I'm not 100% sure on my maths for this question, so I figured I'd double check it here.
A box with a square base and open top must have a volume of $32,000 cm^3$. Find the dimension of the box that minimize the amount of material used.
The constraint is that $V=32,000cm^3$, the size of the box will be $=x^2+4xy$
Putting the size into one variable will become, $=x^2+4x(\frac{32000}{x^2})$
Calculating the derivative will result in $2x-\frac{128000}{x^2}$
The critical points...
$2x-\frac{128000}{x^2} = 0$
$\frac{-128000}{x^2}=-2x$
$\frac{128000}{2}=\frac{2x^3}{2}$
$64000=x^3$
$40=x$
$f(x) = x^2+4x(\frac{32000}{x^2})$
$f(40) = 40^2+4(40)(\frac{32000}{40^2})$
$=1600+160(\frac{32000}{1600})$
$=1600+3200$
$=4800$
Did I do this properly? I'm not sure as I've always struggled with these problems, if not can someone point out what was wrong? Thanks!