The diagram shows a greenhouse built on a rectangular base. $140\text{ m}^2$ of polyethene is used to build the curved roof and the two identical semicircular sides. The radius of each semicircle is $r\text{ m}$ and the length of the greenhouse is $y\text{ m}$.

1. Formulate an equation for the total area of polyethene used.

Area of two semicircular sides = $2 \times \frac{1}{2}\pi r^2 = \pi r^2$
Area of curved roof = $(\frac{1}{2} \times 2\pi r) \times y = \pi r y$
Total area = $\pi r^2 + \pi r y = 140$

2. Express the length $y$ in terms of $r$.

$\pi r y = 140 - \pi r^2$
$y = \frac{140 - \pi r^2}{\pi r}$

3. Formulate the expression for the volume of the greenhouse.

Volume $V$ = Cross-sectional area $\times$ length
$V = \frac{1}{2}\pi r^2 \times y$

4. Substitute $y$ into the volume expression and simplify to obtain the given result.

$V = \frac{1}{2}\pi r^2 \left( \frac{140 - \pi r^2}{\pi r} \right)$
$V = \frac{1}{2}r (140 - \pi r^2)$
$V = 70r - \frac{\pi r^3}{2}$ (shown)

1. Differentiate the volume function with respect to $r$

Given $V = 70r - \frac{\pi r^3}{2}$
$\frac{\text{d}V}{\text{d}r} = 70 - \frac{3\pi r^2}{2}$

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

For a stationary value, $\frac{\text{d}V}{\text{d}r} = 0$
$70 - \frac{3\pi r^2}{2} = 0$
$\frac{3\pi r^2}{2} = 70$

3. Solve for $r$

$r^2 = \frac{140}{3\pi}$
$r = \sqrt{\frac{140}{3\pi}}$ (since $r > 0$)
$r \approx 3.8542 \approx 3.86$ (to 3 s.f.)

1. Find the second derivative of V with respect to r

$\frac{dV}{dr} = 70 - \frac{3\pi r^2}{2}$
$\frac{d^2V}{dr^2} = -3\pi r$

2. Evaluate the sign of the second derivative to justify the maximum

Since $r > 0$ (as it is a physical length), $\frac{d^2V}{dr^2} = -3\pi r < 0$.
Therefore, the stationary value is a maximum, giving the largest possible volume.