Question

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

Hint:

In optimization problems involving geometric shapes, always try to express the quantity to be maximized or minimized (like volume or area) as a function of a single variable (like an angle or a side length). Use the given constraints (like fixed slant height or surface area) to establish relationships between the variables.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem where we need to find the semi-vertical angle of a cone that maximizes its volume, given that its slant height is constant. We will use calculus, specifically the first and second derivative tests, to find the maximum.
Step 2: Key Formula or Approach:
Let \(l\) be the given slant height, \(h\) be the height, \(r\) be the radius, and \(\theta\) be the semi-vertical angle of the cone.
The volume of the cone is given by the formula:
\[ V = \frac{1}{3}\pi r^2 h \] The relationships between \(r\), \(h\), \(l\), and \(\theta\) are:
\[ r = l \sin\theta \] \[ h = l \cos\theta \] We need to express the volume \(V\) as a function of a single variable, \(\theta\), and then find the value of \(\theta\) for which \(V\) is maximum.
Step 3: Detailed Explanation:
First, substitute the expressions for \(r\) and \(h\) in terms of \(l\) and \(\theta\) into the volume formula:
\[ V(\theta) = \frac{1}{3}\pi (l \sin\theta)^2 (l \cos\theta) \] \[ V(\theta) = \frac{1}{3}\pi l^3 \sin^2\theta \cos\theta \] Since \(l\) is a constant, to maximize \(V\), we need to maximize the function \(f(\theta) = \sin^2\theta \cos\theta\).
Now, we differentiate \(V\) with respect to \(\theta\):
\[ \frac{dV}{d\theta} = \frac{1}{3}\pi l^3 \frac{d}{d\theta}(\sin^2\theta \cos\theta) \] Using the product rule \((uv)' = u'v + uv'\), let \(u = \sin^2\theta\) and \(v = \cos\theta\).
\[ u' = 2\sin\theta\cos\theta \] \[ v' = -\sin\theta \] \[ \frac{d}{d\theta}(\sin^2\theta \cos\theta) = (2\sin\theta\cos\theta)(\cos\theta) + (\sin^2\theta)(-\sin\theta) \] \[ = 2\sin\theta\cos^2\theta - \sin^3\theta \] So,
\[ \frac{dV}{d\theta} = \frac{1}{3}\pi l^3 (2\sin\theta\cos^2\theta - \sin^3\theta) \] For maximum or minimum volume, we set \(\frac{dV}{d\theta} = 0\):
\[ \frac{1}{3}\pi l^3 (\sin\theta)(2\cos^2\theta - \sin^2\theta) = 0 \] Since \(l \neq 0\) and for a cone, \(\theta \in (0, \pi/2)\), \(\sin\theta \neq 0\). Thus, we must have:
\[ 2\cos^2\theta - \sin^2\theta = 0 \] \[ 2\cos^2\theta = \sin^2\theta \] Dividing both sides by \(\cos^2\theta\) (which is non-zero for \(\theta \in (0, \pi/2)\)):
\[ \tan^2\theta = 2 \] \[ \tan\theta = \sqrt{2} \] This gives the critical point \(\theta = \tan^{-1}(\sqrt{2})\).
To confirm this is a maximum, we use the second derivative test. It's often easier to check the sign of the first derivative around the critical point. Or, we can analyze the derivative of \(f(\theta) = \sin^2\theta \cos\theta\). Let's find the second derivative of V.
\[ \frac{d^2V}{d\theta^2} = \frac{1}{3}\pi l^3 \frac{d}{d\theta}(2\sin\theta\cos^2\theta - \sin^3\theta) \] \[ \frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [2\cos^3\theta - 4\sin^2\theta\cos\theta - 3\sin^2\theta\cos\theta] \] \[ \frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} [2\cos^3\theta - 7\sin^2\theta\cos\theta] \] \[ \frac{d^2V}{d\theta^2} = \frac{\pi l^3}{3} \cos\theta [2\cos^2\theta - 7\sin^2\theta] \] At \(\tan^2\theta = 2\), we have \(\sin^2\theta = 2\cos^2\theta\). Substituting this into the bracket:
\[ 2\cos^2\theta - 7(2\cos^2\theta) = 2\cos^2\theta - 14\cos^2\theta = -12\cos^2\theta \] Since \(\theta \in (0, \pi/2)\), \(\cos\theta>0\), so \(-12\cos^2\theta<0\).
Therefore, \(\frac{d^2V}{d\theta^2}<0\) at \(\tan\theta = \sqrt{2}\), which confirms that the volume is maximum at this angle.
Step 4: Final Answer:
The volume of the cone is maximized when the semi-vertical angle \(\theta\) satisfies \(\tan\theta = \sqrt{2}\).
Hence, the semi-vertical angle for the cone of maximum volume is \(\theta = \tan^{-1}(\sqrt{2})\).