Extrema/LaGrange in Vector Calc

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SUMMARY

The discussion focuses on finding the extrema of the function f(x,y) = x - y, subject to the constraint x² - y² = 2. The solution involves using the method of Lagrange multipliers, leading to the equations ∇f = λ∇g, where ∇f = (1, -1) and ∇g = (2x, -2y). The conclusion reached is that there are no extrema for this set of equations, as the condition x = y contradicts the constraint, indicating that the function is unbounded. The participant also notes confusion regarding the intersection of the equations in 3D and the nature of the function being a plane parallel to the xy-axis.

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Homework Statement



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


Homework Equations



[tex]\nabla[/tex]f=[tex]\lambda[/tex][tex]\nabla[/tex]g


The Attempt at a Solution



[tex]\nabla[/tex]f=(1,-1)
[tex]\nabla[/tex]g=(2x, -2x)

(1,-1)=[tex]\lambda[/tex](2x, -2x)

1 = [tex]\lambda[/tex](2x) -> [tex]\lambda[/tex]=[tex]\frac{1}{2x}[/tex]

-1 = [tex]\lambda[/tex](-2y) -> [tex]\lambda[/tex]=[tex]\frac{1}{2y}[/tex]

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!
 
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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:

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