Square Pyramids!

$s = \dfrac{\sqrt{4H^2 + x^2}}{2} \implies$ the surface area $A = 2xs = x\sqrt{4H^2 + x^2}$

The volume $V = \dfrac{1}{3}x^2H = K \implies H = \dfrac{3K}{x^2} \implies$ $A = \dfrac{\sqrt{36K^2 + x^6}}{x} \implies$

$\dfrac{dA}{dx} = \dfrac{2x^6 - 36k^2}{x^2\sqrt{36k^2 + x^6}} = 0 \:\ x \neq 0 \implies$ $x^6 - 18k^2 = 0 \implies$

$x = (18K^2)^{\frac{1}{6}} = 3^{\frac{1}{3}} 2^{\frac{1}{6}} k^{\frac{1}{3}}$ $\implies H = (\dfrac{3}{2})^{\frac{1}{3}} K^{\frac{1}{3}}$

$\implies \boxed{\tan(\lambda) = \dfrac{2H}{x} = \sqrt{2}}$ and $\boxed{\tan(\gamma) = 1}$ .

$\vec{OP} = x\vec{i} + 0\vec{j} + 0\vec{k}$

$\vec{OR} = 0\vec{i} - x\vec{j} + 0\vec{k}$

$\vec{OS} = \dfrac{x}{2}\vec{i} -\dfrac{x}{2}\vec{j} + H\vec{k}$

$\vec{U} = \vec{OP} \:\ X \:\ \vec{OS} = 0\vec{i} - xH\vec{j} - \dfrac{x^2}{2}\vec{k}$

and

$\vec{V} = \vec{OR} \:\ X \:\ \vec{OS} = -xH\vec{i} + 0\vec{j} + \dfrac{x^2}{2}\vec{k}$

$\vec{U} \circ \vec{V} = -\dfrac{x^4}{4}$

$|\vec{U}| = \dfrac{x}{2}\sqrt{4H^2 + x^2} = |\vec{V}|$

$\implies \boxed{\cos(\theta) = \dfrac{\vec{U} \circ \vec{V}}{|\vec{U}| |\vec{V}|} = -\dfrac{x^2}{4H^2 + x^2} = -\dfrac{1}{3}}$

$\implies \dfrac{\tan^2(\lambda)\tan(\gamma)}{\cos(\theta)} = \boxed{-6}$ .

Note: $\dfrac{d^2A}{dx^2}|_{x = (18 k^2)^{\frac{1}{6}}} = 4\sqrt{3} > 0 \implies$ min occurs at $x = (18 k^2)^{\frac{1}{6}}$