下图所示的封闭容器由一个高为 $h\text{ cm}$ 的圆柱体和一个圆锥体拼接而成。圆柱体和圆锥体具有相同的底面半径 $r\text{ cm}$，且圆锥的母线与垂直方向成 $30^\circ$ 角。容器的体积为 $70\text{ cm}^3$。

1. Find the height and slant height of the cone in terms of $r$.

Let $h_1$ be the height of the cone and $l$ be its slant height.
Using trigonometry on the right-angled triangle formed by the cone's vertical axis, its base radius, and its slant edge:
$\sin(30^\circ) = \frac{r}{l} \implies l = \frac{r}{\sin(30^\circ)} = 2r$
$\tan(30^\circ) = \frac{r}{h_1} \implies h_1 = \frac{r}{\tan(30^\circ)} = r\sqrt{3}$

2. Express the height of the cylinder, $h$, in terms of $r$ using the given container volume.

The total volume $V$ is the sum of the volume of the cylinder and the volume of the cone:
$V = \pi r^2 h + \frac{1}{3}\pi r^2 h_1 = 70$
Substitute $h_1 = r\sqrt{3}$:
$\pi r^2 h + \frac{1}{3}\pi r^2 (r\sqrt{3}) = 70$
$\pi r^2 h = 70 - \frac{\sqrt{3}}{3}\pi r^3$
$h = \frac{70}{\pi r^2} - \frac{\sqrt{3}}{3}r$

3. Formulate the total surface area $A$ and simplify to the required expression.

The total surface area $A$ consists of the circular base, the curved surface of the cylinder, and the curved surface of the cone:
$A = \text{Base Area} + \text{Curved SA}_{\text{cylinder}} + \text{Curved SA}_{\text{cone}}$
$A = \pi r^2 + 2\pi r h + \pi r l$
Substitute $h = \frac{70}{\pi r^2} - \frac{\sqrt{3}}{3}r$ and $l = 2r$ into the equation:
$A = \pi r^2 + 2\pi r \left( \frac{70}{\pi r^2} - \frac{\sqrt{3}}{3}r \right) + \pi r (2r)$
$A = \pi r^2 + \frac{140}{r} - \frac{2\sqrt{3}}{3}\pi r^2 + 2\pi r^2$
$A = 3\pi r^2 - \frac{2\sqrt{3}\pi r^2}{3} + \frac{140}{r} \quad \text{(shown)}$

1. Differentiate the given area formula with respect to $r$.

$A = \left(3 - \frac{2\sqrt{3}}{3}\right)\pi r^2 + 140r^{-1}$
$\frac{dA}{dr} = 2\left(3 - \frac{2\sqrt{3}}{3}\right)\pi r - 140r^{-2}$
$\frac{dA}{dr} = \left(6 - \frac{4\sqrt{3}}{3}\right)\pi r - \frac{140}{r^2}$

2. Equate the derivative to zero and solve for $r$.

For a stationary point, $\frac{dA}{dr} = 0$.
$\left(6 - \frac{4\sqrt{3}}{3}\right)\pi r - \frac{140}{r^2} = 0$
$\left(6 - \frac{4\sqrt{3}}{3}\right)\pi r^3 = 140$
$r^3 = \frac{140}{\pi\left(6 - \frac{4\sqrt{3}}{3}\right)}$
$r \approx 2.2941$

3. Verify that this value of $r$ gives a minimum surface area using the second derivative test.

$\frac{d^2A}{dr^2} = \left(6 - \frac{4\sqrt{3}}{3}\right)\pi + \frac{280}{r^3}$
Since $6 > \frac{4\sqrt{3}}{3}$ and $r > 0$, we have $\frac{d^2A}{dr^2} > 0$.
Therefore, the surface area is a minimum.
$r = 2.29 \text{ (to 3 s.f.)}$