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 perpendicular height and the length of the non-parallel sides of the trapezoid $ABCD$.

Let the perpendiculars from $D$ and $C$ to $AB$ meet $AB$ at $P$ and $Q$ respectively.
Since $ABCD$ is an isosceles trapezoid, $AP = \frac{AB - DC}{2} = \frac{5x - 3x}{2} = x$.
In $\triangle APD$, $\angle DAP = 60^{\circ}$.
The perpendicular height of the trapezoid, $DP = AP \tan 60^{\circ} = x\sqrt{3}$.
The length of the non-parallel side $AD = \frac{AP}{\cos 60^{\circ}} = \frac{x}{0.5} = 2x$.
By symmetry, $BC = 2x$.

2. Calculate the area of the trapezoid $ABCD$.

$\text{Area of } ABCD = \frac{1}{2}(AB + DC) \times DP$
$= \frac{1}{2}(5x + 3x)(x\sqrt{3})$
$= 4x^2\sqrt{3}\text{ units}^2$.

3. Use the given volume to express $y$ in terms of $x$ and $k$.

The volume of the prism is given by the base area multiplied by its vertical height $y$.
$\text{Volume} = (\text{Area of } ABCD) \times y$
$4k\sqrt{3} = (4x^2\sqrt{3})y$
$y = \frac{4k\sqrt{3}}{4x^2\sqrt{3}} = \frac{k}{x^2}$

4. Calculate the total external surface area by summing the areas of the two bases and the four rectangular sides, then substitute $y$.

The total external surface area $A$ is the sum of the two trapezoidal bases and the four rectangular lateral faces.
$A = 2 \times (\text{Area of } ABCD) + (\text{Perimeter of } ABCD) \times y$
$A = 2(4x^2\sqrt{3}) + (5x + 2x + 3x + 2x)y$
$A = 8x^2\sqrt{3} + 12xy$

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

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

$\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} \left( 8x^2\sqrt{3} + 12kx^{-1} \right) = 16x\sqrt{3} - 12kx^{-2}$

2. Set the first derivative to zero and solve for $x$ in terms of $k$.

$16x\sqrt{3} - \frac{12k}{x^2} = 0$
$16x^3\sqrt{3} = 12k$
$x^3 = \frac{12k}{16\sqrt{3}}$
$x^3 = \frac{3k}{4\sqrt{3}} = \frac{\sqrt{3}k}{4}$
$x = \left( \frac{\sqrt{3}k}{4} \right)^{\frac{1}{3}}$

3. Use the second derivative test to verify that the value of $x$ gives a minimum area.

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x} \left( 16x\sqrt{3} - 12kx^{-2} \right) = 16\sqrt{3} + 24kx^{-3}$
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 16\sqrt{3} + \frac{24k}{x^3}$
Since the volume $4k\sqrt{3} > 0 \implies k > 0$, and length $x > 0$, we have $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$.
Therefore, $A$ is a minimum when $x = \left( \frac{\sqrt{3}k}{4} \right)^{\frac{1}{3}}$.

1. Use the given volume to find the constant $k$ and substitute it into the area equation.

Equating the general volume to 1000:
$4k\sqrt{3} = 1000 \implies k = \frac{1000}{4\sqrt{3}} = \frac{250}{\sqrt{3}}$

Substituting $k$ into the total surface area equation given in part (a):
$A = 8x^2\sqrt{3} + \frac{12}{x}\left(\frac{250}{\sqrt{3}}\right)$
$A = 8x^2\sqrt{3} + \frac{3000}{x\sqrt{3}}$
Rationalising the denominator:
$A = 8x^2\sqrt{3} + \frac{1000\sqrt{3}}{x}$

2. Set the area to 800, form a polynomial equation, and solve for $x$ using a graphing calculator.

$8x^2\sqrt{3} + \frac{1000\sqrt{3}}{x} = 800$
Multiply throughout by $x$ to form a polynomial equation:
$8\sqrt{3}x^3 - 800x + 1000\sqrt{3} = 0$
Divide by $8$:
$\sqrt{3}x^3 - 100x + 125\sqrt{3} = 0$
Using a graphing calculator to solve for $x$:
$x \approx 2.406 \quad \text{or} \quad x \approx 6.104 \quad \text{or} \quad x \approx -8.510$
Since $x$ represents a physical length, $x > 0$. We reject $x \approx -8.510$.
$x = 2.41 \quad \text{or} \quad x = 6.10 \text{ (to 3 s.f.)}$