Maximum revenue problem.

**craigmain** · March 8th 2009, 01:06 AM

Hi,

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

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

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

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

The answer is 749 units.

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

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

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

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

Calculating that out doesn't get me near 749.

I make it $x\approx{3431.3998}$

Thanks
Craig.