# 022 Exam 2 Sample A, Problem 7

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

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

Moreover, we have several important important functions:

• ${\displaystyle C(x)}$, the total cost to produce ${\displaystyle x}$ units;
• ${\displaystyle R(x)}$, the total revenue (or gross receipts) from producing ${\displaystyle x}$ units;
• ${\displaystyle P(x)}$, the total profit from producing ${\displaystyle x}$ units.
In particular, we have the relations
${\displaystyle P(x)=R(x)-C(x),}$
and
${\displaystyle R(x)=x\cdot p(x).}$
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
${\displaystyle R(x)\,=\,x\cdot p(x)\,=\,x\cdot (90-3x)\,=\,90x-3x^{2}.}$
From this,
${\displaystyle P(x)\,=\,R(x)-C(x)\,=\,90x-3x^{2}-\left(200-30x+x^{2}\right)\,=\,120x-4x^{2}-200.}$
Step 2:
Find the Maximum: The equation for marginal revenue is
${\displaystyle P(x)\,=\,120x-4x^{2}-200.}$
Applying our power rule to each term, we find
${\displaystyle P'(x)\,=\,120-8x\,=\,8(15-x).}$
The only root of this occurs at ${\displaystyle x=15}$, and this is our production level to achieve maximum profit.