Optimization Problem 2

For area of $4n - gon$ :

Let $PQ = x$ be a side of the $n - gon$ , $PB = QB= m$ , $AB = h^*$ , 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 $A = nx^2\cot(\dfrac{\pi}{4n}) = a \implies x = \sqrt{\dfrac{a\tan(\dfrac{\pi}{4n})}{n}} \implies m = \dfrac{1}{2}\sqrt{(\dfrac{a}{n})\csc(\dfrac{\pi}{4n})\sec(\dfrac{\pi}{4n})}$ .

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The volume $V = aH = 1 \implies H = \dfrac{1}{a} \implies D(a) = d^2(a) = 4m^2 + H^2 =$ $(\dfrac{a}{n})\csc(\dfrac{\pi}{4n})\sec(\dfrac{\pi}{4n}) + \dfrac{1}{a^2} \implies$ $\dfrac{dD}{da} = \dfrac{\csc(\dfrac{\pi}{4n})\sec(\dfrac{\pi}{4n})a^3 - 2n}{n a^3} = 0$

$a > 0 \implies a(n) = (n\sin(\dfrac{\pi}{2 n}))^{\dfrac{1}{3}}$ and $\dfrac{d^2D}{da^2} = \dfrac{6}{a^4} > 0$ for $a(n) = (n\sin(\dfrac{\pi}{2 n}))^{\dfrac{1}{3}} \implies$ min at $a(n) = (n\sin(\dfrac{\pi}{2 n}))^{\dfrac{1}{3}}$ .

$D(n) = d^2(n) = \dfrac{3}{(n\sin(\dfrac{\pi}{2 n}))^{\dfrac{2}{3}}} \implies$ $d(n) = \dfrac{\sqrt{3}}{(n\sin(\dfrac{\pi}{2 n}))^{\dfrac{1}{3}}}$ .

Let $j_{n} = n\sin(\dfrac{\pi}{2 n}).$

Using the inequality: $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies \dfrac{\pi}{2}\cos(\dfrac{\pi}{2n}) < j_{n} < \dfrac{\pi}{2} \implies$ $\lim_{n \rightarrow \infty} j_{n} = \dfrac{\pi}{2} \implies$ $\lim_{n \rightarrow \infty} d(n) = \dfrac{\sqrt{3}}{(\dfrac{\pi}{2})^{\dfrac{1}{3}}} \approx \boxed{1.49}$

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To show $\lim_{n \rightarrow \infty} V_{n} = V_{cylinder}$ :

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

$x = 2r\sin(\dfrac{\pi}{n})$ and $h^{*} = r\cos(\dfrac{\pi}{n})$ .

Let $H$ be the height of the prism $\implies V_{n} = \dfrac{n}{2}\sin(\dfrac{2\pi}{n})r^2H.$

Using the inequality: $\cos(x) < \dfrac{\sin(x)}{x} < 1 \implies$ $\pi\cos(\dfrac{2\pi}{n}) < \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) < \pi$ $\implies \lim_{n \rightarrow \infty} \dfrac{n}{2}\sin(\dfrac{2\pi}{n}) = \pi \implies$ $\lim_{n \rightarrow \infty} V_{n} = \pi r^2 H = V_{cylinder}.$

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Doing the problem using a cylinder:

$\pi r^2 = a \implies r = \sqrt{\dfrac{a}{\pi}}$ and the volume $V = ah = 1 \implies h = \dfrac{1}{a} \implies$

$D(a) = d^2(a) = 4r^2 + h^2 = \dfrac{4a}{\pi} + \dfrac{1}{a^2} \implies$ $\dfrac{dD}{da} = \dfrac{2}{a^3\pi}(2a^3 - \pi) = 0 \implies a = (\dfrac{\pi}{2})^{\dfrac{1}{3}}$

$\dfrac{d^2D}{da^2} = \dfrac{6}{a^4} > 0$ for $a = (\dfrac{\pi}{2})^{\dfrac{1}{3}} \implies$ min at $a = (\dfrac{\pi}{2})^{\dfrac{1}{3}}$ and $d = \dfrac{\sqrt{3}}{(\dfrac{\pi}{2})^{\dfrac{1}{3}}} \approx \boxed{1.49}$

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