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Extrema/LaGrange in Vector Calc

  1. Sep 29, 2008 #1
    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


    3. The attempt at a solution

    [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 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. jcsd
  3. Sep 29, 2008 #2


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    Science Advisor
    Homework Helper

    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
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