# Extrema/LaGrange in Vector Calc

1. Sep 29, 2008

### DougD720

1. The problem statement, all variables and given/known data

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

2. Relevant equations

$$\nabla$$f=$$\lambda$$$$\nabla$$g

3. The attempt at a solution

$$\nabla$$f=(1,-1)
$$\nabla$$g=(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 paralell 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!

2. Sep 29, 2008

### 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.

Last edited: Sep 29, 2008