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 sphere and cylinder,we ha

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

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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 sphere and cylinder,we have the relation: $r^{2} + (h/2)^{2} = R^{2}$.This implies $(h/2)^{2} = R^{2} - r^{2}$,so $h = 2\sqrt{R^{2} - r^{2}}$.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 $r$:$\frac{dV}{dr} = 2\pi \left[ 2r \sqrt{R^{2} - r^{2}} + r^{2} \cdot \frac{1}{2\sqrt{R^{2} - r^{2}}} \cdot (-2r) ight] = 2\pi \left[ \frac{2r(R^{2} - r^{2}) - r^{3}}{\sqrt{R^{2} - r^{2}}} ight] = 2\pi \left[ \frac{2rR^{2} - 3r^{3}}{\sqrt{R^{2} - r^{2}}} ight]$.Setting $\frac{dV}{dr} = 0$,we get $2rR^{2} - 3r^{3} = 0$.Since $r 
eq 0$,we have $2R^{2} = 3r^{2}$,which gives $r^{2} = \frac{2}{3}R^{2}$,or $r = \sqrt{\frac{2}{3}}R$.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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