Kitty litter box

Let the dimensions be $a,b,c$ , then the volume is $abc = 32$ and surface area $bc + 2ac +2ab$ . (note the top is removed)

Let $a = A', b = 2B', c = 2C'$ . then $A' B' C' = 8$

And we want to minimize $4(B' C' + A'C' + A'B')$ .

Apply AM-GM: $4(B' C' + A'C' + A'B') \geq 4\cdot 3 \cdot \sqrt[3]{(A'B'C')^2} = 4\cdot 3 \cdot 8^{2/3} =\boxed {48}$ .

I Just assumed that when abc = 32 , so a=4 , b=4 ,and c=2 .Then I get the minimum internal surface area . I feel stupid after seeing your complicated answers.

You can solve this by some inequality or other but I bashed my way through with my newfound powers of Langrange multipliers!!!11!1 Yeah, that wasn't as exciting as we might have expected.

Let the box have dimensions $x \times y \times z$ . Our constraint here is that $g = xyz-32 = 0$ and the function to be minimised is $f = 2xy+2yz+xz$ . Now, to minimise, we just set

$\nabla f = \lambda \nabla g$

for $\lambda$ some constant. Taking the partial derivatives of $f$ and $g$ , we obtain the three equations

$2y+z = \lambda yz; \quad 2x+2z = \lambda xz; \quad 2y+x = \lambda xy$

Combining those with our constraint $g$ we get a system of simultaneous equations. Solving it yields $2y = x = z = 4$ ; hence,

$\min f = \boxed{48}$