The semi vertical angle of a right circular cone of maximum volume of a given slant height is

$\cos^{-1}(\frac{1}{\sqrt{3}})$

The correct answer is Option (3) → $\cos^{-1}(\frac{1}{\sqrt{3}})$

Let the slant height of the right circular cone be $l$ (constant).

Let the semi-vertical angle be $\theta$.

Radius of base, $r = l \sin \theta$

Height, $h = l \cos \theta$

Volume of cone:

$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (l \sin \theta)^2 (l \cos \theta) = \frac{1}{3} \pi l^3 \sin^2 \theta \cos \theta$

Maximize $V$ w.r.t $\theta$:

$\frac{dV}{d\theta} = \frac{1}{3} \pi l^3 \frac{d}{d\theta} (\sin^2 \theta \cos \theta)$

Calculate derivative:

$\frac{d}{d\theta} (\sin^2 \theta \cos \theta) = 2 \sin \theta \cos \theta \cdot \cos \theta + \sin^2 \theta \cdot (-\sin \theta) = 2 \sin \theta \cos^2 \theta - \sin^3 \theta$

Set derivative equal to zero for maximum:

$2 \sin \theta \cos^2 \theta - \sin^3 \theta = 0$

$\sin \theta (2 \cos^2 \theta - \sin^2 \theta) = 0$

Either:

$\sin \theta = 0$ (which gives trivial solution $\theta = 0$) or

$2 \cos^2 \theta - \sin^2 \theta = 0$

Using $\sin^2 \theta + \cos^2 \theta = 1$, substitute:

$2 \cos^2 \theta - (1 - \cos^2 \theta) = 0$ $\Rightarrow 2 \cos^2 \theta - 1 + \cos^2 \theta = 0$ $\Rightarrow 3 \cos^2 \theta = 1$ $\Rightarrow \cos^2 \theta = \frac{1}{3}$ $\Rightarrow \cos \theta = \frac{1}{\sqrt{3}}$

Therefore:

The semi-vertical angle $\theta = \cos^{-1} \frac{1}{\sqrt{3}}$