Exercise 6.3 · Q21: Closed can of 100 cm³ with least surface: $h = 2r$

Basics you need

Solution

1. Name the variables and use the volume

   Let the can have radius $r$ cm and height $h$ cm. The volume is fixed:

   $$\pi r^2 h = 100, \qquad\text{so}\qquad h = \frac{100}{\pi r^2}.$$

   Every positive $r$ gives a positive $h$ and hence a genuine can, so the allowed range is $r > 0$⁠, that is $r \in (0, \infty)$⁠.

   The surface area depends on both $r$ and $h$⁠, but we cannot vary them independently: making the can wider forces it to be shorter to keep 100 cm³. Solving the volume equation for $h$ expresses that link, and leaves a single variable to work with.
   $r = 0$ is excluded (no can at all), and there is no upper limit on $r$⁠: a very wide can is simply very flat. Basics: interval notation

2. The surface area as a function of $r$

   Two ends plus the curved side:

   $$\begin{aligned} S &= 2\pi r^2 + 2\pi r h \\ &= 2\pi r^2 + 2\pi r \cdot \frac{100}{\pi r^2} \\ &= 2\pi r^2 + \frac{200}{r}, \qquad r > 0. \end{aligned}$$
   This is the quantity to make as small as possible. The two terms pull in opposite directions: $2\pi r^2$ (the lids) grows as $r$ grows, while $\frac{200}{r}$ (the side) shrinks. The best can is where neither wins outright.
   Closed can: both circular ends count, so $2\pi r^2$⁠, not $\pi r^2$ (Basic 1). Basics: building the function to optimise

3. Differentiate and find the critical point

   $$\begin{aligned} S'(r) &= 4\pi r - \frac{200}{r^2} \\ &= \frac{4\pi r^3 - 200}{r^2} \\ &= \frac{4\pi}{r^2}\left(r^3 - \frac{50}{\pi}\right). \end{aligned}$$

   Since $\frac{4\pi}{r^2} > 0$⁠, $S'(r) = 0$ exactly when

   $$r^3 = \frac{50}{\pi}, \qquad r = \left(\frac{50}{\pi}\right)^{1/3}.$$

   Call this value $c$⁠. A real number has exactly one real cube root, so $c$ is the only critical point in $(0, \infty)$⁠; $S$ is differentiable at every $r > 0$⁠, so there are no others.

   Points of local maxima or minima inside an open interval can only be where the derivative is zero or does not exist (Theorem 2, Section 6.4). Writing $S'$ as a positive factor times $\left(r^3 - \frac{50}{\pi}\right)$ makes both the zero and, in Step 5, the sign easy to read.
   Power rule: $\frac{d}{dr}(2\pi r^2) = 4\pi r$ and $\frac{d}{dr}(200 r^{-1}) = -200 r^{-2}$⁠.

4. Second derivative test

   $$S''(r) = 4\pi + \frac{400}{r^3}.$$

   For every $r > 0$ both terms are positive, so in particular $S''(c) > 0$⁠. By the second derivative test, $r = c$ is a point of local minima of $S$⁠.

   $S'' > 0$ means the slope $S'$ is increasing: the graph of $S$ bends upwards like a valley, so the flat point is the bottom of a valley.
   Theorem 4 (ii), Section 6.4. $\frac{d}{dr}(-200 r^{-2}) = 400 r^{-3}$⁠. Lesson: first and second derivative tests

5. It is the least value among all cans

   From Step 3, $S'(r)$ has the sign of $r^3 - \frac{50}{\pi}$⁠, and $r^3$ increases with $r$⁠:

   • for $0 < r < c$⁠: $r^3 < c^3 = \frac{50}{\pi}$⁠, so $S'(r) < 0$⁠;
   • for $r > c$⁠: $r^3 > \frac{50}{\pi}$⁠, so $S'(r) > 0$⁠.

   Take any allowed $r \ne c$⁠. By the Mean Value Theorem on the interval between $r$ and $c$⁠, $S(r) - S(c) = S'(\xi)(r - c)$ for some $\xi$ strictly between them. If $r < c$⁠, then $S'(\xi) < 0$ and $r - c < 0$⁠; if $r > c$⁠, then $S'(\xi) > 0$ and $r - c > 0$⁠. Either way the product is positive, so

   $$S(r) > S(c) \quad\text{for every } r > 0,\ r \ne c.$$

   The ends of the range agree: as $r \to 0^+$⁠, $\frac{200}{r} \to \infty$⁠; as $r \to \infty$⁠, $2\pi r^2 \to \infty$⁠. So $S$ falls all the way down to $r = c$ and rises all the way after it.

   The second derivative test only says $c$ beats its neighbours. The question asks for the minimum among all cans, so we need the sign of $S'$ on the whole range $(0, \infty)$⁠. One critical point, with $S$ decreasing before it and increasing after it, means it is the lowest point of the entire graph.
   This is the argument in the proof of Theorem 1, Section 6.3 (a negative derivative means decreasing, a positive one increasing), which rests on the Mean Value Theorem. (The book's proof calls it "Theorem 8 in Chapter 5"; that theorem is not printed in this reprint's Chapter 5, but the statement used is exactly the one above.) Lesson: increasing and decreasing functions

6. The height of the best can

   From $c^3 = \frac{50}{\pi}$ we get $\pi c^3 = 50$⁠, so $100 = 2\pi c^3$⁠. Then

   $$h = \frac{100}{\pi c^2} = \frac{2\pi c^3}{\pi c^2} = 2c = 2\left(\frac{50}{\pi}\right)^{1/3}.$$

   So the height is twice the radius: the height equals the diameter of the base.

   Using $\pi c^3 = 50$ avoids messy fractional powers and shows the shape of the answer at once. This matches Q20 just before, where the same $h = 2r$ appears for a fixed surface and maximum volume: the two problems are two views of the same best shape.
   Substitute only after differentiating: the formula $h = \frac{100}{\pi r^2}$ from Step 1 now gives the height for the chosen $r = c$⁠.

7. The numbers, and the least surface area

   $\frac{50}{\pi} \approx 15.9155$⁠, so $c \approx 2.5154$ cm and $h = 2c \approx 5.0308$ cm. Since $200 = 4\pi c^3$⁠, the side term is $\frac{200}{c} = 4\pi c^2$⁠, and

   $$S(c) = 2\pi c^2 + 4\pi c^2 = 6\pi c^2 \approx 119.27 \text{ cm}^2.$$
   The question asks only for the dimensions; the area is a useful extra and is what the numerical check below compares against.

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