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प्रश्न
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is `8/27` of the volume of the sphere.
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उत्तर
Let VAB be the volume of the largest cone contained in the sphere.
Obviously, for maximum volume, the axis of the cone should be along the height of the sphere.
Let, ∠AOC = θ
∴ AC, the radius of the base of the cone = R sin θ, where R
is the radius of the sphere.
Height of the cone VC = VO + OC = R + R cos θ
Volume of a cone; V = `1/3 pi (AC)^2 xx (VC)`
`=> V = 1/3 piR^2 sin^2 θ (R + R cos theta)`
`=> V = 1/3 piR^3 sin^2 theta (1 + cos theta)`
On differentiating
`therefore dV/(d theta) = 1/3 piR^3 [sin^2 theta (- sin theta) + (1 cos theta) * 2 sin theta cos theta]`
`= 1/3 piR^3 [- sin^3 theta + 2 sin theta cos theta + 2 sin theta (1 - sin^2 theta)]`
`= 1/3 piR^3 [- sin^3 theta + 2 sin theta cos theta + 2 sin theta - 2sin^2 theta]`
`= 1/3 pi R^3 [- 3 sin^3 theta + 2 sin theta + 2 sin theta cos theta]`
For minimum and maximum, `(dV)/(d theta) = 0`
`=> 1/3 pi"R"^3 (- 3 sin^3 theta + 2 sin cos theta + 2 sin theta)` = 0
= - 3 sin3 θ + 2 sin θ cos θ + 2 sin θ = 0
= sin θ (- 3 sin2 θ + 2 cos θ + 2) = 0
= - 3 sin2 θ + 2 cos θ +2 = 0 ...[∵ sin θ ≠ 0]
= -3 (1 - cos2 θ) + 2 cos θ + 2 = 0
= - 3 + cos2 θ + 2 cos θ + 2 = 0
⇒ 3 cos2 θ + 2 cos θ - 1 = 0
⇒ (3 cos θ - 1)(cos θ + 1) = 0
⇒ cos θ = `1/3` cos θ = - 1
But cos θ ≠ 1 because cos θ = - 1 ⇒ θ = π which is not possible.
`therefore cos theta = 1/3`
When `cos theta = 1/3`, then `sin theta = sqrt(1 - cos^2 theta) = sqrt(1 - 1/9)`
`= sqrt(8/9)`
`= (2 sqrt2)/3`
Now `(dV)/(d theta) = 1/3 piR^3 sin theta [- 3 sin^2 theta + 2 + 2 cos theta]`
`= 1/3 piR^3 sin theta (3 cos theta - 1)(cos theta + 1)`
The sign of `cos theta = 1/3, (dV)/(d theta)` changes from +ve to -ve.
∴ V is highest at `theta = cos^-1 (1/3)`.
On decreasing θ, cos θ increases.
Now cos θ = `1/3` then V is maximum.
The height of the cone for this value of cos θ is
VC = R + R cos θ
`= R + R xx 1/3 = (4R)/3`
Radius of cone = AC = R sin θ = `R * (2sqrt2)/3 = (2 sqrt2)/3 R`
∴ Maximum volume of the cone is V
`= 1/3 pi (AC)^2 (VC)`
`= 1/3 pi ((2 sqrt(2R))/3)^2 ((4R')/3)`
`= 1/3 pi xx 8/9 R^2 xx (4R)/3`
`= 8/27 (4/3 piR^3)`
`= 8/27 xx` Volume of sphere
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Based on the above information, answer the following questions:
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