Maximizing the volume of an open-top box with fixed surface area

The volume to maximize is $V = (\text{length})(\text{width})(\text{height}) = (2w)(w)(h) = 2w^2h$. The surface area consists of the base $2w\cdot w = 2w^2$ and four sides: two of area $wh$ and two of area $2wh$, giving the constraint $2w^2 + 2(wh) + 2(2wh) = 2w^2 + 6wh = 108.$ Solving for $h$: $h = \dfrac{108 - 2w^2}{6w} = \dfrac{18}{w} - \dfrac{w}{3}$, valid for $0 < w < \sqrt{54}$ so that $h>0$.

Substitute into $V$: $V(w) = 2w^2\left(\frac{18}{w} - \frac{w}{3}\right) = 36w - \frac{2}{3}w^3.$ Differentiate: $V'(w) = 36 - 2w^2$, which is zero when $w^2 = 18$, i.e. $w = \sqrt{18} = 3\sqrt2$ (taking the positive root). Since $V''(w) = -4w < 0$ for $w>0$, this critical point is a maximum.

Compute $h$ at $w = 3\sqrt2$: $h = \dfrac{18}{3\sqrt2} - \dfrac{3\sqrt2}{3} = \sqrt2\cdot 3 - \sqrt2 = 2\sqrt2$. So the base is $w = 3\sqrt2$ ft by $2w = 6\sqrt2$ ft, and the height is $h = 2\sqrt2$ ft. The maximum volume is $V = 2w^2h = 2(18)(2\sqrt2) = 72\sqrt2 \approx 101.8$ ft$^3$.

The optimal width was three times the square root of two feet, approximately four point two four feet, while the height was two times the square root of two feet, approximately two point eight three feet. Since the height is smaller than the width, the box is shorter than it is wide. This can also be seen directly from the ratio of height to width, which equals two thirds, a value less than one.