一位农民刚刚买了一块土地来种植作物。这块土地由一个边长为 $b\text{ m}$ 的等边三角形、一个边长为 $b\text{ m}$ 和 $l\text{ m}$ 的矩形以及一个直径为 $b\text{ m}$ 的半圆形组成，如下图所示。

1. Formulate an equation for the perimeter of the composite shape

The perimeter consists of two sides of the equilateral triangle, two horizontal sides of the rectangle, and the arc of the semi-circle.
Perimeter $= b + b + l + l + \frac{1}{2}\pi b$
$2b + 2l + \frac{\pi b}{2} = 600$

2. Make $l$ the subject of the equation

$2l = 600 - 2b - \frac{\pi b}{2}$
$l = 300 - b - \frac{\pi b}{4}$

1. Find the expression for $l$ in terms of $b$ from the perimeter given in part (i).

The perimeter consists of two sides of the triangle, two lengths of the rectangle, and the arc of the semi-circle.
Perimeter $= 2b + 2l + \frac{1}{2}\pi b = 600$
$2l = 600 - 2b - \frac{\pi}{2}b$
$l = 300 - b - \frac{\pi}{4}b$

2. Formulate the total area of the land in terms of $b$ and $l$.

Area of equilateral triangle $= \frac{1}{2}(b)(b)\sin 60^\circ = \frac{\sqrt{3}}{4}b^2$
Area of rectangle $= bl$
Area of semi-circle $= \frac{1}{2}\pi \left(\frac{b}{2}\right)^2 = \frac{\pi}{8}b^2$
Total Area $A = \frac{\sqrt{3}}{4}b^2 + bl + \frac{\pi}{8}b^2$

3. Substitute $l$ into the area equation and simplify to obtain the required expression.

$A = \frac{\sqrt{3}}{4}b^2 + b\left(300 - b - \frac{\pi}{4}b\right) + \frac{\pi}{8}b^2$
$A = \frac{\sqrt{3}}{4}b^2 + 300b - b^2 - \frac{\pi}{4}b^2 + \frac{\pi}{8}b^2$
$A = 300b + b^2\left(\frac{\sqrt{3}}{4} - 1 - \frac{\pi}{4} + \frac{\pi}{8}\right)$
$A = 300b + b^2\left(\frac{2\sqrt{3}}{8} - \frac{8}{8} - \frac{2\pi}{8} + \frac{\pi}{8}\right)$
$A = 300b + \left(\frac{2\sqrt{3} - 8 - \pi}{8}\right)b^2$ (Shown)

1. Differentiate $A$ with respect to $b$

Given $A = 300b + \left(\frac{2\sqrt{3} - 8 - \pi}{8}\right)b^2$,
$\frac{\text{d}A}{\text{d}b} = 300 + 2\left(\frac{2\sqrt{3} - 8 - \pi}{8}\right)b$
$\frac{\text{d}A}{\text{d}b} = 300 + \left(\frac{2\sqrt{3} - 8 - \pi}{4}\right)b$

2. Equate the derivative to zero to find the stationary value

For a stationary value, $\frac{\text{d}A}{\text{d}b} = 0$:
$300 + \left(\frac{2\sqrt{3} - 8 - \pi}{4}\right)b = 0$

3. Solve for $b$

$\left(\frac{8 + \pi - 2\sqrt{3}}{4}\right)b = 300$
$b = \frac{1200}{8 + \pi - 2\sqrt{3}}$
$b \approx 156.30$
$b = 156 \text{ (to 3 s.f.)}$

1. Find the second derivative of A with respect to b

From (ii), $A = 300b + \left(\frac{2\sqrt{3} - 8 - \pi}{8}\right)b^2$.
$\frac{dA}{db} = 300 + \left(\frac{2\sqrt{3} - 8 - \pi}{4}\right)b$.
$\frac{d^2A}{db^2} = \frac{2\sqrt{3} - 8 - \pi}{4}$

2. Determine the sign of the second derivative to find the nature of the stationary value

Since $2\sqrt{3} \approx 3.464$, $2\sqrt{3} - 8 - \pi < 0$.
Therefore, $\frac{d^2A}{db^2} < 0$.
This means the stationary value is a maximum.

3. Explain why the farmer will be happy

The farmer will be happy because a maximum stationary value means that for the fixed perimeter of $600\text{ m}$, the area of the land he can use to grow crops is maximised.