图 1 显示了一个无盖容器，其形状为一个等腰梯形柱体 $ABCDEFGH$，其正方形底面 $ABCD$ 的边长为 $x\text{ cm}$，其中 $x$ 是一个常数。面 $ABFE$ 和 $DCGH$ 垂直于底面 $ABCD$。然后将该容器沿 $BC$ 倾斜，使得面 $BCGF$ 和底面 $ABCD$ 分别与水平面成 $30^\circ$ 的角，如图 2 所示。接着，以 $x^3\text{ cm}^3/\text{min}$ 的恒定速率将水注入这个空容器中。假设在此问题中容器的厚度可以忽略不计。

1. Identify the lowest point of the container's opening from which water will overflow.

The container is open at the top face $EFGH$. When tilted, the water will overflow from the lowest point on the rim $EFGH$. Since the container is tilted along $BC$, the edge $FG$ is horizontal and lower than $EH$. Thus, water overflows over the edge $FG$.

2. Calculate the vertical height of the edge $FG$ above the horizontal surface.

The face $BCGF$ makes an angle of $30^\circ$ with the horizontal. The length of $BF$ is $x$ (since $ABFE$ is an isosceles trapezoid and $AE=x$, by symmetry $BF=x$).

The height of $F$ (and $G$) above the horizontal surface at $B$ is given by:
$\text{Height} = BF \sin 30^\circ$
$\text{Height} = x \left(\frac{1}{2}\right) = 0.5x$

1. Find the cross-sectional area of the water (triangle ABF)

Just before overflowing, the horizontal water surface passes through $A$ and $F$, both at height $\tfrac{x}{2}$ (from part (a)). The water cross-section is the isosceles triangle $ABF$ with $AB = BF = x$ and included angle $\angle ABF = 30^\circ+30^\circ = 120^\circ$.

Using the surface $AF$ as base with height $\tfrac{x}{2}$, where $AF = 2\bigl(x\cos 30^\circ\bigr) = x\sqrt{3}$:
$\text{Area} = \tfrac{1}{2}\times AF \times \text{height} = \tfrac{1}{2}\times x\sqrt{3}\times \tfrac{x}{2} = \frac{\sqrt{3}}{4}x^2$
(Equivalently $\tfrac{1}{2}x\cdot x\sin120^\circ = \tfrac{\sqrt3}{4}x^2$.)

2. Find the volume of water in the container

The container is a prism of depth $BC = x$, so the volume of water is the cross-sectional area multiplied by the depth:
$V = \frac{\sqrt{3}}{4}x^2 \times x = \frac{\sqrt{3}}{4}x^3 \text{ cm}^3$

3. Find the filling time and convert to seconds

Water is added at a constant rate of $x^3\text{ cm}^3/\text{min}$, so
$\text{Time} = \frac{V}{\text{rate}} = \frac{\tfrac{\sqrt{3}}{4}x^3}{x^3} = \frac{\sqrt{3}}{4}\text{ min}$
Converting to seconds:
$\frac{\sqrt{3}}{4}\times 60 = 15\sqrt{3} \approx 25.98\text{ s} \approx 26\text{ s}$