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Question
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is `(4r)/3.`
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Solution
Let the radius of the sphere = r
Radius of cone = R
Height of the cone = AM
= OA + OM
= r + r cos θ
= r(1 + cosθ)
where ∠BOM = θ
BC = diameter of the base of the cone
∴ Radius of cone = r sin θ
Volume of cone V = `1/3 pi (r sin theta)^2 xx r (1 + cos theta)` ....`[because "volume of cone" = 1/3 pir^2 h]`
`= 1/3 pir^3 sin^2 theta (1 + cos theta)`
On differentiating,
`(dV)/(d theta) = 1/3 pir^3 [2 sin theta cos theta (1 + cos theta) + sin^2 theta (- sin theta)]`
`= 1/3 pir^3 [2 sin theta cos theta (1 + cos theta) - sin^3 theta]`
`= 1/3 pir^3 sin theta [2 cos theta (1 + cos theta) - sin^2 theta]`
`= 1/3 pir^3 sin theta [2 cos theta + 2 cos^2 theta - 1+ cos^2 theta]`
`= 1/3 pir^3 sin theta [3 cos^2 theta + 2 cos theta - 1]`
`= 1/3 pir^3 sin theta (cos theta + 1)(3 cos theta - 1)`
For maximum and minimum, `(dV)/(d theta) = 0`
⇒ cos θ ≠ - 1
⇒ θ ≠ π
∴ (3 cos θ - 1) = 0
⇒ `cos theta = 1/3`
In the interval `(0, pi/2)` cos θ is decreasing, cos θ increases as θ decreases and decreases as θ increases.
⇒ at cos θ = `1/3`
The sign of `(dV)/(d theta)` changes from positive to negative as θ passes through this point.
Hence V is highest at this point.
Height of the cone = `r (1 + cos theta) = r(1 + 1/3)`
`= r xx 4/3`
= `(4r)/3`
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