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Exercises Systems of equations

Example 10 from Algebra and Trigonometry, 11.1 Systems of Linear Equations: Two Variables

Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function $C (x) = 0.85 x + 35,000$ and the revenue function $R (x) = 1.55 x,$ find the break-even point and the profit function.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Write the system of equations using $y$ to replace function notation.
$\begin{array}{l} \begin{array}{l} y = 0.85 x + 35,000 \end{array} \\ y = 1.55 x \end{array}$
Substitute the expression $0.85 x + 35,000$ from the first equation into the second equation and solve for $x .$
$\begin{matrix} 0.85 x + 35,000 = 1.55 x \\ 35,000 = 0.7 x \\ 50,000 = x \end{matrix}$
Then, we substitute $x = 50,000$ into either the cost function or the revenue function.
$1.55 (50,000) = 77,500$
The break-even point is $(50,000, 77,500) .$
The profit function is found using the formula $P (x) = R (x) - C (x) .$

$\begin{array}{l} P (x) = 1.55 x - (0.85 x + 35, 000) \\ = 0.7 x - 35, 000 \end{array}$
The profit function is $P (x) = 0.7 x −35,000.$

How did it go?

PreviousFinding a Solution to a Dependent System of Linear Equations Next Writing and Solving a System of Equations in Two Variables

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 11.1, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.