在一部科幻电影中，为了保护一个正在建造的用于存放致命武器的圆柱形筒仓，创建了一个半球形的力场护盾（一个开半球）。筒仓顶部的圆周恰好位于半球护盾内部，如下图所示。电力供应使得生成的护盾半径 $r$ 固定为 0.5 km。设筒仓的半径和高度分别为 $x$ km 和 $y$ km。

1. Relate the radius $x$ and height $y$ of the silo using the hemisphere's radius.

From the vertical cross-section, the radius of the silo $x$, its height $y$, and the radius of the hemisphere $r$ form a right-angled triangle.
$x^2 + y^2 = r^2$
Given the hemisphere radius is $r = 0.5$ km,
$x^2 + y^2 = 0.5^2 = \frac{1}{4}$
$x^2 = \frac{1}{4} - y^2$

2. Formulate the volume of the silo in terms of $y$ to show the given expression.

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 equate to zero to find the stationary point.

To find the maximum volume, we differentiate $V$ with respect to $y$:
$\frac{\text{d}V}{\text{d}y} = \pi \left( \frac{1}{4} - 3y^2 \right)$
At the maximum volume, $\frac{\text{d}V}{\text{d}y} = 0$:
$\pi \left( \frac{1}{4} - 3y^2 \right) = 0$
$3y^2 = \frac{1}{4}$
$y^2 = \frac{1}{12}$
Since the height $y > 0$, we have $y = \frac{1}{\sqrt{12}} = \frac{1}{2\sqrt{3}}$.

4. Use the second derivative to verify that the volume is a maximum.

$\frac{\text{d}^2V}{\text{d}y^2} = \pi ( -6y ) = -6\pi y$
Since $y = \frac{1}{2\sqrt{3}} > 0$, $\frac{\text{d}^2V}{\text{d}y^2} < 0$.
Thus, the volume is maximized at this value of $y$.

5. Calculate the exact maximum volume by substituting $y$ into the volume expression.

Substitute $y = \frac{1}{2\sqrt{3}}$ into $V$:
$V_{\text{max}} = \pi \left[ \frac{1}{4}\left(\frac{1}{2\sqrt{3}}\right) - \left(\frac{1}{2\sqrt{3}}\right)^3 \right]$
$V_{\text{max}} = \pi \left( \frac{1}{8\sqrt{3}} - \frac{1}{24\sqrt{3}} \right)$
$V_{\text{max}} = \pi \left( \frac{3 - 1}{24\sqrt{3}} \right) = \pi \left( \frac{2}{24\sqrt{3}} \right) = \frac{\pi}{12\sqrt{3}}$
Rationalising the denominator:
$V_{\text{max}} = \frac{\sqrt{3}\pi}{36}$

1. Express the volume $V$ of the hemispherical dome in terms of its radius $r$, and use the chain rule to find an expression for $\frac{dV}{dt}$.

Since the dome is a hemisphere, its volume is half that of a sphere:
$V = \frac{1}{2} \left( \frac{4}{3}\pi r^3 \right) = \frac{2}{3}\pi r^3$
Differentiating with respect to $t$:
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt}$

2. Substitute the given rates to solve for the radius $r$ at the given instant.

Given $\frac{dV}{dt} = 0.75 = \frac{3}{4}$ and $\frac{dr}{dt} = \frac{3}{4\pi}$:
$\frac{3}{4} = 2\pi r^2 \left( \frac{3}{4\pi} \right)$
$\frac{3}{4} = \frac{3}{2} r^2$
$r^2 = \frac{1}{2} \implies r = \frac{1}{\sqrt{2}} \quad (\text{since } r > 0)$

3. Express the surface area $A$ of the open hemispherical dome in terms of $r$ and differentiate with respect to $t$.

Since the dome is an open hemisphere, its surface area is half the surface area of a sphere:
$A = \frac{1}{2}(4\pi r^2) = 2\pi r^2$
Differentiating with respect to $t$:
$\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} = 4\pi r \frac{dr}{dt}$

4. Substitute the values of $r$ and $\frac{dr}{dt}$ to find the exact rate of increase of the surface area.

$\frac{dA}{dt} = 4\pi \left( \frac{1}{\sqrt{2}} \right) \left( \frac{3}{4\pi} \right)$
$\frac{dA}{dt} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$