1. ## Maximum profit

find the price per unit p that produces the max profit P.

Cost function is C = 35x + 500
Demand Function is p = 50 - 0.1(sqrt x)

2. Originally Posted by meli3000
find the price per unit p that produces the max profit P.

Cost function is C = 35x + 500
Demand Function is p = 50 - 0.1(sqrt x)

So, you have a Cost function $C = 35x + 500$ and a Demand function $p = 50 - 0.1\sqrt x = 50 - \frac{{\sqrt x }}{{10}}$.

Find a function of Total revenue $(TR)$

$TR = x \cdot p = x\left( {50 - \frac{{\sqrt x }}{{10}}} \right) = 50x - \frac{{{{\left( {\sqrt x } \right)}^3}}}{{10}}.$

Find a Profit-function $(P)$

$P = TR - C = 50x - \frac{{{{\left( {\sqrt x } \right)}^3}}}{{10}} - \left( {35x + 500} \right) = 15x - \frac{{{{\left( {\sqrt x } \right)}^3}}}{{10}} - 500.$

Now find the derivative of the Profit-function $(P)$ and $\frac{{dP}}{{dx}} = 0$ and find $x$

$\frac{{dP}}{{dx}} = \frac{d}{{dx}}\left( {15x - \frac{{{{\left( {\sqrt x } \right)}^3}}}{{10}} - 500} \right) = 15 - \frac{{3\sqrt x }}{{20}} = 0 \Leftrightarrow \sqrt x = 100 \Leftrightarrow \boxed{x = 10000}$

${P_{\max }} = P\left( {10000} \right) = 15 \cdot 10000 - \frac{{{{\left( {\sqrt {10000} } \right)}^3}}}{{10}} - 500 = 49500$

Finally, find the price per unit that produces the max profit

$p = p\left( {10000} \right) = 50 - 0.1\sqrt {10000} = 50 - \frac{{\sqrt {10000} }}{{10}} = \boxed{40}$