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

Let height of cone AO = h and ` angleBAO = alpha`
Let radius of cylinder
O'P = x
In ` Delta APO', tan alpha = (PO')/(AO')`
` rArr AO' = x cot alpha`
`:. O'O = AO - AO'`
`= h-x cot alpha`
Now, volume of cylinder
`V = pi x^(2) * (OO')`
` = pi x^(2)(h-x cot alpha)`
` = pi (hx^(2) - x^(3) cot alpha)`
` rArr (dV)/(dx) = pi (2hx- 3x^(2) cot alpha)`
and ` (d^(2)V)/(dx^(2)) = pi (2h-6x cot alpha)`
For maxima/minima, `(dV)/(dx) = 0`
` rArr 2hx= 3x^(2) cot alpha rArr x = (2h)/3 tan alpha`
at ` x = (2h) / 3 tan alpha`
` (d^(2)V)/(dx^(2))= pi (2h-4h)=- 2pi h lt 0 `
` rArr` Volume V is maximum.
Now, maximum volume
`V=pi {h((2h)/3 tan alpha)^(2) - ( (2h)/3 tan alpha)^(3) cot alpha}`
`= pi h^(3)tan^(2) alpha(4/9 - 8/27 ) = 4/27 pi h^(3) tan^(2) alpha` .