一家工厂制造微型容器，该容器由一个封闭的圆柱体和一个半径相同的半球组成。圆柱体的半径为 $r\text{ cm}$，高为 $h\text{ cm}$。容器的体积为 $60\pi\text{ cm}^3$。

1. Formulate an equation for the total volume of the container

Volume of cylinder $= \pi r^2 h$
Volume of hemisphere $= \frac{2}{3}\pi r^3$
Total volume $= \pi r^2 h + \frac{2}{3}\pi r^3 = 60\pi$

2. Make $h$ the subject of the formula

Divide the entire equation by $\pi$:
$r^2 h + \frac{2}{3}r^3 = 60$
$r^2 h = 60 - \frac{2}{3}r^3$
$h = \frac{60}{r^2} - \frac{2r}{3}$

1. Recall the expression for $h$ from part (a)

From part (a), the volume is $V = \pi r^2 h + \frac{2}{3}\pi r^3 = 60\pi$.
Solving for $h$ gives:
$h = \frac{60}{r^2} - \frac{2}{3}r$

2. Formulate the total cost equation in terms of $r$ and $h$

Area of cylinder cover = $\pi r^2$
Area of cylinder curved surface = $2\pi r h$
Area of hemisphere curved surface = $2\pi r^2$

Total cost $C = 3(\pi r^2 + 2\pi r h) + 4(2\pi r^2)$
$C = 3\pi r^2 + 6\pi r h + 8\pi r^2$
$C = 11\pi r^2 + 6\pi r h$

3. Substitute $h$ into the cost equation and simplify to show the result

Substitute $h = \frac{60}{r^2} - \frac{2}{3}r$ into $C$:
$C = 11\pi r^2 + 6\pi r \left(\frac{60}{r^2} - \frac{2}{3}r\right)$
$C = 11\pi r^2 + \frac{360\pi}{r} - 4\pi r^2$
$C = 7\pi r^2 + \frac{360\pi}{r}$ (Shown)

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

$C = 7\pi r^2 + 360\pi r^{-1}$

$\frac{\text{d}C}{\text{d}r} = 14\pi r - 360\pi r^{-2}$

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

For a stationary value, $\frac{\text{d}C}{\text{d}r} = 0$.

$14\pi r - \frac{360\pi}{r^2} = 0$

$14\pi r^3 = 360\pi$

$r^3 = \frac{360}{14} = \frac{180}{7}$

$r = \sqrt[3]{\frac{180}{7}} \approx 2.9515 \approx 2.95$ (to 3 s.f.)

1. Find the second derivative of $C$ with respect to $r$ to determine the nature of the stationary value.

From part (c), $\frac{dC}{dr} = 14\pi r - 360\pi r^{-2}$.

$\frac{d^2C}{dr^2} = 14\pi + 720\pi r^{-3}$

Since $r > 0$, $\frac{d^2C}{dr^2} > 0$.
Therefore, the stationary value is a minimum.

2. Calculate the minimum cost using the value of $r$ found in part (c).

From part (c), $r = \sqrt[3]{\frac{180}{7}} \approx 2.9515\text{ cm}$.

Substitute $r = 2.9515$ into $C$:
$C = 7\pi (2.9515)^2 + \frac{360\pi}{2.9515}$
$C \approx 191.55 + 383.10 = 574.65\text{ cents}$

3. Convert the cost to dollars and conclude whether production should continue.

Minimum cost $= \char36{}5.75$ (to 2 d.p.)

Since the minimum cost of $\char36{}5.75$ is greater than $\char36{}5.60$, the factory owner should not continue to produce this container.