## Calculus: Early Transcendentals 8th Edition

The box should be $40\times 40 \times 20$ to minimize the volume.
Let $b$ be the length of the base and $h$ the height. The volume is $32,000=b^{2}h$, so $h=\frac{32,000}{b^{2}}$. The surface area of the open box is $S=b^{2}+4hb=b^{2}+4\frac{32,000}{b}$. $S'(b)=2b-\frac{128,000}{b^{2}}$, so $S'(b)$ is $0$ when $b=40$. This gives an absolute minimum, so the two bases should be $40$ and the height should be $20$.