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The radius of a sphere is $9 c m$ . Find the height of the right circular cone that can be inscribed in the sphere so that the volume of the cone is maximum.

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An Intiative by Sri Chaitanya

h = 12 cm

h = 15 cm

h = 10 cm

h = 11 cm

answer is A.

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Detailed Solution

Let us first draw the diagram.

Let the height of the cone be

h

and the radius of the cone be

r

. From the diagram, we can see that

O A B

forms a right-angled triangle. Since

O B

represents the radius of the sphere, therefore

O B = 9 c m

.
Also,

O C

represents the radius of the sphere, therefore

O C = 9 c m

.
As we can see that

A C = h

and

A C = A O + O C

, therefore

h = A O + 9 ⇒ A O = h − 9

In right angles triangle

Δ O A B, A O = h − 9, B O = 9 c m

.
Therefore, using Pythagoras’ Theorem,

B O 2 = A O 2 + A B 2 ⇒ 92 = h − 9 2 + A B 2 ⇒ A B 2 = 81 − h − 9 2

⇒ A B 2 = 81 − h 2 − 81 + 18 h ⇒ A B 2 = 18 h − h 2

As we can see ABAB represents the radius of the base of the cone, therefore

r 2 = 18 h − h 2

.
Now the volume of the cone is given by ,

V = 13 π r 2 h

where

r

is the radius of the base of the cone and

h

is the height of the cone.
Therefore,

V = 13 π 18 h − h 2 h ⇒ V = 13 π 18 h 2 − h 3

Let us take the derivative of

V

with respect to

h

. We will get,

d V d h = 13 π 36 h − 3 h 2

Taking

d V d h = 0

will give us two values of

h

one of them gives the maximum volume of the cone and the other one gives the minimum volume of the cone.
Therefore,

13 π h 36 − 3 h = 0

gives

h = 0

and

36 − 3 h

that is

h = 12 c m

.
Taking derivative of

d V d h

, we get

d 2 V d h 2 = 13 π 36 − 6 h

When we put

h = 12 c m

d 2 V d h 2

, we get

d 2 V d h 2 h = 12 c m = 13 π 36 − 6 (12) d 2 V d h 2 h = 12 c m = − 13 π 36

d 2 V d h 2 h = 12 c m = − 12 π

,which is negative.
Therefore,

h = 12 c m

represents the maximum volume of cones inscribed in the sphere.

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