Exercise 6.3 · Q24: Cone of given volume, least curved surface: $h = \sqrt{2}\,r$

Basics you need

Solution

1. Variables, the fixed volume, and the range

   Let the cone have base radius $r$⁠, height $h$ and slant height $l$⁠. The volume is fixed at $V$⁠:

   $$\frac{1}{3}\pi r^2 h = V, \qquad\text{so}\qquad h = \frac{3V}{\pi r^2}.$$

   Every $r > 0$ gives a positive height and hence a cone of volume $V$⁠, so the range is $r \in (0, \infty)$⁠.

   The fixed volume ties $h$ to $r$⁠: a wide cone must be low, a narrow one tall. Solving for $h$ leaves one free variable. Here $V$ is a given constant, not a variable.
   Volume of a cone (Basic 1). Basics: interval notation

2. The curved surface, and its square

   The curved surface is $S = \pi r l = \pi r\sqrt{r^2 + h^2}$⁠. Squaring and substituting $h^2 = \frac{9V^2}{\pi^2 r^4}$⁠:

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

   Let $f(r) = S^2 = \pi^2 r^4 + \frac{9V^2}{r^2}$ for $r > 0$⁠. Since $S > 0$⁠, the cone with the least $S$ is exactly the cone with the least $f$ (Basic 2).

   Differentiating $\pi r\sqrt{r^2 + h^2}$ directly would need the chain rule on a square root and then a messy equation. Squaring removes the root, and because squaring keeps the order of positive numbers, minimising $f$ answers the same question.
   $\left(\frac{3V}{\pi r^2}\right)^2 = \frac{9V^2}{\pi^2 r^4}$⁠. Basics: building the function to optimise

3. Differentiate and find the critical point

   $$\begin{aligned} f'(r) &= 4\pi^2 r^3 - \frac{18V^2}{r^3} \\ &= \frac{4\pi^2 r^6 - 18V^2}{r^3} \\ &= \frac{2}{r^3}\left(2\pi^2 r^6 - 9V^2\right). \end{aligned}$$

   Since $\frac{2}{r^3} > 0$ for $r > 0$⁠, $f'(r) = 0$ exactly when

   $$2\pi^2 r^6 = 9V^2, \qquad r^6 = \frac{9V^2}{2\pi^2}.$$

   A positive number has exactly one positive sixth root, so there is exactly one critical point in $(0, \infty)$⁠; call it $c$⁠, with $2\pi^2 c^6 = 9V^2$⁠.

   Writing $f'$ as a positive factor times $\left(2\pi^2 r^6 - 9V^2\right)$ shows the zero and, in Step 5, the sign. We keep the condition in the form $2\pi^2 c^6 = 9V^2$ rather than taking a sixth root: Step 6 uses it exactly in this form.
   Power rule: $\frac{d}{dr}(\pi^2 r^4) = 4\pi^2 r^3$ and $\frac{d}{dr}(9V^2 r^{-2}) = -18V^2 r^{-3}$⁠.

4. Second derivative test

   $$f''(r) = 12\pi^2 r^2 + \frac{54V^2}{r^4} > 0 \quad\text{for every } r > 0.$$

   So $f''(c) > 0$⁠, and by the second derivative test $r = c$ is a point of local minima of $f$⁠.

   Both terms of $f''$ are positive, so the graph of $f$ bends upwards everywhere and the flat point is the bottom of a valley.
   Theorem 4 (ii), Section 6.4. $\frac{d}{dr}(-18V^2 r^{-3}) = 54V^2 r^{-4}$⁠. Lesson: first and second derivative tests

5. It is the least curved surface of all

   From Step 3, $f'(r)$ has the sign of $2\pi^2 r^6 - 9V^2$⁠, which increases with $r$ and is zero at $c$⁠: negative for $0 < r < c$ and positive for $r > c$⁠. So $f$ decreases on $(0, c]$ and increases on $[c, \infty)$⁠, and $f(r) > f(c)$ for every other $r > 0$⁠. Since $S = \sqrt{f}$ and the square root keeps order, $S(r) > S(c)$ too.

   The ends agree: $f \to \infty$ as $r \to 0^+$ (the term $\frac{9V^2}{r^2}$⁠) and as $r \to \infty$ (the term $\pi^2 r^4$⁠).

   "Least curved surface" means least among all cones of volume $V$⁠. One critical point, with $f$ falling before it and rising after it, is the lowest point of the whole graph, not only of its neighbourhood.
   Theorem 1, Section 6.3 (negative derivative: decreasing; positive: increasing). Lesson: increasing and decreasing functions

6. At the best cone, $h = \sqrt{2}\,r$

   For the best cone, $r = c$ and $V = \frac{1}{3}\pi c^2 h$⁠, so $9V^2 = \pi^2 c^4 h^2$⁠. Put this into $2\pi^2 c^6 = 9V^2$⁠:

   $$\begin{aligned} 2\pi^2 c^6 &= \pi^2 c^4 h^2 \\ h^2 &= 2c^2 \\ h &= \sqrt{2}\,c. \end{aligned}$$

   (We divided by $\pi^2 c^4 \ne 0$⁠, and took the positive square root because $h > 0$ and $c > 0$⁠.) So the altitude equals $\sqrt{2}$ times the radius of the base.

   The question asks for a relation between $h$ and $r$⁠, not for their values (which depend on $V$⁠). Replacing $9V^2$ by its expression in $c$ and $h$ makes $V$ disappear and leaves exactly that relation.
   $9V^2 = \left(3V\right)^2 = \left(\pi c^2 h\right)^2$⁠.

7. What the best cone looks like

   Its slant height is $l = \sqrt{r^2 + 2r^2} = \sqrt{3}\,r$⁠, and its semi-vertical angle $\theta$ (between the axis and the slant side) has $\tan\theta = \frac{r}{h} = \frac{1}{\sqrt{2}}$⁠.

   These extras are not asked for, but they describe the shape. Compare Q25: a cone of given slant height and greatest volume has $\tan\theta = \sqrt{2}$⁠, the reciprocal ratio.

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