一个体积为 $250\text{ cm}^3$ 的盒子由厚度可忽略不计的硬纸板制成。盒子的高为 $y\text{ cm}$，底面为边长 $x\text{ cm}$ 的等边三角形。其盒盖的深度为 $ky\text{ cm}$，其中 $0 < k \le 1$（见图）。

1. Express the volume of the box to find $y$ in terms of $x$.

Let the volume of the box be $V$. The base is an equilateral triangle with side $x$.
$V = \text{Base Area} \times \text{height}$
$V = \left(\frac{1}{2} x^2 \sin 60^\circ \right) y = \frac{\sqrt{3}}{4} x^2 y$
Given $V = 250$:
$\frac{\sqrt{3}}{4} x^2 y = 250$
$y = \frac{1000}{\sqrt{3}x^2}$

2. Formulate the expression for the total external surface area in terms of $x$ and $y$.

The total external surface area $A$ consists of the base of the box, the top of the lid, the 3 rectangular sides of the box, and the 3 rectangular sides of the lid.
$\text{Area of box} = \frac{\sqrt{3}}{4} x^2 + 3xy$
$\text{Area of lid} = \frac{\sqrt{3}}{4} x^2 + 3x(ky)$
$A = \left( \frac{\sqrt{3}}{4} x^2 + 3xy \right) + \left( \frac{\sqrt{3}}{4} x^2 + 3kxy \right)$
$A = \frac{\sqrt{3}}{2} x^2 + 3xy(1+k)$

3. Substitute $y$ into the area expression to show the required result.

Substitute $y = \frac{1000}{\sqrt{3}x^2}$ into $A$:
$A = \frac{\sqrt{3}}{2} x^2 + 3x(1+k) \left( \frac{1000}{\sqrt{3}x^2} \right)$
$A = \frac{\sqrt{3}}{2} x^2 + \frac{3000(1+k)}{\sqrt{3}x}$
Rationalising the denominator $\frac{3000}{\sqrt{3}} = 1000\sqrt{3}$:
$A = \frac{1000\sqrt{3}(1+k)}{x} + \frac{\sqrt{3}}{2} x^2$
(Shown)

1. Differentiate the given surface area expression with respect to $x$.

$A = 1000\sqrt{3}(1+k)x^{-1} + \frac{\sqrt{3}}{2}x^2$
$\frac{\mathrm{d}A}{\mathrm{d}x} = -1000\sqrt{3}(1+k)x^{-2} + \sqrt{3}x$

2. Equate the first derivative to zero and solve for $x$.

$-\frac{1000\sqrt{3}(1+k)}{x^2} + \sqrt{3}x = 0$
$\sqrt{3}x = \frac{1000\sqrt{3}(1+k)}{x^2}$
$x^3 = 1000(1+k)$
$x = 10\sqrt[3]{1+k}$

3. Verify that this value of $x$ gives a minimum area using the second derivative test.

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 2000\sqrt{3}(1+k)x^{-3} + \sqrt{3}$
Substitute $x^3 = 1000(1+k)$:
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = \frac{2000\sqrt{3}(1+k)}{1000(1+k)} + \sqrt{3} = 3\sqrt{3}$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$, the surface area is a minimum at $x = 10\sqrt[3]{1+k}$.

1. Express the height $y$ in terms of the base side $x$ using the volume of the box.

The volume of the box is the area of its equilateral triangular base multiplied by its height $y$.
$V = \left(\frac{1}{2}x^2 \sin 60^\circ\right) y = \frac{\sqrt{3}}{4}x^2y$
Given that $V = 250\text{ cm}^3$,
$\frac{\sqrt{3}}{4}x^2y = 250$
$y = \frac{1000}{\sqrt{3}x^2}$

2. Set up the ratio $\frac{y}{x}$ and substitute the value of $x^3$ found in part (ii) for the minimum surface area.

Dividing the expression for $y$ by $x$ gives:
$\frac{y}{x} = \frac{1000}{\sqrt{3}x^3}$
From part (ii), the value of $x$ that gives the minimum surface area satisfies the condition derived from $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$, which yields:
$x^3 = 1000(1+k)$
Substituting this into the ratio expression:
$\frac{y}{x} = \frac{1000}{\sqrt{3}[1000(1+k)]}$
$\frac{y}{x} = \frac{1}{\sqrt{3}(1+k)} \quad \text{or} \quad \frac{\sqrt{3}}{3(1+k)}$

1. Express the ratio $\frac{y}{x}$ in terms of $x$ using the given volume.

The area of the equilateral triangular base is $\frac{1}{2}(x)(x)\sin 60^\circ = \frac{\sqrt{3}}{4}x^2$.
Using the given volume, $V = \frac{\sqrt{3}}{4}x^2 y = 250 \implies y = \frac{1000}{\sqrt{3}x^2}$.
Hence, the ratio is $\frac{y}{x} = \frac{1000}{\sqrt{3}x^3}$.

2. Find $\frac{y}{x}$ in terms of $k$ for the minimum surface area.

From the equation stated in part (i), $A = \frac{1000\sqrt{3}(1+k)}{x} + \frac{\sqrt{3}}{2}x^2$.
At the minimum surface area, $\frac{dA}{dx} = -\frac{1000\sqrt{3}(1+k)}{x^2} + \sqrt{3}x = 0$, which gives $x^3 = 1000(1+k)$.
Substituting this into our ratio:
$\frac{y}{x} = \frac{1000}{\sqrt{3}[1000(1+k)]} = \frac{1}{\sqrt{3}(1+k)}$

3. Determine the range of values for $\frac{y}{x}$ based on the given bounds of $k$.

We are given $0 < k \le 1$.
Adding 1:
$1 < 1+k \le 2$

Taking the reciprocal reverses the inequality:
$\frac{1}{2} \le \frac{1}{1+k} < 1$

Multiplying by $\frac{1}{\sqrt{3}}$:
$\frac{1}{2\sqrt{3}} \le \frac{1}{\sqrt{3}(1+k)} < \frac{1}{\sqrt{3}}$

Rationalising the denominators gives the final range:
$\frac{\sqrt{3}}{6} \le \frac{y}{x} < \frac{\sqrt{3}}{3}$