Question

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 α is one-third that of the cone and the greatest volume of cylinder is `4/27 pih^3` tan²α.

Answer 1

Let VAB be a cone.

Height of the cone = h

Semivertex angle = α

Cylinder A'B'DC is inscribed in a cone whose radius = x.

OO' = Height of cylinder = VO - VO'

= h - x cot α

V, volume of the cylinder = πx² (h - x cot α)

On differentiating with respect to x,

`(dV)/dx = 2 pi xh - 3 pix^2 cot alpha`

For maximum and minimum, `(dV)/dx = 0`

⇒ 2πxh - 3πx² cot α = 0

⇒ πx (2h - 3x cot α) = 0

⇒ 2h - 3 x cot α = 0

⇒  3x cot α = 2h

`therefore x = (2h)/3 tan alpha        ...[x ne 0]`

Now, `(d^2 v)/(dx^2) = 2 pih - 6 pih cot alpha`

When x = `(2h)/3 tan alpha`

`therefore (d^2V)/dx^2 = 2pih - 6pi (2h)/3 tan alpha cot alpha`

`= 2pih - 4pih`

`= pi(2h - 4h)`

`= - 2 pih < 0`

⇒ V is maximum when x = `(2h)/3 tan alpha`

Height of cylinder, OO' = VO - VO' = h - x cot α

`= h - ((2h)/3  tan alpha) cot alpha        ...[because x = (2h)/3 tan alpha]`

`= h - (2h)/3 = h/3`

`= 1/3` height of the cone

Maximum volume of cylinder  = πx² (h - x cot α)

`= pi - ((2h)/3 tan alpha)^2 * (h - (2h)/3)`

`= pi((2h)/3 tan alpha)^2 xx h/3`

`= 4/27 pih^3  tan^2  alpha`