一位建筑师用 40 cm 长的铁丝为房间模型设计 2 个窗框。他希望他的设计能让尽可能多的光线穿过窗口。一个窗框是边长为 $x$ cm 的等边三角形，另一个是半径为 $r$ cm 的半圆形。

1. Form an equation for the total length of the wire in terms of $x$ and $r$.

Perimeter of equilateral triangle = $3x$
Perimeter of semi-circle = $\pi r + 2r = r(\pi + 2)$
Total length of wire = $3x + r(\pi + 2) = 40$

2. Express $r$ in terms of $x$.

$r(\pi + 2) = 40 - 3x$
$r = \frac{40 - 3x}{\pi + 2}$

3. Find the area of the equilateral triangle in terms of $x$.

Area of triangle = $\frac{1}{2}(x)(x)\sin 60^\circ = \frac{1}{2}x^2\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}x^2$

4. Find the area of the semi-circle, substitute $r$, and sum the areas to show the given formula.

Area of semi-circle = $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi \left(\frac{40 - 3x}{\pi + 2}\right)^2 = \frac{\pi(40 - 3x)^2}{2(\pi + 2)^2}$
Total Area $A = \frac{\pi(40 - 3x)^2}{2(\pi + 2)^2} + \frac{\sqrt{3}x^2}{4}$ (shown)

1. Differentiate the expression for $A$ with respect to $x$.

$A = \frac{\pi(40-3x)^2}{2(\pi+2)^2} + \frac{\sqrt{3}x^2}{4}$

$\frac{dA}{dx} = \frac{\pi}{2(\pi+2)^2} \cdot 2(40-3x)(-3) + \frac{\sqrt{3}}{4} \cdot 2x$

$\frac{dA}{dx} = \frac{-3\pi(40-3x)}{(\pi+2)^2} + \frac{\sqrt{3}x}{2}$

2. Set the derivative to zero to find the stationary value of $x$.

For a stationary value, $\frac{dA}{dx} = 0$:

$\frac{\sqrt{3}x}{2} = \frac{3\pi(40-3x)}{(\pi+2)^2}$

$\sqrt{3}x(\pi+2)^2 = 6\pi(40-3x)$

$\sqrt{3}(\pi+2)^2 x = 240\pi - 18\pi x$

$x[\sqrt{3}(\pi+2)^2 + 18\pi] = 240\pi$

$x = \frac{240\pi}{\sqrt{3}(\pi+2)^2 + 18\pi} \approx 7.3676$

3. Substitute the value of $x$ back into the equation for $A$ to find the stationary value.

$A = \frac{\pi(40-3(7.3676))^2}{2(\pi+2)^2} + \frac{\sqrt{3}(7.3676)^2}{4}$

$A \approx 19.034 + 23.504 \approx 42.538$

$A \approx 42.5$ cm$^2$ (to 3 s.f.)

1. Find the second derivative of $A$ with respect to $x$.

From part (b), the first derivative is:
$\frac{\text{d}A}{\text{d}x} = \frac{-3\pi(40-3x)}{(\pi+2)^2} + \frac{\sqrt{3}x}{2}$
Differentiating again with respect to $x$:
$\frac{\text{d}^2A}{\text{d}x^2} = \frac{-3\pi(-3)}{(\pi+2)^2} + \frac{\sqrt{3}}{2}$
$\frac{\text{d}^2A}{\text{d}x^2} = \frac{9\pi}{(\pi+2)^2} + \frac{\sqrt{3}}{2}$

2. Determine the nature of the stationary value using the sign of the second derivative.

Since $\pi > 0$, $\frac{9\pi}{(\pi+2)^2} > 0$ and $\frac{\sqrt{3}}{2} > 0$.
Therefore, $\frac{\text{d}^2A}{\text{d}x^2} > 0$ for all $x$.
This means the stationary value is a minimum.

3. Explain why the architect may not be happy with the design.

The architect wants "as much light as possible", which means he wants to maximise the total area $A$. However, the stationary value gives a minimum area, so the design does not achieve his goal.