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

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

From the net, the lateral edges of the folded pyramid have length $a$.
The distance from the centre of the regular hexagonal base to any of its vertices is $x$ (since the base is formed by 6 equilateral triangles of side $x$).
Considering the right-angled triangle formed by the vertical height $h$, the distance from the centre to a vertex $x$, and the lateral edge $a$:
$h^2 + x^2 = a^2$
$h^2 = a^2 - x^2$

2. Square the given volume equation and substitute $h^2$ to obtain the target expression.

Given the volume formula:
$V = \frac{\sqrt{3}}{2}x^{2}h$
Squaring 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:
$V^2 = \frac{3}{4}x^{4}(a^2 - x^2)$
$V^2 = \frac{3}{4}\left(a^{2}x^{4} - x^{6}\right)$

1. Differentiate $V^2$ with respect to $x$ to find the stationary point.

To maximise the volume $V$, we can maximise $V^2$ since $V > 0$. Differentiating the expression from part (a) with respect to $x$:
$\frac{\mathrm{d}(V^2)}{\mathrm{d}x} = \frac{3}{4}\left(4a^2x^3 - 6x^5\right)$

2. Equate the derivative to zero and solve for $x^2$ in terms of $a$.

Set the derivative to zero to find the stationary point for maximum volume:
$\frac{3}{4}\left(4a^2x^3 - 6x^5\right) = 0$
$\frac{3}{4}x^3(4a^2 - 6x^2) = 0$
Since the edge length $x > 0$, we have:
$4a^2 - 6x^2 = 0 \implies x^2 = \frac{2}{3}a^2$

3. Substitute $x^2 = \frac{2}{3}a^2$ back into the expression for $V^2$.

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

4. Take the positive square root to find the maximum possible volume $V$.

Since the volume $V$ must be positive, we take the positive square root:
$V = \sqrt{\frac{1}{9}a^6} = \frac{1}{3}a^3$

1. Recall or determine the value of $x$ for which the volume is maximum.

From part (b), to find the maximum volume, we set the derivative of $V^2$ with respect to $x$ to $0$:
$\frac{\mathrm{d}}{\mathrm{d}x} \left[ \frac{3}{4}(a^2x^4 - x^6) \right] = 0$
$\frac{3}{4}(4a^2x^3 - 6x^5) = 0$
Since $x > 0$, we divide by $2x^3$:
$2a^2 - 3x^2 = 0 \implies x^2 = \frac{2}{3}a^2 \implies x = a\sqrt{\frac{2}{3}}$

2. Calculate the area of the regular hexagonal base at maximum volume.

The regular hexagonal base consists of 6 equilateral triangles of side $x$. Using the area formula $\frac{1}{2}ab\sin C$:
$\text{Base Area} = 6 \left( \frac{1}{2}x^2 \sin 60^\circ \right) = 3x^2 \left( \frac{\sqrt{3}}{2} \right) = \frac{3\sqrt{3}}{2}x^2$
Substituting $x^2 = \frac{2}{3}a^2$:
$\text{Base Area} = \frac{3\sqrt{3}}{2} \left( \frac{2}{3}a^2 \right) = \sqrt{3}a^2$

3. Find the expression for the lateral surface area of the six triangular faces.

The lateral surface consists of 6 isosceles triangles, each with base $x$ and two equal sides $a$.
By Pythagoras' theorem, the height of one such triangle is:
$h_t = \sqrt{a^2 - \left(\frac{x}{2}\right)^2} = \sqrt{a^2 - \frac{x^2}{4}}$
The total lateral surface area is:
$\text{Lateral Area} = 6 \left( \frac{1}{2} \times x \times h_t \right) = 3x \sqrt{a^2 - \frac{x^2}{4}}$

4. Substitute the condition for maximum volume into the lateral area expression and simplify.

Substituting $x = a\sqrt{\frac{2}{3}}$ and $x^2 = \frac{2}{3}a^2$:
$\text{Lateral Area} = 3 \left( a\sqrt{\frac{2}{3}} \right) \sqrt{a^2 - \frac{1}{4}\left(\frac{2}{3}a^2\right)}$
$= 3a\sqrt{\frac{2}{3}} \sqrt{a^2 - \frac{1}{6}a^2}$
$= 3a\sqrt{\frac{2}{3}} \sqrt{\frac{5}{6}a^2} = 3a^2 \sqrt{\frac{2}{3} \times \frac{5}{6}}$
$= 3a^2 \sqrt{\frac{10}{18}} = 3a^2 \left( \frac{\sqrt{5}}{3} \right) = \sqrt{5}a^2$

5. Calculate the final total surface area.

Total Surface Area = Base Area + Lateral Area
$\text{Total Surface Area} = \sqrt{3}a^2 + \sqrt{5}a^2 = (\sqrt{3} + \sqrt{5})a^2$