Pyramid Mania

In right $\triangle{QPR}, \:\ \overline{PR} = 3a$ and $QP = h$ .

$\implies QR = \sqrt{9a^2 + h^2} \implies A = A_{\triangle{QPR}} = 2a\sqrt{9a^2 + h^2}$

The volume of the pyramid $V = \dfrac{16}{3} a^2 h = k \implies h = \dfrac{3k}{16a^2} \implies$ $A(a) = \dfrac{6\sqrt{256a^6 + k^2}}{a} \implies \dfrac{dA}{da} = \dfrac{6(512a^6 - k^2)}{a^2\sqrt{256a^6 + k^2}} = 0 \:\ a \neq 0 \implies a = \dfrac{k^{\frac{1}{3}}}{2\sqrt{2}} \implies$ $h = \dfrac{3 k^{\frac{1}{3}}}{2}$ .

$\vec{u} = 4a\vec{i} + 0\vec{j} + 0\vec{k}$

$\vec{v} = 3a\vec{i} + 3a\vec{j} + h\vec{k}$

$\implies |\vec{u} X \vec{v}| = 4a\sqrt{h^2 + 9 a^2}, \:\ |\vec{u}| = 4a, \:\ |\vec{v}| = \sqrt{18a^2 + h^2}$

$\implies \sin(\theta) = \dfrac{\sqrt{h^2 + 9a^2}}{\sqrt{18a^2 + h^2}} = \dfrac{\sqrt{3}}{2}$ $\implies \theta = \boxed{60^\circ}$ .

In right $\triangle{QPS}, \:\ \overline{PS} = a$ and $QP = h$

$\implies QS = \sqrt{a^2 + h^2} \implies A = A_{\triangle{QPS}} = 2a\sqrt{a^2 + h^2}$

The volume of the pyramid $V = \dfrac{16}{3} a^2 h = k \implies h = \dfrac{3k}{16a^2} \implies$ $A(a) = \dfrac{2\sqrt{256a^6 + 9k^2}}{a} \implies \dfrac{dA}{da} = \dfrac{2(512a^6 - 9k^2)}{a^2\sqrt{256a^6 + 9k^2}} = 0 \:\$ $a \neq 0 \implies a = \dfrac{3^{\frac{1}{3}}k^{\frac{1}{3}}}{2\sqrt{2}} \implies$ $h = \dfrac{3^{\frac{1}{3}} k^{\frac{1}{3}}}{2}$ .

$\vec{u} = 0\vec{i} + 4a\vec{j} + 0\vec{k}$

$\vec{v} = -a\vec{i} + 3a\vec{j} + h\vec{k}$

$\implies |\vec{u} X \vec{v}| = 4a\sqrt{h^2 + a^2}, \:\ |\vec{u}| = 4a, \:\ |\vec{v}| = \sqrt{10a^2 + h^2}$

$\implies \sin(\theta) = \dfrac{\sqrt{h^2 + a^2}}{\sqrt{10a^2 + h^2}} = \dfrac{1}{2}$ $\implies \theta = \boxed{30^\circ}$

$\implies \theta + \lambda = \boxed{90^\circ}$ .