Example 3.42. Investigate the function for extremum

$$
z=x^{2}+2 y^{2}-2 x y-x-2 y
$$
Solution. To find the stationary points of the given function, we solve the following system of equations:

$$
\left\{\begin{array}{l}
f_{x}^{\prime}=2 x-2 y-1=0 \\
f_{y}^{\prime}=4 y-2 x-2=0
\end{array}\right.
$$

From which it follows that $x=2, y=\frac{3}{2}$. We will investigate the sufficient conditions for an extremum. For this, we find the second-order partial derivatives. We have: $f_{x x}^{\prime \prime}=2, f_{y y}^{\prime \prime}=4, f_{x y}^{\prime \prime}=f_{y x}^{\prime \prime}=-2$. According to Sylvester's criterion, we find the principal minors: $\Delta_{1}=2>0 ; \Delta_{2}=\left|\begin{array}{cc}2 & -2 \\ -2 & 4\end{array}\right|=8-4>0$. Since all the principal minors are positive, the function has a minimum at the point $\left(2, \frac{3}{2}\right)$.