Systems of Linear Equations: Two Variables: Apply It
Solve systems of equations by graphing.
Solve systems of equations algebraically.
Identify inconsistent and dependent systems of equations containing two variables.
Solve applied systems.
Using Systems of Equations to Investigate Profits
Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price.
The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The [latex]x[/latex] -axis represents quantity in hundreds of units. The y-axis represents either cost or revenue in hundreds of dollars.
The point at which the two lines intersect is called the break-even point. We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.
The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.
Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.
Write the system of equations using [latex]y[/latex] to replace function notation.
The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].
Analysis of the Solution
The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.
We see from the graph that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x-axis. Then, the graph emerges into positive y-values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.
The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?
Let c = the number of children and a = the number of adults in attendance.The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.
[latex]c+a=2,000[/latex]
The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.
[latex]25c+50a=70,000[/latex]
We now have a system of linear equations in two variables.
In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].
