Maximizing volume of a box under a length-plus-girth constraint

From the constraint $h + 4x = 108$, solve for $h$ to get $h = 108 - 4x$. Substituting into the volume formula gives $V(x) = x^2(108-4x) = 108x^2 - 4x^3$, a function of the single variable $x$, defined for $0 < x < 27$ (since $h>0$ requires $x<27$, and $x>0$ for a physical box).

Differentiate: $V'(x) = 216x - 12x^2 = 12x(18 - x)$. Setting $V'(x)=0$ gives $x=0$ or $x=18$; only $x=18$ lies in the open domain. Since $V'(x)>0$ for $x<18$ and $V'(x)<0$ for $x>18$, this is a maximum. Then $h = 108 - 4(18) = 36$, and the maximum volume is $V = 18^2 \cdot 36 = 11664$ cubic inches.

At the optimum, the base side is eighteen inches and the height is thirty six inches. Dividing thirty six by eighteen gives two, so the height is exactly twice the base side length. This ratio comes directly from the critical point found by setting the derivative of the volume function equal to zero. The answer is two, or twice.