
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: $\displaystyle r(x)c(x)=.00001x^3.0021x^2+20x360$
$\displaystyle p'(x)=.00003x^2.0042x+20$.
The answer is 749 units.
I cannot seem to factor p' to get this result.
I have tried completing the square.
$\displaystyle x^2+140x\frac{20}{.00003}=0$
$\displaystyle (x+70)^2=70^2+\frac{20}{.00003}$
$\displaystyle x=70\pm \sqrt{70^2+\frac{20}{.00003}}$
Calculating that out doesn't get me near 749.
I make it $\displaystyle x\approx{3431.3998}$
Thanks
Craig.

$\displaystyle p'(x)=.00003x^2.0042x+20=0$
$\displaystyle 0=3x^2+420x2000000$
$\displaystyle x=\frac{420\pm\sqrt{420^2+24000000}}{6}$
$\displaystyle x=\frac{420\pm4916.95...}{6}$
Gives 749.491...

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?