# Thread: optimization for a box

1. ## optimization for a box

Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.

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.

$V = mmn = m^2n$

Assuming the box has an open top, the surface area is

$m^2 + 4mn = 64$

Solve for n from the surface area, then plug the expressions into the volume formula.

$n = \frac{64-m^2}{4m}$

$V = m^2 \frac{64-m^2}{4m} = \frac{64m-m^3}{4}$

Find the first derivative of volume:
$V' = 8 - 0.75 m^2 = 0
$

Then solve for m.

The story changes if this is a closed top box.

Find the dimensions of a rectangular box of maximum volume if the total surface area is given as 64cm^2.
So the volume of a box is

$V=lwh$

Subject to the surface area

$S=2lw+2lh +2wh=64$

$F(l,w,h,\lambda) =lhw-\lambda(2lw+2lh+2wh-64)$

$\nabla F=0$

$hw-\lambda(2w+2h)=0$

$lh-\lambda(2l+2h)=0$

$lw-\lambda(2l+2w)=0$

$0=(2lw+2lh+2wh-64)$

Now multiply the first equation by l and the 2nd by -w and add them together

$lwh-\lambda(2lw+2lw)=0$
$-lwh+\lambda(2lw+2wh)=0$
$-2lw+2wh=0 \iff 2w(-l+h)=0 \iff l=h$

Doing something similar you can show that $w=l=h$

Plugging into the last equation we get.

$2l^2+2l^2+2l^2-64=0 \iff 6l^2=64 \iff l^2 =\frac{32}{3} \iff l=4 \sqrt{\frac{2}{3}}$

So the maximal volume will be a cube.

I hope this helps

4. Nice going, TheEmptySet... I was way off here. A calculus subforum for single-variable people like me? Just kidding.