maximization using Lagrange

Question

I am maximizing

$f(x,y)=-x$

given the constraint

$g(x,y)=x^2-y^2=0$

To satisfy the non degenerate constraint qualification I have:

$Dg(x,y)= [2x\quad-2y]$

and the set of $(x,y)$ that satisfy it is having $x=y$.

However on setting up the Lagrange multiplier:

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

and getting the first order conditions:

$L_x=-1+ 2\lambda x=0$ and

$L_y=-2\lambda y =0$

$L_\lambda= x^2-y^2=0$

I have a contradiction since for $x=y$

The first equation will give:

$-2 \lambda x= -1$

The second however shows:

$-2 \lambda x=0$

Is there anywhere I have gotten wrong here?

Answer 1

$L_\lambda=x^2+y^2=0$ is missing and you got $L_x=-1+2x\lambda=0$ and $L_y=2y\lambda=0$ in this case you will get $-1+2\lambda x=0$ (1)
$-2\lambda y=0$ (2)
$x^2-y^2=0$ (3) your system has no real solutions