[ Area and volume (extremum problems $)]$

Find the maximum value of the volume of the pyramid $S A B C$ under the following constraints $S A \leqslant 4, S B \geqslant 7, S C \geqslant 9, A B=5, B C \leqslant 6, A C \leqslant 8$.
Let $C K$ be the height of the triangular pyramid $S A B C$. Then $C K \leqslant B C \leqslant 6$. By the cosine theorem from triangle $S A B$, we find that

$$
\begin{aligned}
& \cos \angle S A B=\frac{A S^{2}+A B^{2}-S B^{2}}{2 A S \cdot A B} \leqslant \frac{16+25-19}{2 \cdot 5 \cdot A S} = \\
& \leqslant \frac{22}{10 A S} = \frac{11}{5 A S} \leqslant \frac{11}{5 \cdot 4} = \frac{11}{20} = -\frac{1}{5}.
\end{aligned}
$$

Therefore, $\angle S A B > 90^{\circ}$. If $V$ is the volume of the pyramid $S A B C$, then

$$
\begin{aligned}
& V=\frac{1}{3} \cdot \frac{1}{2} A S \cdot A B \sin \angle S A B \cdot C K \leqslant \frac{1}{6} \cdot 4 \cdot 5 \cdot 6 \cdot \sin \angle S A B = \\
& = 20 \sin \angle S A B = 20 \sqrt{1-\cos ^{2} \angle S A B} \leqslant 20 \sqrt{1-\left(-\frac{1}{5}\right)^2} = 20 \sqrt{1-\frac{1}{25}} = 20 \sqrt{\frac{24}{25}} = 20 \cdot \frac{\sqrt{24}}{5} = 4 \sqrt{24} = 8 \sqrt{6}.
\end{aligned}
$$

The pyramid $S A B C$, where $A S=4, A B=5, \angle S A B=\arccos \left(-\frac{1}{5}\right), B C=6$, and $B C$ is perpendicular to the plane of the base $S A B$, satisfies the condition of the problem, since in this case

$$
\begin{gathered}
S B=7, A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{25+36}=\sqrt{61}.
\end{gathered}
$$

| $[\underline{\text { [ Tangents to spheres ]}}]$ |  |
| :---: | :---: |
|  |  |

and the volume of such a pyramid is $8 \sqrt{6}$.

## Answer

$8 \sqrt{6}$.