Maximizing the volume of an open-top box

The volume is $V(x) = x(12-2x)^2$ since the base has area $(12-2x)^2$ and the height is $x$. Physically, we need $x>0$ and $12-2x>0$, so the domain is $0<x<6$. Expanding, $V(x) = 144x - 48x^2 + 4x^3$.

Differentiating gives $V'(x) = 144 - 96x + 12x^2 = 12(x^2 - 8x + 12) = 12(x-2)(x-6)$. Setting $V'(x)=0$ gives $x=2$ or $x=6$, but only $x=2$ lies in the open interval $(0,6)$. Since $V'(x)>0$ for $x<2$ and $V'(x)<0$ for $2<x<6$, $x=2$ gives a maximum, and $V(2) = 2(12-4)^2 = 2(64) = 128$ cubic inches.

The critical points were x equals two and x equals six. The valid domain required x to be strictly between zero and six, since the cardboard side is twelve inches. Since x equals six is not strictly less than six, it lies outside the valid domain. Therefore x equals six is the excluded critical point.