Minimize Surface Area

Recall the volume and the slant surface area of a cone are given as $V:= \dfrac13 \pi r^2 h$ and $S:= \pi r s$ , where $r, h, s$ denote the radius, height and the slant height of the cone.

By Pythagorean theorem , we can see that $r^2 + h^2= s^2$ .

We essentially want to find the value of $\tan^{-1} \left( \dfrac hr \right)$ .

Since the volume is held constant, we can let $V = \dfrac\pi3$ , then $\dfrac\pi3 = \dfrac13 \pi r^2 h \Leftrightarrow h = \dfrac1{r^2}$ .

We want to find the value of $r$ when $S$ is minimized. If $S$ is minimized, then so is $T:= S^2$ . We have

$T = \pi^2 r^2 (r^2 + h^2) = \pi r^2 \left( r^2 + \dfrac1{r^4} \right) = \pi \left( r^4 + \dfrac1{r^2} \right) .$

When at extrema, $\dfrac{dT}{dr} = 0 \Rightarrow \pi \left( 4r^3 - \dfrac2{r^3} \right) = 0 \Rightarrow r^6 = \dfrac12$ .

We apply the second derivative test to prove that $T$ has a minimum value when $r^6 = \dfrac12$ ,

$\dfrac{d^2 T}{dr^2} = \pi \left( 12r^2 + \dfrac{6}{r^4} \right) > 0 .$

Since the second derivative is positive when $r^6 = \dfrac12$ , $T$ is indeed minimized when at $r = 2^{-1/6}$ . Hence $\tan^{-1} \left( \dfrac hr \right) = \tan^{-1} \left( \dfrac{1/r^2}r \right) = \tan^{-1} \sqrt2 \approx \boxed{54.7356^\circ} .$

Using the diagram of the cone above:

Let $m\angle{ABO} = \theta$

Let $K$ be a constant.

The volume $V = \dfrac{1}{3} \pi r^2 h$ and the lateral surface area $P = 2 \pi r s$ .

Let $m\angle{ABO} = \theta$ and let $K$ be a constant.

$r = s \cos(\theta), h = s \sin(\theta) \implies V = \dfrac{1}{3} \pi \cos^2(\theta) \sin(\theta) s^3 = K \implies s = (\dfrac{3K}{\pi \cos^2(\theta) \sin(\theta)})^{\dfrac{1}{3}} \implies$

$P(\theta) = 2\pi \cos(\theta) \sin(\theta) s^2 = 2\pi \cos(\theta) \sin(\theta) (\dfrac{9 k^2}{\pi^2 \cos^4(\theta) \sin(\theta)})^\dfrac{1}{3}$

Let $j = (9 K^2 \pi)^{\dfrac{1}{3}} \implies P(\theta) = 2 j (\sec(\theta))^\dfrac{1}{3} (\csc(\theta))^\dfrac{1}{3} \implies \dfrac{ds}{d\theta} = \dfrac{2j}{3} (\sec(\theta))^\dfrac{1}{3} (\csc(\theta))^\dfrac{2}{3} (\tan(\theta) - 2 \cot(\theta)) = 0 \implies$

$3 \sin^2(\theta) - 2 = 0 \implies \sin(\theta) = \pm \sqrt{\dfrac{2}{3}}$ choosing $\sin(\theta) = \sqrt{\dfrac{2}{3}} \implies \theta = \boxed{54.7356^\circ}$

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$Bonus:$

The desired angle is independent of $n$ and the work below verifies this.

For area of $n - gon$ :

Let $BC = x$ be a side of the $n - gon$ , $AC = AB= r$ , $AD = h^*$ , and $\angle{BAD} = \dfrac{180}{n}$ .

$\dfrac{x}{2} = r \sin(\dfrac{180}{n}) \implies r = \dfrac{x}{2 \sin(\dfrac{180}{n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) \implies A_{\triangle{ABC}} = \dfrac{1}{4} \cot(\dfrac{180}{n}) x^2 \implies$

$A_{n - gon} = \dfrac{n}{4} \cot(\dfrac{180}{n}) x^2 \implies$ the Volume of the pyramid $V_{p} = \dfrac{n}{12} \cot(\dfrac{180}{n}) x^2 H$

Let $AC = H$ be the height of the pyramid, $BC = h^*$ , and $AB = h'$ be the slant height of the pyramid and $m\angle{CBA} = \theta$

The lateral surface area $S = \dfrac{n}{2} x h'$ .

$h^* = \dfrac{x}{2} \cot(\dfrac{180}{n}) = h' \cos(\theta) \implies x = 2 \tan(\dfrac{180}{n}) \cos(\theta) h'$ and $H = h' \sin(\theta)$ and letting $u(n) = \tan(\dfrac{180}{n}) \implies$ $V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h' = K$ and $S = n * u(n) cos(\theta) h'^2.$

$V_{p} = \dfrac{n}{3} u(n) \cos^2(\theta) \sin(\theta) h' = K \implies h' = (\dfrac{3K}{n * u(n_) cos^2(\theta) sin(\theta)})^{\dfrac{1}{3}} \implies S(\theta) = j(n) * (sec(\theta))^{\dfrac{1}{3}} (csc(\theta))^{\dfrac{2}{3}}$ , where $j(n) = (9 k^2 n * u(n))^{\dfrac{1}{3}}$

$\implies \dfrac{dS}{d\theta} = \dfrac{j(n)}{3} (sec\theta)^\dfrac{1}{3} (\csc\theta)^\dfrac{2}{3} * (\tan\theta - 2 \cot\theta) = 0 \implies$

$3 \sin^2\theta - 2 = 0 \implies \sin\theta = \pm \sqrt{\dfrac{2}{3}}$ . choosing $\sin\theta = \sqrt{\dfrac{2}{3}} \implies \theta = \boxed{54.7356^\circ}$

$\therefore$ The measure of the desired angle $\theta$ is independent of $n$ .