Task B-2.2. For what values of the variables $x$ and $y$ does the expression $\frac{4 x^{2}+2 y^{2}-4 y+4}{2 x^{2}+y^{2}-2 y+5}$ have the smallest value, and what is that smallest value?
## Solution.

$$
\begin{aligned}
& \frac{4 x^{2}+2 y^{2}-4 y+4}{2 x^{2}+y^{2}-2 y+5}=\frac{2\left(2 x^{2}+y^{2}-2 y+5\right)-6}{2 x^{2}+y^{2}-2 y+5}= \\
& =2-\frac{6}{2 x^{2}+y^{2}-2 y+1+4}=2-\frac{6}{2 x^{2}+(y-1)^{2}+4}
\end{aligned}
$$

The given expression has the smallest value when the value of $\frac{6}{2 x^{2}+(y-1)^{2}+4}$ is the largest, which occurs when the value of the expression $2 x^{2}+(y-1)^{2}+4$ is the smallest. This is achieved for

$$
x^{2}=0,(y-1)^{2}=0 \Leftrightarrow x=0, y=1
$$

The smallest value of the given expression is $\frac{1}{2}$.