Example 3.49. Investigate the function $f\left(x_{1}, x_{2}\right)=x_{2}^{2}-x_{1}^{2}$ for the presence of a local extremum subject to the condition $x_{1}-2 x_{2}+3=0$.
Solution. Let's form the Lagrange function

$$
L(x, y, \lambda)=x_{2}^{2}+x_{1}^{2}+\lambda\left(x_{1}-2 x_{2}+3\right)
$$

We will find the points where the first-order partial derivatives of this function are zero. We have

$$
\left\{\begin{array}{l}
\frac{\partial L}{\partial x_{1}}=-2 x_{1}+\lambda=0 \\
\frac{\partial L}{\partial x_{2}}=2 x_{2}-2 \lambda=0 \\
\frac{\partial L}{\partial \lambda}=x_{1}-2 x_{2}+3=0
\end{array}\right.
$$

From this, we obtain the point of possible extremum $P_{1}(1,2,-1)$. Next, we find $d^{2} L\left(x_{1}, x_{2}, \lambda\right)=-2 d x_{1}^{2}+2 d x_{2}^{2}+$ $+2 d x_{2} d \lambda-d x_{1} d \lambda$. This quadratic form is indefinite, as the minors of the matrix corresponding to this quadratic form $\left(\begin{array}{cc}-2 & 1 \\ 1 & 2\end{array}\right), \Delta_{1}=-2, \Delta_{2}=\left|\begin{array}{cc}-2 & 1 \\ 1 & 2\end{array}\right|=$ $=-5-$ are negative. Thus, $L(x, y, \lambda)$ does not have an extremum. However, we will prove that the conditional extremum of the function $f\left(x_{1}, x_{2}\right)$ under the constraint exists.

Indeed, from the constraint, it follows that $d x_{1}=2 d x_{2}$ and $d^{2} L=-6 d x_{2}^{2}$ - is a negative definite quadratic form, and the point $x_{1}=1, x_{2}=2$ is a point of local maximum of the function $f\left(x_{1}, x_{2}\right)$ under the constraint $x_{1}-2 x_{2}+$ $+3=0$.