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:
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 C(x)} , the total cost to produce 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 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 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 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 P(x)} , the total profit from producing 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 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 \overline{C}(x)} , the average cost of producing 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 x} units.
In particular, we have the relations
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 P(x)\,=\,R(x)-C(x),}
while
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)\,=\,x\cdot p(x),}
and
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 \overline{C}(x)\,=\,\frac{C(x)}{x}.}
The marginal profit at 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 x_0} units is defined to be the effective profit of the next unit produced, and is precisely 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 P'(x_0)} . Similarly, the marginal revenue or marginal cost would be 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_0)} or  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 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 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)\,=\, x\cdot p(x)\,=\, x(116-3x)\,=\,116x-3x^{2}} . Thus, the marginal revenue at a production level of 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 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 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}+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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle P(12)\,=\,-4(12)^{2}+96(12)+64\,=\,640.}

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 $640$ 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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