6. In a triangular pyramid $S A B C$, the angle $A S B$ at vertex $S$ is $30^{\circ}$, and the lateral edge $S C$ is inclined to the plane of the face $A S B$ at an angle of $45^{\circ}$. The sum of the lengths of the lateral edges of the pyramid is 9. Find the greatest possible value of the volume of the pyramid under these conditions.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_6cabcf1712d23d5ceb2bg-11.jpg?height=460&width=714&top_left_y=250&top_left_x=771)

Let's introduce the notations: $S A=a, S B=b, S C=c, C K=h, \square A S B=\alpha, \square C S K=\beta$,

$C K$ is the perpendicular to the plane $A S B$. Then

$$
V_{S A B C}=\frac{1}{3} S_{S A B} \cdot h=\frac{1}{6} a b \sin \alpha \cdot c \sin \beta=\frac{1}{6} a b c \sin \alpha \sin \beta
$$

From the inequality of the geometric mean and the arithmetic mean

$$
\sqrt[3]{a b c} \leq \frac{a+b+c}{3}=\frac{p}{3} \rightarrow a b c \leq \frac{p^{3}}{27}
$$

Equality is achieved when $a=b=c$. Then the maximum value of the product $a b c$ is $\frac{p^{3}}{27}$, and the maximum volume of the pyramid is $\frac{p^{3}}{162} \sin \alpha \sin \beta=\frac{9 \sqrt{2}}{8}$.

Answer: $V_{\max }=\frac{9 \sqrt{2}}{8}$.

## Answers for other variants.

## Variant 2

Problem 1. Answer: $x=2 ; 4 ; 5-\sqrt{13} ; 5-\sqrt{17}$.