图示为一个容器，它由一个半径为 $r$ m、高为 $h$ m 的无盖圆柱体与一个半径为 $r$ m 的半球形底部拼接而成。该容器由厚度可忽略不计的薄金属片制成。容器圆柱部分的体积为 $300\ \text{m}^3$。 [半径为 r 的球体表面积 = $4\pi r^2$]

1. Set up the equation for the volume of the cylindrical part

$\pi r^2 h = 300$

2. Make $h$ the subject of the formula

$h = \frac{300}{\pi r^2}$

1. Express the total surface area of the container in terms of $r$ and $h$

The container consists of an open cylinder and a hemispherical bottom.
Curved surface area of cylinder = $2\pi r h$
Curved surface area of hemisphere = $\frac{1}{2}(4\pi r^2) = 2\pi r^2$
Total surface area, $A = 2\pi r h + 2\pi r^2$

2. Substitute $h$ from part (a) to express the surface area entirely in terms of $r$

From part (a), the volume of the cylinder is $300 = \pi r^2 h$, so $h = \frac{300}{\pi r^2}$.
Substitute $h$ into the area equation:
$A = 2\pi r \left(\frac{300}{\pi r^2}\right) + 2\pi r^2$
$A = \frac{600}{r} + 2\pi r^2$

3. Calculate the total cost $C$ by multiplying the area by the cost per square metre

Cost per square metre = $\char36{}60$
$C = 60 \times A$
$C = 60 \left(\frac{600}{r} + 2\pi r^2\right)$
$C = \frac{36000}{r} + 120\pi r^2$
$C = 120\pi r^2 + \frac{36000}{r}$ (Shown)

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

$C = 120\pi r^2 + 36000r^{-1}$
$\dfrac{\mathrm{d}C}{\mathrm{d}r} = 240\pi r - 36000r^{-2}$

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

For stationary value, $\dfrac{\mathrm{d}C}{\mathrm{d}r} = 0$
$240\pi r - \dfrac{36000}{r^2} = 0$
$240\pi r^3 = 36000$
$r^3 = \dfrac{36000}{240\pi} = \dfrac{150}{\pi}$
$r = \left(\dfrac{150}{\pi}\right)^{\frac{1}{3}} \approx 3.6278$

3. Calculate the stationary value of $C$

Substitute $r = \left(\dfrac{150}{\pi}\right)^{\frac{1}{3}}$ into $C$:
$C = 120\pi \left(\dfrac{150}{\pi}\right)^{\frac{2}{3}} + 36000\left(\dfrac{\pi}{150}\right)^{\frac{1}{3}}$
$C = 120\pi^{\frac{1}{3}}(150)^{\frac{2}{3}} + 240(150)^{\frac{2}{3}}\pi^{\frac{1}{3}} = 360(150)^{\frac{2}{3}}\pi^{\frac{1}{3}} \approx 15841.5 \approx 15800$ (to 3 s.f.)

4. Determine the nature of the stationary value using the second derivative test

$\dfrac{\mathrm{d}^2C}{\mathrm{d}r^2} = 240\pi + 72000r^{-3}$
Since $r > 0$, $\dfrac{\mathrm{d}^2C}{\mathrm{d}r^2} > 0$.
Therefore, the stationary value is a minimum.