Let the height of the cylinder be x, and height of the cone be h

Therefore $VO' = xcot30° = x\sqrt{3}$

therefore $ OO' = h - x\sqrt{3} $

$ V= \pi x^2(h-x\sqrt{3}) $

on differentiating w.r.t x we get,

dv/dX = 2πxh - 3πx^2√3

For maximum or minimum V

dv/dX = 0

Therefore 2πxh - 3πx^2√3 = 0

This implies x = (2h/3√3)

Again differentiating w.r.t x we get,

d^2y/dx^2 = 2πh - 6πx√3

Substituting for x we get,

d^2y/dx^2 = π(2h- 4h)

Hence V is maximum when x = 2h/3√3

The maximum volume of the cylinder is given by

V = π[(2h/3√3)^2(h- 2h/3)]

V = (4/81)×πh^3

Hence proved.