图 1 展示了一个从一边长为 30 cm 的矩形纸板上裁剪出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正六棱柱，其底面为边长 a cm 的正六边形，直高为 h cm，如图 2 所示。

1. Determine the horizontal width of a regular hexagon.

A regular hexagon of side length $a$ can be partitioned into 6 equilateral triangles.
The perpendicular height from the centre of the hexagon to a side is $a \sin 60^\circ = \frac{\sqrt{3}}{2}a$.
The total horizontal distance across the hexagon (between two opposite parallel sides) is twice this height:
$\text{Width of hexagon} = 2 \times \left(\frac{\sqrt{3}}{2}a\right) = a\sqrt{3}\text{ cm}$

2. Formulate the equation using the total horizontal span of the net and simplify.

The horizontal span of the entire net from left to right consists of the left-hand rectangle, the central hexagon, the right-hand rectangle, and the second hexagon.
$\text{Total width} = h + a\sqrt{3} + h + a\sqrt{3} = 2h + 2a\sqrt{3}$
Given that the rectangular cardboard has a horizontal width of $30\text{ cm}$:
$2h + 2a\sqrt{3} = 30$
Dividing the entire equation by $2$:
$h + a\sqrt{3} = 15 \text{ (shown)}$

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

A regular hexagon of side $a$ can be divided into 6 congruent equilateral triangles of side $a$.
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. Calculate the total area of the hexagon.

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

1. Formulate the expression for the volume $V$ in terms of $a$.

Using the results from (a) and (b), substitute $h = 15 - a\sqrt{3}$ into the volume formula $V = \text{Base Area} \times h$:
$V = \left(\frac{3\sqrt{3}}{2} a^2\right)(15 - a\sqrt{3})$
$V = \frac{45\sqrt{3}}{2} a^2 - \frac{9}{2} a^3$

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

$\frac{\text{d}V}{\text{d}a} = 2 \left(\frac{45\sqrt{3}}{2}\right) a - 3 \left(\frac{9}{2}\right) a^2$
$\frac{\text{d}V}{\text{d}a} = 45\sqrt{3}a - \frac{27}{2}a^2$

3. Find the value of $a$ that gives a stationary point by equating $\frac{\text{d}V}{\text{d}a}$ to zero.

For a stationary point, $\frac{\text{d}V}{\text{d}a} = 0$:
$45\sqrt{3}a - \frac{27}{2}a^2 = 0$
$a\left(45\sqrt{3} - \frac{27}{2}a\right) = 0$
Since the side length $a > 0$, we have:
$\frac{27}{2}a = 45\sqrt{3}$
$a = \frac{90\sqrt{3}}{27} = \frac{10\sqrt{3}}{3}$

4. Perform the second derivative test to verify 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 is a maximum.

5. Substitute the value of $a$ back into the volume equation to find the exact maximum volume.

$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{10\sqrt{3}}{3}\right)^2 \left(15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}\right)$
$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{300}{9}\right) (15 - 10)$
$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{100}{3}\right) (5)$
$V_{\text{max}} = 50\sqrt{3} (5) = 250\sqrt{3}$

1. Find the values of $a$ and $h$ that give the maximum volume (from part c).

From parts (a) and (b), we have $h = 15 - a\sqrt{3}$ and $V = \frac{3\sqrt{3}}{2} a^2(15 - a\sqrt{3}) = \frac{45\sqrt{3}}{2} a^2 - \frac{9}{2} a^3$.

Setting $\frac{\mathrm{d}V}{\mathrm{d}a} = 0$ for maximum volume:
$45\sqrt{3} a - \frac{27}{2} a^2 = 0$

Since $a > 0$, we divide by $a$:
$\frac{27}{2} a = 45\sqrt{3} \implies a = \frac{90\sqrt{3}}{27} = \frac{10\sqrt{3}}{3}$

Substitute $a$ back to find the corresponding height $h$:
$h = 15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3} = 15 - 10 = 5$

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

The ratio of $h$ to $a$ is:
$\frac{h}{a} = \frac{5}{\frac{10\sqrt{3}}{3}}$
$\frac{h}{a} = \frac{15}{10\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$

Expressed as a ratio, $h : a = \sqrt{3} : 2$.

1. Determine the expression for volume, its intercepts, and the valid domain.

From parts (a) and (b), the volume $V$ is given by:
$V = \left(\frac{3\sqrt{3}}{2}a^2\right)(15 - a\sqrt{3})$
To find the $a$-intercepts, set $V = 0$:
$a = 0$ or $15 - a\sqrt{3} = 0 \implies a = \frac{15}{\sqrt{3}} = 5\sqrt{3}$.
Since length $a$ and height $h$ must be positive, the realistic domain for the sketch is $0 \le a \le 5\sqrt{3}$.

2. Identify the maximum point coordinates.

From part (c), the maximum volume occurs at $a = \frac{10\sqrt{3}}{3}$.
Substitute this back into the volume formula to find the maximum volume:
$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{10\sqrt{3}}{3}\right)^2 \left(15 - \left(\frac{10\sqrt{3}}{3}\right)\sqrt{3}\right)$
$V_{\text{max}} = \frac{3\sqrt{3}}{2} \left(\frac{100}{3}\right) (15 - 10) = 50\sqrt{3} \times 5 = 250\sqrt{3}$
The maximum point to label on the sketch is $\left(\frac{10\sqrt{3}}{3}, 250\sqrt{3}\right)$.

3. Describe the features of the sketch.

The sketch should be a smooth curve (a portion of a cubic graph with a negative leading coefficient) restricted to the first quadrant.

• It starts from the origin $(0, 0)$.
• It increases to the turning point $\left(\frac{10\sqrt{3}}{3}, 250\sqrt{3}\right)$.
• It decreases to cross the $a$-axis at $(5\sqrt{3}, 0)$ where it terminates.