A system of equations is also used in the analysis of profits for different volumes of business. As an example, consider cost and revenue for the poster example discussed previously. Suppose the printing business decided that it should charge $0.17 per poster. Then it would have a revenue function:

$R=\$0.17x$

Together with the cost function, this forms a system of equations:

$C=\$80+\$0.11x\text{and}R=\$0.17x$

You have already seen a table for the cost function. Now prepare a short table for revenue, then graph both functions on the same axes, as follows:

To analyze this system for business purposes, you should use the relationship (from Chapter 1):

$Profit=Revenue-Cost$

You can see from the graph that the printing business would lose money at those values of x (the low values) for which the cost function is above the revenue function. And it would make a profit at those values of x (the high values) where revenue exceeds the cost.

As in most systems of equations, the point at which all the equations are satisfied – the solution of the equations – is of particular interest. In the example above, it is the place at which,

$Revenue=Cost$

That is, the point at which the profit is zero – the break-even point.

The break-even point can be estimated from the graph, but it can also be calculated from the equations in the system.

To find this point, set:

$Revenue=Cost$
$\$0.17x=\$80+\$0.11x$

So, by subtracting 0.11p from both sides, you get:

$\$0.17x-\$0.11x=\$80+\$0.11x-\$0.11x$

And:

$\$0.06p=\$80$

So

$x=1333.33$

But we cannot make a fraction of a poster! So we always round up.

x = 1,334 posters (rounded up)

This analysis shows the printing company that small orders lose money at $0.17 per poster. In real life such companies cannot accept orders on which they lose money; they will either set a minimum order size (say, 1,500) or charge proportionately more for small orders (for example, $100 for the first 100 posters).

As long as the revenue and cost functions are known, you can evaluate what the profits will be at any level of activity. Using the revenue function above $R=\$0.17x$,

$Profit=\$0.17x-(\$80+\$0.11x)=\$0.06x-\$80$

so that for, say, 2,000 posters:

$Profit=\$0.06\times 2,000-\$80=\$40.00$

The above example was for a job in a business. To analyze the relationship between profits and volume for an entire business or product line, do the analysis for a time period, such as a month or a year. As before, the fixed costs are those that do not depend on the level of activity. They include:

• rent
• property taxes
• salaries of employees who will be employed regardless of the level of activity (for instance, managers, sales clerks)
• utilities such as heat and light

The variable costs include:

• the costs of materials
• labor costs which are the result of the level of production (called direct labor costs)
• commissions on sales

Example 2.6.1

Suppose a manufacturer of sports bags examined its accounts and found the following costs.

Thus the cost equation would be

$C=\$70,000+\$7x$

where

C = cost
x = number of bags produced in a month

This manufacturer sells the bags for $12 per bag. Hence the revenue equation is

$R=\$12x$

if we assume all bags produced are sold in the same month as they are produced.

The monthly profit would then be

$Profit=R-C=\$12x-(\$70,000+\$7x)$

$=\$5x-\$70,000$

To break even would require:

$0=\$5x-\$70,000$

$\$5x=\$70,000$

So

x = 14,000 bags (produced and sold)

Below 14,000 bags the company would show a loss; above 14,000 bags it would show a profit. You can find the level required for any given profit by solving for x with that profit. For example, to earn a $10,000 profit:

$\$5x-\$70,000=\$10,000$

thus

$x=16,000\text{bags}$

Knowledge Check 2.4