Example 3.48. Find the extremum of the function

\[
f(x, y)=x^{2}+y^{2}-2 x-y
\]

subject to the condition that the variables \(x\) and \(y\) are related by the equation \(\varphi(x, y)=x+y-1=0\).
Solution. For the Lagrange function

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

considering (3.44), we have a system of three equations:

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

The solution is \( x_{0}=0.75, y_{0}=0.25, \lambda_{0}=0.5 \). We then find the second-order partial derivatives: \( L_{x x}^{\prime \prime}=2=A>0 \), \( L_{y y}^{\prime \prime}=2=C>0 \), \( L_{x y}^{\prime \prime}=0=B \), \( \Delta=A C-B^{2}=4>0 \). Therefore, the conditional minimum of the function \( f(x, y) \) is:

$$
f_{\min }(0.75 ; 0.25)=0.75^{2}+0.25^{2}-2 \cdot 0.75-0.25=-1.125
$$