Miscellaneous · Q15: Greatest cylinder in a cone of height $h$⁠, semi-vertical angle $\alpha$⁠: height $\frac{h}{3}$

Basics you need

Solution

1. The cross-section and the variable

   Cut through the common axis. The cone becomes an isosceles triangle with apex angle $2\alpha$ and height $h$⁠, and the cylinder becomes a rectangle standing on its base with its two top corners on the slant sides. Let

   • $x$ = height of the cylinder,
   • $r$ = radius of the cylinder.
   Once the height $x$ is chosen, the top face is at a definite level, and the slant side fixes how wide the cylinder can be there. So $x$ alone determines the cylinder.
   A cylinder "inscribed" in a cone shares its axis, stands on its base and touches its curved surface. Basics: building the function

2. The allowed range of $x$

   $$0 < x < h.$$
   A cylinder of height 0 is flat, and at height $h$ the top face would reach the apex, where the cone has width 0. Neither holds any volume, so both ends are excluded.
   Open interval: the end points are not included. Basics: interval notation

3. Radius in terms of $x$

   The top face is at height $x$ above the base, so it is at depth $h - x$ below the apex. The top corner of the rectangle lies on the slant side, so the right triangle formed by the apex, the centre of the top face and that corner has angle $\alpha$ at the apex, adjacent side $h - x$ and opposite side $r$⁠:

   $$\tan\alpha = \frac{r}{h-x} \quad\Longrightarrow\quad r = (h-x)\tan\alpha.$$
   This is the key relation. Check the extremes: at $x = 0$ it gives $r = h\tan\alpha$⁠, the radius of the cone's own base, and at $x = h$ it gives $r = 0$⁠, the apex. The small cone above the cylinder is similar to the whole cone, which is the same fact: $\frac{r}{h\tan\alpha} = \frac{h - x}{h}$⁠.
   $\tan\alpha = \frac{\text{opposite}}{\text{adjacent}}$ in a right triangle. Basics: trigonometric ratios

4. The volume as a function of $x$

   $$\begin{aligned} V(x) &= \pi r^2 x = \pi (h-x)^2\tan^2\alpha\cdot x \\ &= \pi\tan^2\alpha\,(h-x)^2\,x, \qquad 0 < x < h. \end{aligned}$$
   $h$ and $\alpha$ are fixed numbers for the given cone, so $\pi\tan^2\alpha$ is a positive constant in front, and only $(h-x)^2x$ varies with $x$⁠.
   $V = \pi r^2 \times$ height (Basic 1). Basics: area and volume formulas

5. Critical point

   Product rule on $(h-x)^2\cdot x$⁠:

   $$\begin{aligned} V'(x) &= \pi\tan^2\alpha\left[(h-x)^2 - 2x(h-x)\right] \\ &= \pi\tan^2\alpha\,(h-x)(h-3x). \end{aligned}$$

   On $0 < x < h$ we have $h - x \ne 0$ and $\tan\alpha \ne 0$⁠, so

   $$V'(x) = 0 \iff h - 3x = 0 \iff x = \frac{h}{3}.$$
   $\frac{d}{dx}(h-x)^2 = 2(h-x)\cdot(-1)$ by the chain rule, which gives the $-2x(h-x)$ term. Inside the bracket, $(h-x)$ is common: $(h-x)\left[(h-x) - 2x\right]$ $= (h-x)(h-3x)$⁠. The root $x = h$ is outside the range, so $x = \frac{h}{3}$ is the only critical point.
   Product rule $(uv)' = u'v + uv'$ with $u = (h-x)^2$⁠, $v = x$⁠. Basics: factorising

6. It is the greatest volume on the whole range

   Second derivative test. Expanding, $V'(x) = \pi\tan^2\alpha\,(h^2 - 4hx + 3x^2)$⁠, so

   $$V''(x) = \pi\tan^2\alpha\,(6x - 4h),$$ $$V''\left(\tfrac{h}{3}\right) = \pi\tan^2\alpha\,(2h - 4h) = -2\pi h\tan^2\alpha < 0.$$

   By Theorem 4 (i), $x = \frac{h}{3}$ is a point of local maxima.

   Whole range. On $0 < x < h$⁠, $\pi\tan^2\alpha > 0$ and $h - x > 0$⁠, so $V'$ has the sign of $h - 3x$⁠: positive for $0 < x < \frac{h}{3}$ and negative for $\frac{h}{3} < x < h$⁠. $V$ increases up to $\frac{h}{3}$ and decreases after, so $V(x) < V\left(\frac{h}{3}\right)$ for every other allowed $x$⁠.

   "Greatest volume" is a statement about all inscribed cylinders, so the local test alone is not enough; the sign of $V'$ on the entire range finishes the proof. The height $\frac{h}{3}$ is one-third of the cone's height, as the question states.
   Theorem 4, Section 6.4, and Theorem 1, Section 6.3. Lesson: derivative tests

7. The greatest volume

   At $x = \frac{h}{3}$⁠: $h - x = \frac{2h}{3}$⁠, so $r = \frac{2h}{3}\tan\alpha$ and

   $$\begin{aligned} V\left(\tfrac{h}{3}\right) &= \pi\tan^2\alpha\cdot\frac{4h^2}{9}\cdot\frac{h}{3} \\ &= \frac{4}{27}\pi h^3\tan^2\alpha. \end{aligned}$$
   $\left(\frac{2h}{3}\right)^2 = \frac{4h^2}{9}$⁠, and $\frac{4}{9}\cdot\frac{1}{3} = \frac{4}{27}$⁠. For comparison, the cone's own volume is $\frac13\pi(h\tan\alpha)^2 h = \frac13\pi h^3\tan^2\alpha$⁠, so the greatest cylinder fills $\frac{4/27}{1/3} = \frac{4}{9}$ of the cone.
   Volume of a cone $\frac13\pi(\text{radius})^2(\text{height})$⁠. Basics: area and volume formulas

Every fact used, and where it comes from