1020. Investigate the function for extremum

$$
f(x, y)=x^{2}+y^{2}-4 y+4
$$
Solution. Let's find the partial derivatives:

$$
f_{x}^{\prime}(x, y)=2 x ; f_{y}^{\prime}(x, y)=2 y-4
$$

Set them equal to zero, we get a system of two equations with two unknowns:

$$
\left\{\begin{array}{l}
2 x=0 \\
2 y-4=0
\end{array}\right.
$$

from which $x=0, y=2$.

The function at the critical point $M_{0}(0 ; 2)$ has a minimum $f(0 ; 2)=4-8+4=0$, since near this point for any points $M(x ; y)$, different from $M_{0}(0 ; 2)$, the values of the function are greater than zero. For example, let $x=0, y=1$, then $f(0 ; 1)=1-4+4=1>0$; let $x=-1, y=3$, then $f(-1 ; 3)=1+9-12+4=2>0$.