# Difference between revisions of "022 Sample Final A, Problem 9"

Given demand ${\displaystyle p=116-3x}$, and cost  ${\displaystyle C=x^{2}+20x+64}$, 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, ${\displaystyle p(x)}$, 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;
• ${\displaystyle {\overline {C}}(x)}$, the average cost of producing ${\displaystyle x}$ units.
In particular, we have the relations
${\displaystyle P(x)\,=\,R(x)-C(x),}$
while
${\displaystyle R(x)\,=\,x\cdot p(x),}$
and
${\displaystyle {\overline {C}}(x)\,=\,{\frac {C(x)}{x}}.}$
The marginal profit at ${\displaystyle x_{0}}$ units is defined to be the effective profit of the next unit produced, and is precisely ${\displaystyle P'(x_{0})}$. Similarly, the marginal revenue or marginal cost would be ${\displaystyle R'(x_{0})}$ or  ${\displaystyle C'(x_{0})}$, 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
${\displaystyle R(x)\,=\,x\cdot p(x)\,=\,x(116-3x)\,=\,116x-3x^{2}}$.

Thus, the marginal revenue at a production level of ${\displaystyle 7}$ units is simply

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

Our first derivative is then

${\displaystyle {\overline {C}}\,'(x)\,=\,1-{\frac {64}{x^{2}}}\,=\,{\frac {x^{2}-64}{x^{2}}}\,=\,{\frac {(x-8)(x+8)}{x^{2}}}.}$

This has a single positive root at ${\displaystyle x=8}$, which will correspond to the minimum average cost.

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

To find the maximum value, we need to find a root of the derivative:

${\displaystyle 0\,=\,P'(x)\,=\,-8x+96\,=\,-8(x-12),}$

which has a root at ${\displaystyle x=12}$. Plugging this into our function for profit, we have

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