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

\frac{2 R}{3}

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\frac{R}{3}

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

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\frac{R}{\sqrt{3}}

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Solution

The correct option is C $\frac{2 R}{\sqrt{3}}$
$\frac{h^{2}}{4} + r^{2} = R^{2}$
Volume of cylinder $= π r^{2} h$
$= π (R^{2} - \frac{h^{2}}{4}) h$
$\frac{d v}{d h} = π (R^{2} - \frac{3 h^{2}}{4})$
$0 = R^{2} - \frac{3 h^{2}}{4}$
$h = \frac{2 R}{\sqrt{3}}$

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The height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is

The correct option is C 2R√3h24+r2=R2Volume of cylinder =πr2h=π(R2−h24)hdvdh=π(R2−3h24)0=R2−3h24h=2R√3

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

The correct option is C $\frac{2 R}{\sqrt{3}}$
$\frac{h^{2}}{4} + r^{2} = R^{2}$
Volume of cylinder $= π r^{2} h$
$= π (R^{2} - \frac{h^{2}}{4}) h$
$\frac{d v}{d h} = π (R^{2} - \frac{3 h^{2}}{4})$
$0 = R^{2} - \frac{3 h^{2}}{4}$
$h = \frac{2 R}{\sqrt{3}}$