Example 26: Cylinder in a cone: greatest curved surface at radius $\tfrac{r}{2}$

Basics you need

Solution

1. Look at the cross-section through the axis

   Cut the cone and cylinder by a plane through the axis OA. The cone gives the triangle ABC (apex A, base BC with centre O); the cylinder gives the rectangle DEQP, whose top corners P and Q lie on the slanting sides AB and AC.

   The cylinder's radius $x$ can be any number with $0 < x < r$⁠.

   A 3D problem about a cone and a cylinder becomes a flat problem about a triangle and a rectangle. "Inscribed" means the cylinder stands on the cone's base and its top rim touches the cone, so Q is on AC.
   For $x = 0$ or $x = r$ there is no real cylinder (no width or no height), so the range is open. Basics: interval notation

2. Height of the cylinder from similar triangles

   Triangles QEC and AOC are right-angled at E and O and share the angle at C, so they are similar:

   $$\frac{\text{QE}}{\text{OA}} = \frac{\text{EC}}{\text{OC}}, \quad\text{that is}\quad \frac{\text{QE}}{h} = \frac{r - x}{r}.$$ $$\text{QE} = \frac{h(r - x)}{r}.$$
   The height is what ties the cylinder to the cone. EC = OC − OE = $r - x$⁠. The formula behaves sensibly: $x$ close to 0 gives a height close to $h$ (a thin tall cylinder), $x$ close to $r$ gives a height close to 0 (a flat disc).
   Similar triangles have proportional corresponding sides (Basic 2).

3. The curved surface area as a function of $x$

   $$\begin{aligned} S(x) &= 2\pi x \cdot \frac{h(r - x)}{r} \\ &= \frac{2\pi h}{r}\,(rx - x^2), \quad 0 < x < r. \end{aligned}$$
   This is the quantity to maximise. $r$ and $h$ are fixed by the given cone, so $\frac{2\pi h}{r}$ is just a positive constant; only $x$ varies.
   Curved surface area $2\pi x H$ with $H = \text{QE}$ (Basic 1). Basics: building the function to optimise

4. Critical point

   $$S'(x) = \frac{2\pi h}{r}\,(r - 2x), \qquad S'(x) = 0 \iff x = \frac{r}{2}.$$

   $\frac{r}{2}$ lies in $(0, r)$⁠. $S$ is a polynomial in $x$⁠, so there is no other critical point.

   The constant factor $\frac{2\pi h}{r}$ is not zero, so $S' = 0$ exactly when $r - 2x = 0$⁠.
   Constant multiple and power rules: $\frac{d}{dx}(rx - x^2) = r - 2x$⁠.

5. Second derivative test

   $$S''(x) = -\frac{4\pi h}{r} < 0 \text{ for all } x, \text{ so } S''\!\left(\tfrac{r}{2}\right) < 0.$$

   By the second derivative test, $x = \frac{r}{2}$ is a point of local maxima of $S$⁠.

   This is the book's argument: a negative second derivative means the slope is falling, so the graph has a hill at $\frac{r}{2}$⁠.
   Theorem 4(i), Section 6.4. Lesson: first and second derivative tests

6. It is the greatest value on the whole range

   Since $\frac{2\pi h}{r} > 0$⁠, $S'(x)$ has the sign of $r - 2x$⁠: positive for $0 < x < \frac{r}{2}$⁠, negative for $\frac{r}{2} < x < r$⁠. So $S$ increases up to $x = \frac{r}{2}$ and decreases after it, and $S(x) < S\!\left(\frac{r}{2}\right)$ for every other allowed $x$⁠. Equivalently,

   $$S(x) = \frac{2\pi h}{r}\left[\frac{r^2}{4} - \left(x - \frac{r}{2}\right)^2\right] \le \frac{\pi r h}{2}.$$
   "Greatest" in the question means greatest among all inscribed cylinders, not just nearby ones; the sign of $S'$ over the whole range settles that. Near the ends of the range $S$ is close to 0.
   Theorem 1, Section 6.3. Lesson: optimisation problems

7. Conclusion

   The curved surface area is greatest when the cylinder's radius is $x = \frac{r}{2}$⁠, half the radius of the cone. Then its height is $\frac{h(r - r/2)}{r} = \frac{h}{2}$⁠, and the greatest curved surface area is $\frac{\pi r h}{2}$⁠.

   The question only asks us to prove the radius statement; the height and the area are useful extras that also show the best cylinder reaches exactly halfway up the cone.

Every fact used, and where it comes from