# Problem 2.

Find all positive $x, y$ for which the expression

$$
\frac{x y}{2}+\frac{18}{x y}
$$

takes the smallest value, and among all such $x, y$ the expression

$$
\frac{y}{2}+\frac{x}{3}
$$

is also minimal.
# Solution:

According to the Cauchy inequality,

$$
\frac{x y}{2}+\frac{18}{x y} \geq 2 \sqrt{\frac{x y}{2} \cdot \frac{18}{x y}}=6
$$

Equality is possible only when

$$
\frac{x y}{2}=\frac{18}{x y}
$$

From this,

$$
x y=6 \Rightarrow y=\frac{6}{x}
$$

Substituting the obtained value into the expression

$$
\frac{y}{2}+\frac{x}{3}
$$

we arrive at the expression

$$
\frac{3}{x}+\frac{x}{3} \geq 2
$$

The minimum value is achieved when $\frac{x}{3}=1$, i.e., $x=3$. Therefore, $y=2$.

Answer: $x=3, y=2$.

## Criteria: