A farmer just bought a piece of land to grow his crops. This piece of land consists of an equilateral triangle of sides $b\text{ m}$, a rectangle with sides $b\text{ m}$ and $l\text{ m}$ as well as a semi-circle of diameter $b\text{ m}$ as shown in the diagram below.

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.