Square Pyramids 2!

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

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

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

$2x^3\sqrt{36k^2 + x^6} = 36k^2 - 2x^6 \implies$

$144k^2x^6 + 4x^{12} = 36^2k^4 - 144k^2x^6 + 4x^{12} \implies$ $288 k^2 x^6 - 36^2 k^4 \implies$

$x^6 = \dfrac{36^2 k^2}{36 * 8} = \dfrac{9}{2}k^2 \implies$ $x = \dfrac{3^{\frac{1}{3}} k^{\frac{1}{3}}}{2^{\frac{1}{6}}}$ $\implies H = (6k)^{\frac{1}{3}} = 3^{\frac{1}{3}} 2^{\frac{1}{3}} k^{\frac{1}{3}}$

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

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

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

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

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

and

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

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

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

$\implies \boxed{\cos(\theta) = \dfrac{\vec{U} \circ \vec{V}}{|\vec{U}| |\vec{V}|} = -\dfrac{x^2}{4H^2 + x^2} = -\dfrac{1}{9}}$

$\implies \dfrac{\tan^2(\lambda)\tan(\gamma)}{\cos(\theta)} = \boxed{-144}$ .

Note: $\dfrac{d^2A}{dx^2}|_{x = (\dfrac{9}{2} k^2)^{\frac{1}{6}}} = \dfrac{32}{3} > 0 \implies$ min occurs at $x = (\dfrac{9}{2} k^2)^{\frac{1}{6}}$