Question

Show that the right circular cone of least curved surface and given volume has an altitude equal to time the radius of the base.

Answer 1

Let, r be the radius and h be the height of the cone.

Let V be the volume and S be the curved surface area of the cone.

Since the volume of the cone is,

V= 1 3 π r 2 h

So the height can be written as,

h= 3V π r 2

The curved surface area of the cone is,

S=πr r 2 + h 2 =πr r 2 + ( 3V π r 2 ) 2 =πr π 2 r 6 +9 V 2 π 2 r 4 = 1 r π 2 r 6 +9 V 2

Differentiate the surface area with respect to r,

S ′ = −1 r 2 π 2 r 6 +9 V 2 + 1 r ( 1 2 π 2 r 6 +9 V 2 )( 6 π 2 r 5 +0 ) = −( π 2 r 6 +9 V 2 )+3 π 2 r 6 r 2 ( π 2 r 6 +9 V 2 ) = 2 π 2 r 6 −9 V 2 r 2 ( π 2 r 6 +9 V 2 ) (1)

Put S ′ =0,

2 π 2 r 6 −9 V 2 r 2 ( π 2 r 6 +9 V 2 ) =0 2 π 2 r 6 −9 V 2 =0 2 π 2 r 6 =9 V 2 r 6 = 9 V 2 2 π 2

Differentiate equation (1) with respect to r,

S ″ = 2 π 2 r 6 −9 V 2 r 2 ( π 2 r 6 +9 V 2 ) = r 2 ( π 2 r 6 +9 V 2 )( 12 π 2 r 5 )−( 2 π 2 r 6 −9 V 2 )[ ( 2r )( π 2 r 6 +9 V 2 )+ r 2 6 π 2 r 5 2 π 2 r 6 +9 V 2 ] ( r 2 ( π 2 r 6 +9 V 2 ) ) 2

S ″ ( 9 V 2 2 π 2 )>0

This shows that the surface area is minimum when r 6 = 9 V 2 2 π 2 .

At r 6 = 9 V 2 2 π 2 , the height becomes,

h= 3V π r 2 h 3 = 27 V 3 π 3 r 6 = 27 V 3 π 3 × 9 V 2 2 π 2 = 6V π

Substitute V= 1 3 π r 2 h in the above equation,

h 3 = 6× 1 3 π r 2 h π h 2 =2 r 2 h= 2 r

Therefore, the least curved surface area of the right circular cone has an altitude equal to 2 times the radius of the base.