The semi vertical angle of a right circular cone of maximum volume of…

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

Let l be the given slant height, h the height and r the radius of the base, so that \(r^2 = l^2 - h^2\). The volume as a function of h is

    \(\displaystyle V = \frac{1}{3}\pi r^2h = \frac{\pi}{3}\left(l^2h - h^3\right)\)

    \(\displaystyle \frac{dV}{dh} = \frac{\pi}{3}\left(l^2 - 3h^2\right) = 0,\quad h = \frac{l}{\sqrt{3}}\)

    \(\displaystyle \frac{d^2V}{dh^2} = -2\pi h \lt 0\)

so this h gives the maximum volume. The semi-vertical angle θ is the angle between the axis (height) and the slant height:

    \(\displaystyle \cos\theta = \frac{h}{l} = \frac{1}{\sqrt{3}},\quad \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\)

Why the other options are incorrect

• Option (1) \(\tan^{-1}2\) — Incorrect. With \(\cos\theta = \frac{1}{\sqrt{3}}\), \(\sin\theta = \sqrt{\frac{2}{3}}\), so \(\tan\theta = \sqrt{2}\), not 2.

• Option (2) \(\sin^{-1}\left(\frac{1}{3}\right)\) — Incorrect. \(\sin\theta = \sqrt{\frac{2}{3}}\), not \(\frac{1}{3}\).

• Option (4) \(\sin^{-1}\left(\frac{1}{\sqrt{3}}\right)\) — Incorrect. \(\sin\theta = \sqrt{\frac{2}{3}}\). The value \(\frac{1}{\sqrt{3}}\) is the sine of the angle between the slant height and the base, not of the semi-vertical angle.