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Latest revision as of 07:09, 15 May 2015
Find the quantity that produces maximum profit, given the demand function $p\,=\,903x$ and cost function $C\,=\,20030x+x^{2}$.
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

 $R(x)=x\cdot p(x).$

Using these equations, we can find the maximizing production level by determining when the first derivative of profit is zero.

Solution:
Step 1:

Find the Profit Function: We have

 $R(x)\,=\,x\cdot p(x)\,=\,x\cdot (903x)\,=\,90x3x^{2}.$

From this,

 $P(x)\,=\,R(x)C(x)\,=\,90x3x^{2}\left(20030x+x^{2}\right)\,=\,120x4x^{2}200.$

Step 2:

Find the Maximum: The equation for marginal revenue is

 $P(x)\,=\,120x4x^{2}200.$

Applying our power rule to each term, we find

 $P'(x)\,=\,1208x\,=\,8(15x).$

The only root of this occurs at $x=15$, and this is our production level to achieve maximum profit.


Final Answer:

Maximum profit occurs when we produce 15 items.

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