The following diagram shows the dimensions of a trapezoidal prism with fixed volume $4k\sqrt{3}$ units$^3$, with variables $x$ and $y$. The top surface of the prism, $ABCD$, is an isosceles trapezoid with $AB$ of length $5x$ units, $DC$ of length $3x$ units, $AD = BC$ and $\angle ABC = \angle BAD = 60^\circ$. The rectangular sides $ABFE$ and $BCGF$ are perpendicular to both the top surface $ABCD$ and the bottom surface $EFGH$, with $AE = BF = DH = CG = y$ units.

1. Find the height of the trapezoidal face and the length of its non-parallel sides.

Let $h$ be the perpendicular height of the isosceles trapezoid $ABCD$.
Since $AB = 5x$ and $DC = 3x$, dropping perpendiculars from $D$ and $C$ to $AB$ leaves a horizontal segment of $3x$ in the middle, and two segments of length $\frac{5x - 3x}{2} = x$ on either side.
In the right-angled triangle formed at the base,
$\tan 60^\circ = \frac{h}{x} \implies h = x\sqrt{3}$
$\cos 60^\circ = \frac{x}{AD} \implies AD = \frac{x}{0.5} = 2x$
Since the trapezoid is isosceles, $BC = AD = 2x$.

2. Express the volume of the prism to find $y$ in terms of $x$ and $k$.

The area of the trapezoidal base $ABCD$ is:
$\text{Area} = \frac{1}{2}(AB + DC)h = \frac{1}{2}(5x + 3x)(x\sqrt{3}) = 4x^2\sqrt{3}$
The volume of the prism is the base area multiplied by the height $y$:
$V = 4x^2\sqrt{3} \times y = 4x^2y\sqrt{3}$
Given that $V = 4k\sqrt{3}$, we can equate and solve for $y$:
$4x^2y\sqrt{3} = 4k\sqrt{3}$
$y = \frac{k}{x^2}$

3. Formulate the total external surface area $A$ in terms of $x$ and $y$.

The total external surface area consists of the $2$ trapezoidal bases (top and bottom) and the $4$ rectangular vertical sides.
The perimeter of the trapezoid is:
$P = AB + BC + DC + AD = 5x + 2x + 3x + 2x = 12x$
The lateral surface area (sum of the $4$ rectangular sides) is:
$\text{Lateral Area} = P \times y = 12xy$
Therefore, the total surface area $A$ is:
$A = 2(4x^2\sqrt{3}) + 12xy$
$A = 8x^2\sqrt{3} + 12xy$

4. Substitute $y$ into the area equation to obtain the given expression for $A$.

Substitute $y = \frac{k}{x^2}$ into the equation for $A$:
$A = 8x^2\sqrt{3} + 12x\left(\frac{k}{x^2}\right)$
$A = 8x^2\sqrt{3} + \frac{12k}{x} \text{ (shown)}$

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

Given $A = 8x^2\sqrt{3} + 12k x^{-1}$,

$\frac{\mathrm{d}A}{\mathrm{d}x} = 16\sqrt{3}x - 12k x^{-2}$

2. Set the first derivative to zero to locate the stationary point.

For a minimum area, $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$:

$16\sqrt{3}x - \frac{12k}{x^2} = 0$
$16\sqrt{3}x = \frac{12k}{x^2}$
$16\sqrt{3}x^3 = 12k$

3. Solve for $x$ in terms of $k$.

$x^3 = \frac{12k}{16\sqrt{3}}$
$x^3 = \frac{3k}{4\sqrt{3}}$
$x^3 = \frac{\sqrt{3}k}{4}$
$x = \left( \frac{\sqrt{3}k}{4} \right)^{\frac{1}{3}}$

4. Use the second derivative test to confirm it is a minimum point.

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 16\sqrt{3} + 24k x^{-3}$

Since physical dimensions must be positive ($x > 0$) and the volume is given as $4k\sqrt{3}$ units$^3$ implying $k > 0$, we have:
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$.

Hence, the area $A$ is indeed a minimum at this value of $x$.

1. Find the value of $k$ from the given volume and substitute into the expression for surface area.

The volume of the prism is given as $4k\sqrt{3}$. Equating this to $1000$:
$4k\sqrt{3} = 1000 \implies k = \frac{1000}{4\sqrt{3}} = \frac{250}{\sqrt{3}}$
Substitute $k$ and the given external surface area $A = 800$ into the expression from part (a):
$800 = 8x^2\sqrt{3} + \frac{12}{x}\left(\frac{250}{\sqrt{3}}\right)$

2. Simplify the equation into a cubic polynomial form.

$800 = 8x^2\sqrt{3} + \frac{3000}{x\sqrt{3}}$
Rationalise the denominator of the fraction:
$800 = 8x^2\sqrt{3} + \frac{1000\sqrt{3}}{x}$
Multiply throughout by $x$ and rearrange to form a cubic equation:
$800x = 8\sqrt{3}x^3 + 1000\sqrt{3}$
Divide by 8:
$\sqrt{3}x^3 - 100x + 125\sqrt{3} = 0$

3. Solve the cubic equation using a graphing calculator and identify the valid roots.

Using a graphing calculator to solve $\sqrt{3}x^3 - 100x + 125\sqrt{3} = 0$, we get:
$x \approx -8.510$
$x \approx 2.406$
$x \approx 6.104$

Since $x$ represents a physical length, $x > 0$. Therefore, we reject the negative root.
$x = 2.41$ or $x = 6.10$ (to 3 s.f.)