图 1 显示了一个六角锥的展开图，该展开图由一个等边长为 $a$ cm 的星形纸板折叠而成。该展开图由一个边长均为 $x$ cm 的六边形和六个底边为 $x$ cm、腰长为 $a$ cm 的等腰三角形组成。将该展开图折叠后形成一个直六角锥，其六边形底边的边长为 $x$ cm，垂直高度为 $h$ cm，如图 2 所示。该六边形底面由六个边长为 $x$ cm 的等边三角形组成。 底面边长为 $x$ cm、高度为 $h$ cm 的直六角锥的体积由下式给出： $V = \frac{\sqrt{3}}{2} x^{2} h.$

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

The apex of the right pyramid is directly above the centre of the hexagonal base. The distance from the centre of the base to any of its vertices is $x$ (since the hexagon is composed of 6 equilateral triangles of side $x$). The lateral edge of the pyramid is $a$.
By Pythagoras' theorem,
$h^2 + x^2 = a^2$
$h^2 = a^2 - x^2$

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

Given $V = \frac{\sqrt{3}}{2} x^{2} h$
Squaring both sides:
$V^2 = \left( \frac{\sqrt{3}}{2} x^2 h \right)^2 = \frac{3}{4} x^4 h^2$
Substitute $h^2 = a^2 - x^2$ into the expression for $V^2$:
$V^2 = \frac{3}{4} x^4 (a^2 - x^2)$
$V^2 = \frac{3}{4} \bigl(a^2 x^4 - x^6\bigr) \quad \text{(shown)}$

1. Set up a function for the square of the volume and differentiate it with respect to $x$

Let $W = V^2 = \frac{3}{4}(a^2x^4 - x^6)$.
Since $V > 0$, maximizing $V$ is equivalent to maximizing $W$.
Differentiating with respect to $x$:
$\frac{\mathrm{d}W}{\mathrm{d}x} = \frac{3}{4}(4a^2x^3 - 6x^5)$

2. Set the derivative to zero to find the stationary value of $x^2$

At the maximum volume, $\frac{\mathrm{d}W}{\mathrm{d}x} = 0$.
$\frac{3}{4}x^3(4a^2 - 6x^2) = 0$
Since $x > 0$ (it is a length):
$4a^2 - 6x^2 = 0$
$x^2 = \frac{2}{3}a^2$

3. Substitute the value of $x^2$ back into the expression for $V^2$ to find the maximum $V^2$

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

4. Find the maximum possible volume by taking the square root

Taking the positive square root (since $V > 0$):
$V = \sqrt{\frac{1}{9}a^6}$
$V = \frac{1}{3}a^3$

1. Find the condition on $x$ for the maximum volume.

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

2. Calculate the area of the hexagonal base when the volume is maximum.

The regular hexagonal base comprises 6 equilateral triangles of side $x$.
$\text{Base Area} = 6 \left( \frac{1}{2} x^2 \sin 60^\circ \right) = 6 \left( \frac{\sqrt{3}}{4} x^2 \right) = \frac{3\sqrt{3}}{2} x^2$
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. Calculate the area of the 6 lateral faces when the volume is maximum.

The 6 lateral faces are isosceles triangles with base $x$ and equal sides $a$.
Height of each lateral triangle, $l = \sqrt{a^2 - \left(\frac{x}{2}\right)^2} = \sqrt{a^2 - \frac{x^2}{4}}$
Substituting $x^2 = \frac{2}{3}a^2$:
$l = \sqrt{a^2 - \frac{1}{4}\left(\frac{2}{3}a^2\right)} = \sqrt{a^2 - \frac{1}{6}a^2} = \sqrt{\frac{5}{6}}a$
$\text{Total Lateral Area} = 6 \left( \frac{1}{2} x l \right) = 3 x l$
Substituting $x = \sqrt{\frac{2}{3}}a$ and $l = \sqrt{\frac{5}{6}}a$:
$\text{Total Lateral Area} = 3 \left( \sqrt{\frac{2}{3}}a \right) \left( \sqrt{\frac{5}{6}}a \right) = 3 \sqrt{\frac{10}{18}} a^2 = 3 \left(\frac{\sqrt{5}}{3}\right) a^2 = \sqrt{5}a^2$

4. Combine the areas to get the total surface area.

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