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.`

Answer 1

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`