### Instruction: Find the minimum of the function
\[\frac{xy}{x^2 + y^2}\]in the domain $\frac{2}{5} \le x \le \frac{1}{2}$ and $\frac{1}{3} \le y \le \frac{3}{8}.$.### Response: I want to find the minimum of a fractional function, so I think about the relationship between the numerator and the denominator.
If the numerator is small and the denominator is large, then the fraction is small.
So I want to minimize the numerator and maximize the denominator.
The numerator is the product of $x$ and $y$, so I want to make them as small as possible within the given domain.
The smallest possible value of $x$ is $\frac{2}{5}$ and the smallest possible value of $y$ is $\frac{1}{3}$.
So the minimum value of the numerator is $\frac{2}{5} \cdot \frac{1}{3} = \frac{2}{15}$.
The denominator is the sum of the squares of $x$ and $y$, so I want to make them as large as possible within the given domain.
The largest possible value of $x$ is $\frac{1}{2}$ and the largest possible value of $y$ is $\frac{3}{8}$.
So the maximum value of the denominator is $\frac{1}{2}^2 + \frac{3}{8}^2 = \frac{1}{4} + \frac{9}{64} = \frac{25}{64}$.
Therefore, the minimum value of the fraction is $\frac{\frac{2}{15}}{\frac{25}{64}} = \frac{2 \cdot 64}{15 \cdot 25} = \frac{128}{375}$.
This value occurs when $x = \frac{2}{5}$ and $y = \frac{1}{3}$.