Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
Let the base be m and height be n. I am assuming the problem indicates square base is used.
$\displaystyle V = mmn = m^2n$
Assuming the box has an open top, the surface area is
$\displaystyle m^2 + 4mn = 64$
Solve for n from the surface area, then plug the expressions into the volume formula.
$\displaystyle n = \frac{64-m^2}{4m}$
$\displaystyle V = m^2 \frac{64-m^2}{4m} = \frac{64m-m^3}{4}$
Find the first derivative of volume:
$\displaystyle V' = 8 - 0.75 m^2 = 0
$
Then solve for m.
The story changes if this is a closed top box.
So the volume of a box is
$\displaystyle V=lwh$
Subject to the surface area
$\displaystyle S=2lw+2lh +2wh=64$
$\displaystyle F(l,w,h,\lambda) =lhw-\lambda(2lw+2lh+2wh-64)$
$\displaystyle \nabla F=0 $
$\displaystyle hw-\lambda(2w+2h)=0$
$\displaystyle lh-\lambda(2l+2h)=0$
$\displaystyle lw-\lambda(2l+2w)=0$
$\displaystyle 0=(2lw+2lh+2wh-64)$
Now multiply the first equation by l and the 2nd by -w and add them together
$\displaystyle lwh-\lambda(2lw+2lw)=0$
$\displaystyle -lwh+\lambda(2lw+2wh)=0$
$\displaystyle -2lw+2wh=0 \iff 2w(-l+h)=0 \iff l=h$
Doing something similar you can show that $\displaystyle w=l=h$
Plugging into the last equation we get.
$\displaystyle 2l^2+2l^2+2l^2-64=0 \iff 6l^2=64 \iff l^2 =\frac{32}{3} \iff l=4 \sqrt{\frac{2}{3}}$
So the maximal volume will be a cube.
I hope this helps