A circular cylinder is inscribed in a sphere of radius r. If the volume of the cylinder is maximum, then its height is.

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$r$

$\sqrt{2 r}$

$\frac{2 r}{\sqrt{3}}$

$\frac{3 r}{2}$

answer is A.

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Detailed Solution

$r^{2} = R^{2} + \frac{h^{2}}{4} \Rightarrow R^{2} = r^{2} - \frac{h^{2}}{4} volume V = {πR}^{2} h V = π (r^{2} - \frac{h^{2}}{4}) h = π (r^{2} h - \frac{h^{3}}{4}) V^{1} = π (r^{2} - \frac{3 h^{2}}{4}) V is \min or \max \Rightarrow V^{1} = 0 r^{2} = \frac{3 h^{2}}{4} \Rightarrow h = \frac{2 r}{\sqrt{3}} ∴ h = \frac{2 r}{\sqrt{3}}$

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