# 022 Exam 1 Sample A, Problem 5

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Find the marginal revenue and marginal profit at ${\displaystyle x=4}$, given the demand function

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

and the cost function

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

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

Foundations:
Recall that the demand function, ${\displaystyle p(x)}$, relates the price per unit ${\displaystyle p}$ to the number of units sold, ${\displaystyle x}$.

Moreover, we have several important important functions:

• ${\displaystyle C(x)}$, the total cost to produce ${\displaystyle x}$ units;
• ${\displaystyle R(x)}$, the total revenue (or gross receipts) from producing ${\displaystyle x}$ units;
• ${\displaystyle P(x)}$, the total profit from producing ${\displaystyle x}$ units.
In particular, we have the relations
${\displaystyle P(x)=R(x)-C(x),}$
and
${\displaystyle R(x)=x\cdot p(x).}$
Finally, the marginal profit at ${\displaystyle x_{0}}$ units is defined to be the effective profit of the next unit produced, and is precisely ${\displaystyle P'(x_{0})}$. Similarly, the marginal revenue or marginal cost would be ${\displaystyle R'(x_{0})}$ or ${\displaystyle C'(x_{0})}$, respectively.

Solution:

Step 1:
Find the Important Functions: We have
${\displaystyle R(x)\,\,=\,\,x\cdot p(x)\,\,=\,\,x\cdot {\frac {200}{\sqrt {x}}}\,\,=\,\,200{\sqrt {x}}.}$
From this,
${\displaystyle P(x)\,\,=\,\,R(x)-C(x)\,\,=\,\,200{\sqrt {x}}-\left(100+15x+3x^{2}\right).}$
Step 2:
Find the Marginal Revenue and Profit: The equation for marginal revenue is
${\displaystyle R'(x)\,\,=\,\,\left(200{\sqrt {x}}\right)'\,\,=\,\,200\cdot {\frac {1}{2}}\cdot {\frac {1}{\sqrt {x}}}\,\,=\,\,{\frac {100}{\sqrt {x}}},}$
while the equation for marginal profit is
${\displaystyle P'(x)\,\,=\,\,R'(x)-C'(x)\,\,=\,\,{\frac {100}{\sqrt {x}}}-(15+6x).}$
At ${\displaystyle x=4}$, we find the marginal revenue is
${\displaystyle R'(4)\,\,=\,\,{\frac {100}{\sqrt {4}}}\,\,=\,\,50.}$
On the other hand, marginal profit is
${\displaystyle P'(4)\,\,=\,\,{\frac {100}{\sqrt {4}}}-(15+6(4))\,\,=\,\,50-39\,\,=\,\,11.}$
Thus, it is profitable to produce another item.
${\displaystyle R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,11.}$