Berkeley Math Circle: Monthly Contest 2

Solution:

The answer is $12$. This is achieved when $x = 3$; to see it is optimal, note that
$$2x + \frac{18}{x} \geq 12 \Longleftrightarrow x^2 - 6x + 9 \geq 0,$$
which is obviously true since the left-hand side is $(x-3)^2 \geq 0$.

Alternatively, those who know the so-called AM-GM inequality may apply it directly.