In a futuristic movie, a hemispherical force field dome (an open hemisphere) is created to protect a cylindrical silo under construction to house deadly weapons. The circumference of the top of the silo lies just within the dome as shown in the diagram below. The power supply is such that the radius $r$ of the dome generated is fixed at 0.5 km. Let $x$ km and $y$ km be the radius and height of the silo respectively.

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.

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