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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\overline {C}}\,'(x)\,=\,1-{\frac {64}{x^{2}}}\,=\,{\frac {x^{2}-64}{x^{2}}}\,=\,{\frac {(x-8)(+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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}P(x)&=&{\displaystyle {\displaystyle R(x)-C(x)}}\\\\&=&116x-3x^{2}-(x^{2}+20x+64)\\\\&=&-4x^{2}+136x+64.\end{array}}} To find the maximum value, we need to find a root of the derivative: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 0\,=\,P'(x)\,=\,-8x+136\,=\,-8(x-17),} which has a root at Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle x=17} . Plugging this into our function for profit, we have $P(17)\,=\,-4(17)^{2}+136(17)+64\,=\,1220.$

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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 1220} occurs at a production level of Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 17} units.
Note that monetary units were not provided in the statement of the problem.

Return to Sample Exam