Maximizing box volume from cut-corner squares

Since $x$ must be positive and the folded base side $12-2x$ must also be positive, the domain is $0 < x < 6$. Expanding, $V(x) = x(144 - 48x + 4x^2) = 4x^3 - 48x^2 + 144x$. Differentiating gives $V'(x) = 12x^2 - 96x + 144 = 12(x^2 - 8x + 12) = 12(x-2)(x-6)$.

Setting $V'(x) = 0$ gives $x = 2$ or $x = 6$, but $x=6$ is excluded from the domain since it makes the base side zero. Testing sign changes of $V'(x)$ around $x=2$ shows $V'$ goes from positive to negative, so $x=2$ is a local maximum. Since it's the only critical point in $(0,6)$, it gives the global maximum volume on that interval.

The critical point at six is excluded because it lies on the boundary of the domain, not strictly inside it. At x equals six, the folded base side becomes twelve minus twelve, which equals zero, so the box has no base and cannot physically exist. Since the domain requires x strictly between zero and six, this value is not a valid candidate. Only x equals two remains as the interior critical point, giving the true maximum volume.