Question

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is of the volume of the sphere.

Answer 1

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

Since the triangle AOD is the right angle triangle. So, by Pythagoras theorem,

OD 2 = OA 2 − AD 2 OD= R 2 − r 2 OD= R 2 − r 2

The height of the cone is CD,

CD=CO+OD h=R+ R 2 − r 2

Let, V be the volume of the cone,

V= 1 3 π r 2 h 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 the volume with respect to r,

dV dr = 2 3 πrR+ 2 3 πr R 2 − r 2 + 1 3 π r 2 × ( −2r ) 2 R 2 − r 2 = 2 3 πrR+ 2 3 πr R 2 − r 2 − 1 3 π r 3 × 1 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 (1)

Put dV dr =0,

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= 3 r 2 −2 R 2 R 2 − r 2 2R R 2 − r 2 =3 r 2 −2 R 2

Square both sides of above equation,

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

Differentiate equation (1) with respect to r,

d 2 V d r 2 = 2πR 3 + 3 R 2 − r 2 ( 2π R 2 −3π r 3 )× ( −2r ) 6 R 2 − r 2 9( R 2 − r 2 ) V ″ ( r )= 2πR 3 + 9( R 2 − r 2 )( 2π R 2 −9π r 2 )+2π r 2 R 2 +3π r 4 27 ( R 2 − r 2 ) 3 2 V ″ ( 8 9 R 2 )= 2πR 3 + 9( R 2 − 8 9 R 2 )( 2π R 2 −9π( 8 9 R 2 ) )+2π( 8 9 R 2 ) R 2 +3π ( 8 9 R 2 ) 2 27 ( R 2 − 8 9 R 2 ) 3 2 <0

Since V ″ is negative, so, the maximum volume of cone is when r 2 = 8 9 R 2 .

When r 2 = 8 9 R 2 , the height becomes,

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

And the volume becomes,

V= π 3 ( 8 9 R 2 )( 4 3 R ) = 8 27 ( 4 3 π R 3 ) = 8 27 ×Volume of sphere

Therefore, 8 27 of the volume of sphere is the maximum volume of the cone inscribed in a sphere.