Exercise 6.3 · Q20: Cylinder of given total surface and maximum volume: height $=$ diameter

Basics you need

Solution

1. The fixed surface, in symbols

   Let the cylinder have base radius $r$ and height $h$⁠, and let $S$ be its given (fixed) total surface area. "Surface" here means the whole surface of the closed cylinder: two circular ends and the curved side. So

   $$S = 2\pi r^2 + 2\pi rh.$$
   Opening out the curved side gives a rectangle of width $h$ and length equal to the circumference $2\pi r$⁠, so its area is $2\pi rh$⁠. Each end is a disc of area $\pi r^2$⁠. Why the total surface of a closed cylinder? The book only says "given surface". If only the curved surface $2\pi rh$ were fixed, the volume would have no maximum at all (Common mistakes, item 1); with one end open the same method gives $h = r$ (What to take away, item 1). Of these three readings, only the closed cylinder gives the stated result $h = 2r$⁠, and it is the usual meaning of the surface of a cylinder.
   Total surface area of a closed cylinder $= 2\pi r^2 + 2\pi rh$⁠. Basics: area and volume formulas

2. Height in terms of radius, and the allowed range

   $$h = \frac{S - 2\pi r^2}{2\pi r}.$$

   A real cylinder needs $r > 0$ and $h > 0$⁠; the second means $S - 2\pi r^2 > 0$⁠, that is $r^2 < \frac{S}{2\pi}$⁠. So

   $$0 < r < \sqrt{\frac{S}{2\pi}}.$$
   With the surface fixed, a wider cylinder must be shorter. At the upper end of the range the two discs alone use up all the surface and the height is 0. Every $r$ in the range gives exactly one cylinder with surface $S$⁠, so maximising over this range covers all such cylinders.
   Subtract $2\pi r^2$ and divide by $2\pi r$ (allowed, since $r > 0$⁠). Basics: interval notation

3. The volume as a function of $r$

   $$\begin{aligned} V(r) &= \pi r^2 h \\ &= \pi r^2 \cdot \frac{S - 2\pi r^2}{2\pi r} \\ &= \frac{r(S - 2\pi r^2)}{2} \\ &= \frac{Sr - 2\pi r^3}{2}. \end{aligned}$$
   Substituting for $h$ leaves $r$ as the only variable; $S$ and $\pi$ are constants. The factor $\pi r$ cancels between numerator and denominator.
   $\frac{\pi r^2}{2\pi r} = \frac{r}{2}$⁠. Basics: building the function

4. Critical point

   $$V'(r) = \frac{S - 6\pi r^2}{2},\qquad V''(r) = -6\pi r.$$

   $V'(r) = 0 \iff S = 6\pi r^2$⁠, that is $r = \sqrt{\frac{S}{6\pi}}$ (taking the positive root). It is inside the range, because $\frac{S}{6\pi} < \frac{S}{2\pi}$⁠.

   Differentiate term by term: $\frac{d}{dr}(Sr) = S$ and $\frac{d}{dr}(2\pi r^3) = 6\pi r^2$⁠. A radius is positive, so the negative square root is rejected.
   Power rule with constants $S$ and $\pi$⁠. Basics: derivatives

5. It gives the largest volume

   Let $r_0 = \sqrt{\frac{S}{6\pi}}$⁠. First, $V''(r_0) = -6\pi r_0 < 0$⁠, so $r_0$ is a point of local maxima (Theorem 4). Second, the sign of $V'(r) = \frac{S - 6\pi r^2}{2}$ on the whole range:

   • for $0 < r < r_0$⁠: $6\pi r^2 < S$⁠, so $V'(r) > 0$ and $V$ is increasing;
   • for $r_0 < r < \sqrt{\frac{S}{2\pi}}$⁠: $6\pi r^2 > S$⁠, so $V'(r) < 0$ and $V$ is decreasing.

   So $V(r) < V(r_0)$ for every other cylinder with the same surface.

   The second derivative test is local; the sign of $V'$ across the whole range is what proves "maximum volume" among all cylinders of the given surface.
   Theorem 4, Section 6.4; Theorem 1, Section 6.3. Lesson: derivative tests

6. At the maximum, height equals diameter

   At $r = r_0$ we have $S = 6\pi r_0^2$⁠. Substitute this into the formula for $h$ from Step 2:

   $$\begin{aligned} h &= \frac{6\pi r_0^2 - 2\pi r_0^2}{2\pi r_0} \\ &= \frac{4\pi r_0^2}{2\pi r_0} = 2r_0. \end{aligned}$$

   So the height equals $2r_0$⁠, the diameter of the base, which is what was to be shown.

   The relation $S = 6\pi r^2$ is the clean form of the optimum condition; using it avoids square roots. The largest volume itself is $V = \pi r_0^2 \cdot 2r_0 = 2\pi r_0^3$⁠.
   Diameter $= 2 \times$ radius. Basics: area and volume formulas

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