Application of Derivatives: Maxima and Minima Word Problems

Let the cone have radius $r$, height $h$, slant height $l = \sqrt{r^2 + h^2}$, and semi-vertical angle $\alpha$. The total surface area is constant: $S = \pi r l + \pi r^2 = \pi r \sqrt{r^2 + h^2} + \pi r^2$ Solving for $l$: $l = \frac{S - \pi r^2}{\pi r} \implies l^2 = \frac{(S - \pi r^2)^2}{\pi^2 r^2}$ Since $h^2 = l^2 - r^2$, we have: $h^2 = \frac{(S - \pi r^2)^2}{\pi^2 r^2} - r^2 = \frac{S(S - 2\pi r^2)}{\pi^2 r^2}$

Volume $V = \frac{1}{3}\pi r^2 h$. Squaring to eliminate radical: $V^2 = \frac{1}{9}\pi^2 r^4 h^2 = \frac{1}{9}\pi^2 r^4 \cdot \frac{S(S - 2\pi r^2)}{\pi^2 r^2} = \frac{S}{9}(S r^2 - 2\pi r^4)$ Let $Z = V^2$. Maximizing $Z$ is equivalent to maximizing $V$ since $V > 0$.

$\frac{dZ}{dr} = \frac{S}{9}(2S r - 8\pi r^3)$ Setting $\frac{dZ}{dr} = 0$ for stationary point ($r \ne 0$): $2Sr - 8\pi r^3 = 0 \implies 2r(S - 4\pi r^2) = 0 \implies S = 4\pi r^2$ Substitute $S = \pi r l + \pi r^2$: $4\pi r^2 = \pi r l + \pi r^2 \implies 3\pi r^2 = \pi r l \implies l = 3r$

$\frac{d^2 Z}{dr^2} = \frac{S}{9}(2S - 24\pi r^2)$ At $S = 4\pi r^2$: $\frac{d^2 Z}{dr^2} = \frac{S}{9}(8\pi r^2 - 24\pi r^2) = -\frac{16\pi r^2 S}{9} < 0$ Hence, Volume $V$ is maximum when $l = 3r$. In right triangle formed by semi-vertical angle $\alpha$: $\sin \alpha = \frac{r}{l} = \frac{r}{3r} = \frac{1}{3} \implies \alpha = \sin^{-1}\left(\frac{1}{3}\right)$ Hence proved.