图 1 展示了一张卡片 $ABCD$，它是一个边长为 $a$ cm 的菱形。该卡片由两个完全相同的等边三角形 $ABD$ 和 $CBD$ 组成。从每个角剪掉筝形，得到图 2 所示的形状。将图 2 中剩余的卡片沿虚线折叠，形成一个高为 $x$ 的无盖菱形棱柱，如图 3 所示。

1. Determine the lengths cut off from the original edges at the corners.

Let the side length of the new base rhombus be $y$. The vertical walls of the prism have height $x$, so the perpendicular distance from the fold lines to the original edges is $x$.
At each $60^\circ$ corner (e.g., $A$ and $C$), the fold lines meet on the angle bisector ($30^\circ$). By using right-angled triangles from the kite cut-outs, the length removed from the edge is:
$x \cot 30^\circ = x\sqrt{3}$
At each $120^\circ$ corner (e.g., $B$ and $D$), the fold lines meet on the angle bisector ($60^\circ$). The length removed from the edge is:
$x \cot 60^\circ = \frac{x}{\sqrt{3}}$

2. Express the side length of the base rhombus in terms of $a$ and $x$.

Subtracting these corner portions from the original side length $a$ gives $y$:
$y = a - x\sqrt{3} - \frac{x}{\sqrt{3}}$
By taking a common denominator of $\sqrt{3}$:
$y = a - \frac{3x}{\sqrt{3}} - \frac{x}{\sqrt{3}} = a - \frac{4x}{\sqrt{3}}$

3. Find the area of the base rhombus.

The base is a rhombus with side length $y$ and interior angles $60^\circ$ and $120^\circ$.
Using the area formula $\text{Area} = y^2 \sin 60^\circ$:
$\text{Area of base} = \left(a - \frac{4x}{\sqrt{3}}\right)^2 \left(\frac{\sqrt{3}}{2}\right)$

4. Formulate the volume of the prism.

The volume of a uniform prism is given by its base area multiplied by its vertical height $x$.
$V = \text{Area of base} \times \text{height}$
$V = \frac{\sqrt{3}}{2} \left(a - \frac{4x}{\sqrt{3}}\right)^2 \times x$
$V = \frac{\sqrt{3}}{2}x\left(a - \frac{4x}{\sqrt{3}}\right)^2 \quad \text{(shown)}$

1. Expand the expression for $V$ and find its first derivative with respect to $x$.

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

2. Set the derivative to zero to find the stationary values of $x$ and reject the invalid length.

For a stationary point, $\frac{\mathrm{d}V}{\mathrm{d}x} = 0$.
$8\sqrt{3}x^2 - 8ax + \frac{\sqrt{3}}{2}a^2 = 0$
Using the quadratic formula to solve for $x$:
$x = \frac{-(-8a) \pm \sqrt{(-8a)^2 - 4(8\sqrt{3})\left(\frac{\sqrt{3}}{2}a^2\right)}}{2(8\sqrt{3})}$
$x = \frac{8a \pm \sqrt{64a^2 - 48a^2}}{16\sqrt{3}} = \frac{8a \pm 4a}{16\sqrt{3}}$
$x = \frac{12a}{16\sqrt{3}} = \frac{\sqrt{3}}{4}a \quad \text{or} \quad x = \frac{4a}{16\sqrt{3}} = \frac{\sqrt{3}}{12}a$
Since the cut size $x$ must leave a positive base dimension, $a - \frac{4x}{\sqrt{3}} > 0 \implies x < \frac{\sqrt{3}}{4}a$. Thus, we reject $x = \frac{\sqrt{3}}{4}a$.

3. Substitute the valid $x$ value into $V$ to determine the maximum volume.

Substitute $x = \frac{\sqrt{3}}{12}a$ into the original expression for $V$:
$V_{\text{max}} = \frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}}{12}a\right)\left(a - \frac{4}{\sqrt{3}}\left(\frac{\sqrt{3}}{12}a\right)\right)^2$
$V_{\text{max}} = \frac{3}{24}a\left(a - \frac{1}{3}a\right)^2$
$V_{\text{max}} = \frac{1}{8}a\left(\frac{2}{3}a\right)^2$
$V_{\text{max}} = \frac{1}{8}a\left(\frac{4}{9}a^2\right) = \frac{1}{18}a^3$

4. Determine the second derivative and verify the nature of the stationary point.

$\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -8a + 16\sqrt{3}x$
When $x = \frac{\sqrt{3}}{12}a$:
$\frac{\mathrm{d}^2V}{\mathrm{d}x^2} = -8a + 16\sqrt{3}\left(\frac{\sqrt{3}}{12}a\right) = -8a + 4a = -4a$
Since $a$ represents a positive side length ($a > 0$), $-4a < 0$.
By the second derivative test, this confirms that the volume $V$ is a maximum.

1. Express the volume of sand in terms of the depth using the prism's (constant) base area

From part (ii) the maximum volume occurs at $x=\dfrac{\sqrt3 a}{12}$, giving base rhombus side $s=a-\dfrac{4x}{\sqrt3}=a-\dfrac{a}{3}=\dfrac{2a}{3}$.

Base area (a rhombus = two equilateral triangles of side $s$):
$A=\frac{\sqrt3}{2}s^2=\frac{\sqrt3}{2}\left(\frac{2a}{3}\right)^2=\frac{2\sqrt3 a^2}{9}.$
Since the prism has a uniform cross-section, the volume of sand at depth $h$ is
$V=A\,h=\frac{2\sqrt3 a^2}{9}\,h.$

2. Set up the connected-rate relationship

Differentiating with respect to $t$ (the base area $A$ is constant):
$\frac{dV}{dt}=A\,\frac{dh}{dt}.$
The sand is added at $\dfrac{dV}{dt}=\sqrt3\ \text{cm}^3/\text{s}$, so
$\sqrt3=\frac{2\sqrt3 a^2}{9}\,\frac{dh}{dt}.$

3. Solve for the rate of increase of depth

$\frac{dh}{dt}=\frac{\sqrt3}{\frac{2\sqrt3 a^2}{9}}=\sqrt3\times\frac{9}{2\sqrt3 a^2}=\frac{9}{2a^2}\ \text{cm/s}.$
Because the cross-section is uniform, this rate is the same at every depth, so it equals $\dfrac{9}{2a^2}\ \text{cm/s}$ regardless of the stated depth.