does not cross the x-axis. (e)

Plot all the predetermined points from steps (a) to (d). Draw a smooth curve passing through the points. Examples: Sketch graphs for each of the following quadratic functions: (i)

f (x ) = x 2 ă 4x

(ii)

f ( x ) = 3 ă 2x ă x

(iii)

f (x ) = 2x 2 + 2x + 1

2

Solutions: (i)

Determine the values of a, b and c from the function. f ( x ) = x 2 ă 4 x , where a = 1, b = 4, c = 0. 

The value of a is positive, hence the parabola opens upward. Copyright © Open University Malaysia (OUM)

42 

TOPIC 2




The turning point (x, y) b 4ac  b 2 x , y 2a 4a 

  4  2 1

4 2 2 

,



,



,

4 1 0    4 

2

4 1

0  16 4  4

Therefore, the turning point is (2, 4). 

The y-intercept is the value of c.

c=0 Then, the y-intercept is (0, 0). 

The x-intercept is when f (x) = 0.

x (x ă 4) = 0 The factored method gives:

x (x ă 4) = 0 x = 0,

x=4 or

The quadratic formula: x

b  b 2  4ac 2a

b 2  4ac   4    4 1 0  2

 16  0  16 x

  4   16 2 1

44 2 44 44 , x x 0 2 x  4, x  0 x


TOPIC 2




 43

Thus, there are two x-intercepts, i.e. (0, 0) and (4, 0). Therefore. The graph is:

Graph 2.11

(ii)

2

f ( x ) = 3 ă 2 x ă x , where a = ă1, b = ă2, c = 3.



The value of a is negative, thus the parabola opens downward.



The turning point (x, y) x x

b 2a   2  2  1

,

y

,

y

2 2

,

x  1

,

x

4ac  b 2 4a 4  1 3   2 

2

4  1

12  4 4 16 y 4 y4 y

Hence, the turning point is (ă1, 4). 

The y-intercept is the value of c. c=3



The x-intercept is when f (x) = 0. 3 ă 2x ă x 2 = 0


44 

TOPIC 2


The factored method gives: (3 + x)(1 ă x) = 0 3 + x = 0, 1 ă x = 4

x = ă3, x = 1 

Therefore, the graph is:

Graph 2.12

(iii) f (x) = 2x 2 + 2x + 1, where a = 2, b = 2, c = 1. 

The value of a is positive, thus the parabola opens upward.



The vertex (x, y) b 2a 2 x 2  2 x

x

,

 b  y f   2a 

,

1 Substitute x   into the function 2

2 4

1 x 2

 1 y  f    2 2

,

 1  1 y  2    2   1  2  2 1 y  2  1 1 4 1 y 2 Copyright © Open University Malaysia (OUM)

TOPIC 2




 1 1 Hence, the turning point is   ,  .  2 2

The y-intercept is the value of c. c=1 

Therefore, the y-intercept is (0, 1). The x-intercept is when f (x) = 0.

2x 2 + 2x + 1 = 0 Use the quadratic formula method.

b  b 2  4ac 2a 2 2 b  4ac  2  4  2 1 x

 48  4

b 2 ă 4ac < 0, so the graph has no x-intercept. 

Therefore, the graph is:

Graph 2.13


 45

46 

TOPIC 2


ACTIVITY 2.3 Why do you think quadratic equation has a parabola shape? Share your answer during tutorial.

EXERCISE 2.3 Sketch graphs for each of the quadratic functions below: (a)

f(x) = x2 ă 6x + 5

(b)

f(x) = x2 + 4

(c)

f(x) = ă x2 ă 2x ă 3

(d)

f(x) = x2 ă 16

(e)

f(x) = (x ă 1)(3 ă x)

2.4

INTERSECTION POINT

The point of intersection between two graphs can be obtained by solving the equations simultaneously. Example: Find the intersection point for lines 2x + y = 4 and x ă y = 2. Solution: Solve the two equations simultaneously. Add the two equations together to eliminate y.

(+)

2x + y = 4 xăy = 2 3x = 6 x = 2

Now, substitute x = 2 into equation x ă y = 2. 2ăy = 2 y = 0


TOPIC 2


 47

Therefore, the point of intersection is (2, 0). Example: Find the intersection point for lines 2x + 4y = 6 and 6x + 3y = 18. Solution: Equate the scalar of x in both equations by multiplying each term in equation 2x + 4y = 6 by 3. Thus, we have 6x + 12y = 18. Then, subtract this equation from the other to eliminate the variable x. (-)

6x + 12y = 18 6x + 3y = 18 9y = 0 y = 0

Substituting y = 0 into equation 2x + 4y = 6. 2x + 4(0) = 6 2x = 6 x = 3 Hence, the point of intersection is (3, 0). Example: Find the intersection point for curves y = 4 x ă x 2 and y = x 2 ă 6. Solution: Solve the equations y = 4 x ă x 2 and y = x 2 ă 6. 4x ă x 2 = x 2 ă 6 4x ă x 2 ă x 2 + 6 = 0 4x ă 2x 2 + 6 = 0 (Divide each term by 2) 2 2x ă x + 3 = 0

a = ă1, b = 2, c = 3

(Apply the quadratic formula)


48 

x

TOPIC 2


2  4  4  1 3 2  1

2  16 2 2  4 x 2 2  4 x 2 2 x 2 x  1 x

, , ,

2  4 2 6 x  2 x  3 x 

Hence, the intersection points are (ă1, ă5) and (3, 3). Example: Find the intersection point for curves x 2 + y ă 3 = 0 and 2x + y = 0. Solution: Solve the equations x 2 + y ă 3 = 0 and 2x + y = 0. Substitute 2x + y = 0, that is y = ă2x into x 2 + y ă 3 = 0 x2 + y ă 3 = 0 x 2 ă2x ă 3 = 0 (Factorise the equation) (x ă 3)(x + 1) = 0 x = 3, x = ă1 Insert x = 3 and x = ă1 into y = ă2x. When x = 3,

y = ă2(3) y = ă6

When x = ă1, y = ă2(ă1) y=2 Thus, the intersection points are (3, ă6) and (ă1, 2). Further discussions on function can be found in the following website: http://mathworld.wolfram.com/Function.html


TOPIC 2


 49

ACTIVITY 2.4 When two graphs cross each other, what can we say about their respective value of x and y at the intersection point?

EXERCISE 2.4 Find the intersection points for each of the following equations: (a)

2x + y = 10 and 6x + y = 14

(b)

3x + y ă 2 = 0 and 3x ă 4y + 8 = 0

(c)

2x ă 3y = 7 and 3x + 2y = 4

(d)

y = 8 ă x2 and 4x ă y + 11 = 0

(e)

y = 2x2 ă 3x and y = x2 ă 2

(f)

y = x2 + 6x + 2 and y = 2x2 + 2x + 5


2.

Form the linear equation for a line that passes through point 3,  1 and having slope ă7. A. y  7 x  21

B. y  7 x  22

C. y  7 x  20

D. y  3 x  7

Which pair of lines are parallel? A. L1 : 2 y  3 x  13 , L2 : 3 y  2 x  3 B. L1 : 2 x  2 y  1

, L2 : y  x  9

C. L1 : y  3 x  7 , L2 : y  3 x  7 D. L1 : 3 x  y  7  0 , L2 : x  y  1  0


50 

TOPIC 2


3.

Determine the graph that represents a line having slope ă3.

4.

Let y = f(x) be a quadratic function. Which of the following is false?

5.

A.

Parabola opens downward when the scalar of x is negative.

B.

The y-intercept is obtained by evaluating f(0).

C.

The turning point is maximum when the parabola opens downward.

D.

The highest power of an independent variable for a quadratic function is two. 2

Which graph represents f(x) = 6 + 5x + x ? A.

B.


TOPIC 2

C.


 51

D.



If y is a function of x and is written as y = f(x), then there exists a unique value for y for every value of x.



A linear function is of the first degree polynomial function, hence it is also known as polynomial linear.



A quadratic function is of the second degree polynomial function.

Function

Parabola

Intersection Point

Quadratic Function

Linear Function


Topic

3



Application of Linear and Quadratic Functions


\

1.

Identify demand and supply functions;

2.

Calculate the market equilibrium point;

3.

Identify Cost, Revenue and Profit functions;

4.

Analyse the break-even point; and

5.

Find the maximum and minimum value of a function.



INTRODUCTION

There are many applications of linear and quadratic functions in our daily lives. However, in this topic only its application in economic field will be discussed. The discussion of market equilibrium point involves the demand and supply functions and their intersection point. Likewise, the discussion of break-even point analysis involves the cost, revenue and profit functions. The shape of quadratic function graph is a parabola. So, the maximum or minimum value of a quadratic function can be obtained from its turning point.

3.1

THE DEMAND AND SUPPLY FUNCTION

For every price level of a product, there exists a matched quantity of the product which is demanded by consumers over a certain period of time. In general, the higher the price is, the lower the quantity desired and when the price goes down, the quantity demanded will go back up. Suppose the price for a unit of a product Copyright © Open University Malaysia (OUM)

TOPIC 3

APPLICATION OF LINEAR AND QUADRATIC FUNCTIONS



53

is p and the matching demanded quantity is q, then the equation which links p and q is called demand equation. This demand function has a negative slope.

Graph 3.1

Generally, the higher the unit price of a product is, the more quantity offered by the producer. When the price goes down, the supplied quantity will be reduced. If the price for a unit of a product is p and the matching supplied quantity is q, then the equation which links p and q is known as supply equation. This supply function has a positive slope.

Graph 3.2

3.1.1

The Market Equilibrium Point

The market reaches its equilibrium level when supply and demand are equal. The equilibrium point can be derived by obtaining the intersection point between the demand and supply equations.


54



TOPIC 3


Graph 3.3

Example: Determine which of the following equations are the demand and supply equations. Then, obtain the market equilibrium point. (a)

p + 2q = 100

(b)

3p = q + 125

Solution: Determine the slope for each equation. (a)

p + 2q = 100 p = ă2q + 100 The slope is ă2 (negative). Thus, the equation p + 2q = 100 is a demand equation.

(b)

3 p  q  125

p

q 3



125 3

1 (positive). 3 Thus, the equation 3p = q + 125 is a supply equation.

The slope is


TOPIC 3




55

Solve both equations to obtain the market equilibrium point.

p + 2q 3p + 6q (ă) 3p ă q 7q

= 100 = 300 = 125 = 175 q = 25

(multiply each term by 3) (subtract this equation from the other to eliminate p)

Insert q = 25 into equation p + 2q = 100

p + 2(25) = 100 p + 50 = 100 p = 50  The market equilibrium point is (25, 50).

Example: Given a demand function, q d = p 2 – 100 p + 2500 and a supply function qs = 0.5 p 2 ă 50. (a)

Determine the price at market equilibrium point if the price domain is 5  p  50.

(b)

Find the quantity for such price.

Solution: (a)

Equating the demand and supply functions to find the value for p. 2

2

p – 100p + 2500 = 0.5p – 50 2

2

p – 0.5p – 100p + 2500 + 50 = 0

0.5p 2 – 100p + 2550 = 0 a = 0.5, b = –100, c = 2550

p

b  b 2  4ac 2a

b 2 – 4ac = (–100) 2 – 4(0.5)(2550)

= 10000 – 5100 = 4900


56

TOPIC 3



p


  100   4900 2  0.5 

100  70 1 p  100  70

,

p  100  70

p  170

,

p  30

p

p = 170 does not lie in the given price domain, hence it is not the solution. Therefore, the price at the market equilibrium point is RM30. (b)

Substitute p = 30 into the supply function. q = 0.5p 2 – 50

= 0.5(30) 2 – 50 = 0.5(900) – 50 = 450 – 50 = 400 The corresponding supply quantity is 400.

ACTIVITY 3.1 If you are an entrepreneur, would you be satisfied if the equilibrium point is met? Elaborate your reason(s).

EXERCISE 3.1 1.

Find the equilibrium point if the supply and demand equations of a product are p 

2.

1 1 q  8 and p   q  12 , respectively. 300 180

Suppose the demand function is

p 2 ă 400 and the supply

function is given as p 2 ă 40p + 2600. Determine the price and quantity at which the market reaches its equilibrium point.


TOPIC 3

3.2




57

COST AND REVENUE FUNCTIONS

Fixed costs are costs that are independent of the production level such as insurance cost. Total Cost = Fixed Costs + Variable Costs Total revenue is the money received from products sold. Total Revenue = (Unit Price)  (Total Quantity Sold) Profit is the difference between total revenue and total cost. Profit = Total Revenue  Total Cost Example: A factory produces a certain type of product worth RM200. The costs of its raw material and labour are RM30 and RM15 per unit, respectively. Fixed costs are RM100,000. (a)

Obtain the function for profit.

(b)

Find the profit if 10,000 units are sold.

Solution: (a)

Suppose q is the quantity of products sold. Total Revenue= (Unit Price)  (Total Quantity Sold) = 200q Total Cost = Variable Costs + Fixed Costs = 30q +15q + 100,000 Hence, Profit = Total Revenue ă Total Cost = 200q ă (30q +15q + 100,000) = 200q ă 30q ă 15q ă 100,000 = 155q ă 100,000


TOPIC 3


58



(b)

Substitute q = 10,000 into 155q ă 100,000 Profit = 155(10,000) ă 100,000 = 1,550,000 ă 100,000 = 1,450,000

Example: Assuming the cost of producing 10 units of a given product is RM40, while that of 20 units is RM70. If the cost C is linearly related to production quantity q, find: (a)

The linear equation that links C to q.

(b)

The costs required to produce 35 units of the product.

Solution: Linear equation: C = mq + k ⁄(1) Substitute q = 10, C = 40 and q = 20, C = 70 into equation (1) to form two equations, i.e. 40 = 10m + k and 70 = 20m + k. Subtract one equation from the other to eliminate k and then obtain the value for m. (ă)

(a)

70 40 30 3

= 20m + k = 10m + k = 10m =m

Substitute m = 3 into 40 = 10m + k to derive the value of k. 40 = 10(3) + k 40 = 30 + k 10 = k So, the linear equation becomes C = 3q + 10

(b)

Substitute q = 35 into C = 3q + 10.

C = 3q + 10 = 105 + 10 = 115


TOPIC 3

3.2.1




59

Break-even Point Analysis

Break-even point of a product is the level of production at which there is no profit made or loss incurred. It is the point where total revenue equals to total cost.

Graph 3.4

TR = TC : Break-even point TR < TC : Business suffering loss TR > TC : Business generating profit Example: A company is selling a product at the price of RM45 per unit. Variable costs per unit is RM33, while fixed costs is RM450,000. How many units have to be sold to break even? Solution: Break-even point: Total Revenue = Total Cost Total Revenue = (Unit Price)  (Total Quantity Sold) = 45q Total Cost = Variable Costs + Fixed Costs = 33q + 450,000 Therefore, 45q = 33q + 450,000 45q  33q = 450,000 12q = 450,000 q = 37,500 Hence, 37,500 units have to be sold to reach the break-even point.


60

TOPIC 3




SELF-CHECK 3.1 Which part of the graph in Graph 3.4 should be avoided by an entrepreneur? Explain.

EXERCISE 3.2 1.

2.

3.3

Identify each of the following equations as a demand or supply equation. Then, find their corresponding price and quantity at which they break even. (a)

2p = 100q  600

(b)

p = 50q + 600

A company is producing a type of product with a selling price of RM50 per unit. To produce one unit of the product, the company has to use raw material, at a cost of RM40. Fixed costs are RM5,000. If q represents the quantity of products sold, determine: (a)

Revenue function;

(b)

Cost function;

(c)

Profit function; and

(d)

Quantity to be sold to obtain break-even point.

MAXIMUM AND MINIMUM

Maximum value can be obtained from the turning point on the parabola which opens downward and minimum value can be obtained from the turning point on the parabola which opens upward. Example: A company learnt that the demand function for its product is p = 48 ă 3q, where p represents the unit price and q is the quantity demanded for the product. (a)

Derive the revenue function.

(b)

Determine the quantity for which the revenue is maximised.

(c)

What is the maximum value of the generated revenue? Copyright © Open University Malaysia (OUM)

TOPIC 3




61

Solution: (a)

(b)

Total Revenue = (Unit Price)  (Total Quantity Sold)

R

=

pq

R

=

(48 ă 3q) q

R

=

48q ă 3q 2

The revenue function is a quadratic, therefore its turning point (x,y) will be used as it provides the maximum point. The x-coordinate is the quantity which the revenue is maximised.

R = 48q ă 3q 2 where a = ă3, b = 48, c = 0 Therefore: b 2a 48  2  3

q

48 6 8 

The quantity that will maximise the revenue is 8. (c)

Substitute q = 8 into the revenue function.

R = 48q ă 3q 2 = 48(8) ă 3(64) = 384 ă 192 = 192 Thus, the maximum revenue is RM192. Example: Given a cost function, C = q 2 ă 6q + 16. (a) (b)

Determine the quantity for which the cost is minimised. What is the minimum value of the cost?


62

TOPIC 3




Solution: (a)

The cost function is a quadratic, therefore its turning point (x, y) will be used as it provides minimum point. The x-coordinate is the quantity which the cost is minimised. C  q   q 2  6q  16 where a  1, b  6, c  16 q q

b 2a   6  2 1

q3

The quantity which the cost is minimised is 3. (b)

Substitute q = 3 into the cost function.

C (q) = q 2 ă 6q + 16 = 3 2 ă 6(3) + 16 = 9 ă18 + 16 = 7

Hence, the minimum cost is RM7.

ACTIVITY 3.2 Is it possible for a quadratic function to have both minimum and maximum values? Explain your answer.

EXERCISE 3.3 Company BC learnt that the demand function for its product is q = 1850 ă 5p, where p represents the unit price and q is the quantity demanded for the product. (a)


(b)

Find the revenue, given the unit price is RM5.

(c)

Determine the price for which the revenue is maximised.

(d)

What is the maximum value of the revenue?


TOPIC 3




63


A manufacturer sells a product at RM200 per unit. The labour cost is RM15 per unit and the material cost is RM50 per unit. The rent of the building is RM10,800 per month. Find the break-even quantity. A. 5

2.

B. 8

C. 46

D. 80

Given the demand function is p 2 – 200 and the supply function is p2 ă 20p + 1,400. What is the equilibrium price? A. RM80

B. RM100

C. RM800

D. RM6,200

For questions 3, 4 and 5 refer to the following information: The demand equation for a certain product is given as p = 2,750 ă 5q. 3.

4.

Find the total revenue function.

A. 2,750q ă 5

B. 2,750p ă 5pq

C. 2,750q ă 5q2

D. 2,750p ă 5p2

Determine the quantity that will maximise the total revenue.

A. 0 5.

B. 25

C. 50

D. 275

What is the maximum total revenue? A. 0

B. RM2,750

C. RM378,125

D. RM753,500



The two applications of the intersection point discussed are determining market equilibrium point and break-even point.



The market equilibrium point is obtained by solving the demand and supply equations simultaneously.



The break-even point is obtained by solving the revenue and cost equations simultaneously. Copyright © Open University Malaysia (OUM)

64



TOPIC 3




In this topic, we have covered the maximum value for the revenue function as well as the minimum value for the cost function by using the turning point method.



This is feasible as the two functions are quadratic and that the method only works for quadratic functions, while differentiation method can be used to determine maximum or minimum values for any type of functions.

Break-even point

Minimum value

Cost function

Profit function

Demand function

Revenue function

Equilibrium point

Supply function

Maximum value


Topic  Exponential

4

and Logarithmic Functions

LEARNING OUTCOMES By the end of this topic, you should be able to: 1.

Identify exponential and logarithmic functions;

2.

Identify the graph of an exponential and a logarithmic function;

3.

Calculate equations using properties of exponentials;

4.

Calculate equations using properties of logarithms; and

5.

Solve application problems.



INTRODUCTION

Logarithmic functions are related to exponential functions. Each logarithmic function is called the inverse of its corresponding exponential function, and that exponential function is the inverse of its corresponding logarithmic function. This topic will discuss the relation between these two functions and their applications.

4.1

PROPERTIES OF EXPONENTIALS

A function f is called an exponential function if it has a form f (x) = ax where the base a is positive, with a  0 and its exponent x is any real number.


66 

TOPIC 4

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

(1)

axa y  ax  a y

(2)

a x b x  ( ab) x

(3)

ax  a x y ay

(4)

ax  a    bx  b 

(5)

(a x ) y  a x y

(6)

a x 

(7)

a0  1

(8)

a1  a

(9)

ay 

x

x

1 ax

 a y

x

Examples: Find the values of (a) (d)

32  33 2

3

3

(b)

(22 )3

(e)

1   2

(b)

2 

(c)

42

(f)

3   2

(c)

42 

3

2

Solutions: (a)

32  33  32  3

2 3

 26  64

3 3

1

(d)

1 32 1  9

32 

 223

(e)

3

1 1 3    2  2  23 8

3

 4

3

 23 8

(f)

2

32 3    22 2 1 1   9 4 4  9


TOPIC 4

 67


Examples: Solve (a)

82 x = 2

(c)

2 x 2 x1 

1 8

1 e

(b)

e 2 x 1 

(d)

3 x ă 9 4- x = 0

2

Solutions: (a)

82 x  2

2 

3 2x

2

2 2 6x

(c)

1

(b) (Equate the base) (Compare the exponent)

6x  1 1 x 6 1 2 x 2 x 1  8 2 x  x 1  23 2 x  1  3 2 x  2

x  1

1 e 2 x 1 e  e 1 2 x  1  1 2 x  2 x  1 e 2 x 1 

3x  9 4  x  0 2

(d)

3x   32  2

4 x

x2  8  2 x x2  2 x  8  0

 x  2  x  4   0 x  2, x  4

ACTIVITY 4.1 Between linear and exponential functions, which one has the most rapid change in its values? Explain.


68 

TOPIC 4


EXERCISE 4.1 Find the values of (a)

3  3ă 4

(d)

 1 3   8

(b)

2ă3  8

(e)

1   5

2

1

(c)

27 3

(f)

42  2ă1

3

EXERCISE 4.2 Solve x

4.2

(a)

1    16 4

(b)

e x 3  1

(d)

2x8x = 2

(e)

5x 

1 25

(c)

4x ă 2x+1 = 0

(f)

e e  x2

x 2



1 e

EQUATIONS AND GRAPHS

There are two general shapes of exponentials graphs. The shapes depend on the base value of the exponential functions. (a)

y  a x where a > 1 y  ax

Graph 4.1


TOPIC 4

(b)

 69


y  a x where 0 < a < 1

y  ax

Graph 4.2 x Following are the properties of the graph of exponential function f ( x)  a

(i)

The y-intercept on the exponential graph is (0,1).

(ii)

There is no x-intercept.

(iii) If a > 1, the graph is increasing from left to right. (iv) If 0 < a < 1, the graph is decreasing from left to right. Example: Sketch a graph of y = 2x. Solution: (i)

Construct a table consisting several values of x and y.

(ii)

Plot the points on a plane.

(iii) Draw a smooth curve through all the plotted points. (i) x

2

1

0

1

2

3

y

1 4

1 2

1

2

4

8


.

70 

TOPIC 4


(ii)

Graph 4.3

Example: x

1 Sketch a graph of y    . 2

Solution: (i) x

3

2

1

0

1

2

y

8

4

2

1

1 2

1 4

(ii)

Graph 4.4


TOPIC 4

4.3


 71

LOGARITHMIC FUNCTIONS

A logarithmic function with base a, is written as loga where a > 0, a  1. y is the logarithm for x with base a, denoted by y = log a x. y = log a x Logarithmic Form



ay = x Exponential Form

Example: Convert the following equations, from logarithmic to exponential forms: (a)

log 3 9 = 2

(b)

log 10 y = 4

(c)

log 2 8 = 3

Solution: (a)

32 = 9

(b)

104 = y

(c)

23 = 8

Example: Convert the following equations, from exponential to logarithmic forms: (a)

25 = 32

(b)

100 = 1

(c)

53 = y

Solution: (a)

log 2 32 = 5

(b)

log 10 1 = 0

(c)

log 5 y = 3

Logarithm with base 10 is known as common logarithm, and is written as log 10 x = log x = lg x.


72 

TOPIC 4


While, logarithm with base e, is called natural logarithm and is denoted by log e x = ln x.

ACTIVITY 4.2 Is logarithmic function a reciprocal operation for exponential function? Why? Explain.

4.4

PROPERTIES OF LOGARITHMS

1.

log a a = 1

2.

log a mx = xlog a m

3.

log a m =

4.

log a M + log a N = log a MN

5.

log a M  log a N  log a

6.

If log a M = log a N then M = N

log b m (Logarithm base interchangeable formula) log b a M N

Examples: Using the above properties, find the value for: 1

(a)

log 3 81

(b)

ln

(d)

log 4 2

(e)

log 4 2  log 4 8

e

(c)

loga 1

(f)

log 6 54  log 6 9


TOPIC 4

 73


Solutions: (a)

log 3 81  log 3 34

(b)

 4log 3 3  4 1

1 1 ln  log e e e  log e e 1

(c)

log a 1  log a a 0

 0  log a a 

0

 1log e e

4

 11  1

(d)

log 4 2  log 4 4

(e)

log 4 2  log 4 8  log 4 16

(f)

 log 4 42

1

 log 4 4 2

 2log 4 4

1  log 4 4 2 1  2

54 9  log 6 6

log 6 54  log 6 9  log 6

 2 1

1

2

Examples: Find the value of x. (a)

log (2x + 1) = log (x + 6)

(b)

logx (6 ă x) = 2

(c)

log3 x = 2

(d)

log x = ă1

(e)

log2 x 4 + log 2 4x = 12

(f)

log x ă log (x ă 1) = log 4

Solutions: Find the value of x. (a)

log  2 x  1  log  x  6 

(b)

2x  1  x  6 2x  x  6 1 x5

log x  6  x   2 x2  6  x x2  x  6  0

 x  3 x  2   0 x  3, x  2 x  3 will be ignored as the base x  0 So, x  2

(c)

log 3 x  2 x3 x9

2

(d)

log x  1 log10 x  1 x  101


74 

(e)

TOPIC 4


log 2 x 4  log 2 4 x  12

(f)

log x  log  x  1  log 4

log 2 x 4  4 x   12

 x  log    log 4  x 1 x 4 x 1 x  4x  4 3 x  4 4 x 3 4 x 3

log 2 4 x  12 5

4 x5  212 212 22 x5  210 x5 

1

x   210  5 x  22 x4

EXERCISE 4.3 1.

Convert the following equations, from logarithmic to exponential forms. (a)

2.

4.

(b)

log2 y = x

(c)

log10 0.1 = 1

Convert the following equations, from exponential to logarithmic forms. (a)

3.

log5 25 = 2

102 = 100

(b)

a0 = 1

(c)

23 =

1 8

Using the properties of logarithm, find the values below: (a)

log2 16

(b)

log8 2

(c)

ln e

(d)

log4 1 4

(e)

ln 1 +1g 100

(f)

log2 1  log2 4 4

(c)

log8 64 = x  1

Find the value of x. (a)

log2 x = 0

(b)

logx 1  1 3

(d)

logx (2x + 8) = 2

(e)

logx + log (x ă 15) = 2

(f)

log3 (x + 1) = log3 (x ă 1) + 1


TOPIC 4


 75

ACTIVITY 4.3

4.5

1.

What is the value for e? What is the significance of e?

2.

How does a logarithmic function simplify the calculations involving exponential function?

EQUATIONS AND GRAPHS

There are two general shapes of logarithmic graphs. They depend on the base value of the logarithmic functions. (a)

y = log a x, where a > 1

Graph 4.5

(b)

y = log a x, where 0 < a < 1

Graph 4.6


76 

TOPIC 4


The following are the properties of the graph of logarithmic function f (x) = log a x. (a)

There is no y-intercept.

(b)

The x-intercept on the logarithm graph is (1,0).

(c)

If a > 1, the graph is increasing from left to right.

(d)

If 0 < a < 1, the graph is decreasing from left to right.

Example: Sketch a graph of y = log 2 x. Solution:



Convert the equation, from logarithmic to exponential form.



Construct a table consisting several values of x and y.



Draw a smooth curve through all the points.

(a)

y = log2 x 2y = x

(b) y

2

1

0

1

2

3

x

1 4

1 2

1

2

4

8

(c)

Graph 4.7


TOPIC 4


 77

Example: Sketch a graph of y = log 1/2 x Solution: y

(a)

1   x 2

(b) y

3

2

1

0

1

2

x

8

4

2

1

1 2

1 4

(c)

Graph 4.8

4.5.1

Application on Growth and Decay Processes

Exponential functions can be applied in growth and decay processes. The formula for total growth is:

P = P0 ert where

P P0 r t

= Number of residents after t years. = Number of original residents. = Percentage (rate) of growth = Time period Copyright © Open University Malaysia (OUM)

78 

TOPIC 4


Example: Suppose the total number of residents in a given town is 20,000 and the rate of growth of the residents is 5% per year. (a)

Determine the total number of residents in this town in the period of 6 years from now.

(b)

How many years will it take for the number of residents to double?

Solution: (a)

Substitute all the given values into the formula to find the value of P.

P = P0 e rt , where P0 = 20,000, r = 5% and t = 6. = 5/100 = 0.05

P = 20,000e 0.05(6) = 20,000e 0.3 = 26,997 Hence, the number of the town residents after six more years is 26,997. (b)

Doubling the number of residents implies P = 2Po. Substitute P with 2Po and r = 0.05 into the formula to find the value for t.

P  Po e rt 2 Po  Po e0.05t 2 Po  e0.05t Po 2  e0.05t log e 2  0.05t ln 2  0.05t ln 2 t 0.05 t  13.863 The numbers of residents will double in about 14 years.


TOPIC 4


 79

The formula for decay process is

P = P0 e -rt Example: Suppose a radioactive element is going through power decay after t days based on exponential function P = 100 e  0.075t. How much of the quantity is left after 20 days? Solution: Substitute t = 20 into the formula to find the value for P.

P = 100 e 0.075(20) = 100 e 1.5 = 100 (0.22313) = 22.313

4.5.2

Investment with Compound Interest

The total amount of money, denoted by S is the compound amount for a sum of money P compounding after n-th year, where the interest is payable k times at the rate of r% per annum, is given by the formula below:

r  S  P 1    k

nk

where: S = Compound amount or the prospective value P = Initial investment or the principal value r = Interest rate per annum k = Number of interest paid (compound) in a year n = Number of year/s Example: If RM1,000 is invested at the rate of 6% per annum, compounding (payable) every quarterly, what would the total amount be in the account after 10 years? Solution:

S = ?, P = 1000,

r = 6% = 0.06,

k = Every quarterly = 4  a year,


n = 10

80 

TOPIC 4


Then

r  S  P 1    k

nk

 0.06  S  1000 1   4   S  1000 1.015 

10  4 

40

S  1000 1.81402  S  1814.02 Example: Determine the principal amount of a loan, given that the prospective amount payable after 10 years is RM21,589.20 and the compound rate of 8% per annum, compounding (payable) on a yearly basis. Solution:

S = 21,589.20,

P = ?,

r = 8% = 0.08,

k = Every year = 1  a year,

n = 10

Then

r  S  P 1    k

nk

 0.08  21589.20  P  1   1  

10 1

21589.20  P 1.08 

10

21589.20  P  2.15892  21589.20 2.15892 P  10000 P

Visit the following website: http://webmath.amherst.edu/qcenter/logarithms/index.html regarding logarithms.

for

questions


TOPIC 4


 81

EXERCISE 4.4 1.

2.

3.

(a)

Given that the price of one acre of land is increasing at the rate of 2% per year. How long will it take for the price to increase to RM30,000, if its current value is RM10,000?

(b)

Due to the economic downfall, the total number of residents in a township drops at the rate of 1% per year. The initial population was 100,000 residents. What is the population after 3 years?

Determine the compound amount, given the following principal values, compound interest rates and time period: (a)

RM5,500; 6% per annum compounding on monthly basis; 18 months.

(b)

RM10,000; 8% per annum compounding yearly; 5 years.

(c)

RM7,600; 7.26% per annum compounding on quarterly basis; 5 years and 8 months.

(d)

RM2,300; 5.75% per annum compounding daily; 150 days. (assume 1 year = 365 days)

Determine the principal amount, given the following compound values, compound interest rates and time period: (a)

RM16,084.82; 6% per annum compounding monthly; 14 months.

(b)

RM10,197.02; 5.3% per annum compounding daily; 135 days. (assume 1 year = 365 days)

(c)

RM6,657.02; 12.6% per annum compounding every 2 months; 10 months.

(d)

RM36,361.63; 7.2% per annum compounding every 3 months; 5 years and 3 months.


82 

TOPIC 4



Is x

1 2



1 x2

?

A. Yes

2.

Given 2x 8x = 4. Solve x. A.

3.

1 2

5.

B. 0

Calculate 1  lg A. 10

4.

B. No

C.

1 4

D.

1 2

1 . 10

B. 1

C. 1

D. 10

Suppose RM5,000 is invested for 10 years at 6% per annum, compounded every four months. Find the compound amount. A. 5000(1.02)30

B. 5000(1.02)40

C. 5000(1.015)30

D. 5000(1.015)40

The population P of a city after two years starting from year 1990 0.06 . What is the growth rate? is given by P = 100 000e A. 1%

B. 2%

C. 3%

D. 6%



Exponential function and its inverse, i.e. logarithm function, form graphs reflection upon a line y = x.



The properties of exponentials and logarithms have to be grasped when tackling problem solving questions.



In addition, the skills to convert equations in the form of exponential to logarithm form, and vice versa are equally significant. Copyright © Open University Malaysia (OUM)

TOPIC 4


Exponential Function

Logarithm Function

Growth

Compound Interest

Decay


 83

Topic

5



Differentiation




1.

Develop basic differentiation rules i.e. formula for the derivative of a constant, of x n, of a constant times a function, and of sums and differences of functions;

2.

Find the derivative by applying the product and quotient rules;

3.

Apply the chain rule; and

4.

Derive the power rule as a special case of a chain rule.

INTRODUCTION

In this topic, you will be exposed to the process of obtaining derivatives for functions which can be differentiated. The process of finding the derivative is called differentiation. Differentiating a function by direct use of the definition of a derivative can be tedious, i.e. applying the limit definition. Thus, a set of differentiation rules has been derived from the limit method to simplify the process. The rules are completely mechanical and efficient procedures for differentiation. Nevertheless, for those who are interested in the proofs of the rules, they may refer to Calculus books for verification. The common notation used to denote differentiation of a function, with respect to dy a variable x, is f  ( x) and is pronounced as f prime x and (pronounced „dee y, dx dee x‰). If f  ( x) can be found, f is said to be differentiable, and f  ( x) is called the derivative of function f with respect to variable x or differentiation of f over x. Copyright © Open University Malaysia (OUM)

TOPIC 5

DIFFERENTIATION

CONSTANT RULE

5.1

RULE I: CONSTANT RULE

If f (x) = c, where c is a constant, then f (x) = 0. Example: (a)

If f (x) = 15, then f (x) = 0.

(b)

If g (x) = 1.4, then g  (x) = 0.

(c)

If y = 3x 0 , then y  

(d)

If y = e 5 , then y  = 0, when e 5 is a constant.

(e)

If y = log 7, then y  = 0.

dy  0 , as x 0 = 1. dx

POWER RULE

5.2

RULE II: POWER RULE

If f (x) = x n where n is a constant, then f (x) = nx nă1 Example: (a)

If f (x)

= x 6 , then

dy dx

= 6 x 6 ă1 = 6x

(b)

If h  w  

1 3

5

, then h  w   w



3 5

w5

3  Hence, h  w    w 5

3 1 5

3 8  w 5 5 Copyright © Open University Malaysia (OUM)

 85

86 

TOPIC 5

DIFFERENTIATION

If y  3 x 4 , then

(c)

y 

d   x dx 

4 x 3 4  x 3 

4 3

  

4 1 3

1 3

CONSTANT TIMES A FUNCTION RULE

5.3

RULE III: CONSTANT TIMES A FUNCTION RULE If f (x) = c ( g ( x )) , where c is a constant and g (x) exists, then f (x) = c(g  (x)) Example: Suppose y = 9 x 4 then

dy  d   9 x 4   dx  dx 

 9  4 x 4 1   9 4 x3   36 x 3

Example: Given f  x   4 x 3 . Find f (x). Solution: Convert f (x) into exponential form or power form. 1

f  x   4  x3  2 3

 4x 2


TOPIC 5

DIFFERENTIATION

 87

Therefore; f   x  4

d  32  x  dx  

 3 2 1   4 x  2  3

3 2   4 x  2  1

1

 6 x 2 or 6 x

THE RULES OF SUMS AND DIFFERENCES OF FUNCTIONS

5.4

RULE IV: RULES OF SUMS AND DIFFERENCES OF FUNCTIONS

(a)

If f (x) = g(x) + h(x), where g (x) and h (x) exist, then f (x) = g (x) + h (x)

(b)

If f (x) = g(x)  h(x) , where g(x) and h (x) exist, then f (x) = g (x) - h (x)

Example: (a)

Suppose f (x) = 3x

2

+ 5. According to Rule IV, f (x) = g(x) + h(x), where

g(x) = 3x 2 and h(x) = 5. f (x) = 3 ( 2 x = 6x (b)

2 ă1

) + 0

Suppose f (x) = 10x 5 ă 6x . According to Rule IV, f (x) = g(x) ă h(x), where

g(x) = 10x 5 and h(x) = 6x. Then; f ' (x )

= 1 0 (5 x 5 ă 1 ) ă 6 (1 x 1 ă 1 ) = 1 0 (5 x 4 ) ă 6 ( x 0 ) = 50x

4

ă 6


88 

(c)

TOPIC 5

DIFFERENTIATION

Suppose f (x) = 5x 4 ă 8x 3 + 3x 2 ă x + 12 . Differentiation is performed on each expression of f (x). f '(x)

= 5(4x 4–1 ) – 8(3x 3–1 ) + 3(2x 2–1 ) – 1 + 0 = 20x 3 – 24x 2 + 6x – 1

Example: Given f  x  

5  x 4  3 2

. Find f (x).

Solution: Simplify f (x): 5 x 4  15 2 4 5 x 15   2 2 5 15  x4  2 2

f  x 

Therefore, f   x  

d  5 x 4  d  15      dx  2  dx  2 

5  4 x41   0 2 20 x 3  2  10 x 3 


7 x3  x . Find f (x). 2 x

Solution: Convert f (x) by writing it in exponential form or power form.


TOPIC 5

f  x 

DIFFERENTIATION

7 x3  x 1

2x 2 x 7 x3  1  1 2x 2 2x 2 7 3 1 1 1 1  x 2 x 2 2 2 5 7 1 1  x2  x2 2 2

7  5 5 1  1  1 1 1  Therefore, f   x    x 2    x 2  2 2  2 2  3 1 35 1   x2  x 2 4 4 Example: 3

Suppose f  x   x 5  x 2  7 x  1 . Find f (x). Solution: Expand f (x) by applying the exponential rule. 13 5

8 5

f  x  x  7x  x f   x  

3 5

 8 8 1  3 3 1 13 135 1  7 x5   x5 x 5 5  5 13 85 56 53 3  52 x  x  x 5 5 5 8

3

13x 5  56 x 5  3x  5



2 5


 89

90 

TOPIC 5

5.5

DIFFERENTIATION

PRODUCT RULE

RULE V: PRODUCT RULE If

f ( x )  g ( x ) h ( x ), where g' ( x ) and h' ( x ) exist, then

f  ( x )  h ( x ) g  ( x )  g ( x ) h ( x ) Example: If f (x) = 2 x (3 x 2 ă 2). Determine the derivative for f (x). Solution: Let g(x) = 2x and h(x) = (3x 2 – 2) . Then g(x) = 2 and h(x) = 6x Therefore f (x)

= (3x 2 – 2)(2) + 2x (6x )

= 6x 2 – 4 + 12x 2 = 18x 2 – 4

Example: If f (x) = (x + 3)( 4x 2 + 2x). Determine the derivative for f (x). Solution: Let g(x) = x + 3 g Ê (x) = 1 Therefore fÊ(x)

and h (x) = 4x 2 + 2x, then and hÊ (x) = 8x + 2 = ( 4x 2 + 2x)(1) + (x + 3)(8x + 2) = 4x 2 + 2x + 8x 2 + 24x + 2x + 6 = 12x 2 + 28x + 6

Example: Given s (t ) = (8 ă 7t )(t 2 - 2) . Determine s'( t ) .


TOPIC 5

DIFFERENTIATION

Solution:

d 2 d t  2    t 2  2   8  7t   dx dx 2   8  7t  2t    t  2   7 

s   t    8  7t 

 16t  14t 2  7t 2  14  21t 2  16t  14 Example: Given y = ( x 2 + 3x ă 2)( 2 x 2 ă x ă 3) Solution:

dy d d   x 2  3x  2   2 x 2  x  3   2 x 2  x  3  x 2  3 x  2  dx dx dx   x 2  3x  2   4 x  1   2 x 2  x  3  2 x  3   4 x3  x 2  12 x 2  3x  8 x  2    4 x3  6 x 2  2 x 2  3x  6 x  9    4 x3  11x 2  11x  2    4 x3  4 x 2  9 x  9   8 x3  15 x 2  20 x  7

5.6

QUOTIENT RULE

RULE VI: QUOTIENT RULE

If y  f  x   f   x 

g  x h x

, where g(x) and h(x) exist, then

h  x  g   x   g  x  h  x   h  x  

2


x x 1


 91

92 

TOPIC 5

DIFFERENTIATION

Solution: f   x 

 x  1

d d  x    x   x  1 dx dx 2  x  1

 x  11   x 1 2  x  1  x  1  x  2  x  1 



1

 x  1

2

Example: y

2x  3 4x  1

Solution: dy   dx   

4 x  1

d d  2 x  3   2 x  3  4 x  1 dx dx 2  4 x  1

 4 x  1 2    2 x  3 4  2  4 x  1 8 x  2  8 x  12

 4 x  1

2

14

 4 x  1

2

Example: y

8x2  2 x  1 x2  5x


TOPIC 5

DIFFERENTIATION

 93

Solution: dy 8 x 2  2 x  1  dx x2  5x  x 2  5x  dxd 8 x 2  2 x  1  8x2  2 x  1 dxd  x 2  5x   2  x2  5x 

x    

2

 5 x  16 x  2    8 x 2  2 x  1  2 x  5 

x

16 x

3

2

 5x 

2

 2 x 2  80 x 2  10 x   16 x 3  40 x 2  4 x 2  10 x  2 x  5 

x

2

 5x 

2

16 x 3  82 x 2  10 x  16 x 3  44 x 2  12 x  5

x

2

 5x 

2

38 x 2  2 x  5

x

2

 5x 

2

CHAIN RULE

5.7

RULE VII: CHAIN RULE If y = f(u), where u = g(x), then y  x  

dy  dy  du      dx  du  dx 

Example: Given y = (1 + x ) 4 . Determine

dy . dx

Solution: Step 1:

Introduce one new variable, u, so that

dy du and are easy to calculate. du dx

Let u = 1 + x, then y = u 4 Step 2:

Calculate

dy du and . du dx Copyright © Open University Malaysia (OUM)

94 

TOPIC 5

DIFFERENTIATION

y = u4 ,

When u = 1 + x, and du =1 and Then dx Step 3:

dy = 4u 3 du

Use the chain rule to calculate y  x  

dy . dx

dy  dy  du      dx  du  dx   4u 3 1  4u 3

Step 4:

dy into expressions of x. dx dy , gives Substitute u = 1 + x into dx dy = 4(1 + x) 3. dx

Calculate

Example: Determine

dy , given y = (3 + x 3 )4. dx

Solution: (a)

Let u = (3 + x 3 ), hence y = u 4

(b)

Then

(c)

By using chain rule:

du = 3x 2 and dx

dy = 4u 3 du dy  dy  du      dx  du  dx 

= 4u3 ( 3x 2 ) (d)

Substitute u = (3 + x 3 ) into

dy . dx

3 dy  4  3  x3   3x 2  dx

 12 x 2  3  x 3 

3


TOPIC 5

DIFFERENTIATION

Example: 3

Given y   2 x 2  1 2 , determine yÊ (x). Solution: 3

(a)

Let u = ( 2x 2 + 1), hence y   u  2

(b)

Then

(c)

(d)

du dy 3 12  u = 4x and du 2 dx

dy  dy  du      dx  du  dx  3 1  u 2  4x 2 dy Substitute u = ( 2x 2 + 1) into . dx

By using chain rule:

1 dy 3   2 x 2  1 2  4 x  dx 2 1

 6 x  2 x 2  1 2

5.8

POWER RULE (SPECIAL CASE OF CHAIN RULE)

RULE VIII: POWER RULE (SPECIAL CASE OF CHAIN RULE) If y   g  x   , then y  n  g  x   n

n 1

g  x  .

Example: Given y = (3x + 4)7. Find y(x). Solution: Let g(x) = 3x + 4, then g(x) = 3 and n = 7. Therefore y(x) = 7(3x + 4)7ă1(3) = 21(3x + 4)6 Copyright © Open University Malaysia (OUM)

 95

96 

TOPIC 5

DIFFERENTIATION

Example: Given y = (13 – x 4 )5 . Find y ' ( x) . Solution: Let g(x) = (13 ă x 4 ), then g(x) = ă 4x 3 and n = 5. Therefore y(x) = 5(13 ă x 4 ) 5ă1 (ă 4x 3 ) = ă20 x 3 (13 ă x 4 ) 4 Example: Differentiate function y = (3x 2 ă 2x + 1)1/2 . Solution: 1 1 d dy 1   3 x 2  2 x  1 2  3x2  2 x  1 dx 2 dx 1  1   3 x 2  2 x  1 2  6 x  2  2

  3x 2  2 x  1



1 2

 3x  1

Example:  2x  2  Differentiate function f  x   3    x3  Solution:  2x  2  f   x   3   x3 

3

31

d  2x  2    dx  x  3  d d  2  x  3  2 x  2    2 x  2   x  3   2x  2   dx dx  3    2  x3    x  3   

 2x  2   3   x3 

2

  x  3 2    2 x  2 1    2    x  3  


TOPIC 5

DIFFERENTIATION

 2x  6  2x  2    2    x  3   2 4   2x  2     3   2  x  3    x  3 

 2x  2   3   x3 

2

EXERCISE 5.1 Find the first degree differentiation for the following functions: (a)

f (x) = 15

(b)

f (x ) = 5x 0

(c)

y = 6e 3

(d)

y = 81n 2

(e)

y = x 3 (x 4 )

(f)

h (s ) = x 5

(g)

p(r )  r

(h)

s (t ) 



2 3

1 3

t5 (i)

y  4 x5

(j)

y  x5

(k)

y  4 8 x2

(l)

f ( x)  5 x 2 

(n)

f ( x)  2 

(p)

f ( x) 

(r)

y = (1 + x ) (1 ă 2x )4

(t)

y

(v)

y = (2x ă 1)2

(x)

y  3x 2  2 x  1

(z)

 2x  2  f ( x)     x3 

(m) f (x) = 3x + 7 (o) (q)

y = x 2 + 4x + 8 f ( x) 

4 x3  7 x  4 x

(s)

y = (2x ă 1) (x + 1)4

(u)

y

4 x3  1 x2  1

(w) y = ( x 2 + 4)5 (y)

f ( x)  x  5 x 2

3 4 x

x 4

5( x 4  3) 2

x 1 x2  2


3

 97

98 

TOPIC 5

DIFFERENTIATION

ACTIVITY 5.1 Join the discussion on rules of differentiation at: http://www.maths.fsnet.co.uk/Maths%20A.htm

MULTIPLE CHOICE QUESTIONS

1.

If f ( x)  x 4  5 x, then f (1) is A. 0

2.

B. 4

3.

4.

If y 



3 2

D. 12x3  x

x



3 2

8(9  3x)5 , then y  = 5

8(9  3 x)5 25

A. 24(9  3 x) 4

B. 

8(9  3x)5 C. 25

D. 24(9  3 x) 4

If y 

A.

5.

B. 3x 3  x

x

C. 12x 

D. 9

2 , then g ( x)  x

If g ( x)  3 x 4  A. 3 +

C. 5

3  2x 2 , then y = x2

3 2 x2

B. 

6 x3

C. 

4 x3

D.

3  4x x4

Given y  5 x 3  x  8, then y (2)  y (1) . A. 14

B. 34

C. 46

D. 60


TOPIC 5

DIFFERENTIATION

 99

You are required to understand and comprehend the following rules of differentiation:



If f (x) = c, where c is a constant, then f (x) = 0.



If f (x) = x n where n is a constant, then f (x) = nx n ă1



If f (x) = c (g (x )) , where c is a constant and g(x) exists, then f (x) = c(g(x))



If f (x) = g(x)  h(x), where g(x) and h(x) exist, then f (x) = g(x)  h(x)



If f (x) = g(x) h(x), where g(x) and h(x) exist, then f (x) = h(x) g(x) + g(x) h(x)



If y  f  x   f   x 

g  x h x

, where g(x) and h(x) exist, then

h  x  g   x   g  x  h  x   h  x  

2



dy  dy  du  If y = f (u), where u = g(x), then y  x       dx  du  dx 



If y   g  x   , then y  n  g  x   n

n 1

g  x  .

Chain Rule

Product Rule

Constant Rule

Quotient Rule

Constant Times a Function Rule

Sums and Differences of Functions Rule

Power Rule


Topic

6



Application of Differentiation




1.

Apply the rules of differentiation when deriving higher degree derivatives of various functions;

2.

Solve the functions of total costs, total revenue and total profit in economic and business world;

3.

Calculate the average functions of total costs, total revenue and total profit in economic and business world;

4.

Derive the marginal or ultimate function of total costs, total revenue and total profit in economic and business world; and

5.

Minimise the total costs function while maximising the total revenue and total profit functions, by using differentiation.

INTRODUCTION

The derivative or differentiation of function y = f (x ) , is denoted by: dy yÊ = f ' ( x) = is the first degree differentiation function with respect to x. dx When differentiation is performed on y', then: d  2 y y   f   x    2 is the second derivative of function with respect to x, which is dx read as „f double prime of x‰. Similarly when differentiation is performed on y‰, the third derivative is d  3 y y   f   x    3 is the third degree differentiation function with respect to x. dx Copyright © Open University Malaysia (OUM)

TOPIC 6

APPLICATION OF DIFFERENTIATION

 101

And subsequently, the higher-order derivative is; d  n y y n  f n  x    n  is the n-th degree differentiation function with respect to x. dx Comprehensive understanding and ability in applying the rules of differentiation together with the knowledge on demand and supply functions, will assist the students in realising the applications of differentiation. The students are expected to appreciate the applications of differentiation in economic and business field, which involve calculating the functions of total costs, total revenue and total profit. Differentiation method will determine how to minimise the total costs function while the total revenue function and total profit function are maximised.

6.1

SECOND AND THIRD DEGREE DIFFERENTIATION

This topic will cover differentiation up to the third degree only. Application of the rules of differentiation repeatedly will support the process of obtaining the required level or degree of differentiation. Example: Given y = 4x 3  12 x 2 + 6x + 2. Derive y  .

Solution: y   4  3x 2   12  2 x   6  12 x 2  24 x  6 y   12  2 x   24 1  24 x  24


102 

TOPIC 6


Example: Determine

d  3 y dx

 3

given y  2 x 1  x 2  1

Solution: dy  2  1x 2   2 x dx  2 x 2  2 x d  2 y dx

 2

 2  2 x 3   2  4 x 3  2

d  3 y dx

 3

 4  3x 4   12 x 4

EXERCISE 6.1 1.

Find the second degree differentiation for the given functions: (a)

2.

6.2

y  4 x3  12 x 2  6 x  2

(b)

y

2 x2

Find the third degree differentiation for the following functions: (a)

y  4 x2

(c)

y  2 x 1  x 2  1

(b) y  4 x3  12 x 2  6 x  24

TOTAL COST FUNCTION (C)

Total cost function is the total cost required to produce x units of a product. In short, it is the cost required to conduct a business.


TOPIC 6


 103

There are two types of costs: (a)

Fixed Costs

:

Unchanged or unvaried costs, i.e. flat although the number of units of a product being produced varies. For example, monthly rental of building.

(b)

Variable Costs

:

The costs which depend on the number of units of a product produced. For example, raw material and part-time workers.

In general, the function for costs can be written as:

C (x) = Fixed Costs + Variable Costs =

Fixed Costs + (Number of units)  (Cost per unit)

Example: If the production costs for one unit of a children toy are RM5 while its fixed costs are RM7,000, (a)

Determine the cost function.

(b)

What is the total cost for producing 100 units of the above toy?

Solution: (a)

The cost function, C (x) = Fixed Costs + x (Per Unit Cost) = 7000 + 5x

(b)

When x = 100, C (x)

= 7000 + 5(100) = 7000 + 500 = 7500

Therefore, the total cost for producing 100 units is RM7,500.

6.2.1

Average Total Cost Function ( C )

The average total cost function, C ( x) is the total cost for producing one unit of a product.

C  x 

C  x x


104 

TOPIC 6


Example: Given the total cost function, C (q) = 2q + 40. What is the average total cost function? Solution: The average total cost function, C  q  

C q

q 2q  40  q 40 2 q

6.2.2

Marginal or Ultimate Total Cost Function (C')

In business, the rate of change for a function is known as marginal function. Marginal or ultimate total cost function, denoted by C (x) is the rate of change for total cost function over quantity. Example: Given, the average cost function, C  x  

1 3 x x 10

(a)

What is the total cost function?

(b)

What is the ultimate total cost function?

(c)

Calculate the rate of change for cost (assuming the cost is in RM) when 4 units of product are produced.

Solution: (a)

Total cost function, C ( x)  C ( x)( x) 3 1   x   x x  10 1  x2  3 10

(b)

The ultimate cost, C   x  

1  2x  0 10

1  x 5


TOPIC 6


 105

The rate of change for cost is the ultimate cost, C '(x).

(c)

1 C '(4)  (4) 5 If four units of product are produced, i.e. x = 4, then 4  5 Hence, the rate of change for cost when 4 units of product are produced is RM0.80 per unit.

6.2.3

Minimising Total Cost

In business and economic, the cost is usually reduced (minimised) to obtain the highest (maximised) production revenue and total profit. Steps to minimise the cost function, C (x): 1.

Find C '(x) and C  ( x)

2.

Let C '(x) = 0 and solve for x. Suppose x = a and a has to be positive.

3.

If C (a )  0 , then x = a is the quantity or level of production which minimises the cost.

Example: If C (q ) = 0.01q 2 + 5q + 100 is the cost function. (a) (b) (c)

Obtain the average cost function. Determine the production level, q which minimises the average cost. What is the minimum value for the average cost?

Solution: (a)

Average cost function, C  q  

C q q

0.01q 2  5q  100 q 100  0.01q  5  q 

(b)

(i)

Find C (q) and C (q ) .


106 

TOPIC 6

C q  


dC  0.01  100q 2 dq

and C   q  

(ii)

d 2C  200q 3 dq 2 200  3 q

dC 0 dq

When

0.01 

100 0 q2 100  0.01 q2 q 2  10000 q  100

(iii) Will q = 100 minimise the cost?

d 2 C 200 d 2C 200 , when q = 100,  0  dq 2 q3 dq 2 1003 Therefore

(c)

d 2C  0 , then C ( x) will have a minimum value when q = 100. dq 2

When q = 100, C  q   0.01q  5 

100 q

C 100   0.01100   5 

100 100

7


TOPIC 6


 107

EXERCISE 6.2 1.

If the production costs for one unit of a product are RM10 while its fixed costs are RM5,000. (a) Find the cost function. (b) What is the total cost for producing 200 units of the above product? (c) Derive the average cost function. (d) Determine the ultimate total cost function.

2.

Suppose the average total cost function is C  q  

100000  1500  0.2q q

(a) (b) (c)

3.

Obtain the total cost function. Derive the marginal total cost function. Determine the rate of change for cost of producing 10 units of the product. q2 Given the total cost function is C  q    3q  400 4 (a) Find the average cost function. (b) Derive the ultimate total cost function. (c) What is the quantity which has to be produced so that the average total cost is minimised?

4.

Elyna Trading supplies sports attire to the supermarkets in North Peninsula. The company's annual cost is given by a function 15 C   0.15q  200 , where q is the quantity (in dozen) and C is the q total cost in a year (in thousands Malaysian Ringgit). (a) What is the quantity which minimises the total cost? (b) What is that minimum total cost?

5.

The total production cost of a cosmetic product is C  2500  75q  0.25q 2 (a) (b) (c)

Find the average total cost function. What is the quantity which has to be produced so that the average total cost is minimised? What is total cost at the production level which minimises the average total cost?


108 

6.3

TOPIC 6


TOTAL REVENUE FUNCTION (R )

Total revenue function, R(x) is the revenue received from production and sales of x unit of the product. If p is the unit price and x is the quantity of the product, then Total Revenue Function, R(x) = Price  Quantity = px

6.3.1

Average Total Revenue Function ( R )

The average total revenue function, R ( x) is the revenue received from selling one unit of a product, i.e. R  x  

6.3.2

R  x x

.

Marginal or Ultimate Total Revenue Function (R' )

Ultimate total revenue function is the rate of change of total revenue over quantity of a product: Ultimate Total Revenue Function = R’(x) Example: The demand function of a product is given by p = 200q + 500. (d)

What is the total revenue function?

(e)

Determine the ultimate total cost function.

Solution: (a)

Total revenue function, R(x) = Quantity  Price = qp = q (200q + 500) = 200q 2 + 500q


TOPIC 6


 109

(b) Ultimate total revenue function, R Ê(x) = 200q 2 + 500q =  400q + 500

SELF-CHECK 6.1 What are the definitions for „ultimate‰ and „marginal?‰ What is the significance of these two terms?

6.3.3

Maximising Revenue Functions

In business and economy, the total revenue is usually maximised to achieve the maximum profit. Steps to maximise the revenue function, R(x): 1. Find R( x) and R( x) . 2. Let RÊ(x) = 0 and solve for x. Suppose x = b and b has to be positive. 3. If R(b)  0 , then x = b is the quantity or level of production which maximises the revenue. Example:

80  q . 4 Determine the quantity which maximises the total revenue. Obtain the price which maximises the total revenue.

The demand function of a product is given by p  q   (a) (b)

Solution: (a)

Total revenue function, R(q) = Quantity  Price  q  p

 80  q   q   4  q2  20q  4


110 

TOPIC 6


To maximise the total revenue, the ultimate total revenue function has to be zero and R  q   0 . Ultimate total revenue function, R  q   20 

q 2

and R  q   

1 2

q 0 2 q   20 2 q  40

When RÊ(q) = 0, then 20 

Does the value q = 40 maximise the revenue? Substitute q = 40 into R  q  . 1 1 Observe that R  q    , therefore R  40     0 . 2 2 Therefore q = 40 is the quantity which maximises the total revenue.

(b)

80  q . At the quantity of q = 40, 4 80  40 p  40   4  10

Given p  q  

Hence, the price has to be fixed at RM10 in order to maximise the revenue. Example: A research has been conducted to determine the import tax of a unit of electronic item made in a foreign country. The demand on that particular item is given by a function D(t) = 8000 ă 20t, where D denote the demand quantity (in hundreds units) and t represent the import tax (in RM unit) (a)

Determine the revenue function for tax, R(t).

(b)

Calculate the import tax which needs to be imposed to maximise the tax revenue.

(c)

What is the maximum tax revenue?

(d)

Obtain the quantity of the required electronic item at the tax level which maximises its revenue.

Solution: (a)

R (t )  ( D )(t )  (8000  20t)t

 8000t – 20t 2 Copyright © Open University Malaysia (OUM)

TOPIC 6

(b)

R (t )


 111

= 8000t – 20t 2

dR  8000  40t dt dR  0 , then 8000 ă 40t = 0 When dt ă 40t = ă 8000 t = 200 R(t ) = ă 40

When t = 200, R(200)  0 then the total import tax which need to be imposed is RM200 to maximise the tax revenue. (c)

R (a ) = 8000t – 20t 2

= 8000(200) – 20(200) 2 = 1600000 – 800000 = 800000 Therefore, the maximum tax revenue is RM800,000. (d)

D (t ) = 8000 – 20t = 8000 – 20(200) = 4,000

To reach the level of tax which maximises its revenue, 4,000 units of electronic item have to be imported.


112 

TOPIC 6


EXERCISE 6.3 1.

The demand function of a health product is given by a function

p = 0.001q 2 + 840.

2.

3.

6.4

(a)

Obtain the total revenue function.

(b)

Derive the average total revenue function.

(c)

Determine the marginal total revenue function.

Given the total demand function, p (x) = 2 ă 0.01 p, where p is the unit price in thousand RM and x is the quantity of the item. (a)

Find the total revenue function.

(b)

Determine the price which maximises the total revenue.

Given the total revenue function, p (x) = 2 ă 0.01 p, where p is the unit price in RM. (a)

Determine the price which will maximise the total revenue.

(b)

Calculate the maximum total revenue.

TOTAL PROFIT FUNCTION ()

The total profit or loss function, (x), is obtained from the production of a single unit product. In general:

 = Total Revenue Function ă Total Cost Function = R(x) ă C (x) Example: The demand function for vehicle spare parts items at ATSAS Enterprise is given by p = 400 ă 2q and the average total cost per unit producing the item is given by 2000 a function C  q   q  160  . Determine the total profit function for ATSAS q Enterprise.


TOPIC 6


 113

Solution: Given p = 400 ă 2q and C  q   q  160 

2000 q

Total cost function, C  q   C  q   q

 2000    q  160  q q    q 2  160q  2000 Total Profit = Total Revenue ă Total Cost = R (q) – C (q) = pq – C ( q) = (400 – 2q ) q – (q 2  160q  2000) = –3q 2 + 240q – 2000

6.4.1

Average Total Profit Function (  )

The average total profit is the profit obtained from the production of a single unit product.

 x 

  x x

Example: Given the demand function for a product is p ( x)  8 – 0.025 x and the total cost function is C ( x)  500  7 x. Find (a)

Total revenue function;

(b)

Total profit function; and

(c)

Functions for the average total cost, average total revenue and average total profit.

Solution: (a)

Total revenue function, R (x) = xp (x) = x (8 – 0.025x )

= 8x – 0.025x 2 Copyright © Open University Malaysia (OUM)

114 

(b)

TOPIC 6


Total profit function, ( x)  R( x)  C ( x)

  8 x  0.025 x 2    500  7 x   0.025 x 2  x  500

(c)

Average total cost function, C  x  

C  x

x 500  7 x  x 500  7 x

R  x 

Average total revenue,

R  x

x 8 x  0.025 x 2  x  8  0.025 x

The demand function is also the average total revenue function. Average total profit,   x  

  x

x 0.025 x 2  x  500  x 500  0.025 x  1  x

6.4.2

Ultimate Total Profit Function (’)

It is the rate of change of total profit over the quantity of a product.

  x   R   x   C   x  Example: Suppose the total cost function, C (x ) = 0.05x 2 – 3x + 500 and the function for total revenue is R (x ) = 3x ă 0.01x 2 . Obtain: (a)

Total profit function; and


TOPIC 6

(b)


 115

Functions for the average total cost, average total revenue and average total profit.

Solution: (a)

Total profit function, (x) = R (x ) – C (x) = (3x – 0.01x 2 ) – (0.05x 2 – 3x + 500) = – 0.06x 2 + 6x – 500

(b)

Ultimate total cost function, C ' (x) = 0.05(2x) ă 3 = 0.10x ă 3 Ultimate total revenue function, R '(x) = 3 ă 0.01(2x) = 3 ă 0.02x Ultimate total profit function,  '(x)

= ă 0.06(2x) + 6 = ă 0.12x + 6

Or

 '(x)

6.4.3

= R '(x) ă C '(x) = (3 ă 0.02x) ă (0.10x ă 3) = ă 0.12x + 6

Maximising Total Profit

As we all know, the purpose of having a business as well as of any economy is to obtain the maximum profit. Steps to maximise the profit: 1. Find ( x) and ( x) . 2. Let ( x) = 0 and solve for x. Suppose x = c and c has to be positive. 3. If (c)  0 , then x = c is the quantity or level of production Example: The demand equation for a travel agency company is p = 40 ă 2q and its function 100 for average cost is given by C  q   4  . q (a)

Determine the total revenue function, R(q). Copyright © Open University Malaysia (OUM)

116 

TOPIC 6


(b)

Determine the total cost function, C (q).

(c)

Determine the total profit function, (q).

(d)

Calculate the price which will maximise the profit. Show that the profit is maximised.

Solution: (a)

R(q) = pq = (40 – 2q) q = 40q – 2q 2

(b)

C q 

C q q

=> C  q   q C  q 

 100   q4   q    4q  100 (c)

(q) = R (q) – C (q) = 40q – 2q 2 – (4q + 100) = – 2q 2 + 36q – 100

(d)

d  4q  36 dq d When 0 dq ă 4q + 36 = 0 4q = 36

q =9

When q = 9,

d2   4  0. dq 2

Substitute q = 9 into p, p (9) = 40 – 2(9)

= 22 Therefore, p = RM22 will maximise the profit.


TOPIC 6


 117

EXERCISE 6.4 1.

2.

3.

A company has an average total cost function of K  4  100q 1 . The demand equation for the company is given by function p = 54 ă q, where p is the unit price (in RM) and q is the quantity. (a)


(b)

Determine the cost function.

(c)

Obtain the profit function.

(d)

Find the price for which the company will maximise its profit, by using differentiation method.

The demand function of a local product is p = 300 ă x and the cost function is C (x ) = 0.1x 2 + 14x + 100 . (a)

Obtain the total cost function.

(b)

Derive the total profit function.

(c)

Determine the quantity which maximises the profit.

(d)

Calculate the price at which the profit is maximised.

(e)

Find the value of the maximum profit.

The demand function for a product based on recycle material is 2 given by p  x 2  5 x  16 and the average total cost function is 3 1 5 C  x   x2  2 x  x 3 (a)

Obtain the total cost function.

(b)

Derive the total revenue function.

(c)

Determine the total profit function.

(d)

Calculate the quantity which maximises the profit.

(e)

Find the value of the maximum profit.


118 

TOPIC 6


ACTIVITY 6.1 Can fixed cost be a variable? Why?


Given f ( x)  3x 2  12 x  9 . (a)

Find its critical point. A. (-6, -189)

(b)

2.

C. (2,0)

D. (2, 3)

Determine the nature of its critical point. A. Minimum point

B. Maximum point

C. Inflection point

D. No conclusion

Given C ( x)  0.05 x 2  3x  500 and R ( x)  3x  0.01x 2 . Find: (a)

(b)

3.

B. (-6,-24)

The total profit function. A. 0.06 x 2  6 x  500

B. 3 x  0.01x 2

C. 0.05 x 2  3 x  500

D. 0.06 x 2  500

The marginal profit function. A. 0.10 x  3

B. 3  0.02x

C. 0.12 x  6

D. 0.06 x 2  6 x  500

The quantity that will maximise the profit. A. 0.5

B. 6

C. 12

D. 50


TOPIC 6


 119

Differentiation can be applied to minimise or maximise a quantity. We can minimise cost and maximise revenue and profit. Some of the significant differentiation formulae which are often used in economic and business world are: total cost C , C .  Average Cost = quantity q dC  Ultimate/Marginal Cost = Rate of change for cost, C   . dq

total revenue R , R . quantity q



Average Revenue =



Ultimate/Marginal Revenue = Rate of change, R 



Average Profit =



Ultimate/Marginal Profit = Rate of change for profit,  

dR . dq

total profit  ,  . quantity q d . dq

Average Total Cost Function

Marginal/Ultimate Total Profit Function

Average Total Profit Function

Marginal/Ultimate Total Revenue Function

Average Total Revenue Function Marginal/Ultimate Total Cost Function

Second and Third Degree Differentation Test


Topic

7



Integration

LEARNING OUTCOMES By the end of this topic, you should be able to: 1.

Apply the rules of integration to solve integration problems;

2.

Calculate the definite integrals; and

3.

Apply the substitution technique for a complex problem.



INTRODUCTION

Calculus is divided into two broad areas-differential calculus, which was discussed in the previous topic, and integral calculus which will be discussed in this topic. Derivative is a rate of change of a function with respect to an independent variable. Therefore, if given a derivative of a function, then the function itself can be determined. The process of obtaining the original function is an inverse process of differentiation, known as integration. Geometrically, a definite integral refers to the area under a curve.

7.1

ANTI-DERIVATIVES

d F  x   f  x  then the integration of a function f (x) is F (x), because of their dx inverse relationship.

If


TOPIC 7

INTEGRATION

 121

The notation for integration:

 f  x  dx  F  x   c where



c

: The constant of integral

f (x) : Integrand

F'(x)

: Anti-derivative

dx

F'(x) = f(x)

: Integral sign

: With respect to variable x

Consider the following derivatives: (a)

d 2  x   2x dx

(b)

d 2  x  4  2x dx

(c)

d 2  x  100   2 x dx

Therefore, 2x is not the only derivative of x 2 , but also a derivative of x 2  4 and x 2  100 . Thus, having an arbitrary number c representing all the different numbers, x 2  c is the anti-derivative of 2x and is written as  2 x dx  x 2  c. Integration of a function f (x) is therefore F (x) + c and the process of finding F ( x) is called integration. The following rules are derived from reversing the process and can be applied to solve integration problems. Integration Rules

1.

 k dx  kx  c, k is a constant.

2.

n  x dx 

x n 1  c , n  1 . n 1

Special case ( when n = 1 ) 1 1  x dx   x dx  ln x  c 3.

k

f ( x) dx  k

 f ( x) dx , k is a constant.


122 

TOPIC 7

INTEGRATION

4.

  f ( x)  g ( x)  dx   f ( x) dx   g ( x) dx

5.

  f ( x)  g ( x)  dx   f ( x) dx   g ( x) dx

6.

e

x

dx  e x  c

kx  e dx 

7.

ek x c k ,

k is a constant.

Example: Applying Rule 1 (a)

1dx =

(b)

 5dx = 5 x  c

(c)

100 dx = 100 x  c

(d)

  dx   x  c

(e)

 e dx  ex  c

xc

Example: Applying Rule 2 (a)

1  x dx   x dx 

(b)

3  x dx 

x2 c 2

x 31 x4 c c 3 1 4 3 1 2

x 2 2 32 x dx   x dx   x c 3 3 2

(c)



(d)

1 x 2 1 3   c c dx x dx  x3  2 2 x 2

(e)

 x dx  ln x  c

1


TOPIC 7

INTEGRATION

 123

Example: Applying Rule 3, 4 and 5 (a)

 3x dx  3 x dx (Take the constant out as stated in Rule 3) 3

3

x 31 x4 3 c 3 c 3 1 4 1

1 1

1

(b)

 5 xdx  5  xdx  5 ln x  c

(c)

  3x

3

(Take the constant out as stated in Rule 3)

 1 dx   3x 3 dx   1dx (Applying Rule 4)

3 4 x  xc 4 (d)

  2x

5

 x 2  dx   2 x 5 dx   x 2 dx (Applying Rule 5)

2 6 1 3 1 1 x  x  c  x 6  x3  c 6 3 3 3 (e)

  2 x  x  x  1 dx   2x

2

 x  1 dx (Expand first as there is no rule for product integration)

=  2 x 2 dx   x dx  1dx (Expand using Rule 3 and 4) 2

(e)

x3 x 2   x  c (Integrate one at a time) 3 2

 x x

2

 1 dx

1 1  5    x 2  x 2  1 dx =   x 2  x 2  (dx Expand first)  

5

1

  x 2 dx   x 2 dx 5

1

1

1

x2 x2    c (Integrate one at a time) 5 1 1 1 2 2


124 

TOPIC 7

7

INTEGRATION

3

7

3

x2 x2 2x 2 2x 2   c  c 7 3 7 3 2 2 x4  x2  x 4 dx

(f)

 1 

1 dx (Simplify the expression, as there is no rule for division integration) x2

=  1dx   x 2 dx (Need to be expressed in the form of  n ) x 2 1 c 2  1 x 1 1  x c x c x 1  x

Example: Applying Rule 6 and 7 x

(a)

 e dx  e

(b)

 e dx 

2x

100 x

e

(c)

x

c

e2 x c 2

e100 x dx  c 100

x

(d)

 e 2 dx 

(e)

 4e

2 4 x

x 2

x e  c  2e 2  c 1 2

dx 

4e2 4 x  c  e 2  4 x  c 4

ACTIVITY 7.1 An inverse of differentiation is integration. Is there a mathematical process which does not have an inverse?


TOPIC 7

INTEGRATION

 125

EXERCISE 7.1 Integrate each of the following:

3.

  dx  8 p dx

4.

5.

 u

6.

7.

 x 1  x  dx

1.

9.

2.

3

3

 3u 2  du 3

8.

 e dt 1  x   2  x  dx  x 0.07 t

7.2

3

2

1  3  4    x 2  dx x x 3 x    5 2  x  2x  1    x 4  dx

10.

e

12.

  x  4  e

2

11.

 e dx  ex dx

3s  4

4

ds

x

x

  dx 

DEFINITE INTEGRALS

Suppose f (x) is a function which is defined between interval [a, b] and F (x) is the anti-differentiation for f (x). The definite integrals for f (x) between interval [a, b] is given by:

 f  x  dx   F  x  b

b

a

a

 F b  F  a 

where a is the lower limit and b is the upper limit of the integration. Example:

1 x 3

3

dx

Solution : 3



3

1

 x4   34 14  x dx        4 1 4 4 80  81 1       20 4  4 4 3


126 

TOPIC 7

INTEGRATION

Example:



1 0

e 2 x dx

Solution:

 e2 x  e 21 e 2 0 e dx      0 2 2  2 0 1

1

2x



e 2 e0 e2 1    2 2 2 2

Example:



e

1

1 dx x

Solution:

  ln x 1  ln e  ln1  1  0  1 e

RULES OF DEFINITE INTEGRALS Suppose

 f dx

and

 g dx

are defined between interval (a, b), where

a, b and c are constants. Hence, b

cf  x  dx  c  f  x  dx ; b

1.



2.

  f  x   g  x   dx   f  x  dx   g  x  dx; and

3.

  f  x   g  x   dx   f  x  dx   g  x  dx

a

a

b

b

b

a

a

a

b

b

b

a

a

a

Example:



3 1

3x3 dx


TOPIC 7

INTEGRATION

 127

Solution: 3

 x4   34 1   81 1    3 x dx 3 3     3     1  4 4  4 1  4 4  80   3    3  20   60  4  3

3

Example:



e 1

1    2 x  dx x 

Solution :



e 1

e1 e 1    2 x  dx  1 dx  1 2 x dx x x 

  ln x 1   x 2  1 e

e

  ln e  ln1   e 2  1  1  0  e 2  1  2  e2

ACTIVITY 7.2 What is the term given to the values of a and b for the interval [a, b]?

EXERCISE 7.2 Find the values of the following integrations: x3 dx

2.

 x 1  x  dx

(t 2  2t  8)dt

4.



 2 y y dy  4 

6.



4  y dy  

8.



1.



2

3.



1

5.



7.



3

2

9

8 1

2

3

1

2 1 4 1

4 1

2 1 4   x 3  x 2  x  dx   e5 dx

 x  1  dx  x 


128 

TOPIC 7

INTEGRATION

INTEGRATION BY SUBSTITUTION

7.3

Integration by substitution is one of the techniques used to integrate more complex functions. It will change the basic variable (usually x) to another variable (usually u). The relationship between these two variables must be specified. Once there is changing of the variable, the integration will be easier to handle. Example:

 2x

x 2  1 dx

Solution: Identify u and du, u = g(x), u is assumed without power. Therefore, u = x2 + 1 and hence, du = 2x dx.

Step 1:

Step 2: Substitute the integration from x to u. 1

1

1

2 2  2 x  x  1 2 dx    x  1 2 2 x dx   u 2 du

Integrate with respect to u.

Step 3:

1 2 32 2 u du  u c  3

Substitute back the u to x.

Step 4:



3 2 2 x  1 2  c  3

Example:



x x2  3

dx

Solution: Step 1:

Identify u and du,

u  x2  3

1 du  2 x dx  x dx  du 2 Copyright © Open University Malaysia (OUM)

TOPIC 7

Step 2:

Substitute the integration from x to u.

 Step 3:

1 du 1 1 dx   2 1   u 2 du 2 2 x 3 u2 x

Integrate

 1 1  u2   2 1  2 Step 4:

INTEGRATION

1  12 u du with respect to u. 2

 1    c  u2  c  

Substitute back the u to x. 1

 x 3  3 2  c Example:



ln x dx x

Solution:

u  ln x,

du 

dx x

Step 1:

Identify u and du,

Step 2:


 ln x Step 3:

dx  u du x 


u2 c 2 Step 4:


 ln x  2

2

c Copyright © Open University Malaysia (OUM)

 129

130 

TOPIC 7

INTEGRATION

Example:

x

2

x dx 1

Solution: Step 1:

Identify u and du.

u  x2  1 Step 2:

du  x dx 2

du  2 x dx 


du x x dx 1 du  x2  1 dx   x2  1   u2  2  u Step 3:


1 ln u  c 2 Step 4:


1 ln  x 2  1  c 2

ACTIVITY 7.3 1.

What do you think needs to be done in order to be an expert in recognising the u and du?

2.

Visit: http://www.sosmath.com/calculus/calculus.html for more examples on integration by substitution.


TOPIC 7

INTEGRATION

 131

EXERCISE 7.3 Integrate the following: 1.

 t

3.

e

5.



3

x

 2  t 4  8t  2  dx





e x  2 dx



4.



ds

6.

x

s 3

3x

2.

1  2s

2

dx 4x2  5 x2  x

 4  3x 2

2

 2 x3 

4

dx

x 1 dx  2x  1


x dx x



x2

A.

x 2.

3

e

5x

3

3

c

2

5 3

3 32 D. x  c 2

B. 

3 e 5 x 1  c 5x  1

3 C.  e 5 x  c 5

4 D.  e 5 x  c 5

3x 2  ( x3  8)2 dx

1 A.  c 8  x3 4.

2 12 C. x c 3

dx

A.  e 5 x  c

3.

2 B. x 2  c 3

 3x

2

x3 c B.  (8  x3 )3

x3 1 c C.  c D.  3(8  x3 )3 8  x3

( x 3  5) 20 dx

( x 3  5) 21 A. c 21 ( x 3  5) 21 c C. 63

x3 ( x3  5) 21 B. c 21 x 3 ( x 4  5) 21 D. c 21 Copyright © Open University Malaysia (OUM)

132 

TOPIC 7

3

5.

INTEGRATION

1

 6 x  3dx 1

1 15

A.

B. In 6

C.

1 1  15 2

D.

1 ln 6 6



Integration may be applied to determine a function given by its rate of change.



Basic integration rules include the following:



 k dx  kx  c,



n  x dx 

k is a constant.

x n 1  c , n  1 . n 1

A special case when n = 1 is as follows:

x



1

dx  

1 dx  ln x  c x



k



  f ( x)  g ( x)  dx   f ( x) dx   g ( x) dx



  f ( x)  g ( x)  dx   f ( x) dx   g ( x) dx



e



kx  e dx 

f ( x) dx  k

x

 f ( x) dx , k is a constant.

dx  e x  c

ek x  c , k is a constant. k

Integration by substitution is one of the techniques used to integrate more complex functions.

Indefinite Integrals

Integration by Substitution


Topic

8



Application of Integration




1.

Calculate the area of a region between a graph and x-axis;

2.

Calculate the area between two graphs;

3.

Apply integration to determine producersÊ surplus and consumerÊs surplus; and

4.

Apply integration to determine the cost, revenue and profit function from its marginal functions respectively.

INTRODUCTION

One of the applications of integration is to find the area of a region. This topic will specifically focus on finding the area between a graph and the x-axis and the area between two graphs. Later, integration is used to determine consumersÊ and producersÊ surplus. The applications of integration in economic and business, includes finding the cost, revenue and profit functions from their respective marginal functions.


134 

TOPIC 8

APPLICATION OF INTEGRATION

FINDING AREA UNDER A GRAPH

8.1

Suppose f (x) is continuous and that f (x)  0 in the interval [a, b]. Then, the area under the graph f (x) and above the x-axis, from x = a to x = b is A   f  x  dx . b

a

Graph 8.1

Example: Find the area under y = 2x from x = 0 to x = 2. Solution:

Graph 8.2

The sketch shows that the graph y = 2x is always above the x-axis when x is positive, therefore the area A is:



2 0

2

2 x dx   x 2   4  0  4 unit 2 . 0


TOPIC 8


 135

Suppose f (x) and g (x) are continuous in the interval [a, b] where f (x)  g (x), i.e. f (x) is always above g (x). Then, the area between the graphs f (x) and g (x) in the interval [a, b] is given by: A    f  x   g  x   dx a b

Graph 8.3

Example: Find the area between graphs y = x 2 and

y = x.

Solution: Step 1:

Sketch the graphs to determine which one is above and which is below.

Graph 8.4

Step 2:

Obtain the intersection points between graphs y = x 2 and y = x.

x2  x x2  x  0 x  x  1  0



x  0 and 1

 (0,0) and (1,1) are the intersection points. Copyright © Open University Malaysia (OUM)

136 

Step 3:

TOPIC 8


Determine the function above and function below.

f a = Function above fb = Function below Function above minus function below: Step 4:

x  x2

Determine the integration and obtain its value. 1

 x 2 x3  2 x  x dx    2  3   12  13  16 unit 2 0   0 1

Example: Find the area between graphs y = 2 ă x 2 and y = ă 2x ă 1.

Solution: Step 1:

Sketch the two graphs.

Graph 8.5

Step 2:

Obtain the intersection points between graphs y = 2 – x 2 and y = ă 2x ă 1.

2  x 2  2 x  1 x2  2 x  3  0

 x  3 x  1  0



x  3,  1

 (ă1, 1) and (3, ă7) are the intersection points. Step 3:

Graph above minus graph below  (2 – x ) – (– 2x – 1) = 3 + 2x – x 2 . 2


TOPIC 8

Step 4:


 137

Determine the integration and obtain its value. 3

 x3  2      3 2 x x dx 3 x x      1 3  1  1 2   9  9  9   3  1    10 3 3  3

2

ACTIVITY 8.1 1.

What is the purpose of finding an area under a graph? Explain.

2.

What will you get if you integrate in the interval [-3,3] the function y= 1? Explain.

EXERCISE 8.1 Find the area above or below for the following graphs: 1.

y = x 2 ă 1; x = ă1 to x = 1

2. y = 3 x 2 + 1; x = ă1 to x = 2

3.

y = 9 ă x 2 ; between [ă2, 1]

4. y = 2x + x 2 ; between [0, 3]

Find the area between the followings two curves: 5.

y = 2x + 4 and y = x 2 + 2x + 3.

6.

y = 2x ă 5 and y = ă x 2 + 6x ă 5.

7.

y + x = 6 and y = x 2 + 4.

8.

y=

x and y = x.


138 

TOPIC 8


APPLICATION OF INTEGRATION IN ECONOMICS AND BUSINESS

8.2

In economics and business, integration may be applied in the following areas: (a)

ConsumersÊ and producersÊ surpluses

(b)

Finding function from its marginal function

Let us discuss each of these areas in detail.

8.2.1

Consumers’ and Producers’ Surpluses

Recall the demand function, p = D (q) and the supply function, p = S (q). The intersection point of the two equations is called the market equilibrium (qe, pe), which exists in an ideal competition market.

ACTIVITY 8.2 What is meant by ideal competition? Does a competition which is not ideal exist?

(a)

ConsumersÊ Surplus There are consumers who are willing to pay more than the equilibrium price, so they benefit from the lower equilibrium price. The area of A represents the consumersÊ surplus. It is the total profit gained from the consumersÊ willingness to pay more than the market equilibrium price.

Graph 8.6 Copyright © Open University Malaysia (OUM)

TOPIC 8


 139

From Subtopic 8.1 we noticed this area can be found by finding areas between two graphs. ConsumersÊ surplus,

A

D(q )   qe

0



 The graph below

 The above graph qe

A   D ( q)dq  Pe qe  0   

or, in simpler form

 Area under curve D ( q )

(b)

Pe  dq 

Area of the rectangle

ProducersÊ Surplus On the other hand, some suppliers would have offered the product at a price below the equilibrium price, so they too can gain from the equilibrium price. The total of the suppliers gain is called the producersÊ surplus and is represented by the area B. ProducersÊ surplus,

B

p  S (q )  dq     qe

0

e

 graph above straigh tline

or, in simpler form

B

 graph below

pq  ee  Area of the rectangle

qe

 0

S (q) dq   Area under the below graph

Example: Given that the demand function and the supply function for a company are p = 200 ă q 2 and p = 6q + 160, respectively. Determine the consumersÊ and producersÊ surpluses for the company. Solution: Sketch the graph in the first quadrant only. Obtain the market equilibrium point.


140 

TOPIC 8


Graph 8.7

6q  160  200  q 2

p  200  q 2  200  16  184

q 2  6q  40  0

 q  10  q  4   0 q  10 and q  4 Hence, (4, 184) is the market equilibrium point. ConsumersÊ Surplus: CS  

4 0

 200  q  dq   4184 2

4

 q3    200q     736 3  0  4

3  4        736  200q  3    0 128  3

ProducersÊ Surplus: PS   4 184     6q  600  dq 4

0

 736   3q 2  160q    736   48  640

4 0

 48

ACTIVITY 8.3 Why would a consumer be willing to pay a higher price than the market equilibrium?


TOPIC 8


 141

Example: The demand and supply functions for a company are ps = 1 + q and pd  49  6q , respectively. Determine the consumersÊ and producersÊ surpluses for the company. Solution: Obtain the market equilibrium point.

pd  ps 49  6q  1  q 49  6q  1  q 

2

49  6q  1  2q  q 2 0  q 2  2q  1   49  6q  0  q 2  8q  48 0   q  4  q  12  We will consider only the point q = 4. P = 1 + q= 1 + 4 = 5 (4,5) is the market equilibrium point. Note: The value q = 4 is the integration limit. ConsumersÊ Surplus:

 D  q  dq   4 5 4

0

Graph 8.8


142 

TOPIC 8




4 0

49  6q dq  20

u  49  6q

u  49  6(4)  25

du  6 dq du  dq 6

u  49  0  49

49

 3 49 49 du 1 1 u2   25 u 6  20   6  25 u du  20  6  3   20    2  25 3 3 1 3 1     49  25   20   73  53   20    6 2  9 218 38   20  9 9 1 2

1 2



 



ProducersÊ Surplus: pe qe   S  q  dq   4  5    1  q  dq 4

0

4

 q2   16   20   q    20   4    20  12  8 2 0 2  

ACTIVITY 8.4 Which is better, a situation where there is consumersÊ surplus or producersÊ surplus?

8.2.2

Finding Function From Its Marginal Function

Previously, if given cost, revenue and profit function, we were asked to find the marginal functions namely marginal cost, marginal revenue and marginal profit function respectively. Now , in this topic, given a marginal function, we will be asked to find the cost, the revenue and the profit function. This can be done by integrating their corresponding marginal functions. Example: Suppose the marginal cost function for a company which produces x thousand 50 units of books, is given by C   x   , while its fixed cost is RM25,000. x Copyright © Open University Malaysia (OUM)

TOPIC 8


 143

(a)

Find the cost function.

(b)

Determine the increment in total cost for an increase in production from 100 to 121 units.

Solution: (a)

Given marginal cost function, we need to integrate the marginal cost function to obtain the cost function.

C  x    C   x  dx 1 1 1 1  50 x2 dx   50 x 2 dx  50  c  50  2  x 2  c  100 x 2  c  1 x 2

Given that the fixed cost is RM25,000. Cost when x  0 Solve C (0) = 25000. 1

Therefore, 100x 2 + c = 25000 100(0) + c = 25000 c = 25000 1

Hence, C (x) = 100x 2 + 25000 (b)



121 100

C   x  dx

 C 121  C 100   100 121  25000   100 100  25000   100 11  25000   100 10   25000   100 Thus, the required cost increment is RM100. Example: Suppose that a companyÊs marginal cost function is given by C (x) = 2x 3 + 6x ă 5 , and its fixed cost is RM8,000. Find the companyÊs cost function.


144 

TOPIC 8


Solution: Given the marginal cost function C (x) = 2x 3 + 6x ă 5 , we integrate to get its cost function. C  x    C '  x  dx   2 x 3  6 x  5 dx 2 x4 6 x2   5x  c 4 2 x4 C  x   3 x 2  5 x  c since C(0)=8000 2 C  0  0  0  0  c 

8000  c

Therefore, the cost function is

C  x 

x4  3x 2  5 x  8000 2

Example: A furniture factory has a marginal cost function, C ' ( x )  3 x  15 and marginal revenue function R ' ( x )  150  3 x with x is the quantity produced. Fixed cost for the factory is RM8,000. Determine: (a)

Revenue function

(b)

Demand function

(c)

Total cost function

(d)

Profit function

Solution: (a) R ( x)   R( x) dx   (150  3 x)dx

3x 2 C 2 3x 2  R (0)  0 =150x – 2  150 x –

(b)

The demand function is p  f ( x) and can be obtained from the revenue function. R  p.x


TOPIC 8

Therefore

p


 145

R x

3x 2 150 x  R 2  150  3x p  x x 2 3x The demand function is p= 150  2 (c)

C ( x)   C ( x)dx   3 x  15dx

3x 2  15 x  C 2 3x 2 =  15 x  8000  C (0)  8000 2 

(d)

P ( x)  R( x)  C ( x)  3x 2  3x 2  150 x   15 x  8000) ( 2  2   165 x – 3x 2  8000 EXERCISE 8.2 1.

2.

The marginal cost function for a company producing school bags is given by C '(q ) = 0.003q 2 – 0.03q , where q is the number of school bags produced. The companyÊs fixed cost is RM176. (a)

Find the cost function.

(b)

Determine the increment in total cost when the production increased from 260 to 400 units.

1000 . If R is 100q the revenue in RM, determine the change/increment in revenues when the sales increased from 400 to 900 units.

The marginal revenue function of a factory is R  =


146 

TOPIC 8


A furniture factory has a marginal cost function, C ' ( x )  3 x  15 and marginal revenue function R ' ( x )  150  3 x with x is the quantity produced. Fixed cost for the factory is RM8,000.

3.

Determine: (a)

Total cost function.

(b)

Profit function.

4.

The demand and supply functions for a company are p = 400  q and p = q + 100, respectively. Determine the consumersÊ and producersÊ surpluses for the company.

5.

The marginal cost function for a company is given by c(q) = 0.001 q 2

 0.02q, where q is in units. Fixed cost is RM1,000. Determine the cost function. 6.

The demand and supply functions for a particular product are p = 100 ă 0.05q and p = 10 + 0.1q, respectively. Determine the consumersÊ and producersÊ surpluses of the product.

7.

The demand and supply functions for a company are p = 0. 01 q 2  1.1q + 30 and p = 0.01 q 2 + 8, respectively. Determine the consumersÊ and producersÊ surpluses.


Find the area of the region enclosed by y  4 x 2 , xăaxis, x = ă 4 and x = 6. 1 B. 370 C. 333 D. 333.3 A. 373 3

2.

If a marginal revenue function is R '( x)  10  9 x  x 2 , find the total revenue function. A. TR = ă 9 + 2x

B. TR = ă 9+2x+c

9 2 x3 C. TR= 10 x  x  2 3

9 2 x3 D. TR = 10 x  x  +C 2 3 Copyright © Open University Malaysia (OUM)

TOPIC 8


 147

The following information is for Q3, Q4 and Q5. For the demand function D ( x)  200  x 2 , and supply function S ( x )  6 x  160 . 3.

4.

Determine the equilibrium point. A. x = 4, y = 184

B. x = 184, y = 4

C. x = ă 4, y = 184

D. x =184, y = -4

Determine the consumersÊ surplus. A. 872

5.

B. 48

C. 128/3

D. 248

Determine the producersÊ surplus. A. 48

B. 248

C. 148

D. 872



You should now realise that integration is an inverse of differentiation.



The applications of integration are not limited to only finding the area of a region, but can also be applied to many business and economic problems.

ConsumersÊ Surplus

ProducersÊ Surplus


Topic



9

Partial Differentiation




1.

Identify functions of one, two, three or more variables;

2.

Calculate values for multi-variable functions;

3.

Perform the first order partial differentiation on z = f (x , y ) ; and

4.

Perform the second order partial differentiation.

INTRODUCTION

Up to this point, we have discussed functions of only one independent variable y  f ( x) . We have discussed cost functions, which depend only on quantity, i.e C  C (q) and population functions, which depend only on time, i.e. P = P (t ) . However, in real life, many mathematical models depend on more than one variable. For example, cost function for a company depends on labour and material cost, while wolves' population in the forest depends on the population of rabbits, squirrels and other food resources. In this topic, we will look into the mathematics of functions of multi-variable. We will see that ideas applied to single variable functions, will also apply to functions of multi-variable, for instance, the idea of differentiation. Previously we have seen functions of one variable, y depends on x, written as y  f ( x) . In this topic, we will see functions of two variables, z depends on two variables x and y and hence written as z = f (x, y). We will also encounter Copyright © Open University Malaysia (OUM)

TOPIC 9

PARTIAL DIFFERENTIATION

 149

function of three variables, where w depends on x , y , and z , therefore the function can be written as w  f ( x, y, z ) .

9.1

FUNCTIONS OF MULTI-VARIABLE

Suppose z depends on the values of x and y, based on the following relationship:

z = 3x + 3xy + 3y Since the function of z depends on two variables, hence it can be written as

z = f (x ,y ) For every selected pair of x and y, there is only one value of z. For example, if x = 1 and y = ă 4, then

z

= 3x + 3xy + 3y = 3(1) + 3(1)(– 4) + 3(– 4) = 3 – 12 – 12 = –21

Therefore, if x = 1 and y = ă 4, then z = ă21, which can be written as f (1, – 4) = –21 . To obtain f (3,ă2) , we let x = 3 and y = ă2, then

f (x,y )

= 3x + 3xy + 3y

f (3,–2) = 3(3) + 3(3)(–2) + 3(–2) = 9 – 18 – 6 = –15

In general, z = f (x ,y ) is a function of two variables and each pair of x and y produced only one value of z. Therefore, x and y are called independent variables while z is said to be a dependent variable. Similarly, if w = f (x,y , z ) is a function of three variables and each triple of x and y

z produced only one value of w . Therefore, x , y and z are called independent variables, while z is said to be a dependent variable.


150 

TOPIC 9


Example:  4x

Suppose g  x, y   (a)

g (3,0)

(b)

g (ă 4,ă3)

x2  y 2

. Find the value for:

Solution: (a)

g (3,0) Now, x = 3 and y = 0.

g  3,0   (b)

 4  3 3 0

g (ă 4,ă3) g   4, 3 

2

2



12 9



12  4 3

 4   4

  4

2

  3

2

16 16 16   16  9 25 5



Example: Suppose h  x, y   x 2  2 y 2 . Find the values of: (a)

h(5, 3)

(b)

h(2, 4)

(c)

h(ă1, ă3)

Solution: (a)

h (5, 3) Then, x = 5 and y = 3. h  5,3  52  2  3  25  2(9)  43 2

(b)

h(2, 4) Then, x = 2 and y = 4. h  2, 4   22  2  4   4  2 16   36  6 2


TOPIC 9

(c)


 151

h(ă1, ă3) Then, x = ă1 and y = ă3.

h  1, 3 

 1

2

 2  3  1  2  9   19 2

Example: Function of three variables Suppose f (x, y , z ) = 4 x 2 – 3xz + 2yz 2  1 . Find (a)

f (3, 0, 1)

(b)

f (–2, 1, 0)

Solution: (a)

f (3,0,1) = 4(3) 2 – 3(3)(1) + 2(0) 1  1 2

 36  9  0  1  26 (b)

f (–2,1,0) = 4(–2) 2 – 3(–2)(0) + 2(1)  0   1 2

 16  6  0  1  21 Example: The

labour

charge

for

assembling a type of car is given by L( x, y )  12 x  6 y  2 xy  40 where x is the number of hours required by a skilled worker and y is the number of hours required by a semi-skilled worker. Find

(a)

L (3, 5)

(b)

L (5, 2)

(c)

If a skilled worker requires 7 hours to assemble a car and a semi-skilled worker requires 9 hours, find the total labour cost.

Solution: (a)

L( x, y )  12 x  6 y  2 xy  40

L(3,5)  12(3)  6(5)  2(3)(5)  40  136 (b)

L(5, 2)  12(5)  6(2)  2(5)(2)  40  132

(c)

L(7,9)  12(7)  6(9)  2(7)(9)  40  1304 Copyright © Open University Malaysia (OUM)

152 

TOPIC 9


ACTIVITY 9.1 If f (x, y , z ) = x + 4y 2 + y 3 , where variable z is not in the function. Can the function be expressed as f (x, y ) = x + 4y 2 + y 3 ? Why?

EXERCISE 9.1 1.

2.

3.

4.

Given f (x, y ) = 4x + 5y + 3 . Find the values of (a)

f (2, ă1)

(b)

f (ă2, ă3)

Given g (x, y ) = – x 2 – 4xy + y 3 . Find the values of (a)

g (ă2, 4)

(b)

g (ă2, 3)

Given g (x, y , z ) = – x 2 – 4xy + y 3 . Find the values of (a)

g (3, 0, 1)

(b)

g (ă2, 1, 4)

The population of cats in an area is assumed by a mathematical model, C (x , y ) = x 2 + 200y ă 1200 , where x is the population of rats measured in hundreds and y is the population of mole rats measured in tens. (a)

Determine C (50, 0)

(b)

Determine C (30, 4)

(c)

Find the number of cats if there are 1,400 rats and 150 mole rats.


TOPIC 9


 153

PARTIAL DERIVATIVES

9.2

dy , where it measures the rate at dx which a function y = f(x) is changing with respect to changes in variable x. Let us now look into the rate of change of function f(x,y) with respect to one of its variables.

In Topic 6 we have discussed about derivatives

Let f(x,y) be a function of two variables x and y. Since we want to know how f(x,y) changes both with respect to x and with respect to y, we shall now define partial derivatives with respect to the two variables. Definition 1. The partial derivative of f with respect to x is the derivative of f obtained by treating x as a variable and y as a constant. fx or f/x. Notation: 2. The partial derivative of f with respect to y is the derivative of f obtained by treating y as a variable and x as a constant. Notation:

fy or f/ y

Recall from the differention:

d d (k )  0 (3)  0  dx dx d d   kx   k  2x  2 dx dx d d   kf ( x)   kf ( x)  2 x2   2(2 x)  4 x dx dx d d 2 x 2  6   2(2 x)  0  4 x   kf ( x)  C   kf ( x)  dx dx Example: (a)

f ( x, y )  4 x 2  6 y 3 f x  8x remember y is a constant f y  18 y 2

remember x is a constant


154 

(b)

TOPIC 9


f ( x, y )  2 xy  x 2

fx  2 y  2x

remember y is a constant

f y  2x

remember x is a constant

Example: Let f (x, y ) = 4x 2 – 9xy + 6y 3 . Find fx and fy . Solution: f (x, y ) = 4x 2 – 9xy + 6y 3

(a)

(Remember to find fx , y is a constant)

f 4 x 2   8 x,  x

f  9 xy   9 y, x

f 6 y3   0  x

Hence, fx = 8x ă 9y. (b)

(Remember to find fy , x is a constant)

f 4 x 2   0,  y

f  9 xy   9 x, y

f 6 y 3   18 y 2  y

Example: Suppose f (x, y ) = 2x 2 + 3xy 3 + 2y + 5 . Find (a)

fx (1, 2)

(b)

fy (4, 3)

Solution: (a)

f (x, y ) = 2x 2 + 3xy 3 + 2y + 5

To differentiate with respect x means y is a constant

 

f 2 x 2  4 x, x

f 3xy 3   3 y 3 ,  x

f f 2 y  0  5  0 x x

f x = 4x + 3y 3

f x  1, 2   4(1)  3(2)3  4  3(8)  20 Copyright © Open University Malaysia (OUM)

TOPIC 9

(b)


 155

f (x, y ) = 2x 2 + 3xy 3 + 2y + 5 To differentiate with respect to y means x is a constant

 

f 2 x 2  0, y

f  3xy3   9 xy 2 , y

f f 2y  2  5  0 y y

fx = 9xy 2 + 2 fx (4,3) = 9( ă 4)( ă3)2 + 2 = 9(– 4)(9) 2 + 2 = 324 + 2 = 322

EXERCISE 9.2 1.

Given f (x, y ) = 2x 4 + 6x3 y 3 + 5y 2 – 8 . Find fx and fy .

2.

Given f (x, y ) = 8x 2 + 5xy 2 – 9y + 4 . Obtain

3.

(a)

fx (0, 4)

(b)

fx (1, 2)

(c)

fy (2, 5)

(d)

fy (0, 0)

Given the temperature of the water at the point on a river, where a nuclear power plant discharges its hot water, is approximated by a function T(x, y) = 2x + 5y + xy ă 40 where x represent the temperature of the river water measured in C before it reaches the power plant and y is the number of megawatt (in hundreds) of electricity being produced annually by the plant. (a)

Find the temperature of the water around the factory if the water temperature before reaching the factory is 10C and 400 megawatt of electric is produced a year.

(b)

Find and interpret Tx (5, 4).

(c)

Find and interpret Ty (8, 3).


156 

TOPIC 9


HIGHER ORDER PARTIAL DERIVATIVES

9.3

The idea of second order differentiation for functions of single variable y=f(x) can be applied to functions of multi-variable, where the first degree partial differentiation will be differentiated again with respect to x and y. If z = f (x , y ) z  fx x

and

z = fy y

the first order differentiation

  z    f x   f xx    xx  x 

  z   f y   f yy      y y   y   the second order partial diferentiation  z      f y   f yx   x   x   x   z     f   f xy   y   x   y x 

Example: Find all the second degree 3 2 3 2 f (x , y ) = ă 4x ă 3x y + 2y .

partial

derivatives

for

Solution: First we find the first order partial differentiation, fx and fy . Given f (x , y ) = ă 4x 3 ă 3x 2 y 3 + 2y 2 .

Then,

f x (x , y ) = ă 12x 2 ă 6xy 3 f y (x , y ) = ă 9x 2 y 2 + 4y To get fxx , perform partial differentiation with respect to x on fx

   f x    12 x 2  6 xy 3  x x  24 x  6 y 3 Copyright © Open University Malaysia (OUM)

function

TOPIC 9


 157

To get fyy, perform partial differentiation with respect to y,on fy   fy     9 x 2 y 2  4 y  y y  18 x 2 y  4

To get fyx , perform partial differentiation with respect to x on fy

  fy   9 x 2 y 2  4 y    x y  18 xy 2 To get fxy , perform partial differentiation with respect to y, on fx    f x    12 x 2  6 xy 3  a y y  18 xy 2

EXERCISE 9.3 1.

Obtain all the second order partial derivatives for f (x, y ) = 4x 2 y 2 – 9xy + 8x 2 – 3y 4

2.

Obtain all the second order partial derivatives for the following functions: (a)

f (x, y ) = 6x 3 y – 9y 2 + 2x

(b)

R (x, y ) = 4x 2 – 5xy + 12y 2 x 2

(c)

r  x, y  

4x x y

ACTIVITY 9.2 For examples and application of partial differentiation, visit: http://www.math.hmc.edu/calculus/tutorials/partialdifferentiation/ To observe the difference between ordinary and partial differentiations, visit: http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/intro.html


158 

TOPIC 9



Find the partial derivative of f x and f y A.

2.

3.

f x  4 x  3 y; f y  3 xy

B.

(B) f x  4 x  3 xy; f y  3 x

C.

(C) f x  4 x  3 y; f y  3 x

D.

(D) f x  4 x  3 xy; f y  3 xy

Find the partial derivative of g ( w, z )  3 w2  z 2 .

2w

gw 

B.

g w  2 w  z 2 ; g z  w2  2 w

C.

gw 

D.

gw 



3 w2  z



2 2/3

2w



3 2w  z



2 2/3

1



3 w2  z 2



2/3

; gz 

2z

A.

; gz  ; gz 



3 w2  z 2



2/3

2z



3 w2  2 z



2/3

1



3 w2  z 2



2/3

Find fx(2, 1) if f ( x, y )  3 x 3 y 2 A. 36

4.

of f ( x, y )  2 x 2  3 xy .

B. ă1152

C. 9x 2 y 2

Find all second-order f ( x, y )  6 x 3 y  9 y 2  2 x .

partial

D. 6x3 y derivatives

A.

f xx  18 x 2 y; f yy  18; f xy  f yx  18 x 2

B.

f xx  36 xy; f yy  18 y; f xy  f yx  18 x 2

C.

f xx  36 xy; f yy  18; f xy  f yx  18 x 2

D.

f xx  18 xy; f yy  18 y; f xy  f yx  18 x 2


of

TOPIC 9

5.


 159

Given f ( x, y )  x 3 y  7 x 2 y 2  8 , Find fyy. B. 14x 2

A. 14xy

C. 14 y 2

D. 14x 2 y



The discussion in this topic is only limited to two-variable functions:



The first order partial differentiation of z = f (x , y ) with respect to x is fx or f . Here x is a variable while y is regarded as a constant. x



The first order partial differentiation of z = f (x , y ) with respect to y is fy or f . Here y is a variable while x is regarded as a constant. y



The second order partial differentiations of z = f (x , y ) are:



 fxx

or

( fx ) x

 fyy

or

( fy ) y

 fxy

or

( fx ) y

 fyx

or

( fy ) x

Note that fxy = fyx .

Dependent Variable

Multi-variable Function

Independent Variable

Partial Differentiation


Topic

10



Application of Partial Differentiation




1.

Find the critical points of functions of two variables;

2.

Determine the properties of critical points, whether the points are maximum, minimum or saddle points, using the M Test; and

3.

Apply the Lagrange multiplier method to solve maximum and minimum problems involving constraints.

INTRODUCTION

One of the applications of partial differentiation which will be discussed in this topic is to obtain the maximum and minimum points of a function multi-variable.

Figure 10.1

Figure 10.1 shows that points A, B and C are the maximum points, while points D and E are the minimum points. This topic will discuss the application of partial differentiation. In the business world, we want to maximise profit but with a certain limitations called constraints. This can be solved using the Lagrange multiplier technique. Copyright © Open University Malaysia (OUM)

TOPIC 10

10.1

APPLICATION OF PARTIAL DIFFERENTIATION

 161

MAXIMUM AND MINIMUM FOR FUNCTIONS OF TWO VARIABLES

Similar to single variable functions which was discussed earlier, if a is a critical point (maximum or minimum point), then fÊ(a) = 0. Now in functions of two variables, if (a, b) is a critical point (maximum/minimum point) of a two-variable function, then fx (a, b) = 0 and fy (a, b) = 0. Besides a maximum and minimum point, (a,b) can also be a saddle point where a saddle point is neither a maximum nor a minimum point. Figure 10.2 shows that a critical point which is a saddle point. Approaching a point from one direction (on the x-axis) indicates a maximum point, while approaching a point from another direction (on the y-axis) indicates a minimum point.

Figure 10.2

M Test to determine maximum or minimum point or a saddle point. Given z =

f (x, y), if a point (a, b) exists, where fx (a, b) = 0 and fy (a, b) = 0

Define M = [fxx (a, b)  fyy (a, b)]  [fxy (a, b)] 2 Then

f (a , b ) is a maximum point if M > 0 and fxx (a, b) < 0.

f (a , b ) is a minimum point if M > 0 and fxx (a, b) > 0. f (a , b ) is a saddle point if M < 0. If M = 0, this test does not provide any conclusion.


162 

TOPIC 10


Example: Given f (x, y ) = 6x 2 + 6y 2 + 6xy + 36y – 5 (a)

find the critical point(s) of f (x, y ) .

(b)

determine whether the point(s) is maximum, minimum or a saddle point.

(c)

calculate the minimum/maximum value.

Solution: (a)

To obtain the critical point(s), differentiate with respect to x and y respectively. f (x, y ) = 6x 2 + 6y 2 + 6xy + 36x – 5

fx = 12x + 6y + 36

fy = 12y + 6x

fx = 0 and fy = 0 for critical points 12x +6y + 36 = 0  (1) 6x + 12y = 0

 (2)

Solve the equations simultaneously. From (2), 6x = ă 12y

x = ă 2y

 (3)

Substitute (3) into equation (1). 12(ă2y) + 6y + 36 = 0 ă24y + 6y + 36 = 0 ă18y = ă36

y= 2 Substitute y = 2 into equation (3).

x = ă 2(2) = ă4. Therefore (ă4, 2) is a critical point.


TOPIC 10

(b)


 163

To determine whether it is a maximum or minimum, we need to obtain all the second degree partial differentiation.

fx = 12x + 6y + 36 fxx = 12 fxy = 6

fy = 12y + 6x fyy = 12 fyx = 6

M = fxx fyy ă (fxy) 2 = (12)(12) ă (6) 2 = 144 ă 36 = 108 Since M = 108 > 0, fxx = 12 > 0, then the point (ă4,2) is the minimum point. (c)

To obtain the minimum value, we have to substitute the values of x and y into the function f (x, y ) = 6x 2 + 6y 2 + 6xy + 36x – 5 f (x, y ) = 6x 2 + 6y 2 + 6xy + 36x – 5 f (–4,2) = 6(–4) 2 + 6(2) 2 + 6(–4)(2) + 36(–4) – 5

= ă77 The minimum value is ă77. Example: Find the critical point for f (x, y ) = 50 + 4x - 5y + x 2 + y 2 + xy and determine whether it is the maximum or minimum point. Solution:

f (x, y ) = 50 + 4x – 5y + x 2 + y 2 + xy Differentiate with respect to x and y respectively.

fx = 4 + 2x + y fy = ă5 + 2y + x To obtain the critical point(s),

fx = 0 and fy = 0 Then, 4 + 2x + y = 0 and ă5 + 2y + x = 0 Copyright © Open University Malaysia (OUM)

164 

TOPIC 10


Rearranging gives, 2x + y = ă 4 (1)

x + 2y = 5 (2) Solve using elimination method. Eliminate x as follows: 2x + y = ă 4 2x + 4y = 10

(2)  2

ă 3y = ă14 14 y= 3

(1) ă (3)

Substitute y 

(1) (3)

14 into equation (1). 3

 14  2x      4  3 14 3 12  14 26 2x   3 3 13 x 3 2x   4 

 13 14  Therefore,   ,  is the critical point.  3 3

To determine whether it is a maximum or minimum, we need to obtain all the second degree partial differentiation. fx  4  2x  y

f xx  2

f y  5  2 y  x

f yy  2 f xy  f yx  1

M  f xx f yy   f xy 

2

  2  2   12 3


TOPIC 10


 165

 13 14 

M = 3 > 0, fxx = 2 > 0, then the point   ,  is the minimum point.  3 3 Example: The cost function for a company producing carbonated drinks is approximated by: C (x, y ) = 2200 + 27x 3 – 72xy + 8y 2

where x is the total amount of the sugar in kilograms and y is the amount of flavours in grams. Obtain: (a)

The total amount of sugar and flavours which will minimise the cost; and

(b)

The minimum cost.

Solution: C  x, y   2200  27 x3  72 xy  8 y 2 C x  81x 2  72 y C y  72 x  16 y

(a)

To obtain the critical point(s), Cx = 0 and Cy = 0.

81x 2 ă 72y = 0 ă 72x + 16y = 0

(1) (2)

Solve for y and hence, equate the solutions. From (1),

81x 2 = 72y 81 2 x y 72

 y

9 2 x 8

(3)

 y

9 x 2

(4)

ă72x = ă16y From (2),

72 x y 16

(3) = (4),

9 2 9 x  x 8 2


166 

TOPIC 10


x2 x  8 2 x2  4 x x  x  4  0 x  0 or x  4

When x = 0,

y

9 9 x   0  0 2 2

 (0,0) is the critical point 9 9 y  x   4   18 When x = 4, 2 2  (4, 18) is another critical point

However, the result (0, 0) is irrelevant as we are looking for the total amount of sugar and flavours, thus we will only use the point (4, 18). We need to determine whether it is actually a minimum point using the M test.

C x = 81x 2 ă 72y

Cxx = 162x

C y = ă 72x + 16y

C yy = 16 C xy = C yx = ă72

M = Cxx Cyy ă (Cxy) 2 (x still exist, therefore we substitute x = 4 as the = (162x)(16) ă (ă72) 2 and the critical point is when x = 4) = (162)(4)(16) ă (ă72) 2 = 5184

Cxx = 648 > 0, M > 0, then the cost is minimised. (b)

The cost is minimised when x = 4 and y = 18,

C (x , y ) = 2200 + 27x 3 ă 72xy + 8y 2 C (4,18)

= 2200 + 27(4) 3 ă 72(4)(18) + 8(18)2 = 1336

Thus, the minimum cost is RM1,336.


TOPIC 10


 167

ACTIVITY 10.1 After observing the Figures 10.1 and 10.2, and learning the techniques of differentiation for two-variable functions, what would the application of this topic be in the field of architecture?

EXERCISE 10.1 Find the critical points and determine whether they are maximum, minimum or saddle points. 1.

f (x, y ) = xy + x – y

2.

f (x, y ) = x 2 – 2xy + 2y 2 + x – 5

3.

f (x, y ) = x 2 – xy + y 2 + 2x +2y+ 6

4.

f (x, y ) = x 2 + 3xy + 3y 2 – 6x + 3y

5.

f (x, y )  4 xy – 10 x 2 – 4 y 2  8 x  8 y  9

6.

f ( x, y )  x 2  xy – 2 x – 2 y  2

7.

f (x, y ) = x 2 – y 2 – 2x + 4y – 7

8.

f (x, y ) = 2x 3 – 3y 2 – 12xy + 4

9.

f (x, y ) = x 2 + 4y 3 – 6xy – 1

10.

The profit function for a company is given by P (x, y ) = 1000 + 24x – x 2 + 80y – y 2

where x is the labour cost and y is the cost of raw material. Find the number of x and y which will maximise the profit. What is the maximum profit?

10.2

LAGRANGE MULTIPLIER

In subtopic 10.1, we saw how to find a relative maximum and relative minimum for a function of two variables. In practice, many functions come along with a constraint. Problems involving constraints can be solved by using Lagrange multiplier method. Copyright © Open University Malaysia (OUM)

168 

TOPIC 10


Theorem: Lagrange Multiplier Method The maximum or minimum value for a function z = f (x,y) over a constraint g (x, y) = 0 can be determined between the points (x, y), where there exists , i.e. Fx (x, y, ) = 0 Fy (x, y, ) = 0 F (x, y, ) = 0 given F (x, y, ) = f (x, y) +  g(x, y) Example: Find the minimum value for f (x , y ) = 5x 2 + 6y 2 ă xy over a constraint of x + 2y = 24 using the steps below. (a)

Express the constraint in terms of an equation equals to zero and in the form of

g (x, y) = 0. The constraint x + 2y = 24 then becomes x + 2y ă 24 = 0. (b)

Form the Lagrange function, f (x , y ,  ) where

f  x, y ,    f  x, y    g  x , y   5 x 2  6 y 2  xy    x  2 y  24   5 x 2  6 y 2  xy   x  2 xy  24 (c)

Find Fx , Fy and F and then equate to zero.

0

 1

Fy  12 y  x  2  0

  2

F  x  2 y  24

  3

Fx  10 x  y  

(d)

0

Solve the three simultaneous equations.

10 x  y    0

 1

12 y  x  2  0

  2

x  2 y  24  0

  3


TOPIC 10


 169

Using (1) and (2), equate the  to obtain the relation between x and y.

 4   (5)

because 2  2

2 y  20 x  x  12 y 14 y  21x 2 y  3x Substitute this relationship into (3). Insert 3x = 2y into (3) x  2 y  24  0



 3

x  3 x  24  0 4 x  24 x6

Substitute x = 6 into the equation 3x = 2y 3x  2 y 3 6  2 y 18  2 y y 9

Therefore, x = 6 and y = 9 will minimise the function f (x, y) = 5x 2 + 6y 2 ă xy . The minimum value is f (x, y ) = 5x 2 + 6y 2 – xy is

f (6, 9)  5(6) 2  6(9) 2 – (6)(9)  612 There are four steps for solving problems using the Lagrange multiplier method. 1.

Express the constraint in an equation that equals to zero, g(x, y) = 0.

2.

Form a function F (x, y, ) = f (x, y) +  g(x, y)

3.

Obtain the partial derivatives and then equate them to zero.

Fx = Fy = F = 0 4.

Solve the equations simultaneously.


170 

TOPIC 10


Example: Find two numbers whose sum is 50 and that their product is maximum. Solution: Suppose the two numbers are x and y. We need to maximise the multiplication (product) of the two numbers, i.e. f (x, y) = xy with the constraint x + y = 50 1.

x + y = 50; therefore g (x, y) = x + y ă 50

2.

F (x, y, ) = f (x, y) +  g(x,y) = xy +  (x + y ă 50) = xy + x + y ă 50

3.

Fx = y + 

y+ =0

 (1)

Fy = x + 

x+ =0

 (2)

x + y ă 50 = 0

 (3)

F = x + y ă 50 4.

Solve all the three equations simultaneously by equating the , From (1)  = -y From (2)  = - x Equating the  we note that the relationship is y=x Substitute this relationship y = x into (3),

x  x  50  0 2 x  50 x  25, y  25 Hence, y = 25. Therefore, xy = 25  25 = 625 Thus, the two numbers which maximise their product are 25 and 25 with their product equals to 625.


TOPIC 10


 171

ACTIVITY 10.2 1.

Why does constraint exist? What causes a constraint to occur? Explain.

2.

For further discussion on the Lagrange multipliers, visit: http://www.mathworld.wolfram.com/LagrangeMultiplier.html

EXERCISE 10.2 1.

Maximise f (x, y) = 2xy over a constraint of x + y = 12.

2.

Maximise f (x, y) = x 2 y over a constraint of 2x + y = 4.

3.

Maximise f (x, y) = x 2 + y 2 – xy over a constraint of x + y = 8.

4.

Maximise f (x, y) = x 2 – 10y 2 over a constraint of x ă y = 18.

5.

Find two numbers (say, x and y) which totals up to 12 and that the product of x 2 y is maximised.

6.

Find two numbers which totals up to 20 and that the product of the two numbers is maximised.

7.

Find two numbers, x and y which give x + y = 20 and xy 2 is maximised.

8.

A farmer has 200 metres of fencing. Find the dimensions of the rectangular field of maximum area that can be enclosed by the amount of fencing.


Find the critical points for f ( x, y )  x 2  2 xy  2 y 2  x  5 . A. (ă1, 1/2)

2.

B. (ă1, ă1/2)

C. ( 1, ă1/2)

D. (1, 1/2)

For the function given in question 1 above, the critical point is A. maximum

B. minimum

C. saddle

D. infimum


172 

3.

TOPIC 10


Using Lagrange multipliers, the critical points of f ( x, y )  2 xy , subject to x  y  12 A. (6, 6)

B. (ă 6, ă 6)

C. (4, 4)

D. (3, 3)

The next information is for question 4 and question 5. Determine the dimension of the largest rectangular field that can be enclosed with 600 meters of fencing. Assume that no fencing is needed along one side of the field. 4.

5.

What is the constraint for the problem? A. 2 x  2 y - 600  0

B. 2 x  y - 600  0

C. xy - 600  0

D. x  600  0

What is the dimension of the field? A. (150, 150)



B. (300, 300)

C. (150, 300)

D. (150, 200)

The M test is applied to determine whether a critical point is a maximum, minimum or saddle point.

M Test to determine maximum or minimum point Suppose z = f (x , y ) , and if a point (a, b) exists, where fx(a, b) = 0 and fy(a, b) = 0, then the M Test = M   f xx  a, b   f yy  , b     f xy  a, b  

2

Therefore, f (a , b ) is a maximum point if M > 0 and fxx(a, b) < 0. f (a , b ) is a minimum point if M > 0 and fxx(a, b) > 0. f (a , b ) is a saddle point if M < 0. If M = 0, this test does not provide any conclusion. Copyright © Open University Malaysia (OUM)

TOPIC 10


 173



The Lagrange multiplier method is a technique to obtain the maximum or minimum value of a function z = f (x, y) with a constraint g (x, y) = 0.



The maximum or minimum point (x, y) can be determined when F (x , y ,  ) = f (x, y) + g(x, y) where there exists , i.e. Fx (x, y, ) = 0, Fy (x, y, ) = 0 and F (x, y, ) = 0.

Lagrange Multiplier Method

Minimum Point

Maximum Point

Saddle Point


174  ANSWERS

Answers TOPIC 1: MATRIX Exercise 1.1 1.

(a)

3x2

(b)

1x 3

(c)

3x1

2.

a0

3.

No. Identity matrix only exists for square matrix.

4.

(a)

Zero Matrix (Null)

(b)

Column Matrix

(c)

Row Matrix

Exercise 1.2 1.

(a)

 6 12 

3 0

9 3 

(b)

2 9 

2

1 4 

(c)

 6 3  3 1     2 5 

(d)

Not possible

(e)

 130 140   110 60   

(f)

 27 35  4 26   

(g)

 20 2 2   2 1 3    2 3 10 

1


ANSWERS

2.

 2 1 A  5 2

Exercise 1.3 1.

2.

(a)

11

(b)

a2 + b2

(c)

27

(d)

27

(a)

1

(b)

The determinant does not exist.

Exercise 1.4 1.

(a)

1 7  1  7

2   7  3  14 

(b)

 2 11 3    1 6 2   0 1 0 

(c)

Does not exist

(d)

 13   10   2  5   7  10

7 5 1  5 3 5

1 2  0   1 2 


 175

176  ANSWERS

2.

 5  A1   2   2

3   2  1 

3.

(a)

B 1 

(b)

ad ă bc  0

1  d b  ad  bc  c a 

Exercise 1.5 24 23 , y 5 5 (b) x  1, y  3, z  2

1. and 2. (a) x 

(3)

(a) (b)

2 10 60 , y , z 17 17 17 x  1, y  0, z  5

x

Multiple Choice Questions 1.

(a)

B

(b) D

2.

D

3.

C

4. D

TOPIC 2: LINEAR AND QUADRATIC FUNCTIONS Exercise 2.1 1.

2. 3. 4. 5.

(a)

1 m  , c  1 2

(b)

m = 5, c = 5

(c)

m = 3, c = 0

(d)

m

2 5 ,c  3 3

y = x + 5 3y = 8x  4 2y + x = 4 3y + x + 3 = 0 Copyright © Open University Malaysia (OUM)

ANSWERS

Exercise 2.2 (a)

(b)

(c)

Exercise 2.3 (a)

The parabola opens upward. The turning point is (3, ă 4). The y-intercept is (0, 5). The x-intercepts are (1, 0) and (5, 0).

(b)

The parabola opens upward. The turning point is (ă2, ă4). The y-intercept is (0, 0). The x-intercepts are (0, 0) and (ă4, 0).

(c)

The parabola opens downward. The turning point is (ă1, ă2). The y-intercept is (0, ă3). There is no x-intercept.

(d)

The parabola opens upward. The turning point is (0, ă16). The y-intercept is (0, ă16). The x-intercepts are (4, 0) and (ă 4, 0). Copyright © Open University Malaysia (OUM)

 177

178  ANSWERS

(e)

The parabola opens downward. The turning point is (2, 1). The y-intercept is (0, ă3). The x-intercepts are (1, 0) and (3, 0).

Exercise 2.4 (a)

(1, 8)

(b)

(d)

(ă3, ă1) and (ă1, 7)

(e)

(1, ă1) and (2, 2)

(f)

(1, 9) and (3, 29)

(0, 2)

(c)

(2, ă1)

Multiple Choice Questions 1. C

2. B

3. C

4. A

5. A

TOPIC 3: APPLICATION OF LINEAR AND QUADRATIC FUNCTIONS Exercise 3.1 1.

(450, 9.50)

2.

p = RM75, q = 5225

Exercise 3.2 1.

2.

2p = 100q ă 600 is a supply equation p = -50q + 600 is a demand equation p = RM150, q=9 (a) (b) (c) (d)

50q 40q + 5000 10q ă 5000 500


ANSWERS

 179

Exercise 3.3 (a)

1850p ă 5p 2

(b)

RM9,125

(c)

RM185

(d)

RM171,125

Multiple Choice Questions 1. D

2. A

3. C

4. D

5. C

TOPIC 4: EXPONENTIAL AND LOGARITHMIC FUNCTIONS Exercise 4.1 (a)

1 27

(b)

1

(c)

3

(d)

1 4

(e)

125

(f)

8

(c)

1

Exercise 4.2 (a)

2

(b)

3

(d)

1 4

(e)

2

(f) 1

Exercise 4.3 1.

(a)

52 = 25

(b)

2x = y

(c)

101 = 0.1

2.

(a)

log10 100 = 2

(b)

10ga 1 = 0

(c)

1 log2 = 3 8

3.

(a)

4

(b)

(c)

1

(d)

1

(e)

(f)

4

1 3 2


180  ANSWERS

4.

(a) (d)

1 2 4

(b)

3

(c)

3

(e)

20

(f)

20

Exercise 4.4 1.

(a) (b)

55 years 97,045 people

2.

(a) (b) (c) (d)

RM6,016.61 RM14,693.28 RM11,373.99 RM2,354.99

3.

(a) (b) (c) (d)

RM15,000 RM10,000 RM6,000 RM25,000

Multiple Choice Questions 1. B

2. D

3. B

4. A

5. C

TOPIC 5: DIFFERENTIATION Exercise 5.1 (a)

0

(b)

0

(c)

0

(d)

0

(e)

7x 6

(f)

5x 4

(g)

2 5  r 3 3

(h)

3 8  t 5 5

(i)

5 14 x 4

(j)

5 32 x 2

(k)

x4

(l)

3  10 x x2

3


ANSWERS

(m) 3 (o)

2x + 4

(q)

8x 

(s)

 x  1 10 x  2 

(u)

(w) (y)

4 x2 3

2 x  2 x 3  6 x  1

x

2

 1

2

10x (x 2 + 4) 4 2x  5 2 x2  5x

1 4

(n)



(p)

10x3

(r)

1  2 x   10  7 x 

(t)

 181

3

 x2  2 x  2

x

2

 2

2

(v)

4(2x ă 1)

(x)

 3x

(z)

 2x  2  3   x3 

2

 2 x  1 2



1 2

 3x  1

 4      x  3 2   

Multiple Choice Questions 1. D

2. D

3. A

4. B

5. C

TOPIC 6: APPLICATION OF DIFFERENTIATION Exercise 6.1 1.

(a)

24 x  24

(b)

12 x4

2.

(a)

0

(b)

24

(c)

12x  4

Exercise 6.2 1.

(a)

10x + 5000

(c)

10 

5000 x

(b)

RM7,000

(d)

10


182  ANSWERS

(a)

100 000 + 1500q + 0.2q 2

(c)

1504

(a)

q 400 3 4 q

(c)

40 unit

4.

(a)

5.

2.

3.

(b)

1500 + 0.4q

(b)

1 400  4 q2

10 dozens

(b)

RM203,000

(a)

2500 75  0.25q q

(b)

100

(c)

RM12,500

(b)

0.001q 2 + 840

Exercise 6.3 (a)

0.01q 3 + 840

(c)

0.003q 2 + 840

2.

(a)

2x ă 0.1x 2

(b)

RM1,000

3.

(a)

RM200

(b)

RM400

1.

Exercise 6.4 1.

2.

3.

(a)

R (q ) = 54q ă q 2

(b)

K(q) = 4q + 100

(c)

  q    q 2  50q  100

(d)

RM29

(a)

300x ă x 2

(b)

ă1.1x 2 + 286x ă 100

(c)

130

(d)

RM170

(e)

RM18,490 1 3 x  2x2  5 3

(b)

2 3 x  5 x 2  16 x 3

(d)

8

(a) (c)

1 3 x  3 x 2  16 x  5 3


ANSWERS

(e)

RM101.67

Multiple Choice Questions 1. (a) D

(b) C

2.(a) A

(b) C

3. D

TOPIC 7: INTEGRATION Exercise 7.1 1.

xc

2.

ex  c

3.

8p3x + c

4.

ex 3 c 3

5.

u4 3  c 4 u

6.

1 2 32 12 54 x 3  x  x  c 5 3 2 x 2 3

7.

x 2 x5  c 2 5

8.

x2 2 1   3 c 2 x 3x

9.

e 0.07 t c 0.07

10.

e3 s  4 c 3

11.



12.

4ln x 

1 1  2ln x   x  c 2 x x

x2  ex  c 8

Exercise 7.2 1.

65 4

2.

87 10

3.

22 3

4.



5.

211 5

6.

3e5

7.

4 ln 8

8.

20 3

1 3


 183

184  ANSWERS

Exercise 7.3 1.

2 1 4 t  8t  2   c  8

2.

1 3 2 4 2 x   2  c 4

3

3.

5.

2  ex  2 2 3



c

4.

2 3 1  2s 2  3  c  8

6.

1 18  4  3 x 2  2 x 3 

3

c

1 ln  x 2  2 x  1  c 2

Multiple Choice Questions 1.B

2.C

3.A

4.A

5.D

TOPIC 8: APPLICATION OF INTEGRATION Exercise 8.1 1.

4 3

2.

12

3.

4

4.

18

5.

4 3

6.

32 3

7.

9 2

8.

1 6

Exercise 8.2 1.

(a)

0.001q 3 ă 0.015q 2 + 176

(b)

RM45,038

2.

RM2,000

3.

(a)

3x 2  15 x  8000 2

(b) 165 x  3 x 2  8000


ANSWERS

4.

Consumers' surplus:

11,250

Producers' surplus:

11,250

5.

0.001 3 q  0.01q 2  1000 3

6.

CS = 9,000,

PS = 18,000

7.

CS = 166.66,

PS = 53.33

 185

Multiple Choice Questions 1. A

2. C

3. A

4. C

5. A

TOPIC 9: PARTIAL DIFFERENTIATION Exercise 9.1 1.

(a) 6 (b) 20

4.

(a) 1300 (b) 500 (c) 1996

2.

(b) 92 (b) 47

3.

(a) -8 (b) 68

Exercise 9.2 1.

fx

= 8x 3 + 18x 2 y 3

fy

= 18x 3 y 2 + 10y

(b) 4

(c) 109

(d) 9

2.

(a) 80

3.

(a)

40C

(b)

6. The water temperature around the factory will increase by 6C if the temperature of the water approaching the factory increases from 5C to 6C, while the factory generated 400 megawatt.

(c)

13. The water temperature around the factory will increase by 13C if the electrical power generated by the factory increases from 300


186  ANSWERS

megawatt to 400 megawatt, while the temperature of the water approaching the factory is 8C.

Exercise 9.3 1.

f xx  8 y 2  16 f yy  8 x 2  36 y 2 f xy  16 xy  9  36 y 3  f yx

2.

(a)

f xx  36 xy

f yy  18

f xy  f yx  18 x 2

(b)

Rxx  8  24 y 2

R yy  24 x 2

Rxy  R yx  5  28 xy

(c )

rxx 

8 y

 x  y

ryy 

3

8x

 x  y

3

rxy  ryx 

4x  4 y

 x  y

3


2. A

3. A

4. C

5. B

TOPIC 10: APPLICATION OF PARTIAL DIFFERENTIATION Exercise 10.1 1.

(1, 1)

saddle point

2.

1   1,   2 

minimum point

3.

(2, 2)

minimum point

4.

(15, 8)

minimum point

5.

2 4  ,  3 3

maximum point

6.

(2, 2)

saddle point

7.

(1, 2)

saddle point Copyright © Open University Malaysia (OUM)

ANSWERS

8.

(0, 0)

saddle point and (4,8) minimum point

9.

(0, 0)

saddle point and (9/2, 3/2) minimum point

10.

P (12,40) = 2744

Exercise 10.2 1. 2.

f (6, 6) = 72  4 4  64 f  ,   3 3  27

3.

f (5, 3) = 28

4.

f (20, 2) = 360

5.

x=8 y=4

6.

(10, 10)

7.

x = 20/3 y = 40/3

8.

(50, 50)


2. B

3. A

4. B

5. C


 187

BBMP1103 Mathematics for Management

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