图中显示了一个棱柱，其横截面为梯形 $ABCD$。
$AB = 4x\text{ cm}$，$BC = DC = AD = 2x\text{ cm}$。该棱柱的长度为 $d\text{ cm}$。

1. Find the horizontal projection of the slanted sides.

Let the perpendiculars from $C$ and $D$ to $AB$ meet $AB$ at $F$ and $E$ respectively.
$EF = DC = 2x$
Since the trapezium is isosceles ($AD = BC = 2x$), $AE = FB = \frac{AB - EF}{2} = \frac{4x - 2x}{2} = x$.

2. Calculate the height of the trapezium using Pythagoras' theorem.

In $\triangle ADE$, using Pythagoras' theorem:
$DE^2 + AE^2 = AD^2$
$h^2 + x^2 = (2x)^2$
$h^2 = 4x^2 - x^2 = 3x^2$
$h = \sqrt{3}x$

3. Calculate the area of the trapezium.

Area of trapezium $ABCD = \frac{1}{2} \times (AB + DC) \times h$
$= \frac{1}{2} \times (4x + 2x) \times \sqrt{3}x$
$= \frac{1}{2} \times 6x \times \sqrt{3}x$
$= 3\sqrt{3}x^2\text{ cm}^2$ (shown)

1. Set up the equation for the volume of the prism

$\text{Volume} = \text{Base Area} \times \text{Length}$
$3888\sqrt{3} = (3\sqrt{3}x^2) \times d$

2. Solve for $d$ in terms of $x$

$d = \frac{3888\sqrt{3}}{3\sqrt{3}x^2}$
$d = \frac{1296}{x^2}$

1. Find the length of the prism, $d$, in terms of $x$ using the volume from part (ii).

$\text{Volume} = \text{Base Area} \times d$
$3\sqrt{3}x^2 \times d = 3888\sqrt{3}$
$d = \frac{3888\sqrt{3}}{3\sqrt{3}x^2} = \frac{1296}{x^2}$

2. Express the total surface area $A$ in terms of $x$ and $d$.

The total surface area consists of 2 trapezoidal faces and 4 rectangular faces.
$A = 2 \times (\text{Area of } ABCD) + (AB + BC + CD + DA) \times d$
$A = 2(3\sqrt{3}x^2) + (4x + 2x + 2x + 2x)d$
$A = 6\sqrt{3}x^2 + 10xd$

3. Substitute $d$ into the surface area equation to show the final expression.

$A = 6\sqrt{3}x^2 + 10x\left(\frac{1296}{x^2}\right)$
$A = 6\sqrt{3}x^2 + \frac{12960}{x}$ (shown)

1. Differentiate $A$ with respect to $x$.

$A = 6\sqrt{3}x^2 + 12960x^{-1}$

$\frac{dA}{dx} = 12\sqrt{3}x - 12960x^{-2}$

2. Set the derivative to zero to find the stationary value.

For a stationary value, $\frac{dA}{dx} = 0$:

$12\sqrt{3}x - \frac{12960}{x^2} = 0$

$12\sqrt{3}x^3 = 12960$

3. Solve for $x$.

$x^3 = \frac{12960}{12\sqrt{3}}$

$x^3 = \frac{1080}{\sqrt{3}} = 360\sqrt{3}$

$x = \sqrt[3]{360\sqrt{3}} \approx 8.5433$

$x \approx 8.54$ (to 3 s.f.)

1. Find the second derivative of $A$ with respect to $x$.

From (iv), $\frac{dA}{dx} = 12\sqrt{3}x - 12960x^{-2}$.

$\frac{d^2A}{dx^2} = 12\sqrt{3} - (-2)(12960)x^{-3}$
$\frac{d^2A}{dx^2} = 12\sqrt{3} + \frac{25920}{x^3}$

2. Use the second derivative test to justify the minimum.

Since $x$ is a length, $x > 0$.
Therefore, $\frac{25920}{x^3} > 0$, which means $\frac{d^2A}{dx^2} > 0$.
Since the second derivative is positive, this value of $x$ gives a minimum (smallest possible) surface area.