Example 3.43. Investigate the function for extremum

$$
u=x^{2}+2 y^{2}+4 z^{2}+4 x+2 y-8
$$
Solution. We find the stationary points of the function from the system of equations

$$
\begin{aligned}
& u_{x}^{\prime}=2 x+4=0 \\
& u_{y}^{\prime}=4 y+2=0 \\
& u_{z}^{\prime}=8 z-8=0
\end{aligned}
$$

From which $x=-2, y=-\frac{1}{2}, z=1$. We find the second-order partial derivatives:

$$
\begin{gathered}
u_{x x}^{\prime \prime}=2, u_{x y}^{\prime \prime}=u_{y x}^{\prime \prime}=0, u_{x z}^{\prime \prime}=u_{z x}^{\prime \prime}=0, u_{y y}^{\prime \prime}=4 \\
u_{z y}^{\prime \prime}=u_{y z}^{\prime \prime}=0, u_{z z}^{\prime \prime}=8
\end{gathered}
$$

According to Sylvester's criterion, we form the principal minors:

$$
\Delta_{1}=2>0, \Delta_{2}=\left|\begin{array}{ll}
2 & 0 \\
0 & 4
\end{array}\right|=8>0, \Delta_{3}=\left|\begin{array}{lll}
2 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 8
\end{array}\right|=64>0
$$

Since all the principal minors are positive, the function has a local minimum at the point $\left(-2, -\frac{1}{2}, 1\right)$.