Miscellaneous · Q7: Circle and square with total perimeter $k$⁠: least total area when $s=2r$

Basics you need

Solution

1. One variable, and its range

   From $2\pi r + 4s = k$⁠:

   $$s = \frac{k - 2\pi r}{4}.$$

   Range. Both shapes must exist: $r > 0$ and $s > 0$⁠. The second means $k - 2\pi r > 0$⁠, that is $r < \frac{k}{2\pi}$⁠. So

   $$0 < r < \frac{k}{2\pi}.$$
   Any length of wire $k$ can be cut into a piece for the circle and a piece for the square; $r$ decides how much goes to the circle, and the rest fixes $s$⁠. The range records that each piece must have positive length.
   Perimeters (Basic 1). Basics: building the function to optimise

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

   $$S(r) = \pi r^2 + s^2 = \pi r^2 + \frac{(k - 2\pi r)^2}{16}.$$
   $s^2 = \left(\frac{k - 2\pi r}{4}\right)^2 = \frac{(k - 2\pi r)^2}{16}$⁠. Now $S$ depends on $r$ only ($k$ is a constant).
   Areas (Basic 1).

3. Differentiate

   Power rule for $\pi r^2$⁠; chain rule for the second term, inner function $k - 2\pi r$ with derivative $-2\pi$⁠:

   $$\begin{aligned} S'(r) &= 2\pi r + \frac{2(k - 2\pi r)(-2\pi)}{16} \\ &= 2\pi r - \pi\cdot\frac{k - 2\pi r}{4} \\ &= 2\pi r - \pi s = \pi(2r - s). \end{aligned}$$
   $\frac{2 \cdot 2\pi}{16} = \frac{\pi}{4}$⁠, and $\frac{k - 2\pi r}{4}$ is exactly $s$ (Step 1). Writing the derivative back in terms of $s$ makes the condition in the question appear by itself.
   Chain rule, Theorem 4, Section 5.3.1 (Basic 3).

4. The only critical point is where $s = 2r$

   $$S'(r) = 0 \iff s = 2r \iff \frac{k - 2\pi r}{4} = 2r \iff r = \frac{k}{2\pi + 8}.$$

   So $r^* = \frac{k}{2(\pi + 4)}$ and $s^* = 2r^* = \frac{k}{\pi + 4}$⁠. This $r^*$ lies in the range, since $0 < \frac{k}{2\pi + 8} < \frac{k}{2\pi}$⁠.

   Multiply $\frac{k - 2\pi r}{4} = 2r$ by 4: $k - 2\pi r = 8r$⁠, so $k = (8 + 2\pi)r$⁠. A critical point outside the range would be useless, so we check that this one is inside.

5. It gives the least area on the whole range

   Differentiate again: $S'(r) = 2\pi r - \frac{\pi k}{4} + \frac{\pi^2 r}{2}$⁠, so

   $$S''(r) = 2\pi + \frac{\pi^2}{2} > 0 \quad\text{for every } r.$$

   So $S'$ is strictly increasing (Theorem 1 applied to $S'$⁠). It is zero at $r^*$⁠, hence $S' < 0$ on $(0, r^*)$ and $S' > 0$ on $\left(r^*, \frac{k}{2\pi}\right)$⁠. By Theorem 1, $S$ is decreasing on $(0, r^*]$ and increasing on $\left[r^*, \frac{k}{2\pi}\right)$⁠, so $S(r) > S(r^*)$ for every other $r$ in the range. In particular $S''(r^*) > 0$⁠, so Theorem 4 (ii) also names $r^*$ a point of local minima.

   Therefore the sum of the areas is least exactly when $s = 2r$: the side of the square is double the radius of the circle.

   The second derivative test alone would give "local minimum". Because $S''$ is positive everywhere, the graph of $S$ bends upward along the whole range, so its single lowest point is the lowest of all.
   Theorem 1, Section 6.3; Theorem 4, Section 6.4 (Basic 4). Lesson: derivative tests

6. The least total area

   At $s = 2r$⁠: $S = \pi r^2 + 4r^2 = (\pi + 4)r^2$⁠. With $r^* = \frac{k}{2(\pi + 4)}$⁠:

   $$S_{\min} = (\pi + 4)\cdot\frac{k^2}{4(\pi + 4)^2} = \frac{k^2}{4(\pi + 4)}.$$
   Not asked for, but useful for the numerical check. Compare the extremes: all wire to the square gives about $\frac{k^2}{16}$⁠, all to the circle gives about $\frac{k^2}{4\pi}$⁠; both are larger, since $4(\pi + 4) > 16 > 4\pi$⁠.

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