A gardener plans to construct two flower beds, one a circle of radius $r$ m, the other a square of side $x$ m. To protect the young flowers, the two beds are to be surrounded by wire netting. The total length of wire netting to be used is 30 m.

1. Formulate an equation for the total length of the wire netting

The perimeter of the circle is $2\pi r$ and the perimeter of the square is $4x$.
Total length = $2\pi r + 4x = 30$

2. Make $x$ the subject of the formula

$4x = 30 - 2\pi r$
$x = \frac{30 - 2\pi r}{4}$
$x = \frac{15 - \pi r}{2}$

1. State the expression for the combined area in terms of $r$ and $x$, and recall $x$ from part (i)

From part (i), the total perimeter is $2\pi r + 4x = 30$, so $x = \frac{15 - \pi r}{2}$.
The combined area $A$ is the sum of the area of the circle and the area of the square:
$A = \pi r^2 + x^2$

2. Substitute $x$ into the area equation and expand

$A = \pi r^2 + \left(\frac{15 - \pi r}{2}\right)^2$
$A = \pi r^2 + \frac{225 - 30\pi r + \pi^2 r^2}{4}$

3. Simplify and group terms to match the required form

$A = \pi r^2 + \frac{225}{4} - \frac{30\pi r}{4} + \frac{\pi^2 r^2}{4}$
$A = \pi r^2 + \frac{1}{4}\pi^2 r^2 - \frac{15}{2}\pi r + \frac{225}{4}$
$A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$

1. Differentiate the expression for $A$ with respect to $r$.

Given $A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$,
$\frac{\mathrm{d}A}{\mathrm{d}r} = 2\left(\frac{1}{4}\pi^2 + \pi\right)r - \frac{15}{2}\pi$
$\frac{\mathrm{d}A}{\mathrm{d}r} = \left(\frac{1}{2}\pi^2 + 2\pi\right)r - \frac{15}{2}\pi$

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

For a stationary value, $\frac{\mathrm{d}A}{\mathrm{d}r} = 0$.
$\left(\frac{1}{2}\pi^2 + 2\pi\right)r - \frac{15}{2}\pi = 0$

3. Solve for $r$.

$\left(\frac{1}{2}\pi^2 + 2\pi\right)r = \frac{15}{2}\pi$
Multiply by 2:
$(\pi^2 + 4\pi)r = 15\pi$
$r = \frac{15\pi}{\pi^2 + 4\pi}$
$r = \frac{15}{\pi + 4}$

1. Recall the expression for $x$ from part (i) and the stationary value of $r$ from part (iii).

From (i), $x = \frac{15 - \pi r}{2}$.
From (iii), the stationary value occurs when $r = \frac{15}{\pi + 4}$.

2. Substitute $r$ into the expression for $x$ and simplify to show it equals $2r$.

$x = \frac{15 - \pi\left(\frac{15}{\pi + 4}\right)}{2}$
$x = \frac{\frac{15(\pi + 4) - 15\pi}{\pi + 4}}{2}$
$x = \frac{15\pi + 60 - 15\pi}{2(\pi + 4)}$
$x = \frac{60}{2(\pi + 4)} = \frac{30}{\pi + 4}$
Since $r = \frac{15}{\pi + 4}$, the diameter of the circle is $2r = \frac{30}{\pi + 4}$.
Therefore, $x = 2r$ (shown).

1. Find the second derivative of $A$ with respect to $r$

From part (ii), $A = \left(\frac{1}{4}\pi^2 + \pi\right)r^2 - \frac{15}{2}\pi r + \frac{225}{4}$.

$\frac{dA}{dr} = 2\left(\frac{1}{4}\pi^2 + \pi\right)r - \frac{15}{2}\pi$

$\frac{d^2A}{dr^2} = 2\left(\frac{1}{4}\pi^2 + \pi\right) = \frac{1}{2}\pi^2 + 2\pi$

2. Determine the nature of the stationary value using the sign of the second derivative

Since $\pi > 0$, $\frac{d^2A}{dr^2} = \frac{1}{2}\pi^2 + 2\pi > 0$.

Therefore, the stationary value of $A$ is a minimum.