# Find the volume of the largest right circular cone that can be inscribed in a sphere of radius r.

Volume is basically the measure of the capacity that a body holds. Various bodies have various formulae of volumes.

## Answer: The volume of the largest right circular cone that can be inscribed in a sphere of radius r is (32/81)πR^{3} cubic units or (8/27) times the volume of the sphere.

Let's find the volume of the largest right circular cone.

**Explanation:**

Let us consider the right circular cone, inscribed in a sphere.

The figure is given below:

The radius of the right circular cone is 'r'

'x' is the distance from the base of the cone to O.

O is the center of the sphere.

V be the volume of the sphere

The volume of a cone (V) = (1/3)πr^{2}h

Also, the height of a cone is given in the figure is AC = AO + OC,

Hence, AO + OC = R + x

Since it is a right circular cone then, r^{2} = R^{2} - x^{2} by Pythagoras theorem.

Put the value of height h = (R + x) and r^{2} = (R^{2} - x^{2}) in the volume of a cone formula, (V) = (1/3)πr^{2}h

V = (1/3) π (R^{2} − x^{2}) (R + x)

= (1/3) π (R^{3 }+ R^{2}x − Rx^{2 }− x^{3 })

Now, differentiate the volume w.r.t. x

dV/dx = d[(1/3)π( R^{3 }+ R^{2}x − Rx^{2 }− x^{3 })]/dx

= (π/3)[0 + R^{2 }− 2Rx − 3x^{2}]

= (π/3)[R^{2} − 2Rx − 3x^{2}]

In order to find the volume, double differentiation w.r.t. x,

d^{2}V/ dx^{2} = π/3[−2R − 6x] (eq 1)

For a maximum or minimum value of V.

VdV/dx = 0

R^{2 }− 2Rx − 3x^{2 }= 0

on solving we have,

⇒(R + x) (x − 3x) = 0

x = −R, R/3

but x ≠ −R (As a length R can not be zero)

So, when x = R/3 , (d^{2}V)/(dx^{2}) < 0

Volume is maximum only when x = R/3

Max Volume = (π/3) [R^{2 }− R^{2}/9][ R + R/3]

= (π/3) [8R^{2}/9][4R/3]

= (32/81)πR^{3}

= (8/27) [(4/3) π R^{3})

or

= (8/27) × (volume of a sphere)