在一部科幻电影中，创建了一个半球形的能量保护罩（一个敞开的半球），以保护一个正在建造的用于存放致命武器的圆柱形筒仓。筒仓顶部的圆周刚好位于半球罩内，如下图所示。由于电力供应限制，生成的半球罩半径 $r$ 固定为 $0.5\text{ km}$。设圆柱形筒仓的半径和高度分别为 $x\text{ km}$ 和 $y\text{ km}$。

1. Establish the relationship between $x$ and $y$ using the geometry of the hemisphere and the silo.

Since the silo stands on the same flat base as the hemisphere and the circumference of its top touches the dome, we can form a right-angled triangle with the radius of the dome ($r$), the radius of the silo ($x$), and the height of the silo ($y$).
By Pythagoras' Theorem:
$x^2 + y^2 = r^2$
Given $r = 0.5$,
$x^2 + y^2 = 0.5^2 = \frac{1}{4}$
$x^2 = \frac{1}{4} - y^2$

2. Express the volume of the silo in terms of $y$ to show the given formula.

The volume of the cylindrical silo, $V$, is given by:
$V = \pi x^2 y$
Substituting $x^2 = \frac{1}{4} - y^2$:
$V = \pi \left(\frac{1}{4} - y^2\right) y$
$V = \pi \left(\frac{1}{4}y - y^3\right) \text{ (Shown)}$

3. Differentiate the volume with respect to $y$ and find the value of $y$ for stationary points.

To find the maximum volume, we first find $\frac{dV}{dy}$:
$\frac{dV}{dy} = \pi \left(\frac{1}{4} - 3y^2\right)$
For stationary points, $\frac{dV}{dy} = 0$:
$\pi \left(\frac{1}{4} - 3y^2\right) = 0$
$3y^2 = \frac{1}{4}$
$y^2 = \frac{1}{12}$
Since $y > 0$ (the height must be positive):
$y = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}$

4. Use the second derivative test to confirm that this value of $y$ gives a maximum volume.

Find the second derivative $\frac{d^2V}{dy^2}$:
$\frac{d^2V}{dy^2} = \pi (0 - 6y) = -6\pi y$
Substitute $y = \frac{\sqrt{3}}{6}$:
$\frac{d^2V}{dy^2} = -6\pi \left(\frac{\sqrt{3}}{6}\right) = -\pi\sqrt{3}$
Since $\frac{d^2V}{dy^2} < 0$, the volume is a maximum when $y = \frac{\sqrt{3}}{6}$.

5. Calculate the exact maximum volume.

Substitute $y^2 = \frac{1}{12}$ and $y = \frac{\sqrt{3}}{6}$ into the volume formula:
$V_{\text{max}} = \pi \left(\frac{1}{4}y - y^3\right) = \pi y \left(\frac{1}{4} - y^2\right)$
$V_{\text{max}} = \pi \left(\frac{\sqrt{3}}{6}\right) \left(\frac{1}{4} - \frac{1}{12}\right)$
$V_{\text{max}} = \pi \left(\frac{\sqrt{3}}{6}\right) \left(\frac{3}{12} - \frac{1}{12}\right)$
$V_{\text{max}} = \pi \left(\frac{\sqrt{3}}{6}\right) \left(\frac{2}{12}\right) = \pi \left(\frac{\sqrt{3}}{6}\right) \left(\frac{1}{6}\right)$
$V_{\text{max}} = \frac{\sqrt{3}\pi}{36}$

1. Relate the rate of change of volume to the rate of change of radius and solve for the radius $r$ at the given instant.

Let $V$ be the enclosed volume of the hemispherical dome.
$V = \frac{2}{3}\pi r^3$
Differentiating with respect to time $t$:
$\frac{\mathrm{d}V}{\mathrm{d}t} = 2\pi r^2 \frac{\mathrm{d}r}{\mathrm{d}t}$
Given $\frac{\mathrm{d}V}{\mathrm{d}t} = 0.75 = \frac{3}{4}$ and $\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{3}{4\pi}$:
$\frac{3}{4} = 2\pi r^2 \left(\frac{3}{4\pi}\right)$
$\frac{3}{4} = \frac{3}{2} r^2 \implies r^2 = \frac{1}{2}$
Since $r > 0$, $r = \frac{1}{\sqrt{2}}$.

2. Set up the related rate equation for the surface area of the open hemisphere.

Let $A$ be the surface area of the open hemisphere (curved surface area).
$A = 2\pi r^2$
Differentiating with respect to time $t$:
$\frac{\mathrm{d}A}{\mathrm{d}t} = 4\pi r \frac{\mathrm{d}r}{\mathrm{d}t}$

3. Calculate the exact rate of increase of the surface area.

Substitute $r = \frac{1}{\sqrt{2}}$ and $\frac{\mathrm{d}r}{\mathrm{d}t} = \frac{3}{4\pi}$ into the rate equation for $A$:
$\frac{\mathrm{d}A}{\mathrm{d}t} = 4\pi \left(\frac{1}{\sqrt{2}}\right) \left(\frac{3}{4\pi}\right)$
$\frac{\mathrm{d}A}{\mathrm{d}t} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$