A farmer uses 160 m of fencing to enclose a plot of his land in a shape that comprises an isosceles triangle and a rectangle, with the dimensions shown.

1. Find the length of the base of the isosceles triangle (which is also the top width of the rectangle).

Let the base of the triangle be $w$.
Using the cosine rule on the triangle:
$w^2 = x^2 + x^2 - 2(x)(x)\cos 120^\circ$
$w^2 = 2x^2 - 2x^2(-\frac{1}{2}) = 3x^2$
$w = \sqrt{3}x$

2. Express the perimeter of the plot in terms of $x$ and $l$, and solve for $l$.

The perimeter consists of the two sloping sides of the triangle, the two vertical sides of the rectangle, and the bottom horizontal side of the rectangle.
Perimeter $= x + x + l + l + w = 2x + 2l + \sqrt{3}x$
Given perimeter is 160 m:
$2x + 2l + \sqrt{3}x = 160$
$2l = 160 - (2 + \sqrt{3})x$
$l = 80 - \frac{2 + \sqrt{3}}{2}x$

3. Find the area of the isosceles triangle.

Area of triangle $= \frac{1}{2}(x)(x)\sin 120^\circ$
$= \frac{1}{2}x^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}x^2$

4. Find the area of the rectangle and the total area, then simplify to the required form.

Area of rectangle $= w \times l = \sqrt{3}x \left(80 - \frac{2 + \sqrt{3}}{2}x\right)$
$= 80\sqrt{3}x - \frac{2\sqrt{3} + 3}{2}x^2$
Total Area $A = \text{Area of triangle} + \text{Area of rectangle}$
$A = \frac{\sqrt{3}}{4}x^2 + 80\sqrt{3}x - \frac{2\sqrt{3} + 3}{2}x^2$
$A = 80\sqrt{3}x + \left(\frac{\sqrt{3}}{4} - \frac{4\sqrt{3} + 6}{4}\right)x^2$
$A = 80\sqrt{3}x - \frac{3\sqrt{3} + 6}{4}x^2$
$A = \frac{320\sqrt{3}x - (3\sqrt{3} + 6)x^2}{4}$ (Shown)

1. Differentiate the area function with respect to $x$

Let $A = \frac{320\sqrt{3}\,x - (3\sqrt{3} + 6)\,x^2}{4} = 80\sqrt{3}\,x - \frac{3\sqrt{3} + 6}{4}\,x^2$.

$\frac{dA}{dx} = 80\sqrt{3} - \frac{3\sqrt{3} + 6}{2}\,x$

2. Set the derivative to zero to find the stationary point

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

$80\sqrt{3} - \frac{3\sqrt{3} + 6}{2}\,x = 0$

$\frac{3\sqrt{3} + 6}{2}\,x = 80\sqrt{3}$

3. Solve for $x$

$x = \frac{160\sqrt{3}}{3\sqrt{3} + 6}$

$x \approx 24.742 \approx 24.7$ (to 3 s.f.)

1. Find the second derivative of the area function to determine the nature of the stationary point.

From (i), $A = \frac{320\sqrt{3}}{4}x - \frac{3\sqrt{3} + 6}{4}x^2 = 80\sqrt{3}x - \frac{3\sqrt{3} + 6}{4}x^2$.
$\frac{dA}{dx} = 80\sqrt{3} - \frac{3\sqrt{3} + 6}{2}x$.
$\frac{d^2A}{dx^2} = -\frac{3\sqrt{3} + 6}{2}$.

2. Explain why the area is a maximum.

Since $\frac{d^2A}{dx^2} = -\frac{3\sqrt{3} + 6}{2} < 0$ for all $x$, the stationary point is a maximum. Thus, this value of $x$ gives the largest possible area.

3. Calculate the maximum area using the value of $x$ from (ii).

From (ii), $\frac{dA}{dx} = 0 \implies x = \frac{160\sqrt{3}}{3\sqrt{3} + 6} \approx 24.7436$.
Substitute $x = \frac{160\sqrt{3}}{3\sqrt{3} + 6}$ into $A$:
$A = 80\sqrt{3}\left(\frac{160\sqrt{3}}{3\sqrt{3} + 6}\right) - \frac{3\sqrt{3} + 6}{4}\left(\frac{160\sqrt{3}}{3\sqrt{3} + 6}\right)^2$
$A = \frac{38400}{3\sqrt{3} + 6} - \frac{19200}{3\sqrt{3} + 6} = \frac{19200}{3\sqrt{3} + 6} \approx 1140.39$
$A \approx 1140\text{ m}^2$ (to the nearest square metre).