# Maximum revenue problem.

• Mar 8th 2009, 01:06 AM
craigmain
Maximum revenue problem.
Hi,

I have revenue:
$\displaystyle r(x)=100x-.0001x^2$

and cost:
$\displaystyle c(x)=360+80x+.002x^2+.00001x^3$

I must maximise profit.
profit is: $\displaystyle r(x)-c(x)=-.00001x^3-.0021x^2+20x-360$

$\displaystyle p'(x)=-.00003x^2-.0042x+20$.

I cannot seem to factor p' to get this result.
I have tried completing the square.

$\displaystyle x^2+140x-\frac{20}{.00003}=0$

$\displaystyle (x+70)^2=70^2+\frac{20}{.00003}$

$\displaystyle x=-70\pm \sqrt{70^2+\frac{20}{.00003}}$

Calculating that out doesn't get me near 749.

I make it $\displaystyle x\approx{3431.3998}$

Thanks
Craig.
• Mar 8th 2009, 04:21 AM
Thomas154321
$\displaystyle p'(x)=-.00003x^2-.0042x+20=0$
$\displaystyle 0=3x^2+420x-2000000$
$\displaystyle x=\frac{-420\pm\sqrt{420^2+24000000}}{6}$
$\displaystyle x=\frac{-420\pm4916.95...}{6}$

Gives 749.491...
• Mar 8th 2009, 04:23 AM
Thomas154321
Might be going crazy but can't find the edit button suddenly. (Rofl) Anyway your answer is fine, perhaps you're typing it in wrong?