# 022 Exam 2 Sample A, Problem 7

Find the quantity that produces maximum profit, given the demand function $p\,=\,90-3x$ and cost function $C\,=\,200-30x+x^{2}$ .

Foundations:
Recall that the demand function, $p(x)$ , relates the price per unit $p$ to the number of units sold, $x$ .

Moreover, we have several important important functions:

• $C(x)$ , the total cost to produce $x$ units;
• $R(x)$ , the total revenue (or gross receipts) from producing $x$ units;
• $P(x)$ , the total profit from producing $x$ units.
In particular, we have the relations
$P(x)=R(x)-C(x),$ and
$R(x)=x\cdot p(x).$ Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.

Solution:

Step 1:
Find the Profit Function: We have
$R(x)\,=\,x\cdot p(x)\,=\,x\cdot (90-3x)\,=\,90x-3x^{2}.$ From this,
$P(x)\,=\,R(x)-C(x)\,=\,90x-3x^{2}-\left(200-30x+x^{2}\right)\,=\,120x-4x^{2}-200.$ Step 2:
Find the Maximum: The equation for marginal revenue is
$P(x)\,=\,120x-4x^{2}-200.$ Applying our power rule to each term, we find
$P'(x)\,=\,120-8x\,=\,8(15-x).$ The only root of this occurs at $x=15$ , and this is our production level to achieve maximum profit.