Maximizing cone volume from a circular sector

Since $V(r) \geq 0$, maximizing $V(r)$ is equivalent to maximizing $g(r) = V(r)^2 = \frac{\pi^2}{9} r^4(144 - r^2) = \frac{\pi^2}{9}(144r^4 - r^6)$. Differentiating with respect to $r$: $g'(r) = \frac{\pi^2}{9}(576r^3 - 6r^5) = \frac{\pi^2}{9} r^3(576 - 6r^2)$. Setting $g'(r) = 0$ gives $r = 0$ (excluded, endpoint) or $576 - 6r^2 = 0$, so $r^2 = 96$, giving $r = \sqrt{96} = 4\sqrt{6} \approx 9.80$ cm, which lies in $(0,12)$.

Checking that this is a maximum: $g'(r) > 0$ for $r$ slightly less than $4\sqrt{6}$ and $g'(r) < 0$ for $r$ slightly greater, confirming a maximum (also $V\to 0$ at both endpoints $r=0$ and $r=12$). Substituting $r^2 = 96$ into $V(r)$: $\sqrt{144 - 96} = \sqrt{48} = 4\sqrt{3}$, so $V = \frac{\pi}{3}(96)(4\sqrt{3}) = 128\sqrt{3}\,\pi \approx 696.5$ cm$^3$.