1. (mathematics) Find the values of $a$ for which the coordinates ( $x ; y$ ) of any point in the square $2 \leq x \leq 3,3 \leq y \leq 4$ satisfy the inequality $(3 x-2 y-a)\left(3 x-2 y-a^{2}\right) \leq 0$.
Answer: $a \in(-\infty ;-4]$

Solution: Points with coordinates $(x ; y)$ of the square $2 \leq x \leq 3, 3 \leq y \leq 4 A B C D$ lie in the strip

$$
(3 x-2 y-a)\left(3 x-2 y-a^{2}\right) \leq 0
$$

between the lines $L_{1}: 3 x-2 y-a=0$ and $L_{2}: 3 x-2 y-a^{2}=0$ (see figure), if this strip contains all the vertices of the square:

$$
\begin{aligned}
& A(2 ; 4) \rightarrow(-2-a)\left(-2-a^{2}\right) \leq 0 \rightarrow a \leq-2 \\
& B(2 ; 5) \rightarrow(-4-a)\left(-4-a^{2}\right) \leq 0 \rightarrow a \leq-4 \\
& C(3 ; 5) \rightarrow(-1-a)\left(-1-a^{2}\right) \leq 0 \rightarrow a \leq-1 \\
& \mathrm{D}(3 ; 4) \rightarrow(-3-a)\left(-3-a^{2}\right) \leq 0 \rightarrow a \leq-3
\end{aligned}
$$

Intersecting the sets of solutions of inequalities (A) - (D), we get

![](https://cdn.mathpix.com/cropped/2024_05_06_f912147981c7262840e0g-1.jpg?height=622&width=557&top_left_y=934&top_left_x=1292)
the desired $a \in(-\infty ;-4]$.