Miscellaneous · Q14: Cylinder of maximum volume in a sphere of radius $R$⁠: height $\frac{2R}{\sqrt3}$

Basics you need

Solution

1. The cross-section and the variable

   Every point of either rim is at distance $R$ from the centre $O$ of the sphere, so $O$ lies on the cylinder's axis, and by symmetry it is halfway between the two ends. Cut through the axis: the sphere becomes a circle of radius $R$ and the cylinder a rectangle inscribed in it.

   Let $h$ be the height of the cylinder and $r$ its radius.

   Both rims have the same radius $r$⁠, so each lies in a plane at the same distance $d = \sqrt{R^2 - r^2}$ from $O$⁠. Two different parallel planes at the same distance from $O$ must lie on opposite sides of it, so $O$ is halfway between the faces and $d = \frac{h}{2}$⁠. Choosing $h$ therefore fixes everything else.
   Points of a sphere are all at distance $R$ from the centre. Basics: building the function

2. The allowed range and the radius

   Pythagoras in the right triangle formed by $O$⁠, the centre of the top face and a point on the top rim:

   $$r^2 + \left(\frac{h}{2}\right)^2 = R^2 \quad\Longrightarrow\quad r^2 = R^2 - \frac{h^2}{4}.$$

   A genuine cylinder needs $h > 0$ and $r > 0$⁠, that is $h^2 < 4R^2$⁠:

   $$0 < h < 2R.$$
   At $h = 2R$ the cylinder would collapse to a diameter ($r = 0$⁠), and at $h = 0$ to a flat disc; neither holds any volume, so both ends are excluded.
   Pythagoras' theorem; $R^2 - \frac{h^2}{4} > 0 \iff h < 2R$ for $h > 0$⁠. Basics: interval notation

3. The volume as a function of $h$

   $$\begin{aligned} V(h) &= \pi r^2 h = \pi\left(R^2 - \frac{h^2}{4}\right)h \\ &= \pi R^2 h - \frac{\pi h^3}{4}, \qquad 0 < h < 2R. \end{aligned}$$
   Substituting $r^2$ leaves a polynomial in the single variable $h$⁠; $R$ is a fixed constant.
   $V = \pi r^2 h$ (Basic 1). Basics: area and volume formulas

4. Critical point

   $$V'(h) = \pi R^2 - \frac{3\pi h^2}{4}.$$ $$\begin{aligned} V'(h) = 0 &\iff h^2 = \frac{4R^2}{3} \\ &\iff h = \frac{2R}{\sqrt{3}} \quad (h > 0). \end{aligned}$$

   Since $\frac{2}{\sqrt3} \approx 1.155 < 2$⁠, this $h$ lies in $(0, 2R)$⁠.

   The negative root $-\frac{2R}{\sqrt3}$ is not a height. $V$ is a polynomial, differentiable everywhere, so this is the only critical point in the range.
   Critical point (Note after Theorem 2, Section 6.4). Lesson: maxima and minima

5. It is a maximum, and the largest on the whole range

   $$V''(h) = -\frac{3\pi h}{2} < 0 \quad\text{for every } h > 0,$$

   so at $h = \frac{2R}{\sqrt3}$ the Second Derivative Test (Theorem 4 (i)) gives a point of local maxima. For the whole range, factor $V'$⁠:

   $$V'(h) = \frac{3\pi}{4}\left(\frac{2R}{\sqrt3} - h\right)\left(\frac{2R}{\sqrt3} + h\right).$$

   The last bracket is positive, so $V' > 0$ for $0 < h < \frac{2R}{\sqrt3}$ and $V' < 0$ for $\frac{2R}{\sqrt3} < h < 2R$⁠. $V$ increases up to $\frac{2R}{\sqrt3}$ and decreases after it, so $V(h) < V\left(\frac{2R}{\sqrt3}\right)$ for every other allowed $h$⁠.

   The second derivative test is local; the factorised $V'$ gives its sign everywhere in the range, which proves the critical point is the overall maximum. (Here $V'' < 0$ on the whole range is a second reason: $V'$ is decreasing, so it is positive before its only zero and negative after.)
   $\frac{3\pi}{4}\left(\frac{4R^2}{3} - h^2\right) = \pi R^2 - \frac{3\pi h^2}{4}$⁠, and a difference of squares factors. Theorem 1, Section 6.3. Lesson: increasing and decreasing

6. The maximum volume

   At $h = \frac{2R}{\sqrt3}$⁠: $\frac{h^2}{4} = \frac{R^2}{3}$⁠, so $r^2 = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}$⁠. Then

   $$\begin{aligned} V_{\max} &= \pi\cdot\frac{2R^2}{3}\cdot\frac{2R}{\sqrt3} \\ &= \frac{4\pi R^3}{3\sqrt{3}} = \frac{4\sqrt3}{9}\,\pi R^3. \end{aligned}$$
   Computing $r^2$ first keeps the arithmetic clean. The two forms are equal: multiply top and bottom of $\frac{4}{3\sqrt3}$ by $\sqrt3$ to get $\frac{4\sqrt3}{9}$⁠. The best cylinder has radius $r = \sqrt{\frac23}\,R \approx 0.816R$⁠.
   Rationalising the denominator (Basic 4). Basics: algebra

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