022 Exam 1 Sample A, Problem 5

Find the marginal revenue and marginal profit at $x=4$, given the demand function

$p={\frac {200}{\sqrt {x}}}$

and the cost function

$C(x)=100+15x+3x^{2}.$

Should the firm produce one more item under these conditions? Justify your answer.

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.
In particular, we have the relations
$P(x)=R(x)-C(x),$
and
$R(x)=x\cdot p(x).$
Finally, 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.

 Solution:

Step 1:
Find the Important Functions: We have
$R(x)\,\,=\,\,x\cdot p(x)\,\,=\,\,x\cdot {\frac {200}{\sqrt {x}}}\,\,=\,\,200{\sqrt {x}}.$
From this,
$P(x)\,\,=\,\,R(x)-C(x)\,\,=\,\,200{\sqrt {x}}-\left(100+15x+3x^{2}\right).$

Step 2:
Find the Marginal Revenue and Profit: The equation for marginal revenue is
$R'(x)\,\,=\,\,\left(200{\sqrt {x}}\right)'\,\,=\,\,200\cdot {\frac {1}{2}}\cdot {\frac {1}{\sqrt {x}}}\,\,=\,\,{\frac {100}{\sqrt {x}}},$
while the equation for marginal profit is
$P'(x)\,\,=\,\,R'(x)-C'(x)\,\,=\,\,{\frac {100}{\sqrt {x}}}-(15+6x).$
At $x=4$, we find the marginal revenue is
$R'(4)\,\,=\,\,{\frac {100}{\sqrt {4}}}\,\,=\,\,50.$
On the other hand, marginal profit is
$P'(4)\,\,=\,\,{\frac {100}{\sqrt {4}}}-(15+6(4))\,\,=\,\,50-39\,\,=\,\,11.$
Thus, it is profitable to produce another item.

Final Answer:
$R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,11.$
Thus, it is profitable to produce another item.

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