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)πR3 cubic units or (8/27) times the volume of the sphere.
Let's find the volume of the largest right circular cone.
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)πr2h
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, r2 = R2 - x2 by Pythagoras theorem.
Put the value of height h = (R + x) and r2 = (R2 - x2) in the volume of a cone formula, (V) = (1/3)πr2h
V = (1/3) π (R2 − x2) (R + x)
= (1/3) π (R3 + R2x − Rx2 − x3 )
Now, differentiate the volume w.r.t. x
dV/dx = d[(1/3)π( R3 + R2x − Rx2 − x3 )]/dx
= (π/3)[0 + R2 − 2Rx − 3x2]
= (π/3)[R2 − 2Rx − 3x2]
In order to find the volume, double differentiation w.r.t. x,
d2V/ dx2 = π/3[−2R − 6x] (eq 1)
For a maximum or minimum value of V.
VdV/dx = 0
R2 − 2Rx − 3x2 = 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 , (d2V)/(dx2) < 0
Volume is maximum only when x = R/3
Max Volume = (π/3) [R2 − R2/9][ R + R/3]
= (π/3) [8R2/9][4R/3]
= (8/27) [(4/3) π R3)
= (8/27) × (volume of a sphere)