图 1 展示了一张边长为 $a$ cm 的菱形纸板 $ABCD$。该纸板由两个完全相同的等边三角形 $ABD$ 和 $CBD$ 组成。从每个角剪掉风筝形，得到图 2 所示的形状。将图 2 中剩余的纸板沿虚线折叠，形成一个高为 $x$ 的无盖菱形柱体，如图 3 所示。

1. Determine the interior angles of the rhombus and the perpendicular distance from the folded edge to the original edge.

Since the original rhombus is formed by two identical equilateral triangles, its interior angles are $60^\circ$ and $120^\circ$.
When the card is folded along the dotted lines to form vertical walls of height $x$, the perpendicular distance from each edge of the original rhombus to the corresponding parallel edge of the new inner base is exactly $x$.

2. Calculate the reduction in side length at each corner using trigonometry.

At each end of an original side of length $a$, the length is reduced by the portions forming the cut kites (which become the folded flaps).
At a $60^\circ$ corner, the angle bisector is $30^\circ$. The side length reduction is $\dfrac{x}{\tan 30^\circ} = x\sqrt{3}$.
At a $120^\circ$ corner, the angle bisector is $60^\circ$. The side length reduction is $\dfrac{x}{\tan 60^\circ} = \dfrac{x}{\sqrt{3}}$.

3. Find the side length of the inner base rhombus.

The side length $s$ of the new base rhombus is the original side length minus the reductions at both ends:
$s = a - x\sqrt{3} - \frac{x}{\sqrt{3}}$
$s = a - \frac{3x}{\sqrt{3}} - \frac{x}{\sqrt{3}} = a - \frac{4x}{\sqrt{3}}$

4. Calculate the area of the base and the volume of the prism to complete the proof.

The base is a rhombus with interior angles $60^\circ$ and $120^\circ$.
$\text{Base Area} = s^2 \sin 60^\circ = \left(a - \frac{4x}{\sqrt{3}}\right)^2 \left(\frac{\sqrt{3}}{2}\right)$
The volume $V$ is the base area multiplied by the height $x$:
$V = \text{Base Area} \times \text{height}$
$V = \frac{\sqrt{3}\,x}{2}\left(a-\dfrac{4x}{\sqrt{3}}\right)^{2} \quad \text{(shown)}$

1. Expand the volume expression into a polynomial in terms of $x$.

$V = \frac{\sqrt{3}x}{2} \left(a - \frac{4x}{\sqrt{3}}\right)^2$
$V = \frac{\sqrt{3}x}{2} \left(a^2 - \frac{8ax}{\sqrt{3}} + \frac{16x^2}{3}\right)$
$V = \frac{\sqrt{3}a^2}{2} x - 4ax^2 + \frac{8\sqrt{3}}{3} x^3$

2. Differentiate $V$ with respect to $x$ and set it to zero to find the stationary points.

$\frac{\text{d}V}{\text{d}x} = \frac{\sqrt{3}a^2}{2} - 8ax + 8\sqrt{3} x^2$
Set $\frac{\text{d}V}{\text{d}x} = 0$:
$8\sqrt{3} x^2 - 8ax + \frac{\sqrt{3}a^2}{2} = 0$
Multiply by $2$ to clear the fraction:
$16\sqrt{3} x^2 - 16ax + \sqrt{3}a^2 = 0$

3. Solve the quadratic equation to find the possible values of $x$.

Using the quadratic formula:
$x = \frac{16a \pm \sqrt{(-16a)^2 - 4(16\sqrt{3})(\sqrt{3}a^2)}}{2(16\sqrt{3})}$
$x = \frac{16a \pm \sqrt{256a^2 - 192a^2}}{32\sqrt{3}}$
$x = \frac{16a \pm \sqrt{64a^2}}{32\sqrt{3}} = \frac{16a \pm 8a}{32\sqrt{3}}$
$x = \frac{24a}{32\sqrt{3}} = \frac{\sqrt{3}a}{4}$ or $x = \frac{8a}{32\sqrt{3}} = \frac{\sqrt{3}a}{12}$

4. Use the second derivative test to prove which value of $x$ gives a maximum.

$\frac{\text{d}^2V}{\text{d}x^2} = -8a + 16\sqrt{3} x$
When $x = \frac{\sqrt{3}a}{4}$:
$\frac{\text{d}^2V}{\text{d}x^2} = -8a + 16\sqrt{3}\left(\frac{\sqrt{3}a}{4}\right) = -8a + 12a = 4a > 0$ (Minimum)

When $x = \frac{\sqrt{3}a}{12}$:
$\frac{\text{d}^2V}{\text{d}x^2} = -8a + 16\sqrt{3}\left(\frac{\sqrt{3}a}{12}\right) = -8a + 4a = -4a$
Since $a$ is a side length, $a > 0$, so $\frac{\text{d}^2V}{\text{d}x^2} < 0$.
This proves that $V$ is a maximum when $x = \frac{\sqrt{3}a}{12}$.

5. Substitute the optimum $x$ back into $V$ to find the maximum volume.

$V_{\text{max}} = \frac{\sqrt{3}}{2} \left(\frac{\sqrt{3}a}{12}\right) \left(a - \frac{4}{\sqrt{3}} \left(\frac{\sqrt{3}a}{12}\right)\right)^2$
$V_{\text{max}} = \frac{3a}{24} \left(a - \frac{a}{3}\right)^2$
$V_{\text{max}} = \frac{a}{8} \left(\frac{2a}{3}\right)^2$
$V_{\text{max}} = \frac{a}{8} \cdot \frac{4a^2}{9} = \frac{a^3}{18}$

1. Expand $V$ and differentiate to obtain $\frac{dV}{dx}$

From part (i), $V=\dfrac{\sqrt{3}}{2}x\left(a-\dfrac{4x}{\sqrt3}\right)^2$.

Expanding:
$V=\frac{\sqrt3}{2}a^2x-4ax^2+\frac{8\sqrt3}{3}x^3$

Differentiating with respect to $x$:
$\frac{dV}{dx}=\frac{\sqrt3}{2}a^2-8ax+8\sqrt3\,x^2$

2. Evaluate $\frac{dV}{dx}$ at the given depth

Substitute $x=\sqrt3\,a$:
$\frac{dV}{dx}=\frac{\sqrt3}{2}a^2-8a(\sqrt3 a)+8\sqrt3(\sqrt3a)^2$
$=\frac{\sqrt3}{2}a^2-8\sqrt3 a^2+24\sqrt3 a^2=\frac{33\sqrt3}{2}a^2$

3. Apply the chain rule and solve for $\frac{dx}{dt}$

By the chain rule:
$\frac{dV}{dt}=\frac{dV}{dx}\cdot\frac{dx}{dt}$

With $\dfrac{dV}{dt}=\sqrt3$:
$\sqrt3=\frac{33\sqrt3}{2}a^2\cdot\frac{dx}{dt}$
$\frac{dx}{dt}=\frac{\sqrt3}{\frac{33\sqrt3}{2}a^2}=\frac{2}{33a^2}\text{ cm/s}$