图1展示了从一张一边长为 30 cm 的矩形硬纸板上剪下的直正六棱柱的展开图。该展开图由两个边长为 $a$ cm 的正六边形和六个边长分别为 $a$ cm 和 $h$ cm 的矩形组成。将该展开图折叠后形成一个底面正六边形边长为 $a$ cm、高（垂直高度）为 $h$ cm 的棱柱，如图2所示。

1. Find the horizontal width of a regular hexagon across its opposite parallel sides.

The distance between two opposite parallel sides of a regular hexagon with side length $a$ is twice its apothem:
$2 \times a \sin(60^\circ) = 2 \times a \left(\frac{\sqrt{3}}{2}\right) = a\sqrt{3}\text{ cm}$

2. Formulate an equation for the total width of the net and simplify to the required form.

The horizontal span of the net perfectly fits the $30\text{ cm}$ side of the rectangular cardboard. This span consists of the leftmost rectangle, the central hexagon, the rightmost rectangle, and the second hexagon.
$h + a\sqrt{3} + h + a\sqrt{3} = 30$
$2h + 2a\sqrt{3} = 30$
$h + a\sqrt{3} = 15 \text{ (shown)}$

1. Find the area of one of the 6 congruent equilateral triangles that make up the regular hexagon.

A regular hexagon of side length $a$ can be divided into 6 congruent equilateral triangles with side length $a$.
Area of one equilateral triangle $= \frac{1}{2}(a)(a)\sin 60^\circ$
$= \frac{1}{2}a^2\left(\frac{\sqrt{3}}{2}\right)$
$= \frac{\sqrt{3}}{4}a^2$

2. Multiply the area of one triangle by 6 to find the total area of the hexagonal base.

Area of the hexagonal base $= 6 \times \frac{\sqrt{3}}{4}a^2$
$= \frac{3\sqrt{3}}{2}a^2\text{ cm}^2 \text{ (Shown)}$

1. Express the volume of the prism, $V$, in terms of $a$.

From part (a), $h = 15 - a\sqrt{3}$.
The volume of the prism is given by $V = \text{Base Area} \times h$.
Using the area from part (b):
$V = \left(\frac{3\sqrt{3}}{2}a^2\right)(15 - a\sqrt{3})$
$V = \frac{45\sqrt{3}}{2}a^2 - \frac{9}{2}a^3$

2. Differentiate $V$ with respect to $a$ and set it to $0$ to find the stationary point.

$\frac{\mathrm{d}V}{\mathrm{d}a} = 45\sqrt{3}a - \frac{27}{2}a^2$

For maximum volume, let $\frac{\mathrm{d}V}{\mathrm{d}a} = 0$:
$45\sqrt{3}a - \frac{27}{2}a^2 = 0$
$a\left(45\sqrt{3} - \frac{27}{2}a\right) = 0$

3. Solve for the non-zero value of $a$.

Since $a$ is a length, $a > 0$.
$\frac{27}{2}a = 45\sqrt{3}$
$a = \frac{90\sqrt{3}}{27} = \frac{10\sqrt{3}}{3}$

4. Prove that this value of $a$ gives a maximum volume.

Using the second derivative test:
$\frac{\mathrm{d}^2V}{\mathrm{d}a^2} = 45\sqrt{3} - 27a$

Substitute $a = \frac{10\sqrt{3}}{3}$:
$\frac{\mathrm{d}^2V}{\mathrm{d}a^2} = 45\sqrt{3} - 27\left(\frac{10\sqrt{3}}{3}\right)$
$\frac{\mathrm{d}^2V}{\mathrm{d}a^2} = 45\sqrt{3} - 90\sqrt{3} = -45\sqrt{3}$

Since $\frac{\mathrm{d}^2V}{\mathrm{d}a^2} < 0$, the volume is a maximum.

5. Calculate the exact maximum volume.

Substitute $a = \frac{10\sqrt{3}}{3}$ into the volume expression:
$V = \left[\frac{3\sqrt{3}}{2}\left(\frac{10\sqrt{3}}{3}\right)^2\right] \left[15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}\right]$
$V = \left[\frac{3\sqrt{3}}{2}\left(\frac{300}{9}\right)\right] \left[15 - 10\right]$
$V = \left(\frac{3\sqrt{3}}{2} \times \frac{100}{3}\right) (5)$
$V = (50\sqrt{3}) (5) = 250\sqrt{3}$

1. Retrieve the value of $a$ that gives the maximum volume from part (c) and calculate the corresponding value of $h$.

From part (c), substituting $h = 15 - a\sqrt{3}$ gives the volume $V = \frac{45\sqrt{3}}{2}a^2 - \frac{9}{2}a^3$.
Setting $\frac{dV}{da} = 45\sqrt{3}a - \frac{27}{2}a^2 = 0$ yields $a = \frac{10\sqrt{3}}{3}$ (for $a > 0$).
Substituting this back to find $h$:
$h = 15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}$
$h = 15 - 10 = 5$

2. Form the ratio $h : a$ and simplify it to exact form.

$\frac{h}{a} = \frac{5}{\frac{10\sqrt{3}}{3}}$
$= \frac{15}{10\sqrt{3}}$
$= \frac{3}{2\sqrt{3}}$
Rationalising the denominator gives:
$\frac{h}{a} = \frac{\sqrt{3}}{2}$
The ratio $h : a$ is $\sqrt{3} : 2$.

1. Determine the intercepts and the practical domain of the volume function.

The volume function is $V = \frac{3\sqrt{3}}{2}a^2(15 - a\sqrt{3})$.
Setting $V = 0$ gives $a = 0$ (repeated root) or $15 - a\sqrt{3} = 0 \implies a = \frac{15}{\sqrt{3}} = 5\sqrt{3}$.
Since $a$ (side length) and $h$ (height) must be positive physical dimensions, we require $a > 0$ and $15 - a\sqrt{3} > 0$, so the practical domain is $0 < a < 5\sqrt{3}$.
The curve intersects the $a$-axis at the origin $(0,0)$ and at $(5\sqrt{3}, 0)$.

2. Determine the maximum point and sketch the curve.

From part (c), the maximum volume occurs at $a = \frac{10\sqrt{3}}{3}$.
Substituting this into the volume equation gives the maximum volume:
$V = \frac{3\sqrt{3}}{2}\left(\frac{10\sqrt{3}}{3}\right)^2\left(15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}\right) = \frac{3\sqrt{3}}{2}\left(\frac{300}{9}\right)(15 - 10) = 250\sqrt{3}$.
The graph of $V$ against $a$ is a section of a cubic curve with a negative $a^3$ coefficient.
Sketching the valid portion yields an inverted U-shaped curve starting from $(0,0)$, peaking at the maximum point $\left(\frac{10\sqrt{3}}{3}, 250\sqrt{3}\right)$, and terminating at the $a$-intercept $(5\sqrt{3}, 0)$.