The radius of the cylinder of maximum volume, | vedclass

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(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R$. From the geometry of the figure,we have the relat

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$\sqrt {{2 \over 3}} R$

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(B) Let $r$ be the radius and $h$ be the height of the cylinder inscribed in a sphere of radius $R$. From the geometry of the figure,we have the relation:${r^2} + {\left( {\frac{h}{2}} \right)^2} = {R^2}$${h^2} = 4({R^2} - {r^2}) \implies h = 2\sqrt {{R^2} - {r^2}}$The volume $V$ of the cylinder is given by:$V = \pi {r^2}h = 2\pi {r^2}\sqrt {{R^2} - {r^2}}$To find the maximum volume,we differentiate $V$ with respect to $r$:$\frac{{dV}}{{dr}} = 2\pi \left[ {2r\sqrt {{R^2} - {r^2}} + {r^2} \cdot \frac{1}{{2\sqrt {{R^2} - {r^2}} }} \cdot ( - 2r)} \right]$$\frac{{dV}}{{dr}} = 2\pi \left[ {2r\sqrt {{R^2} - {r^2}} - \frac{{{r^3}}}{{\sqrt {{R^2} - {r^2}} }}} \right]$Setting $\frac{{dV}}{{dr}} = 0$ for critical points:$2r\sqrt {{R^2} - {r^2}} = \frac{{{r^3}}}{{\sqrt {{R^2} - {r^2}} }}$$2({R^2} - {r^2}) = {r^2}$$2{R^2} - 2{r^2} = {r^2}$$3{r^2} = 2{R^2}$${r^2} = \frac{2}{3}{R^2}$$r = \sqrt {\frac{2}{3}} R$Thus,the radius of the cylinder of maximum volume is $\sqrt {\frac{2}{3}} R$.

The radius of the cylinder of maximum volume,which can be inscribed in a sphere of radius $R$ is

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