1. REVIEW OF BASIC MATHEMATICS

1.0 INTRODUCTION

Perhaps twenty years ago the only mathematics a business student was expected to know was a little basic arithmetic. However with the advent and widespread use of computers, it has become possible to process huge quantities of data and to use sophisticated mathematical techniques to solve complex business problems with relative ease. Consequently, it has become expedient to formulate many business problems mathematically and hence business students in our current age are required to attain at least a basic understanding of the various mathematical techniques that are available.

The mathematics that is presented in this section is by no means sophisticated. It is meant to serve only as a reminder of basic definitions and techniques learned in secondary school. However because of its very basic nature it is essential that the work presented here be completely mastered if you are to fully understand the work of later sections and later years.

Your primary aim for this section is therefore quite simple, to master each of the topics listed in the contents covering the area of number systems, algebra and equation solving. Your guide to a sufficient understanding of these subjects is also simple. If you can correctly solve the problems at the end of each section then you can be fairly confident you have mastered the necessary techniques required for later work. You should attempt all the problems presented unless you are fully confident of your mastery of a topic.

1.1 INTRODUCTION TO NUMBER SYSTEMS

This section provides an introduction to some of the basic notation, concepts and number systems with which you will have to deal.

Integers
The integers consist of all the positive and negative whole numbers including zero, …… -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, ……

Number line
It is sometimes useful to think of the integers as a set of equally spaced points on a straight line of infinite extent. This is called the number line. A point on the line is labelled zero. The points to the right of the zero point are labelled by the positive integers while the points to the left of the zero point are labelled by the negative integers as illustrated below.

-5 -4 -3 -2 -1 0 1 2 3 4 5

In the work that follows, construct your own number line to assist you in understanding the examples and problems.

Order
The number line makes the concept of ordering the integers quite natural. If one number lies to the right of another number on the number line, we say the first number is greater than the second and denote this relationship by the symbol, >, which reads “is greater than”.

For example, since –2 lies to the right of –5 we write –2 > -5, which reads “-2 is greater than –5”.

Remember that while it is true to say –2 > -5, it is not true to say 2 > 5.

If one number lies to the left of another number on the number line we say the first number is less than the second and denote this relationship by the symbol, : greater than : greater than or equal to
< : less than : less than or equal to : not greater than : not less than : not greater than or equal to : not less than or equal to : not equal to
= : approximately equal to

These symbols may be used in combination. Consider the Taxation office’s table of taxable income ranges. If x denotes a taxable income in dollars, then an income range in the table can be expressed as 15,000 x < 20,000 …
This reads as, $15,000 is less than or equal to the taxable income (that is, the taxable income is greater than or equal to $15,000) and the taxable income is less than $20,000.

The symbols must be combined sensibly. For example, 25,000 > x > 30,000 does not make sense as x is required to be less than 25,000 and greater than 30,000.

Rational Numbers
The rational numbers provide an extension of the integers to include all ratios of positive and negative integers with the exception of dividing by zero. In other words, the rational numbers consist of all positive and negative fractions and zero. Some examples of rational numbers are the following:

In general, a rational number is denoted by , where a is any integer and b is any integer except zero. If is a rational number then a is referred to as the numerator and b as the denominator. The rational numbers can be associated not only with the points on the number line associated with the integers but also with many other points. Note that there are infinitely many ways of representing the same rational number.

For example, all represent the same rational number which in the simplest form is
Any rational number of the form equals 0, provided that b is not zero.

Addition and Subtraction of Rational Numbers
When adding or subtracting two rational numbers we must first represent them with the same denominator. Their sum or difference is then found by adding or subtracting their numerators and putting the result over the common denominator.

In general the sum and the difference of two rational numbers is defined as follows:

Sum: …………. (1.1)

Difference: ………….. (1.2)

Example 1.1 Sum of two fractions:

Multiplication and division of rational numbers
To multiply two rational numbers, the procedure is to multiply the two numerators and multiply the two denominators, then cancel any common factors in the numerator and denominator.

The general rule is defined as follows:

………… (1.3)

Example 1.2 Multiplication of two fractions:

To divide an expression by a rational number, multiply the expression by the inverted rational number.

The general rule is defined as follows:

………… (1.4)

Example 1.3 Division of two rational numbers:

= = =

Irrational numbers
The irrational numbers are those numbers which cannot be expressed as the ratio of two integers. Just as there are an infinite number of rational numbers there are an infinite number of irrational numbers.

Some examples of irrational numbers are , - , , , e, ….

e is the base of natural logarithms and has the value 2.71828…

Note that the square root of a number is not always irrational, for example = 2.

The square root of a negative number, for example , is a complex number, but we will not be dealing with that here.

Real numbers
All the rational numbers together with all the irrational numbers make up the real number system. The real numbers can be associated with all the points on the number line.

1.1.1 Basic Operations With Negative Numbers

In a business context, a cash inflow may be represented by a positive number. Changing the sign of a number changes its direction. So, a cash outflow may be represented by a negative number.

Addition of a negative number

Example 1.4 Consider a person who receives $100 with a debt of $70 to then repay:

100 + (-70) = 100 – 70 = 30 $30 would be left

Suppose the debt had been $160

100 + (-160) = 100 – 160 = -60 A debt of $60 is the net result

Subtraction of a negative number

Example 1.5 Consider a person who receives $100 and then this person receives full repayment for $80 previously lent to another party (reversal of a prior cash outflow becomes an inflow):

100 – (-80) = 100 + 80 = 180

$180 is the net result

Multiplication and division of negative numbers
The rules for the sign of the result of two negative numbers that are multiplied, and for the sign when two numbers are divided are

Example 1.6 (a) 3 x (−7) = −21
(b) −24 ÷ (−12) = 2 On some calculators, negative numbers are entered using the +/− key. For example, the calculation 3 x (−7) would use the keystrokes

3 × +/− 7 =

1.1.2 Order of Operations

In mathematics, the order of performing arithmetic operations is represented by the acronym BODMAS (or BOMDAS).
That is, evaluate:
• Bracketed expressions first
• Of next
• Division and Multiplication next, in the order that they occur
• Addition and Subtraction last, in the order that they occur.

Example 1.7 (a) Evaluate = = 14
(b) Evaluate
10% of
=
(Of before Multiplication) (Multiplication before addition)

1.2 DECIMALS, RATIOS, PROPORTIONS AND PERCENTAGES

1.2.1 Decimal Numbers
The decimal notation provides a much simpler formulation for carrying out numerical calculations than working with fractions since all techniques used for computing with integers can be used for decimal numbers.
In the decimal system a decimal point is placed after the units digit of a number. The digits to the right of the decimal point represent integral numbers of tenths, hundredths, thousandths etc. If there are no units digits in a given number, a zero is sometimes placed to the left of the decimal point, for example 0.28.

Example 1.8 (a) 0.5 = 5 tenths

(b) 2.67 = 2 units + 6 tenths + 7 hundredths = 2 + = 2 + =
(c) 46.219 = 46 + =

Any rational number can be expressed in decimal form.

Example 1.9 (a)
(b)

The decimal form of a rational number may have a finite number of digits, as in Example 1.9, or an infinite number of digits, where the digits repeat themselves in groups after a certain stage, as in the following example.

Example 1.10 (a)
(b)
(c)

An irrational number, when represented in decimal form, will have an infinite number of digits after the decimal point and these digits will not have a repeating pattern. Since a recorded number can be expressed to only a finite number of decimal places, all recorded numbers must be rational.

1.2.2 Ratios and Proportions

As the name suggests, a rational number can be used to represent ratios.

Example 1.11 Mr Jardin prepares his 2-stroke fuel by mixing 14 litres of petrol with 4 litres of oil. The ratio of petrol to oil is 14 to 4 and may be written as 14:4 or . Equivalently, the ratio may be described as 7 to 2 or 3.5 to 1, since

One use of the word proportion is to express the size of a part in its relation to the whole.

Example 1.12 Question: In a fuel mixed from petrol and oil in the ratio 7:2, what is the proportion of petrol?

Solution: There are 7 parts petrol and 2 parts oil, that is 7 + 2 = 9 parts. So the proportion of petrol is . In mathematics, the word proportion is also used to mean a statement equating two ratios ………… (1.5)

In a typical problem, one of the terms of these ratios is unknown and has to be calculated.

Example 1.13 Question: The Best Egg superannuation portfolio requires an investment of three thousand dollars in bonds for every ten thousand dollars invested in stocks. If it is proposed to invest eighteen thousand dollars in stocks, how much should be invested in bonds?

Solution: The ratio of investment in bonds to investment in stocks is required to be 3,000 to 10,000, that is . If c represents the proposed amount to be invested in bonds, then is the ratio of bonds to stocks in the proposed investment. These two ratios must be equal so

Therefore

So $5400 must be invested in bonds.

1.2.3 Percentages

A useful way of representing ratios is to use a denominator of 100. The denominator can then be omitted and the symbol "%” placed after the numerator. The ratio is then called a percentage. For example, if 3 out of 4 of the employees in a firm are female, . That is, we mean that 75 out of every 100 employees are female or that the ratio of female employees to all employees is 75 to 100. Equivalently, the proportion of females is 0.75.

A percentage problem is usually expressed as two equal ratios in which one of the four terms is to be determined when the other three are known.

Calculating the percentage one number is of another

Example 1.14 Question: If 9 employees are female in a firm with 48 employees, what percentage of the employees are female?

Solution: If x represents the percentage of female employees then

So

Calculating a given percentage of a number

Example 1.15 Question: A television antenna is priced at $92.00 plus 15% sales tax. Calculate sales tax for the antenna.

Solution: If x is sales tax on the antenna, then

then = 13.8

So the sales tax is $13.80.

Given a value and its percentage of a number, calculate the number

Example 1.16 Question: Your share of the annual bonus to sales staff was $529. What was the total annual bonus?

Solution: Invert both sides Then = 4232

So the total annual bonus was $4,232.

Example 1.17 Question: A shirt with a retail price of $55.95 was marked up 27.5% on the cost price to the retailer. Calculate the cost price.

Solution: $55.95 represents 100 + 27.5 = 127.5% of the cost price. Invert both sides Then

= 43.88

So the cost price if $43.88.

Example 1.18 Question: A property bought for $80,000 is currently valued at $95,000. Calculate the percentage change in the property’s value.

Solution: The change in value is 95,000 – 80,000 = 15,000. Expressing this as a percentage gives a percentage change of = 18.75%.

All problems associated with percentage change are solved similarly. If we let be the final value, the percentage change is given by Percentage change ………… (1.6)

Exercises Basic Mathematics Exercises 1.1-1.17

1.3 ROUNDING, SCIENTIFIC NOTATION AND SIGNIFICANT FIGURES

1.3.1 Rounding

It is often useful, and indeed necessary, to approximate values by using only the first few digits of their decimal representations. For example, when represented in decimal form is equal to 0.3333333…… with the 3’s extending to infinity. Obviously it is quite impractical to try to write this fraction in decimal form to all decimal places and so we often approximate it to 0.333 or 0.33 or 0.3, depending on the level of accuracy required. The symbol is used to mean “approximately equal to” and we can write

The process of rounding off involves more than simply omitting the final digits. Ignoring these digits is referred to as truncating the number. Rounding also involves adjusting the last retained digit according to the following generally accepted rules. If the first digit of the omitted string is 5 or greater then increase the last retained digit by 1, that is, round up. If the first digit of the omitted string is 4 or less then make no change to the digits retained, that is, round down.

Example 1.19 Round the following numbers to 2 decimal places:

(a) 72.63615 72.64 (that is, round up)

(b) 72.63498711 72.63 (that is, round down)

(c) 153.6982 153.70 (that is, round up)

The purpose of rounding off is to reduce the error incurred in approximating the original number by a number with fewer decimal places.

Rounding off is often necessary when a calculation or series of calculations is performed.

Example 1.20 The heights of seven trees distributed around my yard measure as 1.2m, 1.7m, 2.2m, 2.6m, 3.2m, 4.6m and 4.9m. To find the mean height add the individual heights and divide by seven. My calculator gives a result of 2.914285714m. Obviously it is ridiculous to quote this degree of accuracy when the original measurements were made to only one decimal place and we should quote the result only to one decimal place, that is, 2.9m.

1.3.2 Scientific Notation

A number is expressed in scientific notation when it is represented as a number ranging from 1 up to (but not including) 10, and multiplied by 10 raised to a power.

For example:

When a number is multiplied by ten raised to a power, the decimal place is moved to the right for a positive power, or left for a negative power, by the number of places equivalent to the power.

Scientific notation is especially useful for expressing very large or very small numbers.

Example 1.21 The Coles-Myer Company’s reported sales of $11,369,000,000 in 1986/1987 is written in scientific notation as . Enter this number on your calculator by typing 11369 × 1000000 = The calculator displays the number in scientific notation as 1.1369 10. Many computer programs would present this number in scientific notation as 1.1369E10.

Example 1.22 When a communications satellite transmits signals at the speed of light, the signal takes 0.00000000333564 seconds to travel a metre. In scientific notation this is seconds.

1.3.3 Significant Figures

The accuracy to which a measurement is made is conveyed by the number of significant figures in the quoted value. For example, a surveyor quoting a distance as 19.83 metres (4 significant figures) implies that he has measured more accurately than another surveyor quoting 19.8 metres (3 significant figures).

Trailing zeros to the right of the decimal point are important in conveying accuracy. Values quoted as $5 (1 significant figure), $5.0 (2 significant figures), $5.00 (3 significant figures) are not the same as they represent different accuracy of measurement: nearest dollar, nearest 10c and nearest 1c respectively.

The number of significant figures in a value can be made clear when the value is expressed in scientific notation.

Example 1.23 Express the following numbers in scientific notation and specify the number of significant figures: (a) 1,271.062 = 1.271062 : 7 significant figures (b) 0.7610 = 7.610 : 4 significant figures (c) 0.0007610 = 7.610 : 4 significant figures (d) 4,300 = 4.300 : 4 significant figures if 4,300 was measured to the nearest unit.

4,300 = 4.30 : 3 significant figures if 4,300 was measured to the nearest hundred.

4,300 = 4.3 : 2 significant figures if 4,300 was measured to the nearest ten.

1.3.4 Degree of Accuracy

Beware of rounding off too early or too much during a calculation.

Example 1.24 Question:
$100,000 is to be invested in a bank at 14% per year, with the interest to be compounded monthly. Calculate the future value (FV) of the invested amount (PV) after 5 years.

Solution:
The required formula is The theory underlying this formula comes from Financial mathematics.

(a) If were rounded off as 0.012

= 100000(1 + 0.012) = $204,564.73

(b) If were rounded off as 0.011666667

= 100000(1 + 0.011666667) = $200,560.98

If the investment had been paid out on the basis of the calculation in part (a), the bank would have paid $204,564.73 - $200,560.98 = $4,003.75 too much.

The lesson to be learned is that you should keep as many significant figures as possible during a calculation, but round the final result to a suitable degree of accuracy.

There are definite procedures for properly determining this degree of accuracy. The procedures governing two basic circumstances are given below:

• For calculations involving only addition or subtraction the maximum number of decimal places in the answer is the least number of decimal places in the numbers being added or subtracted.
• For calculations involving only multiplication or division the maximum number of significant figures in the answer is the least number of significant figures in the numbers being multiplied or divided.

Example 1.25 Evaluate the following expressions writing the result to an appropriate degree of accuracy.

(a) 3.721 + 46.1 – 11.3514 = 38.4696 = 38.5
The final result is rounded to 1 decimal place since 46.1 has 1 decimal place.

(b) 2.631 0.0017 ÷ 53.046 = 0.000084317 … = 8.4317 … = 8.4
The final result is rounded to 2 significant figures as 0.0017 = 1.7 has only 2 significant figures.

1.4 EXPONENTS AND LOGARITHMS

1.4.1 Exponents

When a number is multiplied by itself several times it can be abbreviated by the use of exponents as shown in the following examples.

Example 1.26 (a) (5 to the power of 3 or 5 cubed)

(b) (2 to the power of 6)

The number being multiplied is referred to as the base and the number of times it occurs in the multiplication is referred to as the exponent or power or index to which the base is raised. For part (a) in Example 1.26, 5 is the base and 3 is the exponent. In general if x is a real number and n is a positive integer then

………… (1.7)

where x appears n times on the right hand side of the equation.

Negative exponents are defined in the following way. If x is a real number, with the restriction that and n is a positive integer, then

………… (1.8)

Example 1.27 (a) 2

(c)

For any real number x, with the restriction that , then

In general, for any real number x, with , and for any integers m and n we have the following result:

………… (1.9)

This is illustrated in the following examples:

Example 1.28 (a) =

(c) = = =

Non-integer exponents
It is possible to have a meaningful interpretation for when m is not an integer. Our discussions will restrict the base x to a positive real number and is always taken to be positive.

Example 1.29 It we let x = 4 and in law (1.9) we have

= 4 Therefore is the positive number which multiplied by itself equals 4, so = 2.

Example 1.30 The law (1.9) can be extended to obtain multiplication of 3 numbers as

= 64

As is the positive number which multiplied by itself twice equals 64, so = 4. Numbers raised to powers can be obtained with by using your calculator’s power key which could look like ax or xy .

Example 1.31 Calculation of powers with a calculator uses the following keystrokes:

Calculate Keystrokes Result

(a) 3 ax 2 = 9 (b) 2 ax 3 +/- 0.125

1.4.2 Logarithms

If the word logarithm strikes fear into your heart, then relax. It just means exponent or power. For example, if you can understand the statement , then you have the basis for understanding logarithms. Logarithms simply provide an alternative way of expressing numbers raised to powers.

that is, when the base 10 is raised to the power of 3, the result is 1000.

The same information expressed in logarithmic form is log That is, the power (log) to which we must raise the base of 10 to obtain a result of 1000 is 3.

Formally, ………… (1.10)

provided that x and a are greater than 0 and a must not equal 1.

Example 1.32 (a) Since then (b) Since then (c) Since then (d) Since then (e) Since then (f) Since then

The two most commonly used bases for logarithms are:
• 10, referred to as common logarithms, or log
• the irrational number e = 2.7182818…, referred to as natural logarithms, or ln

Before calculators and computers were widely available, logarithms to the base 10 (10 because of our decimal numbering system) provided an important method for performing calculations. Although has largely fallen into disuse, logarithms are still needed. Consequently, the logarithm that is important in other areas of mathematics is the one we will use.

Example 1.33 (a) Since

(b) Since

(c) Since

Example 1.34 Calculator keystrokes:

Calculate Key In Displays

(a) In 5 5 ln 1.609

(b) In 0.3 0.3 ln -1.2040

Exercises Basic Mathematics Exercises 1.18-1.25

1.5 INTRODUCTION TO ALGEBRA

Algebra is a system of using symbols or letters to represent numbers either because the value of the numbers is unknown or we may want the symbol to represent a range of possible numbers. It is possible to use algebraic expressions composed of constants and variables to express a relationship between the variables by an equation or a series of equations. In order to develop these relationships and solve the resulting equations, we must know how to manipulate, simplify and transpose algebraic equations. An important point to remember is that the symbols we will be dealing with represent real numbers and are therefore manipulated in exactly the same way as real numbers.

1.5.1 Basic Laws

Listed below are three basic laws used in the manipulation of algebraic expressions. Although the first two may appear self evident they are important and although they hold for real numbers, they do not always hold. For example, if you have ever dealt with matrices, you will recall that if a and b are matrices then That is, the commutative law does not hold for the multiplication of matrices. For most of our purposes however, it is the distributive law which is the least obvious and with which you should take special care.

For any real numbers a, b, c the following laws hold:

Commutative Laws

………… (1.11)

………… (1.12) Associative Laws

………… (1.13) ………… (1.14)

Distributive Law

………… (1.15)

The operations of subtraction and division are defined respectively by

………… (1.16)

………… (1.17)

Example 1.35 Some examples of the use of these laws:

(a) 4 + 6 = 6 + 4 Using 1.11 = 10

(b) Using 1.12 = 21

(c) (4 + 9) + 8 = 4 + (9+8) Using 1.13 = 21

(d) Using 1.14 = 30

(e) Using 1.15 = = 280 +14 = 294

(f) 6a + 13a = (6 + 13)a Using 1.15 = 19a

(g) 2a(6b +3c) = 2a(6b) + 2a(3c) Using 1.15 = 12ab + 6ac

(h) 3 – 3 = 3 + (-3) Using 1.16 = 0

(i) Using 1.17 = 1

(j) Using 1.17

= 4

(k) 7m – 5(m + 3n) = 7m – 5m – 15n Using 1.15 = (7 – 5)m – 15n = 2m – 15n

1.5.2 Laws of Indices

When we multiply n factors of x together and the result is written as that is

n factors of x where x is a real number and n is a positive integer.

The number n is called the index or power or exponent of x, and x is called the base. We also say is the nth power of x.

The basic laws for manipulating indices are as follows:

For any real numbers x, y, m, n.

………… (1.18)

………… (1.19)

, provided x ≠ 0 ………… (1.20)

………… (1.21)

………… (1.22)

………… (1.23)

………… (1.24)

Example 1.36 Some examples involving the use of these laws are given below:

(a) Using 1.18

(b) Using 1.18

(c) Using 1.18 (d) Using 1.19

(e) Using 1.19

(f) Using 1.19 and then 1.20 = 1

(g) Using 1.22 =

(h) Using 1.22 and then 1.21

(i) Using 1.23 and then 1.22

The notation is also used to denote the nth root of x, that is = If you come across this notation simply replace by and proceed using the rules above.

Example 1.37 Simplify the expression

We are now in a position to combine the laws of this section with those of the previous section to expand and simplify more complex algebraic expressions. We will illustrate this with some examples below. Some of these expressions involve algebraic fractions. When adding or subtracting algebraic fractions, the procedure is analogous to the procedure for adding or subtracting numerical fractions as given by Equations 1.1 and 1.2. Both terms must be expressed with a common denominator following which the numerators may be added or subtracted and any simplification performed.

Example 1.38 (a) Simplify

(b) Expand

1.5.3 Laws of Logarithms

Using the definition of a logarithm and the laws of indices of Section 1.5.2, the following basic laws may be derived for manipulating logarithms.

If ………… (1.25)

………… (1.26)

………… (1.27)

………… (1.28)

………… (1.29)

………….(1.30)

………… (1.31)

………… (1.32)

………… (1.33)

Example 1.39 Simplify the following expressions.

(a) Using 1.28

(b) Using 1.29

(c) Using 1.28

(d) Using 1.30

(e) Using 1.30

(f) Using 1.31

(g) Using 1.33

1.5.4 Transformations of Equations

An equation is an equality of two algebraic expressions, a left hand expression and a right hand expression. Solving an equation often involves transforming or transposing an equation to make a particular variable the subject on the left hand side of the equation. Values can then be substituted for the other variables on the right hand side of the equation to obtain the value of the subject. The order of operations must be considered carefully in any such transformation.

Example 1.40 Question: Make x the subject of the following equation, 5x – 11 = 3x + 7

Solution: Add 11 to both sides: 5x – 11 + 11 = 3x + 7 + 11 Simplify: 5x = 3x + 18

Subtract 3x from both sides: 5x – 3x = 3x + 18 – 3x

Simplify: 2x = 18

Divide both sides by 2:

Simplify: x = 9

Example 1.41 Question: The cost (C) of producing n items of a particular product is given by the equation

C = 11 + 13n

If the cost of production is $167, determine the number of items that were produced.

Solution:
Let C = 167 in the equation, giving

167 = 11 + 13n

The task is now to make n the subject of the equation.

Subtract 11 from both sides: 167 – 11 = 11 + 13n – 11

Simplify: 156 = 13n

Divide both sides by 13:

Simplify: 12 = n

The number of items produced was 12.

Example 1.42 Question:
For how many years must $400 be invested at an interest rate of 10 percent per annum with interest compounded annually to reach a value of $484?

Solution:
The formula that gives the future value (FV) after n periods of an initial investment (PV) compounded at 1% per period is FV = PV(1 +
In this problem, PV = 400, FV = 484 and i = 0.10, giving 484 = 400(1 + 0.10)
The problem is to make n the subject of the equation.

Divide both sides by 400:

Simplify:

Take natural logs of both sides:

Simplify:

Divide both sides by ln 1.1:

Simplify:

Evaluate the ln:

Simplify: 2.00 = n

The number of periods required is 2.

If you realised that 1.21 = then the last steps would have been:

Instead of evaluating:

This can be written:

Simplify:

Simplify: 2 = n

1.5.5 Construction of Equations

Many problems we come across are not formulated in mathematical terms but are stated in ordinary language. If the problem can be translated into a mathematical form this will expedite the solution of the problem. In this section, a number of relatively simple examples are given to illustrate the process and allow you to gain some skill in this procedure.

Example 1.43 Question:
Seven people hire a van for the day and they share the expenses consisting of $107 for the van hire and $12 for the petrol. How much does each person need to contribute?

Solution:
First translate the problem into a mathematical expression. The resulting equation can then be transposed to obtain an explicit solution.

Let x be the amount each person contributes, then the problem can be expressed mathematically as

7x = 107 + 12 = 119

Divide both sides by 7

= 17

The amount each person contributes is $17.

Example 1.44 Question:
A consulting group charged $7,000 for a project. The project leader wants to deduct an amount for his services and $3,000 for administration costs. The remainder is to be divided equally amongst the three other members who insist that their share is the same as the project leader. How much does each group member receive?

Solution:
If we again let x be the share given to each member of the group, then the problem translates into the following mathematical equation:

Transforming the equation to make x the subject gives:

4000 – x = 3x 4000 = 4x 1000 = x

Each member, including the leader, receives $1000

Exercises Basic Mathematics Exercises 1.26-1.35

1.6 LINEAR FUNCTIONS

1.6.1 Cartesian Coordinate System

To graph any function relating two variables x and y, we construct what is known as the xy-coordinate plane. This consists of two perpendicular lines, one horizontal and one vertical, drawn in a plane. The point of intersection is known as the origin. These two lines will correspond to number lines with the origin being the zero point on both number lines. The horizontal line will in general be called the x-axis with the positive direction (that is, direction of increasing values of x) to the right. The vertical line will in general be called the y-axis with the positive direction (that is, direction of increasing values of y) upward.

Figure 1.1 Cartesian Coordinate System

4 3

2 P

1

-4 -3 -2 -1 0 1 2 3 4 -1

-2 -3 Q

-4

The position of any point in the plane can be identified by two numbers called the coordinates of the point. The coordinates of the point are expressed as an ordered pair (x,y). The first number in the pair is the x-coordinate and indicates the horizontal distance of the point from the origin. The second number in the pair is the y-coordinate and indicates the vertical distance of the point from the origin. In Figure 1.1, the coordinates of the point P are (4,2) and the coordinates of the point Q are (1,-3). This system for locating points in a plane is known as a cartesian coordinate system.

1.6.2 Linear Functions

A linear function is an expression of the form:

………… (1.34)

where a and b are constants; and x and y are variables

It is one of the simplest algebraic equations expressing a relationship between two variables x and y and expressed in this form we say y is a function of x. In general, we will use this equation to obtain values of the dependent variable y, given values of the independent variable, x. The constants a and b are known as coefficients.

1.6.3 Graphs and Solutions

Given the general linear equation, y = a + bx, one of the most frequent tasks we will be given is to find the value of y for some given value of x. In such cases we simply substitute the value of x into the right hand side of this equation and perform the appropriate arithmetic.

Example 1.45 Question:
Sales for the STEADY-AS-IT-GOES Company were 1000 gizmos in 1994 and have increased by 200 per year.
(a) Write down the equation relating sales (S) to the year t, where t represents the number of years since 1994.
(b) Tabulate the sales for the years 1994 to 1999.
(c) Use the table to graph the sales for the years 1994 to 1997.
(d) View the graph and comment on the coefficients in the equation in part (a).
(e) Use the equation in part (a) to forecast sales for 1999.
(f) Use the equation from part (a) to forecast when sales will reach 2,500 gizmos.
(g) Comment briefly on the forecasts in (e) and (f).

Solution:
(a) S = 1000 + 200t.

This equation is linear since it is of the form y = a + bx, with a = 1000 and b = 200. S is the dependent variable in this problem as the sales vary with (depend on) the year.

(b) In 1994, that is, t = 0, the sales can be calculated by

Sales, S = 1000 + 200 x 0 = 1000

The sales for the other years can be calculated similarly and the values tabulated for each year:

Year 1994 1995 1996 1997 t 0 1 2 3 Sales 1000 1200 1400 1600

(c) The graph can then be drawn using the Excel Chart Wizard

Since the equation is a linear function the graph is a straight line.

(d) a = 1000
The graph cuts the vertical axis at S = 1000, showing that S = 1000 when t = 0. The coefficient a is called the y-intercept.

b = 200
On the graph, note how S increases by 200 for each increase of 1 in x. The coefficient b is called the slope, or gradient, of the line.

(e) If we extrapolate, that is extend, the graph to t = 5 we could read the forecast sales for 1999.
Alternatively, substitute t = 5 in the equation

S = 1000 + 200t = 1000 + 200 x 5 = 2000

The forecast sales volume for 1999 is 2,000 gizmos.

(f) Substitute S = 2500 into the equation from part (a) and solve for x

2500 = 1000 + 200t then 200t = 2500 – 1000 = 1500 and t = = 7.5

The S values are aggregated yearly sales. Sales are forecast to be 2400 in 2001, when t = 7 and 2600 in 2002, when t = 8 In 2002 sales are forecasted to reach, in fact exceed, 2500 gizmos.

(g) The forecasts depend on the assumption that the past growth pattern continues.

1.6.4 Interpreting the Coefficients

In the linear equation y = a + bx

• a is termed the y-intercept, the y value where the line cuts the vertical axis. That is the value of y when x = 0.
• b is termed the slope, or gradient, of the line. That is for each increase of 1 unit in x, y increases by b units.

Example 1.46 Question: Interpret the values of the coefficients of the equations in Example 1.45.

Solution: The equation is Sales, S = 1000 + 200t

a = 2000. The value of S when t = 0 is 1000. In the context of this problem the value of a is the sales volume in 1987.

b = 200. For each increase of t by 1, S increases by 200 units. In the context of this problem, annual sales volume increases by 200 gizmos each year.

Let and be two points with coordinates and respectively. The equation of the line passing through these two points is of the form y = a + bx.

Since and are both on the line and

If (that is, the line is not vertical), then by subtraction

and solving for b we obtain:

………… (1.35)

Figure 1.2 Gradient of a straight line

y

(x2 y2)

Rise = y2-y1 (x1 y1) (0 a) Run = x2 – x1

x

The slope of the line can be positive (b > 0), negative (b < 0) or zero (b = 0).

If b > 0 the line slopes from the lower left of the graph to the upper right. Consequently, as the value of x increases, the value of y also increases.

If b = 0 the line is horizontal. Consequently, the value of y is the same for all values of x.

If b < 0 the line slopes from the upper left to the lower right of the graph. Consequently, as the value of x increases, the value of y decreases.

Figure 1.3 Straight line graph for b > 0, b = 0 and b < 0 y y y

x x x

b > 0 b = 0 b < 0

It is important to note that the interpretation placed on the coefficients will vary according to the context of the problem. In some cases the y-intercept may not have a meaningful interpretation.

Example 1.47 The initial costs to STEADY-AS-IT-GOES for setting up a production run of gizmos is $1,846. After that it costs $22.25 to produce each gizmo. Let C represent the cost of a production run in which N items are produced. C is the dependent variable, since the cost of a run depends on the number produced. Then the relationship between the cost and number of units is expressed by the equation, C = 1846 + 22.25N The fixed cost of production, the production cost if N = 0, is $1,846. $22.25, the marginal cost of production, also known as the variable cost, is the increase in production cost if one extra gizmo is produced.

1.6.5 Graphs

Graphing an equation generally involves plotting a sufficient number of points over a large enough range of the variable in order to observe the equation’s important characteristics. The points are then joined to give the graph. If a spreadsheet is being used to calculate the points, then plotting these points becomes easy.

In the case of a linear function it is sufficient to plot any two points that are calculated from the equation and then to join these two points with a straight line.

Example 1.48 Question: Plot the equation C = 1846 + 22.25N

Solution: For the two points use N = 0 and N = 20 (any two points can be used).

For N = 0, C = 1846 + 22.25 x 0 = 1846. For N = 20, C = 1846 + 22.25 x 20 = 2291.

Plot the two points: N = 0, C = 1846 and N = 20, C = 2291; and draw a straight line through them.

1.6.6 Range of Validity

For many mathematical equations the graph extends infinitely in one, or both, directions. However, when the equation is applied to a business problem, only part of the graph may be relevant to the problem. That is, there is a limited range of validity for the graph.

Example 1.49 The initial costs for STEADY-AS-IT-GOES setting up a production run of gizmos if $1,846. After that it costs $22.25 to produce each gizmo. The relationship between the cost and number of units was expressed by the equation, C = 1846 + 22.25N.

Negative N values do not make sense. Also, the equation is probably inapplicable for large N values. This could be due to economies of scale, or additional expenditures being required, or that the production volume is beyond the company’s capacity. The range of validity is then only for positive values of N, provided N is not too large.

When the relationship between two variables is not linear, a linear function may be used to approximate the relationship over a particular range of values. Provided you are interested only in this range of values, a linear approximation may be reasonable. However, the value of the intercept on the vertical axis may be meaningless.

1.6.7 Finding the equation to a straight line given two points

Given two points and the equation of the straight line that passes through the points can be found by two methods:

• Method A
• Calculate the gradient
• Substitute the coordinates of one of the points and the value for b into the equation y = a + bx and solve for a.
• Method B
• Substitute the coordinates of both points into ………… (1.36)
• Rearrange the equation to obtain an equation of the form y = a + bx.

Example 1.50 Question:
SIMPLICITY TAXIS charge a fare of y dollars for a journey of x kilometres. It is known that a fare is calculated as a flagfall charge plus a fixed charge per kilometre, or part of a kilometre.

If a 6km journey cost $7.50 and an 18km journey cost $16.50:
(a) find the equation relating the fare to the distance using
(1) Method A
(2) Method B
(b) Interpret the coefficients of the equation
(c) Comment on the range of validity for the equation

Solution: y is the dependent variable, as the fare depends on the distance travelled, and y is related to x by an equation of the form y = a + bx.

When x = 6 then y = 7.50 And when x = 18 then y = 16.50

(a) (1) Method A

= 0.75

Substitute the first point and the value for b into

y = a + bx

then 7.50 = a + 0.75 x 6 7.50 = a + 4.50 and a = 3.00 so y = 3.00 + 0.75x

(2) Method B

then

0.75(x – 6) = y – 7.50 0.75x – 0.75 x 6 = y – 7.50 0.75x – 4.50 + 7.50 = y

and y = 3.00 + 0.75x

(b) As y = 3.00 + 0.75x.

y-intercept: a = 3.00. A journey of 0 km costs $3.00. That is a flagfall of $3.00. gradient: b = 0.75. Each extra kilometre travelled increases the fare by $0.75. That is, after the flagfall charge, there is a charge of $0.75 per kilometre.

(c) Range of validity: from 0 to an upper limit on the number of kilometres. For long journeys, for example Melbourne to Sydney, a driver would probably negotiate a special fare rather than use the meter and the equation would not be applicable.

1.6.8 Average Unit Cost (AUC)

At SIMPLICITY TAXIS a fare is related to the number of kilometres travelled by the equation y = 3.00 + 0.75x. If you have just paid $7.50 for a 6km journey, how much did you pay per kilometre?

You paid $0.75 for each extra kilometre, that is the marginal cost was $0.75 per kilometre.

However, on average, you paid per kilometre. That is, the average unit cost was $1.25 per kilometre.

In general

or

Then, for a 6km journey costing $7.50,

= $1.25 per km

Note that as x increases, then AUC decreases. For example, an 18km journey costing $16.50 has

= $0.92 per km

1.6.9 Alternative form of linear equation

The standard form of a linear equation is y = a + bx.

An equation of the form is also linear, although the coefficients and do not have a direct interpretation. For example, 2x + 3y = 48 is a linear equation which may be arranged to give

Exercises Basic Mathematics Exercises 1.36-1.41

1.6.10 Solution of Simultaneous Linear Equations

Important applications of linear functions are to represent supply and demand functions in economics, and to represent cost and revenue functions for break-even analysis in accounting. In these situations, we are dealing with two linear functions and are interested in their intersection, that is, where two lines cross. This point of intersection may be found graphically or algebraically by solving a pair of “simultaneous equations”. To solve a pair of linear equations simultaneously means to find all points (x,y) which satisfy both equations.

Consider the pair of linear equations:

………… (1.37)

Since a pair of linear functions represents two straight lines in a plane, three situations can occur:

(a) If the two lines are not parallel to each other there will be a single point where the two lines cross and the system of equations will have a single, unique solution. (Example 1.51)
(b) If the two lines are parallel, but not coincident, then there is no point of intersection and the system of equations will not have a solution. (Example 1.53)
(c) If the two lines are coincident, that is lie on top of each other, then there are an infinite number of points of intersection and the system of equations has an infinite number of solutions. In this case, both equations represent the same line and one equation can be obtained from the other by multiplication by a constant. (Example 1.52)

Example 1.51 Question:
WILLY-BUY-IT’S DISCOUNT STORE has recently received two shipments of Moonbeam electric kettles. The first shipment of 2 deluxe and 3 standard models had an invoice for $48.00. The second shipment of 4 deluxe and one standard had an invoice for $56.00

(a) How much was the store charged for a deluxe kettle and how much for a standard kettle?
(b) Depict the problem graphically.

Solution:
(a) Let x represent the price of a deluxe model kettle and y represent the price of a standard model.

We must solve the following system of linear equations:

2x + 3y = 48 … (1) and 4x + y = 56 … (2)

There are two methods to solve the system of Equations 1.37, the elimination method and the substitution method. These methods of solution are illustrated in this example:

(1) Elimination Method:

This method involves multiplying each equation by appropriate constants in order to eliminate either the variable x or y when the equations are added or subtracted.

If we multiply (1) by +2 and (2) by –1 we obtain

4x + 6y = 96 .. (3) and -4x – y = -56 ..(4)

Adding these two equations now gives 0 + 5y = 40

and dividing both sides of the equation by 5 gives y = 8

Now, to find x substitute y = 8 in either of the original equations. If we use (2) we obtain:

4x + 8 = 56

then 4x = 48 and x = 12

The solution is therefore (12,8) or x = 12, y = 8

We can check the solution by substitution in (1):

Left hand side, LHS = 2x + 3y = 2 x 12 + 3 x 8 = 48 = Right hand side, RHS

Therefore the solution checks:

(2) Substitution Method:

This method involves solving one of the equations for one variable in terms of the other variable. This solution is then substituted in the remaining equation yielding an equation in only one variable. This equation can now be solved and the solution substituted in the original equation.

Applying this procedure to the current example we can first transpose (2) to give

y = 56 – 4x … (3)

We now substitute this for y in (1) giving

2x + 3(56 – 4x) = 48
Simplifying
2x + 168 – 12x = 48 and then -10x = -120 and solving for x gives x = 12
Substituting x = 12 in equation (3) gives

y = 56 – 4 x 12 and y = 8

Therefore the solution is (12,8).

So the deluxe model cost $12.00 and the standard model cost $8.00.

(b) After rearranging the equations become: y = 16 - and y = 56 – 4x

The graph of the two non-parallel lines shows a unique solution at x = 12 and y = 8.

Example 1.52 Question:
The latest two shipments of Moonbeam electric kettles WILLY-BUY-ITS will be invoiced under the new price schedule. A shipment of one deluxe kettle and one standard kettle was invoiced at $61; and another shipment of three deluxe kettles and three standard kettles was invoiced at $183. What price was WILLY-BUY-ITS store charged for each type of kettle?

Solution:
Let x represent the price of a deluxe model kettle and y represent the price of a standard model kettle.

We have two equations to be solved simultaneously:

x + y = 61 .. (1) and 3x + 3y = 183 .. (2)

The second equation can be obtained by multiplying the first equation by 3. The graphs of these equations are therefore equivalent and they have an infinite number of common points. Finding solutions to either equation provides a solution to both equations, If we try to apply the method of elimination or substitution to this pair of equations we end up with either 0 = 0 or 183 = 183, neither of which will give us values of x and y. We only have one independent equation, where two are needed to obtain a unique solution. This means that, from the available information, we cannot calculate the prices of the two models. After rearrangement, equations (1) and (2) both become the equation y = 61 – x.

Equation 1.5.3 Question:
Last quarter, FREDA’S FISH’N’CHIPPERY had 12 floodlights and 9 fluorescent lights illuminating the front of the shop and the electricity bill for this was $142. This quarter, in order to cut costs, Freda disconnected four of the floodlights and three of the fluorescent lights and the electricity bill was $118. If you assume the usage of the lights was identical for each quarter, how much does it cost to run each type of light?

Solution:
Let x represent the quarterly electricity cost for a floodlight and y represent the quarterly electricity cost for a fluorescent light.

As Freda’s Chippery used 8 floodlights and 6 fluorescent lights in the second quarter, we have two equations to be solved simultaneously:

12x + 9y = 142 ..(1) and 8x + 6y = 118 ..(2)

Solving these equations by the elimination method

Multiplying (1) by 2 and (2) by –3 gives

24x + 18y = 284 ..(3) and -24x – 18y = -354 ..(4)

Adding (3) and (4) gives

0 = -70, which is obviously false.

We can conclude that there is not a pair of x and y that satisfy equations (1) and (2) simultaneously.

For a further explanation, multiply equation (2) by 1.5 to get equation (2a).

12x + 9y = 142 ..(1) and 12x + 9y = 177 ..(2a)

The equations do not have a simultaneous solution as they are inconsistent. Apparently electricity charges were different in the two quarters.

Applications to break-even analysis
The requirement of most business projects is to make a profit, that is, the total revenue for the project should exceed the total costs. If the revenue and the costs are dependent on the same variable, then the aim of break-even analysis is to determine the value for this variable at which the revenue will equal the costs. Determination of the break even point gives management the range of values of the variable for which the project will make a profit and where it will make a loss.

Example 1.54 Question: The cost, C, for producing N UBUTE COMPUTERS is given by C = 135,712 + 142.86N If the computers sell for $260, giving the revenue function as R = 260N, how many computers must Ubute produce before any profit will be made?

Solution:
The break even point tells us how many computers must be produced before a profit is realised, that is, before the revenue exceed costs. Production numbers greater than the break even point will lead to higher profits with increased production. To find the break even point we set costs equal to revenue and solve the resulting equation to find N as follows:

If C = R
Then 135712 + 142.86N = 260N 135712 = 260N – 142.86N and =1,158.54

So Ubute must produce at least 1,159 computers to realise a profit.

Exercises Basic Mathematics Exercises 1.42-1.45

1.7 NON-LINEAR FUNCTIONS

Although many situations can be approximated by the linear function discussed in the last section, there are others which are best represented by some type of curved function. A function whose graphical representation in a plane is not a straight line is called a non-linear function. In this section we will present some types of non-linear functions such as quadratic, cubic, logarithmic and exponential functions.

1.7.1 Quadratic Functions

Many functions associated with business problems are of a form where the function increases to a maximum value and then decreases or decreases to a minimum value then increases. A very useful model for such problems is provided by the quadratic function which is defined by the following equation.

………… (1.38)

where the coefficients a, b and c are real numbers and

The graph of a quadratic function is called a parabola and has one of the two symmetric shapes shown below, depending on whether a is negative or positive.

Figure 1.4 General forms of a quadratic equation

a > 0 a < 0

If a > 0 the function will have a minimum value.

For example, such functions are often used to model:
• Supply, the y dependent variable, as a function of price, the x independent variable
• Cost, the y dependent variable, as a function of number of goods produced, the x independent variable.

If a < 0 the function will have a maximum value. Such functions are often used, for example, to model demand as a function of price or revenue as a function of number of goods sold.

To obtain a graph of a quadratic function it is necessary to substitute a range of values of x into Equation 1.38 and calculate the corresponding values of y. These points are then plotted. At this stage, choosing the range of values of x is somewhat arbitrary, however we will shortly see how to determine the turning point of the graph and this will help in choosing the appropriate values of x.

Example 1.55 Question:
It has been determined that the net profit, y, realised by the Mmm-aah COFFEE company of Brazil is given by the equation where y is measured in millions of dollars and x-the production per month is in the range 0.00 to 4.00 hundred tonnes.

Solution:
Compute a table of coordinates for the equation and plot the points:

x 0.00 0.50 1.00 1.50 1.75 2.00 2.25 2.50 3.00 3.50 4.00 y 3.00 1.25 0.00 -0.75 -0.94 -1.00 -0.94 -0.75 0.00 1.25 3.00

Minimum or Maximum value
The minimum or maximum value of the quadratic function, Equation 1.38, is given by

………… (1.39)

which occurs when Therefore the coordinates of the turning point (i.e., where the maximum and the minimum occurs) are ………… (1.40)

and this will be a maximum if a < 0 and a minimum if a > 0.

X Axis Intercept
If a quadratic function cuts the x-axis then another common problem is to find these points. That is, we want to find the values of x for which y = 0 or, in other words, to find the solutions of the following equation:

………… (1.41)

These solutions are in general given by

………… (1.42)

These formulas provide real solutions only if the value under the square root is positive. This tells us that if

then the graph does not cross the x-axis. then there is only one solution which tells us that the graph just touches the x-axis, that is, the turning point is on the axis.

then the graph cuts the x-axis in two places.

Example 1.56 Question: The Mmm-aah COFFEE company of Brazil, in Example 1.55, can have net profit calculated from the coffee output by the equation

(a) Determine the production value for which profits are a minimum.
(b) Determine the production range for which the company will be suffering a loss.
(c) Comment briefly on the range of validity of the equation.

Solution:
(a) For this function we have a = +1, b = -4 and c = 3.

Since a > 0 the function has a minimum value with coordinates

= 2

= -1

The minimum profit is at the turning point (2, -1).

This tells us that when production is 2 hundred tonnes per month the company experiences its minimum profit, a loss of one million dollars.

(b) To determine the production range for which the company suffers a loss we will need to find the production values which yield zero profit, that is, y is zero.

We first note that

therefore, the equation has two solutions for y = 0, that is crosses the x-axis in two places.

The solutions of the equation for y = 0 are obtained using Equation 1.42:

= 3 and

= 1

Therefore, when the production exceeds one hundred or is below three hundred tonnes per month the company will make a loss.

(c) Negative values for production, x, are not meaningful. According to the equation, if x = 0, y = 3, suggests a profit of $3m when there is zero production. If this cannot be explained, for example by subsidies, then the equation seems to be invalid for small x-values. In practice there would also be an upper limit to the range of validity. This limit would be determined, for example, by the company’s capacity for production or the market’s ability to accept the production.

1.7.2 Exponential Functions

The general form we shall use for an exponential function is

…………(1.43)

where b is a positive real number known as the base, or growth factor, of the exponential function and a is a real number, the value of y when x = 0.

An exponential function increases by a constant factor of b each time x increases by 1. For example, if b = 2, a = 1 then y starts off having a value of 1 when x = 0. By x = 1, y has doubled and has a value of 2; by x = 2, y has doubled again and now has a value of 4 and so on. With this type of “exponential growth” y increases very rapidly and for instance by the time x = 10, y = 1024 and by x = 25, y = 33,554,432. There are many examples of exponential growth, however the most important to us here will be in their application to compound interest and related financial calculations.

There are three basic situations to consider for the function:

(a) If b > 1 then y increases as x increases for all values of x, giving an exponentially growing function as shown in the graph below. Note that y is increasing rapidly as x takes larger and larger positive values. However, since we see that y gets closer and closer to zero as x takes larger and larger negative values.
(b) If b = 1 the situation is extremely simple since 1 raised to any power is equal to 1, that is and y = a for any value of x.
(c) If 0 < b < 1 then y decreases as x increases for all values of x yielding an exponentially decaying function as shown in the graph below.

Note that all of these graphs cut the y-axis at y = a since

a a

a

b > 1 b = 1 0 < b < 1 exponential growth constant exponential decay Two commonly used base numbers are 10 and e = 2.7182818…

The function is often referred to as the exponential function or the natural exponential function. For this function a = 1 and b = e.

Example 1.57 Question: $100 is invested at an interest rate of 20% per year compounded annually.
(a) Calculate and tabulate the value of the investment for the first ten years. Then write an equation showing the value of the investment after n years.
(b) Graph the value of the investment against the year for the first 10 years.
(c) Comment briefly on the range of validity of this equation.

Solution:
(a) Let FV represent the future value of the investment after n years.

At the end of the first year, n = 1:

= 100 (1.2) = 120

At the end of the second year, n = 2:

= 120 (1.2) = 100 (1.2)(1,2) = 144

After n years we have which is of the form of an exponential function with an initial value a = 100 and growth factor b = 1.2.

The values for the first 10 years are shown in the table below.

n 0 1 2 3 4 5 6 7 8 9 10
FV 100 120 144 173 207 249 299 358 430 516 619

(b)

(b) Negative values of n are not relevant as the investment did not begin until n = 0. For n > 0 the investment will grow according to the equation.

A more elaborate use of the exponential function is shown in the next example.

Example 1.58 Question: The production of the GEE-WHIZZ Company’s new blender have been recorded for the product for the first fifteen months. The total production for the blender may be modelled by a Logistic function:

Monthly production,

Where t is the number of months since production commenced

(a) Tabulate the total production for the first fifteen months and graph the function.
(b) Describe briefly the pattern for the monthly production for the blender.

Solution:
(a) The total production for months 0 to 15 is given in the table and the function is graphed from these tabulated values:

Month Total t Production 0 100 1 220 2 475 3 999 4 1980 5 3550 6 5518 7 7353 8 8646 9 9390 10 9767 11 9948 12 10031 13 10069 14 10086
15 10094

(b) From low production in the early months, production increased sharply until about the eighth month. From then on, the rate of increase reduced sharply so that production levelled off at slightly above 10000.
1.7.3 Logarithmic Functions

The general form we shall use for a logarithmic function is

…………(1.44)

where the base, a, is a positive real number and

The logarithmic and exponential functions are the inverses of each other since, if then For the logarithmic function, Equation 1.44, we have two basic cases to consider.

(a) If 0 < a < 1, then y decreases as x increases for all values of x > 0 and y = 0 for x = 1.
(b) If a > 1 then y increases as x increases for all values of x > 0. The rate of increase is inversely proportional to the size of the base, that is, y increases more rapidly for smaller values of a. If x > 1 then y > 0. If x = 1 then y = 0, and if 0 < x < 1 then y < 0. Remember that has no meaning for

Graphs for these two cases are shown below:

The two most commonly used bases are 10 and e. Logarithms to the base 10, are known as common logarithms and are usually written as log x. Logarithms to the base e, are known as natural logarithms and are usually written as ln x. Logarithms to the base 10 were used for calculating before calculators and computers became readily available. Now they are used less frequently and on many calculators only natural logarithms are available.

Values of logarithmic functions may be evaluated on a calculator using the log key for base 10, if one is available, or the ln key for base e.

Example 1.59 Question: Evaluate the following without using a calculator:

(a)
(b)
(c)
(d)

Solution: If If

(a) Let then and since
That is

(b) log 0.001 is understood to mean let then 0.001 and since
That is

(c) Let
Then
And y = 12
That is

(d) By definition, ln
Now by laws of logarithms,
(Recall Law (1.26), p. 23)

Example 1.60 Question: For the new GEE-WHIZZ blender, the relationship between the number of units produced per hour, P, and the number of staff, S is given by the function

P = 10ln(S – 4)

(a) Calculate the number of staff that are needed to produce 45 blenders in an hour.
(b) Tabulate the production for staff levels from 5 to 20 and graph the function.
(c) Comment briefly on the staff productivity at GEE-WHIZZ.

Solution:
(a) For P = 45, solve the equation P = 10ln(S – 4) 45 = 10ln(S – 4) ln(S – 4) =

then

giving = 94.02

About 94 staff are needed.

(b) The production for staff levels from 5 to 20 is given in the table and the function is graphed from these tabulated values:

Staff Production 5 0 6 7 7 11 8 14 9 16 10 18 11 19 12 21 13 22 14 23 15 24 16 25 17 26 18 26 19 27
20 28

(c) The productivity, number of items produced per staff member, is: 0 for 5 staff, presumably non-production staff; the production increases slowly with higher staff numbers.

Exercises Basic Mathematics Exercises 1.46-1.49

1.8 SOLVING SIMPLE NON-LINEAR SYSTEMS OF EQUATIONS

We have shown in Section 1.6.10 how to solve a system of two linear equations. From a graphical viewpoint this involved finding the coordinates of the point where two straight lines intersected. The procedure was relatively straight forward and we were able to write down an explicit series of steps leading to the solution. We now wish to deal with systems of two equations where one or both equations is non-linear. The procedure can become quite complex and for many cases algebraic solutions cannot be found and we must rely on either numerical or graphical techniques to obtain approximate solutions.

Example 1.61 Question: The total revenue function for Never Write biros is given by

………(1)

where N is the number of thousands of biros sold per month. And R is the revenue in hundreds of dollars

The monthly production costs are given by

C = 4N + 24 …………(2) Where C is the cost of production in hundreds of dollars

If all the biros are sold each month then the profit, P = R – C.

Graph the cost and revenue functions and find the break-even points:
(a) graphically
(b) algebraically.

Solution:
Our task is to find the values of N, C and R which satisfy both equations simultaneously. At break-even the revenue will equal the cost, (that is, R = C) so we solve the following system of equations.

…………(3) …………(4)

(a) Graphical Method

To solve the system graphically requires sketching a graph of each question and then identifying the coordinates of the points of intersection. In the current example we will express R as a function of N.

We now substitute a range of values of N into each equation and find the corresponding values of R. These pairs of points may then be plotted. We can also use our knowledge of linear and quadratic equations to identify key features of each graph. Equation (4) is a linear equation with gradient 4 and the intercept on the Revenue-axis at 24. Equation (3) is a quadratic equation with, using Equation 1.39, a maximum turning point at (5, 50) and, using Equation 1.42, intercepts on the N-axis at N = 0 and N = 10. The graphs of these two functions are then drawn.

(b) Algebraic Method
To solve the system algebraically, we use a substitution method similar to that used for the linear systems in Section 1.6.10, Example 1.51.

The method involves solving one of the equations for one variable in terms of the other variable. This solution is then substituted in the remaining equation giving an equation in only one variable. This new equation can be solved and the solution substituted in the original equation.

Substitute the right hand side of this equation (3) for R in (4) giving

simplifying this becomes

This is of the form of a quadratic equation which we can solve using Equation 1.42, taking a = 1, b = -8 and c = 12. This gives

= 2 or 6

Substituting these values of N into (4) gives R = 32 for N = 2 and R = 48 for N = 6. Therefore, the system of equations has two solutions (2, 32) and (6, 48) showing break-even points at production levels of 2,000 and 6,000 biros.

NOTE: When revenue exceeds cost, a profit is made. Hence a profit is made when production levels exceeds 2,000 biros but is less than 6,000 biros.

The next example is a problem involving depreciation. This is a problem which cannot be solved exactly using algebraic methods and hence we must use graphical methods.

Example 1.62 Question: An asset is bought for $1000. It is to be depreciated either on a straight line basis at 10% per annum or on a reducing balance at 15% per annum. After how many years would the straight line book value be less than the reducing book value?

Solution: The equations giving the straight line book value and the reducing balance book value after a time t are:

Straight line depreciation: ………(1) Reducing balance depreciation: ………(2)

The derivation of equations (1) & (2) come from financial mathematics. (Just assume these relationships.) For now all we need be concerned with is that equation (1) is a linear equation with a gradient of –100 and an intercept on the vertical, that is, the BV-axis, at 1000. Equation (2) is an exponential function of the form of Equation 1.45 with a = 1000 and b = 0.85. Since b < 1 this is an exponentially decaying function.

Since this system cannot be solved algebraically to obtain an exact solution, feel free to try if you so wish, we resort to graphical methods. The first step is to determine the book value using each method over a number of years. Graphs of these results may then be plotted and the point of intersection sought. Since it makes no sense to determine the book value for negative times and since the straight line book value reaches zero after ten years we need only consider values of t between zero and ten. The results are given in the table below:

Time (years) 0 1 2 3 4 5 6 7 8 9 10 Straight line 1000 900 800 700 600 500 400 300 200 100 0 ($’s) Reducing 1000 850 723 614 522 444 377 320 272 232 197 Balance ($’s)

From this table we can see that the crossover point comes somewhere during the seventh year. The graph of these results is shown below:

From the graph we see that the straight line book value will be less than the reducing balance book value after approximately years.

Exercises Basic Mathematics Exercises 1.50-1.53