Question

If \(\theta\) is the semi-vertical angle of a cone of maximum volume and given slant height, then \(\tan \theta\) is given by

• A. 2
• B. 1
• C. \(\sqrt{2}\)
• D. \(\sqrt{3}\)

Hint:

For a cone with fixed slant height, volume is maximized when \(\tan \theta = \sqrt{2}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Volume of cone \(V = \frac{1}{3} \pi r^2 h\). Given slant height \(l\) is constant. \(r = l \sin \theta\), \(h = l \cos \theta\).

Step 2: Detailed Explanation:
\(V = \frac{1}{3} \pi (l^2 \sin^2 \theta)(l \cos \theta) = \frac{1}{3} \pi l^3 \sin^2 \theta \cos \theta\).
Let \(f(\theta) = \sin^2 \theta \cos \theta = \cos \theta - \cos^3 \theta\).
\(f'(\theta) = -\sin \theta + 3 \cos^2 \theta \sin \theta = \sin \theta (3 \cos^2 \theta - 1) = 0\).
\(\sin \theta \neq 0\), so \(3 \cos^2 \theta = 1 \implies \cos^2 \theta = \frac{1}{3}\), \(\sin^2 \theta = \frac{2}{3}\).
\(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} = 2 \implies \tan \theta = \sqrt{2}\).

Step 3: Final Answer:
Option (C) \(\sqrt{2}\).