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; Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Final Answer:
Maximum profit occurs when we produce 15 items.

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