12.47 The lateral face of a regular quadrilateral pyramid has a constant given area and is inclined to the base plane at an angle $\alpha$. For what value of $\alpha$ is the volume of the pyramid the greatest?
12.47 The volume of the pyramid \(S_{ABCD}\) (Fig. 12.9) is expressed by the formula \(V = \frac{1}{3} S_{ABCD} \cdot SO\). Let \(AD = a\), then \(S_{ABCD} = a^2\), and from \(\triangle SOK\) we find \(SO = OK \operatorname{tg} \angle SKO = \frac{a}{2} \operatorname{tg} \alpha\).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-397.jpg?height=532&width=467&top_left_y=330&top_left_x=659)

Fig. 12.9

Thus,

\[ V = \frac{1}{3} a^2 \cdot \frac{a}{2} \operatorname{tg} \alpha = \frac{1}{6} a^3 \operatorname{tg} \alpha. \]

Next, let \( S_{\triangle SDC} = Q = \frac{1}{2} a \cdot SK \),

where

\[ SK = \frac{OK}{\cos \alpha} = \frac{a}{2 \cos \alpha}. \] Therefore, \( Q = \frac{a^2}{4 \cos \alpha} \), from which

\[ a = \sqrt{4 Q \cos \alpha}. \]

Substituting (2) into (1), we get

\[ V = V(\alpha) = \frac{1}{6} (\sqrt{4 Q \cos \alpha})^3 \operatorname{tg} \alpha = \frac{4 Q}{3} \sin \alpha \sqrt{Q \cos \alpha} \]

where \(0 < \alpha < \frac{\pi}{2}\). We find \( V'(\alpha) = \frac{4}{3} Q^2 \frac{2 \cos^2 \alpha - \sin^2 \alpha}{2 \sqrt{Q \cos \alpha}} \); thus, \( V'(\alpha) = 0 \) if \( 2 \cos^2 \alpha - \sin^2 \alpha = 0 \); further, we have \( \operatorname{tg}^2 \alpha = 2 \); \( \operatorname{tg} \alpha = \pm \sqrt{2} \); \( \alpha = \pm \operatorname{arctg} \sqrt{2} + \pi n \), \( n \in \mathbb{Z} \). The only value that satisfies the problem's condition is \( \alpha = \operatorname{arctg} \sqrt{2} \). Therefore, the volume of the pyramid is maximized when \( \alpha = \operatorname{arctg} \sqrt{2} \).

Answer: when \( \alpha = \operatorname{arctg} \sqrt{2} \).