Exercise 6.3 · Q26: Given surface area, greatest volume: angle $\sin^{-1}\tfrac{1}{3}$

Basics you need

Solution

1. Slant height in terms of $r$

   Let the cone have radius $r$⁠, height $h$⁠, slant height $l$ and semi-vertical angle $\theta$⁠. The surface area is fixed:

   $$\pi r l + \pi r^2 = S, \qquad\text{so}\qquad l = \frac{S - \pi r^2}{\pi r}.$$
   "Surface area" of a cone means its total surface: the curved part and the circular base. (Step 8 shows that with only the curved surface fixed the answer would be a different angle, so the stated result confirms this reading.) Solving for $l$ leaves $r$ as the only variable.
   Total surface area of a cone (Basic 1).

2. The allowed range of $r$

   A genuine cone needs $h > 0$⁠, that is $l > r$⁠:

   $$\frac{S - \pi r^2}{\pi r} > r \iff S - \pi r^2 > \pi r^2 \iff r^2 < \frac{S}{2\pi}.$$

   So the range is $0 < r < \sqrt{\frac{S}{2\pi}}$⁠, and every $r$ in it gives a cone with surface area $S$⁠.

   The slant side is the hypotenuse of the cone's right triangle, so it must be longer than the radius. If $r$ were too large, the base alone would use up so much of $S$ that no proper cone could be made. Multiplying the inequality by $\pi r > 0$ keeps its direction.
   At $r^2 = \frac{S}{2\pi}$ the slant height equals the radius and the cone is flat ($h = 0$⁠). Basics: interval notation

3. Height squared in terms of $r$

   $$\begin{aligned} h^2 &= l^2 - r^2 = \frac{(S - \pi r^2)^2}{\pi^2 r^2} - r^2 \\ &= \frac{S^2 - 2\pi S r^2 + \pi^2 r^4 - \pi^2 r^4}{\pi^2 r^2} \\ &= \frac{S(S - 2\pi r^2)}{\pi^2 r^2}. \end{aligned}$$
   The volume needs $h$⁠, and Pythagoras gives $h^2$⁠. The $\pi^2 r^4$ terms cancel, which is what makes the next step simple. The numerator is positive exactly on the range of Step 2, as it should be.
   $(S - \pi r^2)^2 = S^2 - 2\pi S r^2 + \pi^2 r^4$⁠; and $r^2 = \frac{\pi^2 r^4}{\pi^2 r^2}$ to use a common denominator.

4. The square of the volume

   $$\begin{aligned} V^2 &= \frac{1}{9}\pi^2 r^4 h^2 = \frac{1}{9}\pi^2 r^4 \cdot \frac{S(S - 2\pi r^2)}{\pi^2 r^2} \\ &= \frac{S}{9}\left(S r^2 - 2\pi r^4\right). \end{aligned}$$

   Let $f(r) = \frac{S}{9}\left(S r^2 - 2\pi r^4\right)$ on $0 < r < \sqrt{\frac{S}{2\pi}}$⁠. Since $V > 0$⁠, the greatest $V$ occurs exactly where $f$ is greatest (Basic 3).

   $V = \frac{1}{3}\pi r^2 h$ involves $h = \sqrt{\dots}$⁠; its square is a plain polynomial in $r$⁠. Squaring is safe because it keeps the order of positive numbers.
   $\left(\frac{1}{3}\pi r^2 h\right)^2 = \frac{1}{9}\pi^2 r^4 h^2$⁠. Basics: building the function to optimise

5. Differentiate and find the critical point

   $$f'(r) = \frac{S}{9}\left(2Sr - 8\pi r^3\right) = \frac{2Sr}{9}\left(S - 4\pi r^2\right).$$

   On the range, $\frac{2Sr}{9} > 0$⁠, so $f'(r) = 0$ exactly when $S = 4\pi r^2$⁠, that is

   $$r = \sqrt{\frac{S}{4\pi}}.$$

   Call this $c$⁠. Since $\frac{S}{4\pi} < \frac{S}{2\pi}$⁠, $c$ is inside the range, and it is the only critical point there.

   The positive factor $\frac{2Sr}{9}$ cannot be zero on the range, so only the bracket matters. Keep the condition as $S = 4\pi c^2$⁠: it is exactly what Step 7 needs.
   Power rule; $S$ is a constant. Theorem 2, Section 6.4 (extrema inside an open interval only at critical points).

6. Second derivative test

   $$\begin{aligned} f''(r) &= \frac{S}{9}\left(2S - 24\pi r^2\right), \\ f''(c) &= \frac{S}{9}(2S - 6S) = -\frac{4S^2}{9} < 0. \end{aligned}$$

   So $r = c$ is a point of local maxima of $f$⁠, by the second derivative test.

   At $c$⁠, $24\pi c^2 = 6 \cdot 4\pi c^2 = 6S$⁠. A negative second derivative means the graph bends downwards: a hilltop.
   Theorem 4 (i), Section 6.4. Lesson: first and second derivative tests

7. It is the greatest volume of all

   From Step 5, $f'(r)$ has the sign of $S - 4\pi r^2$⁠, which decreases as $r$ increases: positive for $0 < r < c$⁠, negative for $c < r < \sqrt{\frac{S}{2\pi}}$⁠. So $f$ increases up to $c$ and decreases after it, and $f(r) < f(c)$ for every other allowed $r$⁠. Since $V = \sqrt{f}$⁠, also $V(r) < V(c)$⁠.

   At the ends, $f = \frac{S}{9}r^2(S - 2\pi r^2) \to 0$⁠: as $r \to 0^+$ the cone is a thin needle, and as $r^2 \to \frac{S}{2\pi}$ it is flat.

   "Maximum volume" means greatest among all cones with this surface area, so the sign of $f'$ over the whole range is needed, not just the local test.
   Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

8. At the best cone $l = 3r$⁠, so $\sin\theta = \frac{1}{3}$

   Use $S = 4\pi c^2$ in the formula for $l$ from Step 1:

   $$l = \frac{S - \pi c^2}{\pi c} = \frac{4\pi c^2 - \pi c^2}{\pi c} = \frac{3\pi c^2}{\pi c} = 3c.$$

   So $\sin\theta = \frac{r}{l} = \frac{c}{3c} = \frac{1}{3}$⁠. The semi-vertical angle is acute, and on $\left(0, \frac{\pi}{2}\right)$ the sine takes the value $\frac{1}{3}$ exactly once; the principal value $\sin^{-1}\frac{1}{3}$ lies in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and is positive, so

   $$\theta = \sin^{-1}\left(\frac{1}{3}\right) \approx 19.47^\circ.$$

   If instead only the curved surface were fixed, $\pi r l = K$⁠, the same method would give $l = \frac{K}{\pi r}$ and

   $$\begin{aligned} V^2 &= \tfrac{1}{9}\pi^2 r^4 (l^2 - r^2) \\ &= \tfrac{1}{9}\left(K^2 r^2 - \pi^2 r^6\right), \end{aligned}$$

   whose derivative $\frac{1}{9}\left(2K^2 r - 6\pi^2 r^5\right)$ is zero when $K^2 = 3\pi^2 r^4$⁠, that is $\pi^2 r^2 l^2 = 3\pi^2 r^4$⁠, $l = \sqrt{3}\,r$⁠. That gives $\sin\theta = \frac{1}{\sqrt{3}}$⁠, not $\frac{1}{3}$⁠; so the question's "surface area" is the total surface area, as used here.

   The question asks for the angle, and $\sin\theta = \frac{r}{l}$ needs only the ratio $l : r$⁠, which the condition $S = 4\pi c^2$ gives directly. Writing the angle as $\sin^{-1}\frac{1}{3}$ is correct because it lies in the principal range of $\sin^{-1}$⁠.
   Range of the principal value branch of $\sin^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2). Basics: signs of sin and cos

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