Question

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is .

Answer 1

It is given that a sphere has a radius r.

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

The volume of the cone will be,

V= 1 3 π R 2 h

In right angled triangle BCD,

BC= r 2 − R 2 h−r= r 2 − R 2 h=r+ r 2 − R 2

So, the volume is,

V= 1 3 π R 2 ( r+ r 2 − R 2 ) V= 1 3 π R 2 r+ 1 3 π R 2 r 2 − R 2

Differentiate volume with respect to R,

dV dR = 2 3 πRr+ 2 3 πR r 2 − R 2 + π R 2 3 × ( −2R ) 2 r 2 − R 2 = 2 3 πRr+ 2 3 πR r 2 − R 2 − π R 3 3 r 2 − R 2 = 2 3 πRr+ 2πR( r 2 − R 2 )−π R 3 3 r 2 − R 2 = 2 3 πRr+ 2πR r 2 −3π R 3 3 r 2 − R 2

Put dV dR =0, then,

2 3 πRr+ 2πR r 2 −3π R 3 3 r 2 − R 2 =0 2 3 πRr= 3π R 3 −2πR r 2 3 r 2 − R 2 2r r 2 − R 2 =3 R 2 −2 r 2

Squaring both sides,

4 r 2 ( r 2 − R 2 )= ( 3 R 2 −2 r 2 ) 2 4 r 4 −4 r 2 R 2 =9 R 4 +4 r 4 −12 r 2 R 2 9 R 2 =8 r 2 R 2 = 8 r 2 9

Differentiate dV dR with respect to R,

d 2 V d R 2 = 2πr 3 + 3 r 2 − R 2 ( 2π r 2 −9π R 2 )−( 2πR r 2 −3π R 3 )( −6R ) 1 2 r 2 − R 2 9( r 2 − R 2 ) = 2πr 3 + 3 r 2 − R 2 ( 2π r 2 −9π R 2 )+( 2πR r 2 −3π R 3 )( 3R ) 1 r 2 − R 2 9( r 2 − R 2 )

When R 2 = 8 r 2 9 , then d 2 V d R 2 <0.

So, the volume is maximum when R 2 = 8 r 2 9 .

The height of the cone will be,

h=r+ r 2 − 8 r 2 9 =r+ r 2 9 =r+ r 3 = 4r 3

Hence, it is proved that the altitude of the right circular cone of the maximum volume that can be inscribed in a sphere of radius r is 4r 3 .