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Show that the height of the cylinder of maximum volume that can be inscribed in a cone of height h is \(\frac{1}{3}h.\)

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Let r and H be the radius and height of inscribed cylinder respectively and θ be the semi - vertical angle of given cone.

If V be the volume of cylinder.

then V = πr2H

∴ V = πr2 ( h - r cot θ)

⇒ V = π( hr2 - r3cot θ)

Differentiating with respect to r, we get

For maxima or minima,

\(\frac{dV}{dr}=0\)

Hence, volume will be maximum when,
\(r = \frac{2h}{3}\,tan\,\theta\)

∴ H (height of cylinder) =

 \(h-\frac{2h}{3}\,tan\,\theta.cot\,\theta\)

\(=\frac{3h-2h}{3}\)\(= \frac{h}{3}.\)

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