Prisms

For area of $4n - gon$ :

Let $PQ = x$ be a side of the $4n - gon$ , $PB = QB= m$ , $AB = h^*$ , and $\angle{QBA} = \dfrac{\pi}{4n}$ .

$\dfrac{x}{2} = m \sin(\dfrac{\pi}{4n}) \implies m = \dfrac{x}{2 \sin(\dfrac{\pi}{4n})} \implies h^* = \dfrac{x}{2} \cot(\dfrac{\pi}{4n}) \implies$ the area of the $4n$ -gon $A = nx^2\cot(\dfrac{\pi}{4n})$ .

The volume $V = nx^2\cot(\dfrac{\pi}{4n}) H = 1 \implies H = \dfrac{\tan(\dfrac{\pi}{4n})}{nx^2}$

The lateral surface area $S = 4nxH = \dfrac{4\tan(\dfrac{\pi}{4n})}{x} = a \implies x = \dfrac{4\tan(\dfrac{\pi}{4n})}{a} \implies m = \dfrac{2\sec(\dfrac{\pi}{4n})}{a}$ and $H = \dfrac{\cot(\dfrac{\pi}{4n})a^2}{16n} \implies$ $D(a) = d^2(a) = 4m^2 + H^2 = \dfrac{16\sec^2(\dfrac{\pi}{4n})}{a^2} + \dfrac{\cot^2(\dfrac{\pi}{4n})a^4}{16^2 n^2} \implies$

$\dfrac{dD}{da} = \dfrac{\cot^2(\dfrac{\pi}{4n}) a^6 - 2^{11}\sec^2(\dfrac{\pi}{4n})n^2}{2^6 n^2 a^3} = 0$ $\implies a^6 = \dfrac{2^{11}\sin^2(\dfrac{\pi}{4n}) n^2}{\cos^4(\dfrac{\pi}{4n})}$

$a > 0 \implies \boxed{a_{n} = 2^{\dfrac{11}{6}}(\dfrac{n\sin(\dfrac{\pi}{4n})}{\cos^2(\dfrac{\pi}{4n})})^{\dfrac{1}{3}}}$

$\dfrac{d^2D}{da^2} > 0$ at $a = 2^{\dfrac{11}{6}}(\dfrac{n\sin(\dfrac{\pi}{4n})}{\cos^2(\dfrac{\pi}{4n})})^{\dfrac{1}{3}} \implies$ min at $a$ . You can check this for yourself.

Let $j_{n} = n\sin(\dfrac{\pi}{4})$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \dfrac{\pi}{4}\cos(\dfrac{\pi}{4n}) < j_{n} < \dfrac{\pi}{4} \implies \lim_{n \rightarrow \infty} a_{n} = 2^{\dfrac{11}{6}} (\dfrac{\pi}{4})^{\dfrac{1}{3}} =$ $\boxed{2^{\dfrac{7}{6}} * \pi^{\dfrac{1}{3}}}$ .

$a_{n} = 2^{\dfrac{11}{6}}(\dfrac{n\sin(\dfrac{\pi}{4n})}{\cos^2(\dfrac{\pi}{4n})})^{\dfrac{1}{3}} \implies m = \dfrac{\sqrt{2}}{(n\sin(\dfrac{\pi}{2n}))^{\dfrac{1}{3}}}$ and $H = \dfrac{1}{(n\sin(\dfrac{\pi}{2n}))^{\dfrac{1}{3}}} \implies \boxed{d_{n} = \dfrac{\sqrt{3}}{(n\sin(\dfrac{\pi}{2n}))^{\dfrac{1}{3}}}}$

Let $j^{*}_{n} = n\sin(\dfrac{\pi}{4})$ .

Using the inequality $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \dfrac{\pi}{2}\cos(\dfrac{\pi}{2n}) < j^{*}_{n} < \dfrac{\pi}{2} \implies \lim_{n \rightarrow \infty} j^{*}_{n} = \dfrac{\pi}{2} \implies$ $\lim_{n \rightarrow \infty} d_{n} = \boxed{\sqrt{3} * (\dfrac{2}{\pi})^{\dfrac{1}{3}}}$ .

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Doing the problem using a cylinder:

$V = \pi r^2 H = 1 \implies H = \dfrac{1}{\pi r^2} \implies S = 2\pi r H = \dfrac{r}{a} \implies H = \dfrac{a^2}{4\pi} \implies D(a) = d^2(a) = 4r^2 + H^2 = \dfrac{16}{a^2} + \dfrac{a^4}{16\pi^2} \implies$

$\dfrac{dD}{da} = \dfrac{a^6 - 2^7\pi^2}{4\pi^2 a^3} = 0 \implies$ $\boxed{a = 2^{\dfrac{7}{6}} \pi^{\dfrac{1}{3}}}$ for $a > 0 \implies \boxed{d = \sqrt{3} * (\dfrac{2}{\pi})^{\dfrac{1}{3}}} \approx \boxed{1.49}$ .

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