Triangular Pyramids

$V_{T} = \dfrac{1}{4\sqrt{3}} x^2 H = K \implies H = \dfrac{4\sqrt{3}K}{x^2}$

Since the base is an equilateral triangle $OB = \dfrac{x}{\sqrt{3}}$ and $OD = \dfrac{x}{2\sqrt{3}}.$

$H = \dfrac{4\sqrt{3}K}{x^2}$ and $OD = \dfrac{x}{2\sqrt{3}} \implies$ The lateral surface area is $A_{S} = \dfrac{3x}{2} s = \dfrac{\sqrt{3}}{4x}\sqrt{x^6 + 576K^2} \implies \dfrac{dA_{s}}{dx} = \dfrac{\sqrt{3}}{4}(\dfrac{2x^6 - 576K^2}{x^2\sqrt{x^6 + 576K^2}}) \implies$ $x = (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} 2^{\frac{1}{6}}$ for $x > 0 \implies$ $H = \dfrac{4 * 3^{\frac{1}{2}} K^{\frac{1}{3}}}{12^{\frac{2}{3}} 2^{\frac{1}{3}}} \implies$

$\tan(\theta) = \dfrac{\sqrt{3} H}{x} = \dfrac{1}{\sqrt{2}}$ and $\tan(\lambda) = \dfrac{2\sqrt{3} H}{x} = \sqrt{2}$ and $\tan(90 - \theta) = \cot(\theta) = \dfrac{1}{\tan(\theta)} = \sqrt{2} = \tan(\lambda)$

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

Testing for a minimum at $x = (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} 2^{\frac{1}{6}}$ :

For $x \neq 0$ . Since the denominator of $\dfrac{dA_{s}}{dx}$ is always positive we just need to test the numerator.

For $0 < x < (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} 2^{\frac{1}{6}}$ pick $x = (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} \implies \dfrac{dA_{s}}{dx} < 0$

For $x > (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} 2^{\frac{1}{6}}$ pick $x = 2 * (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} \implies \dfrac{dA_{s}}{dx} > 0$

$\implies$ minimum at $x = (12)^{\frac{1}{3}} (K)^{\frac{1}{3}} 2^{\frac{1}{6}}$ .