Extrema/LaGrange in Vector Calc

[DougD720]

Homework Statement

Find the extrema of f(x,y)=x-y ; subject to x²-y²=2

Homework Equations

\nablaf=\lambda\nablag

The Attempt at a Solution

\nablaf=(1,-1)
\nablag=(2x, -2x)

(1,-1)=\lambda(2x, -2x)

1 = \lambda(2x) -> \lambda=\frac{1}{2x}

-1 = \lambda(-2y) -> \lambda=\frac{1}{2y}

Which means x = y , but it has to satisfy x²-y²=2 and if x=y then it cannot satisfy this meaning there are no extrema for this set of equations.

Am i right? I tried working it out with other methods but it just keeps not working, however, i plotted the two equations in 3D on Maple and they do intersect so shouldn't there be extrema? Or is the fact that x-y is a plane parallel to the xy-axis mean that all points are extrema?

We never did a problem like this in class, one with no apparent solution, so I'm confused a bit here.

And i just did another problem where I'm coming up with a solution that doesn't satisfy one of the constraints... ugh... what am i doing wrong?

Thanks for the help!

 

[Dick]

Yes, you aren't doing anything wrong. Your constraint says (x-y)=2/(x+y). So (x-y) is unbounded from both above and below. There are no extrema.