Question

Prove that the semi-vertical angle of a right circular cone of given surface and maximum volume is \(\sin^{-1}(1/3)\).

Hint:

In optimization problems, maximizing or minimizing a function \(f(x)\) is equivalent to maximizing or minimizing \((f(x))^2\), provided \(f(x)\) is non-negative (which volume is). This trick often eliminates square roots and makes differentiation much simpler.

Answer 1

Step 1: Understanding the Concept:
This is an optimization problem using calculus (applications of derivatives). We are given a cone with a fixed total surface area and we need to find the semi-vertical angle that maximizes its volume. The key is to express the volume as a function of a single variable (either radius, height, or the semi-vertical angle) and then use differentiation to find the maximum.
Step 2: Key Formula or Approach:
1. Let \(r\) be the radius, \(h\) the height, \(l\) the slant height, and \(\alpha\) the semi-vertical angle of the cone.
2. Write the formulas for total surface area \(S = \pi r^2 + \pi r l\) and volume \(V = \frac{1}{3}\pi r^2 h\).
3. Use the geometric relations: \(l^2 = r^2 + h^2\), \(r = l\sin\alpha\), \(h = l\cos\alpha\).
4. Since S is constant, express one variable (e.g., \(l\)) in terms of another (e.g., \(r\)) from the surface area formula.
5. Substitute this into the volume formula to get V as a function of a single variable. To make differentiation easier, it's better to work with \(V^2\).
6. Differentiate \(V^2\) with respect to the chosen variable, set the derivative to zero to find the condition for maximum volume.
7. Relate this condition back to the semi-vertical angle \(\alpha\).
Step 3: Detailed Explanation:
Let S be the constant total surface area. \(S = \pi r l + \pi r^2\).
From this, we can express the slant height \(l\) in terms of \(r\): \[ l = \frac{S - \pi r^2}{\pi r} \] The volume of the cone is \(V = \frac{1}{3}\pi r^2 h\). We also know \(h = \sqrt{l^2 - r^2}\).
To simplify the differentiation, we will maximize \(V^2\) instead of V. \[ V^2 = \frac{1}{9}\pi^2 r^4 h^2 = \frac{1}{9}\pi^2 r^4 (l^2 - r^2) \] Substitute the expression for \(l\): \[ V^2 = \frac{1}{9}\pi^2 r^4 \left[ \left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2 \right] \] \[ V^2 = \frac{1}{9}\pi^2 r^4 \left[ \frac{(S - \pi r^2)^2 - (\pi r^2)^2}{\pi^2 r^2} \right] = \frac{1}{9} r^2 [(S - \pi r^2)^2 - (\pi r^2)^2] \] Using the difference of squares \(a^2-b^2 = (a-b)(a+b)\): \[ V^2 = \frac{1}{9} r^2 (S - \pi r^2 - \pi r^2)(S - \pi r^2 + \pi r^2) = \frac{1}{9} r^2 (S - 2\pi r^2)(S) \] Let \(Z = V^2\). Then \(Z = \frac{S}{9}(Sr^2 - 2\pi r^4)\).
To find the maximum, we differentiate Z with respect to \(r\) and set it to zero: \[ \frac{dZ}{dr} = \frac{S}{9}(2Sr - 8\pi r^3) \] Setting \(\frac{dZ}{dr} = 0\): \[ \frac{S}{9}(2Sr - 8\pi r^3) = 0 \implies 2Sr = 8\pi r^3 \] Since \(r \neq 0\), we can divide by \(2r\): \[ S = 4\pi r^2 \] This is the condition for maximum volume. Now we relate this back to the semi-vertical angle \(\alpha\).
Substitute \(S = \pi r l + \pi r^2\) back into this condition: \[ \pi r l + \pi r^2 = 4\pi r^2 \] \[ \pi r l = 3\pi r^2 \] \[ l = 3r \] The semi-vertical angle \(\alpha\) is defined by \(\sin\alpha = \frac{r}{l}\).
Substituting \(l=3r\): \[ \sin\alpha = \frac{r}{3r} = \frac{1}{3} \] \[ \alpha = \sin^{-1}\left(\frac{1}{3}\right) \] (One can verify this is a maximum by checking the second derivative, which will be negative).
Step 4: Final Answer:
The semi-vertical angle of the cone with a given surface area and maximum volume is \(\alpha = \sin^{-1}(1/3)\).