Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone.

Explanation

Let the height of cone = h and semivertical angle be `alpha`.
Let the radius of cylinder = x.'
Height of cylinder = OO'
=AO - AO'
=` h - x cot alpha`
`:.` Curved surface of cylinder
`C = 2 pi x*(h- x cot alpha)`
`= 2 pi (hx-x^(2) cot alpha)`
`rArr (dC)/(dx) = 2 pi (h-2x cot alpha)`

For maxima/minima
` (dC)/(dx) = 0`
` rArr h-2x cot alpha = 0`
`rArr cot alpha = h/(2x)`
` rArr h/r = h/(2x)`
(Where 'r' = radius of the cone)
` rArr x=r/2`
and ` (d^(2)C)/(dx^(2)) = 2 pi (0-2 cot alpha ) =-4 pi cot alpha lt 0`
`rArr ` The value of C is maximum.
Hence, the curved surface area of the cylinder will be maximum when `x= r//2 `i.e., the radius of cylinder will be half of the radius of cone.