A farmer has a plot of the land, $OAB$, with adjacent sides $OA$ and $OB$. With reference to $O$ as the origin, $OA$ and $OB$ are parallel to the $x$-axis and $y$-axis respectively. The arc $AB$ is decribed by the parametric equations
$x = \cos\theta, \quad y = \theta - \frac{1}{2}\sin 2\theta, \quad 0 \le \theta \le \frac{\pi}{2}.$

1. Set up the definite integral for the area of the plot $OAB$ using the parametric equations.

The area is bounded by the axes $OA$ (where $y=0$) and $OB$ (where $x=0$).
When $x = 1$, $\cos\theta = 1 \implies \theta = 0$.
When $x = 0$, $\cos\theta = 0 \implies \theta = \frac{\pi}{2}$.
Given $x = \cos\theta$, we have $\frac{dx}{d\theta} = -\sin\theta$.
$\text{Area} = \int_{0}^{1} y \, dx = \int_{\pi/2}^{0} y \frac{dx}{d\theta} \, d\theta$
$= \int_{\pi/2}^{0} \left( \theta - \frac{1}{2}\sin 2\theta \right) (-\sin\theta) \, d\theta$
$= \int_{0}^{\pi/2} \left( \theta\sin\theta - \frac{1}{2}(2\sin\theta\cos\theta)\sin\theta \right) \, d\theta$
$= \int_{0}^{\pi/2} \left( \theta\sin\theta - \sin^2\theta\cos\theta \right) \, d\theta$

2. Evaluate the first part of the integral using integration by parts.

For $\int \theta\sin\theta \, d\theta$, let $u = \theta$ and $\frac{dv}{d\theta} = \sin\theta$.
Then $\frac{du}{d\theta} = 1$ and $v = -\cos\theta$.
$\int_{0}^{\pi/2} \theta\sin\theta \, d\theta = \left[ -\theta\cos\theta \right]_{0}^{\pi/2} - \int_{0}^{\pi/2} (-\cos\theta) \, d\theta$
$= \left( -\frac{\pi}{2}\cos\frac{\pi}{2} - (0) \right) + \int_{0}^{\pi/2} \cos\theta \, d\theta$
$= 0 + \left[ \sin\theta \right]_{0}^{\pi/2} = 1 - 0 = 1$

3. Evaluate the second part of the integral and find the total area.

For $\int \sin^2\theta\cos\theta \, d\theta$, we recognise it as the form $\int [\mathrm{f}(\theta)]^2 \mathrm{f}'(\theta) \, d\theta$.
$\int_{0}^{\pi/2} \sin^2\theta\cos\theta \, d\theta = \left[ \frac{1}{3}\sin^3\theta \right]_{0}^{\pi/2}$
$= \frac{1}{3}(1)^3 - 0 = \frac{1}{3}$
Therefore, the total area is:
$\text{Area} = 1 - \frac{1}{3} = \frac{2}{3} \text{ units}^2 \quad \text{(shown)}$

1. Express the area of the rectangle, $K$, in terms of the parameter $\theta$.

The sides of the rectangle are parallel to the $x$- and $y$-axes, with one vertex at the origin $O(0,0)$ and the opposite vertex $(x,y)$ on the arc $AB$.
The area of the rectangle is given by $K = xy$.
Substituting the given parametric equations $x = \cos\theta$ and $y = \theta - \frac{1}{2}\sin 2\theta$:
$K = \cos\theta \left( \theta - \frac{1}{2}\sin 2\theta \right)$

2. Differentiate $K$ with respect to $\theta$ and simplify.

Using the product rule, $\frac{\mathrm{d}K}{\mathrm{d}\theta} = x \frac{\mathrm{d}y}{\mathrm{d}\theta} + y \frac{\mathrm{d}x}{\mathrm{d}\theta}$:
$\frac{\mathrm{d}x}{\mathrm{d}\theta} = -\sin\theta$
$\frac{\mathrm{d}y}{\mathrm{d}\theta} = 1 - \cos 2\theta = 2\sin^2\theta$
Now, substituting these derivatives:
$\frac{\mathrm{d}K}{\mathrm{d}\theta} = \cos\theta (2\sin^2\theta) + \left( \theta - \frac{1}{2}\sin 2\theta \right)(-\sin\theta)$
Since $\sin 2\theta = 2\sin\theta\cos\theta$:
$\frac{\mathrm{d}K}{\mathrm{d}\theta} = 2\sin^2\theta\cos\theta - \theta\sin\theta + \sin^2\theta\cos\theta$
$\frac{\mathrm{d}K}{\mathrm{d}\theta} = 3\sin^2\theta\cos\theta - \theta\sin\theta$

3. Equate the derivative to zero and solve for the optimal $\theta$.

To find the maximum area, set $\frac{\mathrm{d}K}{\mathrm{d}\theta} = 0$:
$\sin\theta (3\sin\theta\cos\theta - \theta) = 0$
Since the rectangle has a non-zero area, $\theta \neq 0$, which means $\sin\theta \neq 0$.
Thus, we require:
$3\sin\theta\cos\theta - \theta = 0 \implies \frac{3}{2}\sin 2\theta = \theta$
Using a graphing calculator to solve for $\theta$ in the valid range $0 < \theta < \frac{\pi}{2}$:
$\theta \approx 1.13943$

4. Verify that this value of $\theta$ yields a maximum.

We verify the nature of the stationary point using the first derivative test:

• When $\theta = 1$, $\frac{\mathrm{d}K}{\mathrm{d}\theta} \approx 0.306 > 0$
• When $\theta = 1.2$, $\frac{\mathrm{d}K}{\mathrm{d}\theta} \approx -0.176 < 0$

Since the gradient changes from positive to negative, $K$ is indeed a maximum at $\theta \approx 1.13943$.

5. Calculate the maximum value of $K$.

Substitute $\theta \approx 1.13943$ into the equation for $K$:
$K = \cos(1.13943) \left[ 1.13943 - \frac{1}{2}\sin(2 \times 1.13943) \right]$
$K \approx 0.41805 \times (1.13943 - 0.37981)$
$K \approx 0.31756$
$K \approx 0.318 \text{ units}^2 \text{ (to 3 s.f.)}$

1. Set up the equation to find the intersection point between the arc $AB$ and the dividing line $y = 2x + 0.1$.

At the point of intersection, substitute the parametric equations $x = \cos\theta$ and $y = \theta - \frac{1}{2}\sin 2\theta$ into the equation of the line:
$\theta - \frac{1}{2}\sin 2\theta = 2\cos\theta + 0.1$
$\theta - \frac{1}{2}\sin 2\theta - 2\cos\theta - 0.1 = 0$

2. Solve the equation for $\theta$ and find the corresponding $x$-coordinate using a graphing calculator.

Using a graphing calculator, solve the equation for $0 \le \theta \le \frac{\pi}{2}$:
$\theta \approx 1.18956$
The corresponding $x$-coordinate at the intersection point is:
$x = \cos(1.18956) \approx 0.37206$

3. Set up the integrals to find the area of the portion of land below the line.

The line divides the land into two portions. Let's find the area of the portion bounded below by the $x$-axis. This region can be split into a trapezium (under the line from $x=0$ to $x=0.37206$) and the area under the curve (from $x=0.37206$ to $x=1$).
$\text{Area}_{\text{lower}} = \int_{0}^{0.37206} (2x + 0.1) \, \mathrm{d}x + \int_{0.37206}^{1} y \, \mathrm{d}x$
For the second integral, substitute $x = \cos\theta \implies \frac{\mathrm{d}x}{\mathrm{d}\theta} = -\sin\theta$:
When $x = 0.37206$, $\theta = 1.18956$.
When $x = 1$, $\theta = 0$.
$\int_{0.37206}^{1} y \, \mathrm{d}x = \int_{1.18956}^{0} \left(\theta - \frac{1}{2}\sin 2\theta\right) (-\sin\theta) \, \mathrm{d}\theta$
$= \int_{0}^{1.18956} \left(\theta - \frac{1}{2}\sin 2\theta\right) \sin\theta \, \mathrm{d}\theta$

4. Evaluate the areas using a graphing calculator, determine the larger portion, and state the final answer.

Using a graphing calculator to evaluate both integrals:
$\text{Area of trapezium} = \int_{0}^{0.37206} (2x + 0.1) \, \mathrm{d}x \approx 0.17564$
$\text{Area under curve} = \int_{0}^{1.18956} \left(\theta - \frac{1}{2}\sin 2\theta\right) \sin\theta \, \mathrm{d}\theta \approx 0.21907$
$\text{Total Area}_{\text{lower}} = 0.17564 + 0.21907 = 0.39471 \text{ units}^2$
From part (i), the total area of the plot is $\frac{2}{3} \approx 0.66667 \text{ units}^2$.
The area of the upper portion is:
$\text{Area}_{\text{upper}} = \frac{2}{3} - 0.39471 = 0.27196 \text{ units}^2$
Since $0.39471 > 0.27196$, the lower portion is the larger plot.
$\text{Area} \approx 0.395 \text{ units}^2 \text{ (to 3 s.f.)}$