Fig. 1 shows a net of a hexagonal pyramid folded from a star-shaped cardboard of equal edge length $a\text{ cm}$. The net consists of a hexagon with equal sides of $x\text{ cm}$ and six isosceles triangles with base $x\text{ cm}$ and side $a\text{ cm}$. The net is folded to form a right pyramid with a hexagonal base of edge length $x\text{ cm}$ and vertical height $h\text{ cm}$, as shown in Fig. 2. The hexagonal base is made up of six equilateral triangles of side length $x\text{ cm}$. The volume of a right hexagonal pyramid with base edge $x\text{ cm}$ and height $h\text{ cm}$ is given by
$V = \frac{\sqrt{3}}{2}x^2 h.$

1. Relate the vertical height $h$, base radius $x$, and slant edge $a$ using Pythagoras' theorem.

The base of the pyramid is a regular hexagon made up of six equilateral triangles of side length $x$. Thus, the distance from the centre of the base to any of its vertices is exactly $x$.
Consider the right-angled triangle formed by the vertical height $h$, the distance from the centre to a base vertex $x$, and the slant edge $a$:
$h^2 + x^2 = a^2$
$h^2 = a^2 - x^2$

2. Square the volume formula and substitute $h^2$ to obtain the required expression.

Given the volume $V = \frac{\sqrt{3}}{2}x^2 h$,
Square both sides:
$V^2 = \left(\frac{\sqrt{3}}{2}x^2 h\right)^2$
$V^2 = \frac{3}{4}x^4 h^2$
Substitute $h^2 = a^2 - x^2$ into the equation:
$V^2 = \frac{3}{4}x^4(a^2 - x^2)$
$V^2 = \frac{3}{4}\left(a^2 x^4 - x^6\right) \text{ (shown)}$

1. Differentiate the given expression for $V^2$ with respect to $x$.

Let $W = V^2 = \frac{3}{4}(a^2 x^4 - x^6)$.
To find the turning point, we differentiate $W$ with respect to $x$:
$\frac{\mathrm{d}W}{\mathrm{d}x} = \frac{3}{4}(4a^2 x^3 - 6x^5)$

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

For maximum volume, $\frac{\mathrm{d}W}{\mathrm{d}x} = 0$.
$\frac{3}{4}x^3(4a^2 - 6x^2) = 0$
Since $x$ is a length, $x > 0$, so we can divide by $x^3$:
$4a^2 - 6x^2 = 0$
$6x^2 = 4a^2$
$x^2 = \frac{2}{3}a^2$

3. Substitute $x^2$ back into the equation for $V^2$ to find the maximum squared volume.

Substitute $x^2 = \frac{2}{3}a^2$ into the expression for $V^2$:
$V^2 = \frac{3}{4}\left[a^2\left(\frac{2}{3}a^2\right)^2 - \left(\frac{2}{3}a^2\right)^3\right]$
$V^2 = \frac{3}{4}\left[a^2\left(\frac{4}{9}a^4\right) - \frac{8}{27}a^6\right]$
$V^2 = \frac{3}{4}\left[\frac{12}{27}a^6 - \frac{8}{27}a^6\right]$
$V^2 = \frac{3}{4}\left(\frac{4}{27}a^6\right) = \frac{1}{9}a^6$

4. Calculate the maximum volume $V$ by taking the square root.

Since $V > 0$ and $a > 0$,
$V = \sqrt{\frac{1}{9}a^6}$
$V = \frac{1}{3}a^3$

1. Find the relationship between $x$ and $a$ for maximum volume.

Let $U = V^2 = \frac{3}{4}(a^2 x^4 - x^6)$.
To find the maximum volume, we set $\frac{\mathrm{d}U}{\mathrm{d}x} = 0$:
$\frac{\mathrm{d}U}{\mathrm{d}x} = \frac{3}{4}(4a^2x^3 - 6x^5) = 0$
Since $x > 0$, dividing by $x^3$ gives:
$4a^2 - 6x^2 = 0 \implies x^2 = \frac{2}{3}a^2 \implies x = \sqrt{\frac{2}{3}}a$

2. Calculate the area of the hexagonal base in terms of $a$.

The hexagonal base is composed of 6 equilateral triangles of side length $x$.
$\text{Area of base} = 6 \left( \frac{1}{2} x^2 \sin 60^\circ \right) = \frac{3\sqrt{3}}{2}x^2$
Substituting $x^2 = \frac{2}{3}a^2$:
$\text{Area of base} = \frac{3\sqrt{3}}{2} \left( \frac{2}{3}a^2 \right) = \sqrt{3}a^2$

3. Calculate the slant height and the area of the six isosceles triangular faces.

Let $l$ be the perpendicular slant height of each of the 6 isosceles triangular faces.
Using Pythagoras' theorem on a triangular face:
$l = \sqrt{a^2 - \left(\frac{x}{2}\right)^2} = \sqrt{a^2 - \frac{x^2}{4}}$
Substituting $x^2 = \frac{2}{3}a^2$:
$l = \sqrt{a^2 - \frac{1}{4}\left(\frac{2}{3}a^2\right)} = \sqrt{a^2 - \frac{1}{6}a^2} = \sqrt{\frac{5}{6}}a$
$\text{Area of 6 triangular faces} = 6 \left( \frac{1}{2} x l \right) = 3xl$
$= 3 \left( \sqrt{\frac{2}{3}}a \right) \left( \sqrt{\frac{5}{6}}a \right) = 3a^2 \sqrt{\frac{10}{18}} = 3a^2 \left( \frac{\sqrt{5}}{3} \right) = \sqrt{5}a^2$

4. Calculate the total surface area.

$\text{Total Surface Area} = \text{Area of base} + \text{Area of 6 triangular faces}$
$= \sqrt{3}a^2 + \sqrt{5}a^2$
$= (\sqrt{3} + \sqrt{5})a^2$