# 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.
$R'(4)\,\,=\,\,50;\qquad P'(4)\,\,=\,\,11.$ 