Minimizing surface area of an open-top box

The surface area (no top) is $S = x^2 + 4xh$, where $x^2h=32$ gives $h = 32/x^2$. Substituting, $S(x) = x^2 + 4x\left(\dfrac{32}{x^2}\right) = x^2 + \dfrac{128}{x}$, defined for $x>0$ since it is a physical length.

Differentiating, $S'(x) = 2x - \dfrac{128}{x^2}$. Setting $S'(x)=0$ gives $2x^3 = 128$, so $x^3 = 64$, hence $x=4$. Since $S'(x)<0$ for $x<4$ and $S'(x)>0$ for $x>4$, this is a minimum.

Plugging $x=4$ back into $S(x)$: $S(4) = 4^2 + \dfrac{128}{4} = 16 + 32 = 48$. Thus the minimum surface area is $48$ ft$^2$, achieved when the base has side length $4$ ft (and height $h = 32/16 = 2$ ft).