Miscellaneous · Q8: Window of perimeter 10 m: most light when width is $\frac{20}{\pi+4}$ m

Basics you need

Solution

1. Variables and the perimeter

   Let the semicircle have radius $r$ m, so the rectangle is $2r$ m wide; let the rectangle be $h$ m high. The boundary of the whole opening consists of the two vertical sides, the bottom and the arc:

   $$2h + 2r + \pi r = 10.$$
   The top side of the rectangle is shared with the semicircle and lies inside the window, so it is not part of the "total perimeter". Counting it would give $2h + 4r + \pi r$⁠, a different problem.
   Length of a semicircular arc (Basic 1). Basics: area and volume formulas

2. Express $h$ in $r$⁠, and the range of $r$

   $$h = \frac{10 - (\pi + 2)r}{2}.$$

   Range. We need $r > 0$ and $h > 0$⁠; the second means $(\pi + 2)r < 10$⁠. So

   $$0 < r < \frac{10}{\pi + 2} \approx 1.945.$$
   $2r + \pi r = (\pi + 2)r$⁠. A window with $h \le 0$ would have no rectangular part, so it is not the shape described.

3. The area as a function of $r$

   $$\begin{aligned} A &= 2rh + \tfrac12\pi r^2 \\ &= r\big(10 - (\pi + 2)r\big) + \tfrac12\pi r^2 \\ &= 10r - \left(\pi + 2 - \tfrac\pi2\right)r^2 \\ &= 10r - \tfrac{\pi + 4}{2}\,r^2. \end{aligned}$$
   Rectangle area $2r \times h$⁠, with $2rh = r\big(10 - (\pi + 2)r\big)$ from Step 2, plus the semicircle's area. Collecting the $r^2$ terms: $\pi + 2 - \frac\pi2 = \frac{\pi + 4}{2}$⁠.
   Basics: building the function to optimise

4. Differentiate and find the critical point

   $$A'(r) = 10 - (\pi + 4)r, \qquad A'(r) = 0 \iff r = \frac{10}{\pi + 4}.$$

   This lies in the range, because $\pi + 4 > \pi + 2$ gives $\frac{10}{\pi + 4} < \frac{10}{\pi + 2}$⁠.

   Power rule: $\frac{d}{dr}10r = 10$ and $\frac{d}{dr}\frac{\pi + 4}{2}r^2 = (\pi + 4)r$⁠.

5. It gives the greatest area on the whole range

   $A'(r) = 10 - (\pi + 4)r$ is positive for $r < \frac{10}{\pi + 4}$ and negative for $r > \frac{10}{\pi + 4}$⁠. By Theorem 1, $A$ increases up to $r = \frac{10}{\pi + 4}$ and decreases after it, so this $r$ gives a larger area than every other $r$ in the range. The second derivative agrees: $A''(r) = -(\pi + 4) < 0$⁠, a local maximum by Theorem 4 (i).

   $A$ is a quadratic in $r$ with a negative $r^2$ coefficient, a downward parabola, so its vertex is its highest point. The sign of $A'$ is the calculus way of saying the same thing for the whole range at once.
   Theorem 1, Section 6.3; Theorem 4 (i), Section 6.4 (Basic 3). Lesson: optimisation problems

6. The dimensions

   With $r = \frac{10}{\pi + 4}$⁠:

   $$\begin{aligned} \text{width } 2r &= \tfrac{20}{\pi + 4} \approx 2.80 \text{ m}, \\ h &= \tfrac{10}{\pi + 4} \approx 1.40 \text{ m}. \end{aligned}$$

   For $h$⁠, put $r$ into Step 2:

   $$\begin{aligned} 10 - (\pi + 2)r &= 10 - \tfrac{10(\pi + 2)}{\pi + 4} \\ &= \tfrac{10(\pi + 4) - 10(\pi + 2)}{\pi + 4} \\ &= \tfrac{20}{\pi + 4}, \end{aligned}$$

   and $h$ is half of that, $\frac{10}{\pi + 4}$⁠.

   So the rectangle is $\frac{20}{\pi + 4}$ m wide and $\frac{10}{\pi + 4}$ m high, with a semicircle of radius $\frac{10}{\pi + 4}$ m. The greatest area is $A = \frac{100}{\pi + 4} - \frac{50}{\pi + 4} = \frac{50}{\pi + 4} \approx 7.00$ m².

   A neat feature: the height of the rectangle equals the radius, so the whole window is exactly as tall ($h + r = \frac{20}{\pi + 4}$⁠) as it is wide.

Every fact used, and where it comes from