一张先锋卡纸 $ABCDEF$ 呈边长为 $a\text{ cm}$ 的正六边形。从每个角剪掉一个风筝形，得到如图 1 所示的阴影形状。然后将其折叠成一个高为 $h\text{ cm}$ 的无盖正六边形盒子，如图 2 所示。

1. Find the length cut off from each corner of the original hexagon's sides.

Let the original regular hexagon be $ABCDEF$ with side $AB = a$. When the box is unfolded into its net, the top edge of each rectangular side lies on the perimeter of $ABCDEF$. Let the top edge on $AB$ have endpoints $P$ and $Q$.
Let $V$ be the corresponding vertex of the central hexagonal base. The side of the rectangle on the net is $PV$, which is perpendicular to $AB$ and has a length equal to the height of the box, $h$.
In the right-angled $\triangle APV$, $\angle APV = 90^\circ$.
Since the center of the regular hexagon aligns with its vertices, $AV$ bisects the $120^\circ$ interior angle at corner $A$. Thus, $\angle VAP = \frac{120^\circ}{2} = 60^\circ$.
Using trigonometry:
$\tan 60^\circ = \frac{PV}{AP} \implies \sqrt{3} = \frac{h}{AP} \implies AP = \frac{h}{\sqrt{3}}$

2. Express the side length of the central hexagonal base.

By symmetry, the length cut off from the other end of side $AB$ is $QB = \frac{h}{\sqrt{3}}$.
The side length $x$ of the central hexagonal base is equal to the length of the rectangle's top edge $PQ$.
$x = AB - AP - QB = a - \frac{h}{\sqrt{3}} - \frac{h}{\sqrt{3}} = a - \frac{2}{\sqrt{3}}h$

3. Calculate the base area of the box.

The central base is a regular hexagon of side length $x$. It can be divided into $6$ identical equilateral triangles.
The area of one such equilateral triangle is $\frac{1}{2} x^2 \sin 60^\circ = \frac{\sqrt{3}}{4}x^2$.
$\text{Base Area} = 6 \times \frac{\sqrt{3}}{4}x^2 = \frac{3\sqrt{3}}{2}x^2$
Substituting $x = a - \frac{2}{\sqrt{3}}h$:
$\text{Base Area} = \frac{3\sqrt{3}}{2} \left(a - \frac{2}{\sqrt{3}}h\right)^2$

4. Formulate the volume $V$ and algebraically manipulate it to the required form.

The volume $V$ is given by $\text{Base Area} \times \text{height } h$.
$V = \frac{3\sqrt{3}}{2}h \left(a - \frac{2}{\sqrt{3}}h\right)^2$
To match the required expression, factor out $\frac{2}{\sqrt{3}}$ from the terms inside the squared bracket:
$a - \frac{2}{\sqrt{3}}h = \frac{2}{\sqrt{3}} \left(\frac{\sqrt{3}}{2}a - h\right)$
Substitute this back into the volume equation:
$V = \frac{3\sqrt{3}}{2}h \left[ \frac{2}{\sqrt{3}} \left(\frac{\sqrt{3}}{2}a - h\right) \right]^2$
$V = \frac{3\sqrt{3}}{2}h \times \frac{4}{3} \times \left(\frac{\sqrt{3}}{2}a - h\right)^2$
$V = 2\sqrt{3}h \left(\frac{\sqrt{3}}{2}a - h\right)^2 \quad \text{(Shown)}$

1. Find the first derivative of $V$ with respect to $h$.

$V = 2\sqrt{3}h\left(\frac{\sqrt{3}}{2}a - h\right)^2$
Using the product rule and chain rule:
$\frac{\mathrm{d}V}{\mathrm{d}h} = 2\sqrt{3}\left(\frac{\sqrt{3}}{2}a - h\right)^2 + 2\sqrt{3}h\left[2\left(\frac{\sqrt{3}}{2}a - h\right)(-1)\right]$
$\frac{\mathrm{d}V}{\mathrm{d}h} = 2\sqrt{3}\left(\frac{\sqrt{3}}{2}a - h\right) \left[ \left(\frac{\sqrt{3}}{2}a - h\right) - 2h \right]$
$\frac{\mathrm{d}V}{\mathrm{d}h} = 2\sqrt{3}\left(\frac{\sqrt{3}}{2}a - h\right)\left(\frac{\sqrt{3}}{2}a - 3h\right)$

2. Set the first derivative to zero and solve for the stationary values of $h$.

For maximum volume, let $\frac{\mathrm{d}V}{\mathrm{d}h} = 0$:
$2\sqrt{3}\left(\frac{\sqrt{3}}{2}a - h\right)\left(\frac{\sqrt{3}}{2}a - 3h\right) = 0$
$\frac{\sqrt{3}}{2}a - h = 0 \implies h = \frac{\sqrt{3}}{2}a$
$\frac{\sqrt{3}}{2}a - 3h = 0 \implies h = \frac{\sqrt{3}}{6}a$

3. Use the second derivative test to verify which value of $h$ gives a maximum.

Expand $\frac{\mathrm{d}V}{\mathrm{d}h}$ to find $\frac{\mathrm{d}^2V}{\mathrm{d}h^2}$:
$\frac{\mathrm{d}V}{\mathrm{d}h} = 2\sqrt{3}\left(\frac{3}{4}a^2 - 2\sqrt{3}ah + 3h^2\right) = \frac{3\sqrt{3}}{2}a^2 - 12ah + 6\sqrt{3}h^2$
$\frac{\mathrm{d}^2V}{\mathrm{d}h^2} = -12a + 12\sqrt{3}h$

Substitute $h = \frac{\sqrt{3}}{6}a$:
$\frac{\mathrm{d}^2V}{\mathrm{d}h^2} = -12a + 12\sqrt{3}\left(\frac{\sqrt{3}}{6}a\right) = -12a + 6a = -6a$
Since $a > 0$, $\frac{\mathrm{d}^2V}{\mathrm{d}h^2} < 0$, which confirms a maximum volume.
(Note: substituting $h = \frac{\sqrt{3}}{2}a$ gives $\frac{\mathrm{d}^2V}{\mathrm{d}h^2} = -12a + 18a = 6a > 0$, which corresponds to a minimum.)