By Lagrange Multipliers, the function $f$ has no minima or maxima under constraint $g$?

Question

Find the extrema of $f$ subject to the stated condition.

$f(x,y)=x-y$ subject to $g(x,y)=x^2-y^2=2$.

Ok, by Lagrange Multiplier method, we find the points that satisfy $\nabla f(x,y) = \lambda \nabla g(x,y)$ for some $\lambda \in \mathbb R$.

Well then we have $(1,-1)=\lambda(2x,-2y)$ and the system: $$1=\lambda2x \\ -1 = -\lambda2y \implies 1 = \lambda2y$$

This also implies $\lambda,x,y \neq 0$. So we have $\lambda2x = \lambda2y \implies x = y$.

But $x=y \implies x^2-y^2=0$. And the constraint is that $x^2-y^2=2$. Does this mean that there cannot be any points that satisfy the system? And does it mean that there are no maxima/minima under these conditions? Please explain.

Answer 1

The Lagrangian method brings conditionally stationary points to the fore, if there are any. In this example there are none, as you have found out.

Now $x^2-y^2=2$ defines a hyperbola $\gamma$ with apexes at $(\pm\sqrt{2},0)$ and asymptotes $y=\pm x$. The function $f(x,y):=x-y$ essentially measures the distance from the point $(x,y)$ to the ascending asymptote of $\gamma$. This distance is monotonically increasing (resp. decreasing) on each of the two branches of $\gamma$, whence there can be no local maximum or minimum of $f$ on $\gamma$.

Answer 2

As $x^2-y^2=(x-y)(x+y)$ the function becomes $(x,y)\mapsto2/(x+y)$, which clearly has no extrema.