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 tan 2 α .

Answer 1

Let in the triangle PQR the height of the cone is h.

And α be the semi vertical angle of cone.

Let, x be the radius of cylinder ABCD which is inscribed in the cone PQR.

Now, height of cylinder is O O ′ is equal to the difference of PO and P O ′ .

In ΔAP O ′ ,

tanα= A O ′ P O ′ = x P O ′ P O ′ = x tanα =xcotα

Height of the cylinder is,

O O ′ =PO−P O ′ =h−xcotα (1)

We need to maximize the volume of the cylinder.

Calculate the volume of cylinder,

V=π ( A ′ O ′ ) 2 ×( O O ′ ) V=π( x 2 )×( h−xcotα ) V=πh x 2 −πcotα. x 3

Differentiate with respect to x.

dV dx =2πhx−3πcotα⋅ x 2

Equate the derivative of the volume is zero.

dV dx =0 2πhx−3πcotα⋅ x 2 =0 x= 2h 3 tanα

Calculate the value of d 2 V d x 2 ,

dV dx =2πhx−3πcotα. x 2 d 2 V d x 2 =2πh−6πcotα.x

Substitute the value of x= 2h 3 tanα.

d 2 V d x 2 =2πh−4πh d 2 V d x 2 =−2πh <0

So, volume is maximum when x= 2h 3 tanα,

From equation (1),

O O ′ =h−xcotα O O ′ = h 3

Thus, the volume is maximum when the height of cylinder is one third of the height of cone.

Substitute the value to calculate the maximum volume,

V=π ( x ) 2 ( h−xcotα ) =π( 4 h 2 ×h 9 cot 2 α ) = 4 2π π h 3 tan 2 α cubic unit

Thus, the maximum volume is 4 2π π h 3 tan 2 α cubic unit.