Minimizing surface area of an open box with square base

The surface area of the open box consists of the square base plus four rectangular sides: $S = x^2 + 4xh$. From the volume constraint $x^2h = 32000$, solve for $h$ to get $h = \frac{32000}{x^2}$. Substituting this into $S$ gives $S(x) = x^2 + 4x\left(\frac{32000}{x^2}\right) = x^2 + \frac{128000}{x}$, valid for $x>0$.

Differentiate: $S'(x) = 2x - \frac{128000}{x^2}$. Setting $S'(x)=0$ gives $2x = \frac{128000}{x^2}$, so $x^3 = 64000$, which gives $x = 40$. Since $S''(x) = 2 + \frac{256000}{x^3} > 0$ for all $x>0$, this critical point is a minimum. Then $h = \frac{32000}{40^2} = 20$, and the minimum surface area is $S(40) = 40^2 + \frac{128000}{40} = 1600 + 3200 = 4800\text{ cm}^2$.