Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle is one-third that of the cone and the greatest volume of cylinder is `4/(27)pih^3tan^2alphadot`

Explanation

The given right circular cone of fixed height `(h) `and semi-vertical angle `(α) `
Here, a cylinder of radius `R` and height `H` is inscribed in the cone.
We have `r=tanh alpha`
Now since `Delta`AOG is similar to `traingle`CEG, we have
`frac{AO}{OG}=frac{CE}{EG}`
`implies frac{h}{r}=frac{H}{r-R}`
`implies H=frac{h}{r}(r-R)=frac{1}{tan alpha}(h tan alpha-R)`
Now the volume V of the cylinder is given by
`V=piR^2H=piR^2frac{h}{r}(r-R)=frac{1}{tan alpha}(h tan alpha-R)=frac{1}{tan alpha}(htan alpha-R)`
`frac{dV}{dR}=2piRh-frac{3piR^2}{tan alpha}`
Now, `frac{dV}{dR}=0`
`therefore 2piRh-frac{3piR^2}{tan alpha}=0`
`implies R=frac{2h tanalpha}{3}`
`Now,frac{d^2V}{dR^2}=2pih-frac{6piR}{tan alpha}`
`=2pih-frac{6pi(frac{2h tanalpha}{3})}{tan alpha}=-2pih<0`
By second derivative test, the volume of the cylinder is the greatest when `R=frac{2h tanalpha}{3}`
Now, when `R=frac{2h tanalpha}{3} H=frac{1}{tanalpha}(htanalpha-frac{2htanalpha}{3})=frac{h}{3}`
Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder can be obtained as:
`=pi(frac{2h tanalpha}{3})^2(frac{h}{3})=frac{4}{27}pih^3tan^2alpha`