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 and `lg e (alpha)` can be drawn as follows

Here a cylinder of radius R and height H is inscribed in the cone.
Then `angleGAO=alpha, OG =r, OA =h, OE =R and CE=H`
we have
`r=tan alpha`
Now since `triangle AOG` is similar to `triangleCEG`, we have
`(AO)/(OG)=(Ce)/(EG)`
or `(h)/(r)=(H)/(r-R)[EG=OG-OE]`
or `H=(h)/(r)(r-r)=(h)/(h tan alpha)(h tan alpha-R)=(1)/(tanalpha)(htanalpha-R)`
Now the volume (V) of the cylinder is given by
`V=piR^(2)H=(piR^(2))/(tanalpha)(h tan alpha-R)=piR^(2)h-(piR^(3))/(tanalpha)`
`therefore (dV)/(dR)=2piRh-(3piR^(2))/(tan alpha)`
Now `(dV)/(dr)=0`
or `2piRh=(3piR^(2))/(tan alpha)`
or `R=(2h)/(tan alpha)`
or `R=(2h)/(3)tanalpha`
Also for this value of R,
`(d^(2)V)/(dR^(2))=2pih-(6pi)//(tan alpha)(2h)/(3)tanalpha`
`=2pih-4pih`
`=2pih-4pih`
`=-2pihlt0`
Therefore ,R =`(2h)/(3)tan alpha` is point of maxima.
Now the maximum volume of the cylinder is
`pi((2h)/(3)tan alpha)^(2)(h/3)=pi(4h^(2))/(9)tan^(2alpha)(h/3)`
`=(4)/(27)pih^(3)tan^(2)alpha`