# Math Help - revenue/profit maximization

1. ## revenue/profit maximization

The selling price for q units of a product is given by the equation p = 250-25q. The cost of producing q unites of the product is:
C = 100 + 5q
a) what is the max revenue that can be generated?
b) what is the max profit that can be generated?

2. Hello, jamesk486!

You say you solved it.
But did you notice a small snag at the end?

The selling price for $x$ units of a product is: . $P(x) \:=\: 250-25x$

The cost of producing $x$ units of the product is: . $C(x) \:=\: 100 + 5x$

a) What is the max revenue that can be generated?

$R \;=\;x\cdot P(x) \:=\:x(250-25x) \;=\;250x - 25x^2$

Then: . $R' \,=\,0 \quad\Rightarrow\quad 250-50x\:=\:0 \quad\Rightarrow\quad x \:=\:5$

Max Revenue: . $R(5) \:=\:5(250-25\cdot5) \;=\;\625$

b) What is the max profit that can be generated?

$\text{Profit}\;=\;\text{Revenue} - \text{Cost} \;=\;x\cdot P(x) - C(x)$

. . . $P \;=\;x(250-25x) - (100 + 5x)$

. . . $P \;=\;-25x^2 + 245x - 100$

Then: . $P' \:=\:0 \quad\Rightarrow\quad -50x + 245 \:=\:0 \quad\Rightarrow\quad x \:=\:4.9$

Assuming we cannot manufacture nor sell a fraction of a unit,

. . we have: . $\begin{array}{ccccc}P(4) &=& -25(4^2) + 245(4) - 100 &=& \480 \\
P(5) &=& -25(5^2) + 245(5) - 100 &=& \500 \end{array}$

If 5 units are made and sold, the maximum profit is \$500.