Exercise 6.3 · Q22: Wire of 28 m into a square and a circle: least total area

Basics you need

Solution

1. Choose the variable and its range

   Let the piece bent into the square have length $x$ m. The piece for the circle then has length $28 - x$ m. Both pieces must have positive length, so

   $$0 < x < 28.$$
   Everything about the two shapes is decided by where we cut, so one variable is enough. "Cut into two pieces" means neither piece is empty, which is why the ends 0 and 28 are excluded; Step 6 shows that the ends would not be best anyway.
   An open interval $(0, 28)$ contains every number strictly between 0 and 28. Basics: interval notation

2. Side of the square and radius of the circle

   The square's perimeter is $x$⁠, so its side is $\frac{x}{4}$⁠. The circle's circumference is $28 - x$⁠, so its radius $\rho$ satisfies $2\pi\rho = 28 - x$⁠:

   $$\text{side} = \frac{x}{4}, \qquad \rho = \frac{28 - x}{2\pi}.$$
   The wire becomes the boundary of each shape, so the lengths are perimeters. We need the side and the radius because the area formulas are written in terms of them.
   Perimeter of a square $= 4 \times$ side; circumference of a circle $= 2\pi \times$ radius (Basic 1). We write $\rho$ for the radius to keep $r$ free.

3. The combined area as a function of $x$

   $$\begin{aligned} A(x) &= \left(\frac{x}{4}\right)^2 + \pi\left(\frac{28 - x}{2\pi}\right)^2 \\ &= \frac{x^2}{16} + \frac{(28 - x)^2}{4\pi}, \quad 0 < x < 28. \end{aligned}$$
   This is the quantity to make as small as possible. In the circle's term, $\pi \cdot \frac{(28-x)^2}{4\pi^2} = \frac{(28-x)^2}{4\pi}$⁠: one factor $\pi$ cancels.
   Area of a square $= \text{side}^2$⁠; area of a circle $= \pi\rho^2$⁠. Basics: building the function to optimise

4. Differentiate and find the critical point

   By the chain rule, $\frac{d}{dx}(28 - x)^2 = 2(28 - x)\cdot(-1)$⁠. So

   $$\begin{aligned} A'(x) &= \frac{x}{8} - \frac{28 - x}{2\pi} \\ &= \frac{\pi x - 4(28 - x)}{8\pi} \\ &= \frac{(\pi + 4)x - 112}{8\pi}. \end{aligned}$$

   $A'(x) = 0$ exactly when $(\pi + 4)x = 112$⁠, that is

   $$x = \frac{112}{\pi + 4}.$$

   Call this value $c$⁠. Since $\pi + 4 \approx 7.14$⁠, $c \approx 15.68$⁠, which lies in $(0, 28)$⁠. $A$ is a polynomial, so this is the only critical point.

   Inside an open interval, a minimum can only occur where $A' = 0$ (Theorem 2, Section 6.4). Putting $A'$ over the common denominator $8\pi$ turns it into a single straight-line expression whose zero and sign are easy to see.
   Chain rule with inner function $28 - x$ (derivative $-1$⁠) and outer function $u^2$ (derivative $2u$⁠). The common denominator of 8 and $2\pi$ is $8\pi$⁠: multiply the first fraction by $\frac{\pi}{\pi}$ and the second by $\frac{4}{4}$⁠.

5. Second derivative test

   $$A''(x) = \frac{1}{8} + \frac{1}{2\pi} > 0 \quad\text{for every } x.$$

   In particular $A''(c) > 0$⁠, so by the second derivative test $x = c$ is a point of local minima.

   A positive second derivative means the graph bends upwards, so the flat point is the bottom of a valley. Here $A''$ is a positive constant: $A$ is an upward-opening parabola.
   Theorem 4 (ii), Section 6.4. Lesson: first and second derivative tests

6. It is the least value, and the ends are worse

   From Step 4, $A'(x)$ has the sign of $(\pi + 4)x - 112$⁠, which increases with $x$⁠: it is negative for $0 < x < c$ and positive for $c < x < 28$⁠. So $A$ decreases on $(0, c]$ and increases on $[c, 28)$⁠, and $A(x) > A(c)$ for every other allowed $x$⁠.

   For comparison, the two extreme ways of using the wire (not allowed here, since each would leave an empty piece) give

   $$A(0) = \frac{784}{4\pi} = \frac{196}{\pi} \approx 62.39, \qquad A(28) = \frac{784}{16} = 49,$$

   both larger than the value at $c$ found in Step 8.

   The second derivative test alone only compares $c$ with nearby cuts. The sign of $A'$ over the whole range settles that $c$ beats every possible cut: the area falls all the way to $c$ and rises all the way after it.
   Theorem 1, Section 6.3: a negative derivative means decreasing, a positive one increasing (its proof by the Mean Value Theorem gives strict inequalities). Lesson: increasing and decreasing functions

7. Lengths of the two pieces

   $$\begin{aligned} \text{square piece} &= \frac{112}{\pi + 4} \approx 15.68 \text{ m}, \\ \text{circle piece} &= 28 - \frac{112}{\pi + 4} \\ &= \frac{28\pi + 112 - 112}{\pi + 4} \\ &= \frac{28\pi}{\pi + 4} \approx 12.32 \text{ m}. \end{aligned}$$
   The question asks for the lengths of both pieces, so we finish by computing the second one. Writing 28 as $\frac{28(\pi + 4)}{\pi + 4}$ puts both terms over the same denominator.
   $\pi \approx 3.14159$⁠, so $\pi + 4 \approx 7.14159$⁠.

8. The shapes, and the least area

   At $x = c$ the side of the square is $\frac{c}{4} = \frac{28}{\pi + 4}$ and the radius is $\frac{1}{2\pi}\cdot\frac{28\pi}{\pi + 4} = \frac{14}{\pi + 4}$⁠. So the side of the square equals the diameter of the circle. The least combined area is

   $$\begin{aligned} A(c) &= \frac{784}{(\pi + 4)^2} + \frac{196\pi}{(\pi + 4)^2} \\ &= \frac{196(4 + \pi)}{(\pi + 4)^2} = \frac{196}{\pi + 4} \approx 27.44 \text{ m}^2. \end{aligned}$$
   This extra fact is a neat way to remember the answer and gives the minimum area for the check. It also explains the condition $A' = 0$⁠: at the best cut, moving a little wire from one shape to the other changes the two areas by equal and opposite amounts.
   $784 = 28^2$ and $196 = 14^2$⁠; $784 = 4 \times 196$⁠.

Every fact used, and where it comes from