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 (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
${\overline {C}}(x)\,=\,{\frac {C(x)}{x}}.$
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 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'(7)\,=\,116-6x\bigg|_{x=7}\,=\,116-6(7)\,=\,74.}

(b):

We have that the average cost 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 \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 (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)\,=\,1-\frac{64}{x^{2}}\,=\,\frac{x^{2}-64}{x^{2}}\,=\,\frac{(x-8)(+8)}{x^{2}}.} This has a single positive root 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=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 (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 \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 (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 0\,=\,P'(x)\,=\,-8x+136\,=\,-8(x-17),} which has a root 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=17} . Plugging this into our function for profit, we have 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(17)\,=\,-4(17)^{2}+136(17)+64\,=\,1220.}

Final Answer:
(a) 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 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 74} .
(b) The minimum average cost occurs 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 8} units.
(c) The maximum profit 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 1220} occurs 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 17} units.
Note that monetary units were not provided in the statement of the problem.

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