Question

Prove that the semi-vertical angle of a cone with given slant height and maximum volume is \( \tan^{-1}(\sqrt{2}) \).

Hint:

In optimization problems involving geometric shapes, it's crucial to express the quantity to be optimized (like volume or area) as a function of a single variable. Using trigonometric relations is often the most efficient way to do this for problems involving angles.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem where we need to maximize the volume of a cone given a constant slant height. We will express the volume as a function of the semi-vertical angle, then use calculus to find the angle that maximizes this volume.
Step 2: Key Formula or Approach:
Let \( l \) be the given slant height (a constant), \( h \) be the height, \( r \) be the radius, and \( \theta \) be the semi-vertical angle of the cone.
The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3}\pi r^2 h \] The relationships between the variables are:
\[ r = l\sin\theta \text{and} h = l\cos\theta \] We will express \( V \) as a function of \( \theta \), find its derivative \( \frac{dV}{d\theta} \), set it to zero to find critical points, and use the second derivative test to confirm a maximum.
Step 3: Detailed Explanation or Calculation:
1. Express Volume in terms of \( \theta \):
Substitute the expressions for \( r \) and \( h \) into the volume formula:
\[ V(\theta) = \frac{1}{3}\pi (l\sin\theta)^2 (l\cos\theta) \] \[ V(\theta) = \frac{1}{3}\pi l^3 \sin^2\theta \cos\theta \] 2. Differentiate Volume with respect to \( \theta \):
To find the maximum volume, we differentiate \( V \) with respect to \( \theta \) and set the result to zero. Let \( K = \frac{1}{3}\pi l^3 \) (a constant).
\[ V(\theta) = K \sin^2\theta \cos\theta \] Using the product rule \( (uv)' = u'v + uv' \):
\[ \frac{dV}{d\theta} = K \left[ (2\sin\theta\cos\theta)\cos\theta + \sin^2\theta(-\sin\theta) \right] \] \[ \frac{dV}{d\theta} = K [2\sin\theta\cos^2\theta - \sin^3\theta] \] 3. Find Critical Points:
Set \( \frac{dV}{d\theta} = 0 \):
\[ K [2\sin\theta\cos^2\theta - \sin^3\theta] = 0 \] Factor out \( \sin\theta \):
\[ K\sin\theta(2\cos^2\theta - \sin^2\theta) = 0 \] For a cone, \( \theta \in (0, \pi/2) \), so \( \sin\theta \neq 0 \). Therefore, we must have:
\[ 2\cos^2\theta - \sin^2\theta = 0 \] \[ 2\cos^2\theta = \sin^2\theta \] Divide by \( \cos^2\theta \) (which is non-zero for the domain):
\[ 2 = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta \] \[ \tan\theta = \sqrt{2} (\text{since } \theta \text{ is acute, } \tan\theta>0) \] \[ \theta = \tan^{-1}(\sqrt{2}) \] 4. Second Derivative Test:
To confirm this is a maximum, we check the sign of the second derivative.
\[ \frac{d^2V}{d\theta^2} = K \frac{d}{d\theta} [2\sin\theta\cos^2\theta - \sin^3\theta] \] \[ \frac{d^2V}{d\theta^2} = K [2\cos^3\theta - 4\sin^2\theta\cos\theta - 3\sin^2\theta\cos\theta] \] \[ \frac{d^2V}{d\theta^2} = K \cos\theta [2\cos^2\theta - 7\sin^2\theta] \] At the critical point, we know \( \sin^2\theta = 2\cos^2\theta \). Substituting this:
\[ \frac{d^2V}{d\theta^2} = K \cos\theta [2\cos^2\theta - 7(2\cos^2\theta)] = K \cos\theta [-12\cos^2\theta] = -12K\cos^3\theta \] Since \( K>0 \) and \( \cos\theta>0 \) for \( \theta \in (0, \pi/2) \), we have \( \frac{d^2V}{d\theta^2}<0 \).
This confirms that the volume is maximum at \( \theta = \tan^{-1}(\sqrt{2}) \).
Step 4: Final Answer:
The semi-vertical angle of a cone with a given slant height and maximum volume is \( \tan^{-1}(\sqrt{2}) \). Hence proved.