Example 3.41. Find the stationary points of the following functions:

1) $f(x, y)=(x-3)^{2}+(y-2)^{2}$
2) $f(x, y, z)=x^{2}+4 y^{2}+9 z^{2}-4 x+16 y+18 z+1$.
Solution. 1) Find the partial derivatives $f_{x}^{\prime}=2(x-3)$; $f_{y}^{\prime}=2(y-2)$. Setting the partial derivatives to zero, we get the system of equations: $x-3=0$ and $y-2=0$. From this, it follows that $x=3, y=2$. Thus, the stationary point is $(3,2)$. The partial derivatives are

$$
f_{x}^{\prime}=2 x-4, f_{y}^{\prime}=8 y-16, f_{z}^{\prime}=18 z+18
$$

Stationary points are found from the condition that the partial derivatives are equal to zero, leading to the following system of equations: $2 x-4=0, 8 y-16=0, 18 z+18=0$. From which it follows that $x=2, y=2, z=-1$. The stationary point is $(2,2,-1)$.

## Sufficient Conditions for Extremum

Let the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ be defined and have continuous second-order partial derivatives in some neighborhood of the point $P_{0}\left(x_{1}^{0}, x_{2}^{0}, \ldots, x_{n}^{0}\right)$, which is a stationary point of the function $f$. Then, if the quadratic form (second differential)

$$
\Phi\left(d x_{1}, d x_{2}, \ldots, d x_{n}\right)=\sum_{i, j=1}^{n} f_{x_{2} x_{j}}^{\prime \prime}\left(P_{0}\right) d x_{i} d x_{j}
$$

is positive definite (negative definite) at the point $P_{0}$, then the function $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ has a minimum (maximum) at the point. If the quadratic form $\Phi$ takes both positive and negative values (i.e., is indefinite), then there is no extremum at the point $P_{0}$.

Remark. The quadratic form $\Phi$ is called positive (negative) definite if for any point $P\left(z_{1}, z_{2}, \ldots, z_{n}\right) \in E^{n}$, it holds that $\Phi\left(z_{1}, z_{2}, \ldots, z_{n}\right)>0$ $(\Phi<0)$.

Remark. The quadratic form $\Phi$ is positive definite if and only if all the leading principal minors of the matrix of the quadratic form are positive, i.e., $\Delta_{1}>0, \Delta_{2}>0, \ldots, \Delta_{n}>0$.

Remark. If the signs of the leading principal minors alternate starting with a negative, i.e., $\Delta_{1}<0, \Delta_{2}>0, \Delta_{3}<0, \ldots$, then the quadratic form is negative definite.

For a function of two variables, the following criteria can be used to determine the nature of the stationary point:

1) minimum, if $\Delta>0, A>0$ (or $C>0$);
2) maximum, if $\Delta>0, A<0$ (or $C<0$);
3) no extremum, if $\Delta<0$;
4) the question is open, further investigation is required, if $\Delta=0$. Here $\Delta=A \cdot C-B^{2}, A=f_{x x}^{\prime \prime}\left(x_{0}, y_{0}\right)$, $B=f_{x y}^{\prime \prime}\left(x_{0}, y_{0}\right), C=f_{y y}^{\prime \prime}\left(x_{0}, y_{0}\right)$.