Exercise 6.3 · Q25: Given slant height, greatest volume: angle $\tan^{-1}\sqrt{2}$

Basics you need

Solution

1. The variable and its range

   Let $\theta$ be the semi-vertical angle of a cone with slant height $l$⁠. In the right triangle formed by the axis, a base radius and the slant side (Basic 1),

   $$r = l\sin\theta, \qquad h = l\cos\theta.$$

   A genuine cone needs $r > 0$ and $h > 0$⁠, which happens exactly for $0 < \theta < \frac{\pi}{2}$⁠; every such $\theta$ gives a cone of slant height $l$⁠.

   The question asks about the angle, so the angle is the natural variable. With $l$ fixed, $\theta$ alone decides the cone. At $\theta = 0$ the cone collapses to a line, at $\theta = \frac{\pi}{2}$ to a flat disc, so both ends are excluded.
   In a right triangle, the side opposite an angle is hypotenuse × sine and the adjacent side is hypotenuse × cosine; here the hypotenuse is $l$⁠, the opposite side $r$ and the adjacent side $h$⁠. Basics: interval notation

2. The volume as a function of $\theta$

   $$\begin{aligned} V(\theta) &= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\,(l\sin\theta)^2\,(l\cos\theta) \\ &= \frac{\pi l^3}{3}\,\sin^2\theta\,\cos\theta, \quad 0 < \theta < \frac{\pi}{2}. \end{aligned}$$
   This is the quantity to maximise. The factor $\frac{\pi l^3}{3}$ is a positive constant, so all the variation is in $\sin^2\theta\cos\theta$⁠.
   Volume of a cone $\frac{1}{3}\pi r^2 h$⁠. Basics: building the function to optimise

3. Differentiate

   Product rule, with $\frac{d}{d\theta}\sin^2\theta = 2\sin\theta\cos\theta$ (chain rule) and $\frac{d}{d\theta}\cos\theta = -\sin\theta$⁠:

   $$\begin{aligned} V'(\theta) &= \frac{\pi l^3}{3}\left(2\sin\theta\cos^2\theta - \sin^3\theta\right) \\ &= \frac{\pi l^3}{3}\,\sin\theta\left(2\cos^2\theta - \sin^2\theta\right) \\ &= \frac{\pi l^3}{3}\,\sin\theta\cos^2\theta\left(2 - \tan^2\theta\right). \end{aligned}$$
   The last line divides the bracket by $\cos^2\theta$ and multiplies it back outside, which is allowed because $\cos\theta \ne 0$ on $\left(0, \frac{\pi}{2}\right)$⁠. It leaves one factor whose sign can change, $2 - \tan^2\theta$⁠, and that is the form we want for both the zero and the sign.
   Product rule: $(uv)' = u'v + uv'$ with $u = \sin^2\theta$⁠, $v = \cos\theta$⁠. Chain rule: inner $\sin\theta$⁠, outer $u^2$⁠. Basics: the chain rule

4. The critical point

   On $\left(0, \frac{\pi}{2}\right)$⁠, $\sin\theta > 0$ and $\cos^2\theta > 0$⁠. So $V'(\theta) = 0$ exactly when $\tan^2\theta = 2$⁠, and since $\tan\theta > 0$ there,

   $$\tan\theta = \sqrt{2}, \qquad \theta = \tan^{-1}\sqrt{2}.$$

   The principal value $\tan^{-1}\sqrt{2}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and is positive because $\sqrt{2} > 0$⁠, so it is in our range $\left(0, \frac{\pi}{2}\right)$⁠. Because $\tan$ is strictly increasing on this interval, it takes the value $\sqrt{2}$ only once: this is the only critical point. Call it $\theta_0$ (about $54.7356^\circ$⁠).

   We reject $\tan\theta = -\sqrt{2}$ because the tangent of an angle between 0 and $\frac{\pi}{2}$ is positive. Writing the answer as $\tan^{-1}\sqrt{2}$ is valid precisely because $\theta_0$ lies inside the principal range of $\tan^{-1}$⁠.
   Range of the principal value branch of $\tan^{-1}$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ (Chapter 2). Basics: signs of sin and cos

5. Second derivative test

   Differentiate $2\sin\theta\cos^2\theta - \sin^3\theta$ term by term:

   $$\begin{aligned} V''(\theta) &= \frac{\pi l^3}{3}\left(2\cos^3\theta - 7\sin^2\theta\cos\theta\right) \\ &= \frac{\pi l^3}{3}\cos\theta\left(2\cos^2\theta - 7\sin^2\theta\right). \end{aligned}$$

   At $\theta_0$⁠, $\tan^2\theta_0 = 2$ gives $\cos^2\theta_0 = \frac{1}{3}$ and $\sin^2\theta_0 = \frac{2}{3}$⁠, so the bracket is $\frac{2}{3} - \frac{14}{3} = -4$ and

   $$V''(\theta_0) = \frac{\pi l^3}{3}\cdot\frac{1}{\sqrt{3}}\cdot(-4) < 0.$$

   By the second derivative test, $\theta_0$ is a point of local maxima of $V$⁠.

   For the first term, by the product rule, $\frac{d}{d\theta}(2\sin\theta\cos^2\theta)$ equals $2\cos^3\theta - 4\sin^2\theta\cos\theta$⁠; for the second, $\frac{d}{d\theta}\sin^3\theta = 3\sin^2\theta\cos\theta$⁠. Together: $2\cos^3\theta - 7\sin^2\theta\cos\theta$⁠.
   Theorem 4 (i), Section 6.4. $\cos^2\theta = \frac{1}{1 + \tan^2\theta} = \frac{1}{3}$ and $\sin^2\theta = 1 - \cos^2\theta$⁠. Lesson: first and second derivative tests

6. It is the greatest volume of all

   From Step 3, $V'(\theta)$ has the sign of $2 - \tan^2\theta$ on $\left(0, \frac{\pi}{2}\right)$⁠. As $\tan\theta$ increases through $\sqrt{2}$⁠:

   • for $0 < \theta < \theta_0$⁠: $0 < \tan\theta < \sqrt{2}$⁠, so $V'(\theta) > 0$⁠;
   • for $\theta_0 < \theta < \frac{\pi}{2}$⁠: $\tan\theta > \sqrt{2}$⁠, so $V'(\theta) < 0$⁠.

   So $V$ increases on $(0, \theta_0]$ and decreases on $\left[\theta_0, \frac{\pi}{2}\right)$⁠: $V(\theta) < V(\theta_0)$ for every other allowed $\theta$⁠. At both ends $V \to 0$ ($\sin\theta \to 0$ at one end, $\cos\theta \to 0$ at the other).

   "Cone of the maximum volume" means greatest among all cones with this slant height. One critical point, with $V$ rising before it and falling after it, is the top of the whole graph.
   Theorem 1, Section 6.3. Lesson: increasing and decreasing functions

7. Conclusion, and a cross-check with the height

   The volume is greatest when the semi-vertical angle is $\theta = \tan^{-1}\sqrt{2}$. Its value is $V(\theta_0) = \frac{\pi l^3}{3}\cdot\frac{2}{3}\cdot\frac{1}{\sqrt{3}}$ $= \frac{2\sqrt{3}\,\pi l^3}{27}$⁠.

   Cross-check using the height as the variable: $r^2 = l^2 - h^2$⁠, so $V = \frac{\pi}{3}(l^2 h - h^3)$ on $0 < h < l$⁠, and $V' = \frac{\pi}{3}(l^2 - 3h^2) = 0$ gives $h = \frac{l}{\sqrt{3}}$⁠. Then $r = \sqrt{l^2 - \frac{l^2}{3}} = \sqrt{\frac{2}{3}}\,l$⁠, and $\tan\theta = \frac{r}{h} = \sqrt{2}$⁠, the same angle.

   Two different choices of variable lead to the same cone, which is strong evidence that no slip was made. The second route avoids trigonometry until the very end.
   $\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{r}{h}$ in the cone's right triangle.

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