图 1 展示了一个直正六棱柱的展开图，它是从一张边长为 30 cm 的矩形纸板上裁剪下来的。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。将展开图折叠后形成一个直六棱柱，其底面为边长 $a$ cm 的正六边形，直高为 $h$ cm，如图 2 所示。

1. Find the distance across opposite parallel sides of the regular hexagon.

A regular hexagon of side $a$ can be divided into 6 equilateral triangles of side length $a$.
The perpendicular height of one such equilateral triangle is:
$a \sin 60^\circ = a \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{2}a$
The horizontal distance between opposite parallel sides of the hexagon is twice this height:
$\text{Distance} = 2 \left( \frac{\sqrt{3}}{2}a \right) = a\sqrt{3}$

2. Formulate the equation for the width of the cardboard and simplify to obtain the required result.

Based on the layout of the net, the maximum span across the rectangular boundary consists of the length of two rectangles (each $h$) and the width of two hexagons (each $a\sqrt{3}$).
$\text{Total width} = h + a\sqrt{3} + h + a\sqrt{3}$
$\text{Total width} = 2h + 2a\sqrt{3}$
Given that this total width corresponds to the cardboard side length of 30 cm:
$2h + 2a\sqrt{3} = 30$
Dividing the entire equation by 2 gives:
$h + a\sqrt{3} = 15 \quad \text{(shown)}$

1. Find the area of one of the 6 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 $a$.
$\text{Area of one equilateral triangle} = \frac{1}{2}(a)(a)\sin 60^\circ$
$= \frac{1}{2}a^2\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}a^2$

2. Calculate the total area of the regular hexagon.

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

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

$V = \text{Base Area} \times h$
$V = \left( \frac{3\sqrt{3}}{2}a^2 \right) h$
From part (a), $h = 15 - a\sqrt{3}$. Substitute this into the volume equation:
$V = \frac{3\sqrt{3}}{2}a^2 (15 - a\sqrt{3})$
$V = \frac{45\sqrt{3}}{2}a^2 - \frac{9}{2}a^3$

2. Differentiate $V$ with respect to $a$.

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

3. Find the critical value of $a$ by setting the derivative to zero.

Set $\frac{\text{d}V}{\text{d}a} = 0$ for stationary points:
$a \left( 45\sqrt{3} - \frac{27}{2}a \right) = 0$
Since $a > 0$ for a valid physical length, we reject $a=0$:
$45\sqrt{3} - \frac{27}{2}a = 0$
$a = \frac{90\sqrt{3}}{27} = \frac{10\sqrt{3}}{3}$

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

$\frac{\text{d}^2V}{\text{d}a^2} = 45\sqrt{3} - 27a$
Substitute $a = \frac{10\sqrt{3}}{3}$:
$\frac{\text{d}^2V}{\text{d}a^2} = 45\sqrt{3} - 27\left(\frac{10\sqrt{3}}{3}\right)$
$\frac{\text{d}^2V}{\text{d}a^2} = 45\sqrt{3} - 90\sqrt{3} = -45\sqrt{3}$
Since $\frac{\text{d}^2V}{\text{d}a^2} < 0$, the volume $V$ is a maximum.

5. Calculate the exact maximum volume.

When $a = \frac{10\sqrt{3}}{3}$, we first find $h$:
$h = 15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3} = 15 - 10 = 5$
Substitute into $V = \text{Base Area} \times h$:
$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{10\sqrt{3}}{3}\right)^2 (5)$
$V_{\text{max}} = \frac{15\sqrt{3}}{2} \left(\frac{100 \times 3}{9}\right)$
$V_{\text{max}} = \frac{15\sqrt{3}}{2} \left(\frac{100}{3}\right) = 250\sqrt{3} \text{ cm}^3$

1. Find the corresponding value of $h$ for the maximum volume

From part (c), the maximum volume occurs when $\frac{\mathrm{d}V}{\mathrm{d}a} = 0$, which gives $a = \frac{10\sqrt{3}}{3}$.

Substituting this into the equation relating $h$ and $a$ from part (a):
$h = 15 - a\sqrt{3}$
$h = 15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}$
$h = 15 - \frac{10(3)}{3} = 15 - 10 = 5$

2. Calculate and simplify the ratio of $h$ to $a$

Ratio $\frac{h}{a} = \frac{5}{\frac{10\sqrt{3}}{3}}$
$\frac{h}{a} = \frac{15}{10\sqrt{3}}$
$\frac{h}{a} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$

The ratio of $h$ to $a$ is $\sqrt{3} : 2$.

1. Identify the equation for the volume and the practical domain for $a$.

From earlier parts, the volume is $V = \frac{3\sqrt{3}}{2} a^2 (15 - a\sqrt{3})$.
For a physical prism, the side length $a > 0$ and the height $h > 0$.
Since $h = 15 - a\sqrt{3} > 0$, we have $a < \frac{15}{\sqrt{3}} = 5\sqrt{3}$.
Thus, the meaningful domain for the graph is $0 < a < 5\sqrt{3}$.

2. Determine the coordinates of the intercepts and the turning point.

The $a$-intercepts occur where $V = 0$, which gives $a = 0$ and $a = 5\sqrt{3}$.
From part (c), the maximum volume occurs at $a = \frac{10\sqrt{3}}{3}$.
Substituting this into $V$ yields the maximum volume:
$V = \frac{3\sqrt{3}}{2} \left(\frac{10\sqrt{3}}{3}\right)^2 \left(15 - \frac{10\sqrt{3}}{3}\sqrt{3}\right) = \frac{3\sqrt{3}}{2} \left(\frac{100}{3}\right)(15 - 10) = 250\sqrt{3}$.
So the maximum point is $\left(\frac{10\sqrt{3}}{3}, 250\sqrt{3}\right)$.

3. Describe the features of the sketched curve.

The graph is a smooth cubic curve. Since $\frac{dV}{da} = 0$ at $a = 0$, the curve starts tangent to the horizontal axis at $(0,0)$, increases to the maximum turning point at $\left(\frac{10\sqrt{3}}{3}, 250\sqrt{3}\right)$, and then decreases until it hits the $a$-axis at $(5\sqrt{3}, 0)$.