农夫用 160 米长的栅栏围起他的一块土地，该土地的形状由一个等腰三角形和一个矩形组成，其尺寸如图所示。

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).