Minimizing Surface Area

Let $OP$ be the height $H$ of the hexagonal pyramid.

Let $E: (\dfrac{-x}{2}, \dfrac{\sqrt{3}}{2}, 0)$ , $A: (\dfrac{-x}{2}, \dfrac{-\sqrt{3}}{2}, 0)$ , $F: (-x,0,0)$ , $P: (0,0,H)$ .

$\vec{FP} = x\vec{i} + 0\vec{j} + H\vec{k} , \:\ \vec{EP} = \dfrac{x}{2}\vec{i} - \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}, \:\ \vec{AP} = \dfrac{x}{2}\vec{i} + \dfrac{\sqrt{3}}{2} x\vec{j} + H\vec{k}$

$\vec{u} = \vec{FP} \times \vec{EP} = \dfrac{\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} - \dfrac{\sqrt{3}}{2} x^2\vec{k}$ and $\vec{v} = \vec{FP} \times \vec{AP} = \dfrac{-\sqrt{3}}{2} xH \vec{i} -\dfrac{xH}{2}\vec{j} + \dfrac{\sqrt{3}}{2} x^2\vec{k}$

$\vec{u} \cdot \vec{v} = \dfrac{-x^2}{4} (2 H^2 + 3 x^2)$ and $|\vec{u}| = |\vec{v}| = \dfrac{x}{2} \sqrt{4H^2 + 3x^2} \implies \cos(\theta) = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2}$ , where $\theta$ is the angle between the two adjacent slant faces.

The height of each of the six equilateral triangles of the hexagon is $h^{*} = \dfrac{\sqrt{3}}{2} x \implies$
$A_{hexagon} = \dfrac{3\sqrt{3}}{2} x^2 \implies V_{p} = \dfrac{\sqrt{3}}{2} x^2 H$

The slant height of the hexagonal pyramid $s = \sqrt{{h^{*}}^2 + H^2} = \dfrac{\sqrt{3x^2 + 4H^2}}{2} \implies$ the lateral surface $S_{p} = 3xs = \dfrac{3x}{2} \sqrt{3x^2 + 4H^2}.$

$V_{p} = K(constant) \implies H = \dfrac{2K}{\sqrt{3} x^2} \implies S_{p}(x) = \dfrac{\sqrt{3}}{2x} \sqrt{9x^2 + 16K^2} \implies$

$\dfrac{dS_{p}}{dx} = \sqrt{3} (\dfrac{9x^6 - 8K^2}{x^2\sqrt{9x^6 + 16K^2}}) = 0 \implies$ $x = (\dfrac{8 K^2}{9})^\dfrac{1}{6}$

Pick $x = (\dfrac{K^2}{2^6})^\dfrac{1}{6} < (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies \dfrac{dS_{p}}{dx} < 0$

Pick $x = (K^2)^\dfrac{1}{6} > (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies \dfrac{dS_{p}}{dx} > 0 \implies x = \dfrac{dS_{p}}{dx}$ minimizes $S_{p}(x).$

Let $u = (\dfrac{8 K^2}{9})^\dfrac{1}{6} \implies x = u^\dfrac{1}{6} \implies H = \dfrac{2K}{\sqrt{3}u^\dfrac{1}{3}}$

Using the above $\cos(\theta) = \dfrac{-(2H^2 + 3x^2)}{4H^2 + 3x^2} \implies$ $\cos(\theta) = \dfrac{-(8K^2 + 9u)}{16K^2+ 9u} = \dfrac{-2}{3} \implies \theta = \boxed{131.81^\circ}$