Quantitative Methods for Business chapter 2 doc

To plot an equation start by setting out a scale of possible values of one unknown along one axis, or dimension, and a scale of the possible values Example 2.1 A sales agent is paid a ba

Getting linear models

straight

2

Chapter objectives

This chapter will help you to:

■ plot and solve linear equations

■ apply basic break-even analysis

■ interpret inequalities

■ undertake simple linear programming using graphs

■ use the technology: Solver in EXCEL

■ become acquainted with business uses of linear programmingThis chapter is intended to introduce you to the use of algebra in solv-ing business problems For some people the very word algebra conjures

up impressions of abstract and impenetrable jumbles of letters andnumbers that are the preserve of mathematical boffins Certainly parts

of the subject of algebra are complex, but our concern here is with

alge-braic techniques that help to represent or model business situations.

In doing this we are following in the footsteps of the ‘father of algebra’, Mohammed ibn-Musa al-Khwarizmi In the ninth century

al-Khwarizmi wrote Al-jabr wa’l-muqabala, which might be translated as

‘Calculation Using Balancing and Completion’ The first part of theArabic title gives us the word algebra Although al-Khwarizmi was ascholar working at the House of Wisdom in Baghdad, he saw his task invery practical terms, namely to focus on

… what is easiest and most useful in arithmetic, such as men constantlyrequire in cases of inheritance, legacies, partitions, law-suits, and trade,

and in all their dealings with one another… (cited in Boyer,

1968, p 252)

In the course of this chapter we will confine our attention to simplealgebra and how you can use it to solve certain types of business prob-lem We will focus on linear equations, which are those that are straightlines when they are plotted graphically They form the basis of linearmodels that assist business problem-solving

2.1 Linear equations

Central to algebra is the use of letters, most frequently x and y, to

represent numbers Doing this allows us to deal systematically withquantities that are unknown yet of importance in an analysis These

unknown quantities are often referred to as variables, literally things that vary over a range of numbers or values Sometimes the point is to

express a quantitative procedure in a succinct way, and the use of letters merely constitutes convenient shorthand

These types of expression are called equations because of the equalssign, ‘’, which symbolizes equality between the quantity to its left andthe quantity to its right An equation is literally a state of equating orbeing equal

An equation that involves just two unknown quantities can be drawn

as a line or a curve on a graph Each point along it represents a

com-bination of x and y values that satisfies, or fits the equation.

To plot an equation start by setting out a scale of possible values of one

unknown along one axis, or dimension, and a scale of the possible values

Example 2.1

A sales agent is paid a basic wage of £200 per week plus 10% commission on sales

The procedure for working out his/her total wage could be written as:

Total wage 200  10% of sales

It is often more useful to abbreviate this by using letters If y is used to represent the total wage and x to represent sales we can express the procedure as:

y  200  0.1x

Using this we can find the total wage for a week when sales were £1200:

y 200  0.1 * 1200  200  120  £320

of the other unknown along the other axis Ensure the scales cover therange of plausible values, and start them at zero unless interest in the line

is limited to part of it well away from zero Plot the x values along the izontal axis, known as the x axis, and the y values along the vertical axis, known as the y axis This conveys that y depends on x.

hor-Once each axis has been prepared, portraying an equation in itsgraphical form involves finding two points that lay along the line thatwill represent the equation This means you have to identify two pairs of

x and y values both of which satisfy the equation The way to do this is to

specify an x value and use the equation to work out what value y would

have to take in order to satisfy the equation, then repeat the process for

another x value To ensure that your line is accurate it is important to take one x value from the far left hand side of the horizontal axis and

the other from the far right hand side

To help us design the x axis let us suppose that the maximum sales the agent could

achieve in a week is £5000 Using the equation we can use this to find the maximum wage:

y 200  0.1(5000)  700

This means the highest value we need to include in the scale on the y axis is £700.

We are now in a position to construct the framework for our graph, which might looklike Figure 2.1

To plot the line that represents the equation we need to find two points that lie onthe line One of these should be on the left hand side The lowest number on the left

of the horizontal axis is zero, so we could use the equation to work out the wage whensales are zero:

y 200  0.1(0)  200

When sales are £0 the wage is £200, this pair of values gives us the position, or

coord-inates of one point on the line The sales value, 0, positions the point along the

hori-zontal axis and the wage value, 200, positions the point along the vertical axis

To get a second set of coordinates we should take a sales figure from the right handside of the horizontal axis, say the maximum figure of 5000, and work out the wagewhen sales are £5000, again using the equation:

y 200  0.1(5000)  700

When sales are £5000 the wage is £700 The point we plot to represent this pair of ues will be positioned at 5000 along the horizontal axis and at 700 along the vertical axis.

val-We can now plot both points as in Figure 2.2:

If plotting an equation is new to you, or just something you haven’t done for a while,

it is a good idea to plot a third point between the first two A third point should lie inline with the first two so it is a good way of checking that you have plotted the otherpoints correctly A suitable position for our third point in this case might be when sales

0 1000 2000 3000 4000 5000 0

100 200 300 400 500 600 700

x (sales in £)

5000 4000

3000 2000

1000 0

700 600 500 400 300 200 100 0

are £2000 and the wage will be:

y 200  0.1(2000)  400The point that represents these coordinates, sales of £2000 and a wage of £400, hasbeen plotted in Figure 2.3

The final stage in plotting the equation is to draw a straight line linking the plottedpoints This is shown in Figure 2.4:

0 1000 2000 3000 4000 5000 0

100 200 300 400 500 600 700

100 200 300 400 500 600 700

Lines that represent simple linear equations, such as the one plotted

in Figure 2.4, have two defining characteristics: a starting point, or

inter-cept, and a direction, or slope We can think of the intercept as specifying

the point the line begins, and the slope as specifying the way in whichthe line travels In Figure 2.4 the line begins at 200, when sales are zero,and travels upwards at a rate of 0.1 for every one-unit increase in sales,reflecting the fact that the sales agent receives an extra £0.10 for everyadditional £1 of sales

Different lines will have different intercepts and slopes It will helpyou interpret results of this type of analysis if you can associate basictypes of intercept and slope in linear equations with their plotted forms

To illustrate this we can extend the sales agent example to include contrasting approaches to wage determination

Example 2.3

Suppose the basic wage of the sales agent in Example 2.1 is increased to £300 and thecommission on sales remains 10% Express the procedure for determining the wage as

an equation and plot it

The total wage (y) in terms of sales (x) is now:

y  300  0.1x

The line representing this has an intercept of 300 and a slope of 0.1 It is the upperline in Figure 2.5

0 200 400 600 800 1000

You can see two lines plotted in Figure 2.5 The lower is the line ted in Figure 2.4, the original formulation for finding the wage Theupper represents the equation from Example 2.3, where the basicwage is increased to £300 It is higher because the intercept is 300 com-pared to the 200 in the original equation but note that the two lines are parallel since they have exactly the same slope Lines that have thesame slope will be parallel whatever their intercept.

plot-The bottom line in Figure 2.6 represents the equation from Example2.4 It starts from the point where both wage and sales are zero, known

as the origin, since the intercept of the line is zero It is parallel to the

lines above it because it has the same slope as them

Example 2.4

Identify the equation that would express the calculation of the wages of the sales agent in Example 2.1 if there were no basic wage and the commission rateremained 10%

The total wage would be:

y  0  0.1x

This is plotted in Figure 2.6 together with the equations from Examples 2.1 and 2.3

0 100 200 300 400 500 600 700 800 900

The equations plotted in Figure 2.7 have the same intercept, 200, butdifferent slopes This means they start at the same point on the lefthand side of the graph but their paths diverge The upper, steeper linerepresents the equation from Example 2.5 It has a slope of 0.2, twicethe slope of the line representing the equation in Example 2.1, 0.1,reflecting the greater rate at which commission is earned, 20% ratherthan 10% The slope is twice as steep since the same sales will result inthe sales agent earning double the commission.

Example 2.5

The basic wage of the sales agent in Example 2.1 is to remain at £200, but the rate ofcommission increases to 20% Express the procedure for determining the wage as anequation and plot it

The total wage (y) in terms of sales (x) is now:

y  200  0.2x

The line representing this has an intercept of 200 and a slope of 0.2 It is plotted inFigure 2.7 together with the equation from Example 2.1

0 100 200 300 400 500 600 700 800 900

Identify the equation that would express the calculation of the wages of the sales agent

in Example 2.1 if the basic wage is £200 and there is no commission

The equation in Example 2.6 is plotted as the bottom, horizontal line

in Figure 2.8 It has a zero slope; literally it goes neither up nor down.Whatever the level of sales the wage will be unaffected

The slopes in the equations we have looked at so far have been

upward, or positive, and in the case of Example 2.6, zero You will also come across equations that have negative, or downward slopes.

The total wage would be:

y  200  0x

This is plotted in Figure 2.8 together with the equation from Example 2.1

0 200 400 600 800

y  800  10x where y represents the number of units sold and x the price at which they are sold in £.

The equation is plotted in Figure 2.9

0 200 400 600 800 1000

x (price in £)

Figure 2.9

The line of the equation in Example 2.7

In Figure 2.9 the line slopes downwards because the slope,10, is ative It means that for every increase of £1 in the price of the productthe number of units sold will decrease by ten

neg-At this point you may find it useful to try Review Questions 2.1 and

2.2at the end of the chapter

2.2 Simultaneous equations

In the previous section we looked at how linear equations can be used

to show the connection between two variables Such equations sent the relationship in general terms; they are in effect recipes or for-mulae that specify how the value of one quantity can be establishedwith reference to another quantity This is how a wage that consists of

repre-a brepre-asic component plus srepre-ales commission crepre-an be crepre-alculrepre-ated or how repre-aphone bill made up of a fixed charge plus a cost per unit can beworked out In each case a single linear equation provides a clearnumerical definition of the process involved and can be used to work

out the appropriate y value for any given x value.

Sometimes it is necessary to consider two linear equations jointly, orsimultaneously, hence the fact that such combinations of equations are

known as simultaneous equations Typically the aim is to find a pair of specific values of x and y that satisfy both equations You can achieve

this by plotting both equations on the same pair of axes and identifyingthe point where the lines cross

Finding values that fit both of two equations is known as solvingsimultaneous equations In Example 2.8 the point where the lines

Example 2.8

The sales agent in Example 2.1, currently receiving a wage of £200 plus 10% sion on sales, is offered the alternative of receiving 20% commission on sales with nobasic wage What is the minimum level of sales the agent would have to reach to makethe alternative commission-only wage attractive?

commis-The existing arrangement can be represented as:

y  200  0.1x where y represents the wage and x the sales.

The alternative can be expressed as:

y  0  0.2x

Both equations are plotted in Figure 2.10

In Figure 2.10 the lines cross at the point representing sales of £2000 and a wage of

£400 The line representing the current method of determining the wage is the higherline when sales are below £2000, indicating that it would be the better arrangement forthe agent when sales are less than £2000 The line representing the alternative arrange-ment is the higher line when sales are greater than £2000, indicating that it would

be the better arrangement when sales exceed £2000 The minimum level of sales theagent would have to reach to make the alternative commission-only wage attractive istherefore £2000

0 100 200 300 400 500 600 700 800

cross meets the requirements of both equations because it is a point onboth lines The fact that it is the only point where the lines cross tells usthat it represents the only combination of wage level and sales that fitsboth equations.

You can solve simultaneous equations without plotting their lines on

a graph using a method known as elimination As the name implies this

involves removing or eliminating one of the unknown quantities withthe intention of leaving a numerical value for the other

In order to end up with a clear result conventionally simultaneousequations are arranged so that the unknown quantities and the factors

applied to them, their coefficients, are located on the left hand side of the equals sign and the intercept, or constant appears to its right This

may involve rearranging the equations In doing this we need to ensurethat any manipulation preserves the equality inherent in the equation

by balancing every operation performed on one side with the exactsame operation on the other side

Following this we proceed to the removal of one of the unknown

quantities, either x or y This is straightforward if the number of x’s or

y’s in both equations is the same, in which case if you subtract one

equation from the other you will be left with an expression which hasjust one unknown on the left of the equals sign and a number on itsright Using suitable multiplication or division you can then establish

the value of this remaining unknown, one of the pair of x and y values

that fits both equations

Having found one of the pair of values mutually compatible withboth equations you can substitute it into one of the original equationsand find the value of the other unknown that, in combination with thevalue of the first unknown, satisfies the original equations jointly, or

simultaneously.

Example 2.9

In Example 2.8 the sales agent is presented with two possible wage arrangements resented as:

rep-where y represents the wage and x the sales.

We will start by rearranging both equations so the components, or terms, involving x and y are on the left of the equals sign and the ‘stand-alone’ numbers are on the right.

In the case of the first equation, representing the original arrangement, this entails

y  200  0.1x (the original arrangement)

moving the 0.1x from the right of the equals sign over to the left In doing this we have

to reverse the sign in front of it, as strictly speaking we are subtracting 0.1x from both

sides of the equation, and hence preserving the balance:

y  200  0.1x Subtract 0.1x from both sides:

y  0.2x  0  0.2x  0.2x

to get:

y  0.2x  0

We can now set these rearranged equations alongside each another and subtract one

from the other to eliminate y:

We can break this operation down into three parts:

Applying elimination in Example 2.9 was made easier because in both

equations there was only one ‘y’, that is the coefficient on y in each

equation was one If the coefficients on an unknown are different youhave to apply multiplication or division to one or both equations tomake the coefficients on the unknown you wish to eliminate equalbefore you can use subtraction to remove it

Example 2.10

Find the level of wages at which the two procedures for determining the sales agent’s

wage in Example 2.8 result in the same wage by eliminating x, the level of sales.

The equations representing the procedure, as rearranged in Example 2.9, are:

y  0.1x  200

y  0.2x  0

If we multiply the first equation by two we get:

2y  0.2x  400

Subtracting the second equation from this:

Again we find that the wage level at which the two wage determination models produce

the same result is £400 If we substitute this value of y into the equation representing the

original arrangement we can find the level of sales that will yield a wage of £400:



The original approach to establishing the sales agent’s wage will produce a wage of

£400 when sales are £2000 The alternative, commission-only formulation will of courseyield the same wage when sales are £2000:

y 0  0.2 * 2000  400The values of 2000 and 400 for sales and wages respectively therefore satisfy bothwage determination equations simultaneously

Multiply both sides by minus one:

(1) * (0.1x)  (1) * (200)

0.1x 200Multiply both sides by ten:

x 2000The level of sales at which both approaches to wage determination will produce awage of £400 is therefore £2000

Not all pairs of equations can be solved simultaneously These areeither cases where one equation is a multiple of another, such as:

At this point you may find it useful to try Review Question 2.3 at the

end of the chapter

In setting up a break-even analysis we need to make several itions and assumptions First we assume that there are two types of cost,fixed and variable Fixed costs, as the name implies, are those costs thatare constant whatever the level of production These might be the costs

defin-of setting up the operation such as the purchase defin-of machinery as well asexpenses, such as business rates, that do not vary with the level of out-put Variable costs on the other hand are costs that change in relation

to the amount produced, such as the costs of raw materials and labour

We can define the total costs (TC) as the sum of the total fixed costs(TFC) and the total variable costs (TVC):

TC TFC  TVCThe total variable costs depend on the quantity of output We willassume that the variable cost of producing an extra unit is the samehowever many units we produce; in other words, it is linear or varies in

a straight line with the amount produced We can therefore expressthe total variable cost as the variable cost per unit produced, known asthe average variable cost (AVC) multiplied by the quantity produced(Q), so the total cost is:

TC TFC  AVC * QThe total revenue (TR) is the price per unit (P) at which the output

is sold multiplied by the quantity of output (Q):

TR P * QOnce we have defined the total cost and total revenue equations wecan plot them on a graph and look at exactly how total revenue com-pares to total cost This is a key comparison as the total revenue minusthe total cost is the amount of profit made:

Profit TR  TCThe point at which the lines representing the two equations cross isthe point at which total cost is precisely equal to total revenue, thebreak-even point

Example 2.11

The Ackrana Security Company intends to manufacture video security cameras Thecosts of acquiring the necessary plant and machinery and meeting other fixed costs areput at £4.5 million The average variable cost of producing one of their cameras is esti-mated to be £60 and the company plans to sell them at £150 each How many will theyneed to produce and sell in order to break even?

Total cost, TC 4500000  60QTotal revenue, TR 150Q

These equations are plotted in Figure 2.11 Conventionally the money amounts, cost

and revenue are plotted on the vertical or y axis and the output is plotted on the zontal or x axis This arrangement reflects the assumption that the money amounts

hori-depend on the output and makes it easier to interpret the diagram

In Figure 2.11 the steeper line that starts from the origin represents the total revenueequation and the other line represents the total cost equation You can see that the linescross when output is about 50,000 units At this level of production both the total costand total revenue are equal, at about £7.5 million This is the break-even point, at whichcosts precisely match revenues.

We can verify the break-even point by solving the total cost and total revenue tions simultaneously:

equa-When total cost and total revenue are equal, subtracting one from the other will leave

us with an expression in which the only unknown is the level of output, Q:

Dividing both sides by 90 means that the level of output at which total cost and totalrevenue are equal is 50,000:

4500000/90 50000The total cost and total revenue when 50000 units are produced will be:

Break-even analysis can be extended to illustrate the levels of outputthat will yield a loss and those that will yield a profit A level of outputless than the break-even level, and hence to the left of the position ofthe break-even point along the horizontal axis of the graph, will result

in a loss A level of output higher than the break-even level, to the right

of the break-even point on the horizontal axis, will yield a profit

At any point to the left of the break-even point the total cost line isthe higher line indicating that total cost is higher than total revenue;the greater the difference between the two lines, the larger the loss Atany point to the right of the break-even point the total revenue is thehigher line, which means that the total revenue is higher than the totalcost; the bigger the difference between the two lines, the larger theprofit The areas representing loss and profit are shown in Figure 2.12.Using Figure 2.12 you can establish how much profit or loss will beachieved at a particular level of production If for instance productionwere 30,000 units the graph suggests that the total cost would be £6.3million and the total revenue would be £4.5 million resulting in a loss

of £1.8 million

We expect that a company would seek to operate at a level of duction at which they would make a profit The difference between the

pro-output they intend to produce, their budgeted pro-output, and the break-even

level of output is their safety margin This can be expressed as a centage of the budgeted output to give a measure of the extent to whichthey can fall short of their budgeted output before making a loss

Loss

Profit

60 80 100 120 Output (000)

Example 2.12

If the Ackrana Security Company in Example 2.11 aims to produce 80,000 cameraswhat profit should they expect and what is their safety margin?

The break-even analysis we have considered is the simplest case, whereboth costs and revenue are assumed to be linear, that is to form straightlines when plotted graphically In practice companies might find thatwith greater levels of production come economies of scale that meantheir variable cost per unit is not constant for every unit produced butfalls as output increases Furthermore, they may have to reduce theirprice if they want to sell more products so that their total revenuewould not have a linear relationship to their output level.

Despite these shortcomings the basic model can be a useful guide tothe consequences of relatively modest changes in output as well as

a framework for considering different levels of initial investment, pricing strategies and alternative sources of raw materials

At this point you may find it useful to try Review Questions 2.4 to 2.9

at the end of the chapter

2.4 Inequalities

So far we have concentrated on the use of equations to model business

situations Under some circumstances it is appropriate to use

inequal-ities, also known as inequations, expressions of relationships in which

one or more unknowns are not necessarily equal to a specific ical value

numer-The basic forms of inequalities are not equal (), less than ( ), and

greater than (

x x x



x x x

In Example 2.11 we found that their break-even point was 50,000 cameras so theirsafety margin is:

budgeted output break-even output

It may help you to distinguish betweensharp end of each symbol as pointing to the lesser quantity, and the

open end to the greater In x 10 the sharp end points the x so it is the smaller quantity and 10 is the larger quantity whereas in x

sharp end points to 10 so that is the smaller quantity and x is the larger.

There are two types of composite inequality that you will meet later

in this chapter These are less than or equal to () and greater than or

constraint on the business activity An inequality can be used to

repre-sent the relationship between the amounts of the resource requiredfor the products and the quantity in stock so that we can tell whichcombinations of products or services can be produced with the given

materials, in other words what output levels are feasible.

Example 2.13

The Sirdaria Citrus Company produces juices using exotic fruits including two madefrom hoormah, ‘Anelle’ and ‘Emir’ Fruit concentrate for these products is importedand the supply is erratic Sirdaria has 4000 litres of hoormah concentrate in stock The Anelle brand consists of 8% concentrate Emir, the luxury product, consists of 10%concentrate

Represent the fruit concentrate constraint as an inequality

Since we don’t know exactly how much of each product can be produced, indeed wewant an inequality to identify what the possibilities are, we must start by defining the key

variables: the amounts of each product produced We will use x to represent the amount of Anelle produced and y to represent the amount of Emir produced.

Anelle requires 0.08 litres of concentrate per litre, so if we produce x litres of Anelle

we will need 0.08x litres of concentrate Emir requires 0.1 litres of concentrate per litre

so producing y litres of Emir will use up 0.1y litres of concentrate Whatever the volume

of Anelle and Emir produced the amount of concentrate needed will be the amountrequired for Anelle production added to the amount required for Emir production:

Concentrate required 0.08x  0.1y

Inequalities, like equations, represent the connection or relationshipbetween amounts either side of the appropriate sign Like equationsthey can also be represented graphically, but whereas the graphicalform of an equation is a line the graphical form of an inequality is anarea bounded or limited by a line To portray an inequality graphicallyyou have to start by plotting the line that bounds the area If theinequality is a composite type the line to plot is the line for the equa-tion that represents the limits of feasibility, output combinations thatuse up all of the available resource.

The concentrate required must be balanced against the available supply of 4000litres To be feasible the output of products must not give rise to a demand for concen-trate that exceeds the available supply, in other words the demand must be less than orequal to the available supply:

0.08x  0.1y 4000

Example 2.14

Show the inequality in Example 2.13 in graphical form

If all the hoormah concentrate that the Sirdaria Citrus Company have in stock is usedthen the amounts of concentrate required for Anelle and Emir production will equal

4000 litres:

0.08x  0.1y  4000

The best way to plot this equation is to work out how many litres of each productcould be produced using this amount of concentrate if none of the other product weremade You can work out how many litres of Anelle could be produced from 4000 litres

of concentrate by dividing 4000 by the amount of concentrate needed for each litre ofAnelle, 0.08 litres:

4000/0.08 50,000

So if only Anelle were made, 50,000 litres could be produced using the 4000 litres ofconcentrate, but there would be no concentrate remaining for Emir production Thepoint on the graph representing 50,000 litres of Anelle and zero litres of Emir is the

intercept of the line of the equation on the x axis.

If all the concentrate were committed to the manufacture of Emir the number oflitres produced would be:

4000/0.1 40,000

At this level of output there would be no concentrate available for Anelle production.The point on the graph representing 40,000 litres of Emir and zero litres of Anelle is

the intercept of the line on the y axis.

The line can be plotted using these two points

Each point on the line plotted in Figure 2.13 represents a combination of Anelle andEmir output that would use up the entire stock of concentrate It is of course feasible toproduce output combinations that require a lesser amount of concentrate than 4000litres, for instance producing 20,000 litres of each product would require only 3600litres of concentrate We can confirm this by putting these output levels in the expres-sion representing the concentrate requirement in Example 2.13:

Concentrate required 0.08x  0.1y

For 20,000 litres of both products:

Concentrate required 0.08 * 20000  01 * 20000  1600  2000  3600Look at Figure 2.14 and you can see that the point representing this combination liesbelow the line

The line represents all production combinations that use precisely 4000 litres of centrate All the points below it represent combinations that require less than 4000litres of concentrate All the points above it represent combinations that require morethan 4000 litres of concentrate and are therefore not feasible An example of this is themanufacture of 30,000 litres of each product, which would require:

con-0.08 * 30000 0.1 * 30000  2400  3000  5400

0 10 20 30 40 50 60

Clearly 5400 litres of concentrate is considerably higher than the available stock of 4000

so it is simply not possible to produce these quantities

The graphical representation of the inequality that expresses the constraint is fore the line that defines the limits to the production possibilities and the area under-neath it In Figure 2.15 the shaded area represents the output combinations that arefeasible given the amount of concentrate available

there-0 10 20 30 40 50 60

The constraint analysed in Examples 2.13 and 2.14 is an example of

a less than or equal to form of inequality Other constraints might take a

greater than or equal to form of inequality.

Example 2.15

The Sirdaria Citrus Company has a contractual obligation to produce 10,000 litres ofAnelle for an important customer

This is a constraint on production because it obliges Sirdaria to produce at least

10,000 litres of Anelle If x represents the litres of Anelle produced then the inequality for this constraint is that x must be greater than or equal to 10,000 litres:

x 10000Given this limitation any output mix is feasible if it involves producing 10,000 ormore litres of Anelle Were they to produce only 8000 litres, for instance, they wouldhave insufficient to fulfil their commitment to the customer

To represent this constraint on a graph we need to plot the limit to the constraint

(x 10000) and identify which side of it represents production combinations that arefeasible

In Figure 2.16 the vertical line represents the lower limit on Anelle production ing from the contractual commitment Any point to its left would result in too little

result-0 10 20 30 40 50 60

... Figure 2. 16 the vertical line represents the lower limit on Anelle production ing from the contractual commitment Any point to its left would result in too little

result-0 10 20 30... if it involves producing 10,000 ormore litres of Anelle Were they to produce only 8000 litres, for instance, they wouldhave insufficient to fulfil their commitment to the customer

To represent