[Algebra]
Problem: If $2 \leq x \leq 5$ and $10 \leq y \leq 20$, what is the maximum value of $15-\frac{y}{x}$ ?

Solution: Since we want to make $15-\frac{y}{x}$ as large as possible, then we want to subtract as little as possible from 15.

In other words, we want to make $\frac{y}{x}$ as small as possible.

To make a fraction with positive numerator and denominator as small as possible, we make the numerator as small as possible and the denominator as large as possible.

Since $2 \leq x \leq 5$ and $10 \leq y \leq 20$, then we make $x=5$ and $y=10$.

Therefore, the maximum value of $15-\frac{y}{x}$ is $15-\frac{10}{5}=13$.
Since $y$ is positive and $2 \leq x \leq 5$, then $15-\frac{y}{x} \leq 15-\frac{y}{5}$ for any $x$ with $2 \leq x \leq 5$ and positive $y$.

Since $10 \leq y \leq 20$, then $15-\frac{y}{5} \leq 15-\frac{10}{5}$ for any $y$ with $10 \leq y \leq 20$.

Therefore, for any $x$ and $y$ in these ranges, $15-\frac{y}{x} \leq 15-\frac{10}{5}=13$, and so the maximum possible value is 13 (which occurs when $x=5$ and $y=10$ ).

Final answer: 13