# 022 Sample Final A, Problem 9

Given demand $p=116-3x$ , and cost  $C=x^{2}+20x+64$ , find:

a) Marginal revenue when x = 7 units.
b) The quantity (x-value) that produces minimum average cost.
c) Maximum profit (find both the x-value and the profit itself).
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;
• ${\overline {C}}(x)$ , the average cost of producing $x$ units.
In particular, we have the relations
$P(x)\,=\,R(x)-C(x),$ while
$R(x)\,=\,x\cdot p(x),$ and
${\overline {C}}(x)\,=\,{\frac {C(x)}{x}}.$ The marginal profit at $x_{0}$ units is defined to be the effective profit of the next unit produced, and is precisely $P'(x_{0})$ . Similarly, the marginal revenue or marginal cost would be $R'(x_{0})$ or  $C'(x_{0})$ , respectively.

On the other hand, any time they speak of minimizing or maximizing, we need to find a local extrema. These occur when the first derivative is zero.

Solution:

(a):
The revenue function is
$R(x)\,=\,x\cdot p(x)\,=\,x(116-3x)\,=\,116x-3x^{2}$ .

Thus, the marginal revenue at a production level of $7$ units is simply

$R'(7)\,=\,116-6x{\bigg |}_{x=7}\,=\,116-6(7)\,=\,74.$ (b):
We have that the average cost function is
${\begin{array}{rcl}{\overline {C}}(x)&=&{\frac {C(x)}{x}}}}\\\\&=&{\frac {x^{2}+20x+64}{x}}}}\\\\&=&x+20+{\frac {64}{x}}.}\end{array}}$ Our first derivative is then

${\overline {C}}\,'(x)\,=\,1-{\frac {64}{x^{2}}}\,=\,{\frac {x^{2}-64}{x^{2}}}\,=\,{\frac {(x-8)(x+8)}{x^{2}}}.$ This has a single positive root at $x=8$ , which will correspond to the minimum average cost.

(c):
First, we find the equation for profit. Using part of (a), we have
${\begin{array}{rcl}P(x)&=&R(x)-C(x)}}\\\\&=&116x-3x^{2}-(x^{2}+20x+64)\\\\&=&-4x^{2}+96x-64.\end{array}}$ To find the maximum value, we need to find a root of the derivative:

$0\,=\,P'(x)\,=\,-8x+96\,=\,-8(x-12),$ which has a root at $x=12$ . Plugging this into our function for profit, we have

$P(12)\,=\,-4(12)^{2}+96(12)-64\,=\,512.$ (a) The marginal revenue at a production level of $7$ units is $74$ .
(b) The minimum average cost occurs at a production level of $8$ units.
(c) The maximum profit of $512$ occurs at a production level of $12$ units.