Candy is being stored in a closed container in the form of a regular hexagonal prism, with sides $x\text{ cm}$ and height $h\text{ cm}$ (as shown in the figure below). Given that the container has a volume of $972\text{ cm}^3$,

1. Find the area of one of the 6 constituent equilateral triangles that make up the regular hexagon.

A regular hexagon can be divided into 6 congruent equilateral triangles of side length $x$ cm.
$\text{Area of one equilateral triangle} = \frac{1}{2}(x)(x)\sin 60^\circ = \frac{1}{2}x^2 \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}x^2$

2. Calculate the total area of the regular hexagon.

$\text{Base area} = 6 \times \frac{\sqrt{3}}{4}x^2 = \frac{3\sqrt{3}x^2}{2}\text{ cm}^2$

1. Express the height $h$ in terms of $x$ using the given volume.

$V = \text{Base Area} \times h$
$972 = \left( \frac{3\sqrt{3}x^2}{2} \right) h$
$h = \frac{972 \times 2}{3\sqrt{3}x^2} = \frac{648}{\sqrt{3}x^2} = \frac{216\sqrt{3}}{x^2}$

2. Formulate the equation for the total surface area $A$ in terms of $x$.

$A = 2 \times (\text{Base Area}) + (\text{Perimeter of Base}) \times h$
$A = 2 \left( \frac{3\sqrt{3}x^2}{2} \right) + 6x \left( \frac{216\sqrt{3}}{x^2} \right)$
$A = 3\sqrt{3}x^2 + \frac{1296\sqrt{3}}{x}$

3. Differentiate $A$ with respect to $x$ and equate to $0$ to find the stationary point.

$\frac{\mathrm{d}A}{\mathrm{d}x} = 6\sqrt{3}x - \frac{1296\sqrt{3}}{x^2}$
For minimum area, set $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$:
$6\sqrt{3}x - \frac{1296\sqrt{3}}{x^2} = 0$
$6\sqrt{3}x = \frac{1296\sqrt{3}}{x^2}$
$6\sqrt{3}x^3 = 1296\sqrt{3}$
$x^3 = 216$
$x = 6$

4. Verify that the stationary point is a minimum using the second derivative test.

$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 6\sqrt{3} + \frac{2592\sqrt{3}}{x^3}$
When $x = 6$,
$\frac{\mathrm{d}^2A}{\mathrm{d}x^2} = 6\sqrt{3} + \frac{2592\sqrt{3}}{6^3} = 18\sqrt{3} > 0$
Since $\frac{\mathrm{d}^2A}{\mathrm{d}x^2} > 0$, the surface area $A$ is a minimum.

5. Calculate the minimum surface area and leave the answer to 2 decimal places.

$A = 3\sqrt{3}(6)^2 + \frac{1296\sqrt{3}}{6}$
$A = 108\sqrt{3} + 216\sqrt{3}$
$A = 324\sqrt{3} \approx 561.184$
$A \approx 561.18 \text{ cm}^2 \text{ (to 2 d.p.)}$

1. Determine the minimum surface area of the container from part (ii).

From part (ii), the surface area $A$ is the sum of the two hexagonal bases and the 6 rectangular sides. Using the volume $V$ to find height $h$:
$V = \frac{3\sqrt{3}x^2}{2}h = 972 \implies h = \frac{648}{\sqrt{3}x^2}$
$A = 2\left(\frac{3\sqrt{3}x^2}{2}\right) + 6xh = 3\sqrt{3}x^2 + 6x\left(\frac{648}{\sqrt{3}x^2}\right) = 3\sqrt{3}x^2 + \frac{1296\sqrt{3}}{x}$
Differentiating to find the minimum area:
$\frac{\mathrm{d}A}{\mathrm{d}x} = 6\sqrt{3}x - \frac{1296\sqrt{3}}{x^2} = 0 \implies x^3 = 216 \implies x = 6$
$\text{Minimum } A = 3\sqrt{3}(6)^2 + \frac{1296\sqrt{3}}{6} = 324\sqrt{3} \approx 561.184\text{ cm}^2$

2. Calculate the minimum cost required based on the minimum area.

The cost of the material is $\char36{}0.05$ per $100\text{ cm}^2$.
$\text{Cost} = \frac{324\sqrt{3}}{100} \times 0.05$
$\text{Cost} = 0.28059\dots$
$\text{Cost} \approx \text{\$}^0.28$