图中显示了一块场地的轮廓，该场地由一个半径为 $r\text{ m}$ 的半圆弧和直角边长均为 $x\text{ m}$ 的直角三角形的边组成。该场地的周长为 $10\text{ m}$。

1. Express the perimeter of the field in terms of $x$ and $r$.

The perimeter consists of the two sides of the right-angled triangle and the semi-circular arc.
Perimeter $= x + x + \pi r = 2x + \pi r$
Given perimeter $= 10$, so $2x + \pi r = 10$.

2. Express $x$ in terms of $r$.

$2x = 10 - \pi r$
$x = \frac{1}{2}(10 - \pi r)$

3. Formulate the area of the field and substitute $x$ to show the given expression.

Area of field, $A = \text{Area of semi-circle} + \text{Area of triangle}$
$A = \frac{1}{2}\pi r^2 + \frac{1}{2}(x)(x)$
$A = \frac{1}{2}\pi r^2 + \frac{1}{2}x^2$
Substitute $x = \frac{1}{2}(10 - \pi r)$:
$A = \frac{1}{2}\pi r^2 + \frac{1}{2}\left[\frac{1}{2}(10 - \pi r)\right]^2$
$A = \frac{1}{2}\pi r^2 + \frac{1}{2}\left[\frac{1}{4}(10 - \pi r)^2\right]$
$A = \frac{1}{2}\pi r^2 + \frac{1}{8}(10 - \pi r)^2$ (Shown)

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

$A = \frac{1}{2}\pi r^2 + \frac{1}{8}(10 - \pi r)^2$
$\frac{\mathrm{d}A}{\mathrm{d}r} = \pi r + \frac{1}{4}(10 - \pi r)(-\pi)$
$\frac{\mathrm{d}A}{\mathrm{d}r} = \pi r - \frac{5\pi}{2} + \frac{\pi^2 r}{4}$

2. Set the derivative to zero to find the stationary value of $r$.

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$:
$\pi r - \frac{5\pi}{2} + \frac{\pi^2 r}{4} = 0$
Divide by $\pi$:
$r - \frac{5}{2} + \frac{\pi r}{4} = 0$
$r\left(1 + \frac{\pi}{4}\right) = \frac{5}{2}$
$r\left(\frac{4 + \pi}{4}\right) = \frac{5}{2}$
$r = \frac{5}{2} \times \frac{4}{4 + \pi} = \frac{10}{4 + \pi}$

1. Find the value of $r$ for which $A$ is stationary (required from part ii).

$\frac{dA}{dr} = \pi r + \frac{1}{4}(10 - \pi r)(-\pi) = \pi r - \frac{5\pi}{2} + \frac{\pi^2}{4}r$

For a stationary value, $\frac{dA}{dr} = 0$:
$\pi r + \frac{\pi^2}{4}r = \frac{5\pi}{2}$
$r\left(1 + \frac{\pi}{4}\right) = \frac{5}{2}$
$r = \frac{10}{4 + \pi}$

2. Substitute the value of $r$ into the expression for $A$ to find the stationary value.

$A = \frac{1}{2}\pi \left(\frac{10}{4+\pi}\right)^2 + \frac{1}{8}\left(10 - \pi\left(\frac{10}{4+\pi}\right)\right)^2$
$A = \frac{50\pi}{(4+\pi)^2} + \frac{1}{8}\left(\frac{40+10\pi-10\pi}{4+\pi}\right)^2$
$A = \frac{50\pi}{(4+\pi)^2} + \frac{1}{8}\left(\frac{40}{4+\pi}\right)^2$
$A = \frac{50\pi}{(4+\pi)^2} + \frac{200}{(4+\pi)^2}$
$A = \frac{50(\pi+4)}{(4+\pi)^2} = \frac{50}{4+\pi} \approx 7.00$

3. Use the second derivative test to determine the nature of the stationary value.

$\frac{d^2A}{dr^2} = \frac{d}{dr}\left(\pi r - \frac{5\pi}{2} + \frac{\pi^2}{4}r\right) = \pi + \frac{\pi^2}{4}$

Since $\pi + \frac{\pi^2}{4} > 0$, the stationary value is a minimum.
Therefore, this value of $A$ is the smallest possible area.