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 ${\begin{array}{rcl}{\overline {C}}(x)&=&{\displaystyle {\displaystyle {\frac {C(x)}{x}}}}\\\\&=&{\displaystyle {\displaystyle {\frac {x^{2}+20x+64}{x}}}}\\\\&=&{\displaystyle x+20+{\frac {64}{x}}.}\end{array}}$ Our first derivative is then ${\overline {C}}\,'(x)\,=\,1-{\frac {64}{x^{2}}}\,=\,{\frac {x^{2}-64}{x^{2}}}\,=\,{\frac {(x-8)(x+8)}{x^{2}}}.$ 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 ${\begin{array}{rcl}P(x)&=&{\displaystyle {\displaystyle R(x)-C(x)}}\\\\&=&116x-3x^{2}-(x^{2}+20x+64)\\\\&=&-4x^{2}+96x-64.\end{array}}$ To find the maximum value, we need to find a root of the derivative: $0\,=\,P'(x)\,=\,-8x+96\,=\,-8(x-12),$ which has a root at $x=12$. Plugging this into our function for profit, we have $P(12)\,=\,-4(12)^{2}+96(12)-64\,=\,512.$

Final Answer:
(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.
Note that monetary units were not provided in the statement of the problem.

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