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Lagrange multi.

  1. Nov 9, 2003 #1
    We have started to do Lagrange Multi. in my class and my book has a very short section on how to solve these. I was wondering if someone couls help.
    The problem is f(x,y)=x^2-y^2 with the constraint x^2+y^2=1.
    I have found the partial derv. but I am not sure on what else to do. Any help would be sweet, later.
     
  2. jcsd
  3. Nov 9, 2003 #2
    If you have a function f(x,y), and a constraint g(x,y) = 0, then to find the constrained extrema you set the partial derivatives of f + λg to zero, and solve for x and y. Remember to take partial derivatives not only with respect to x and y, but also with respect to λ; otherwise, you won't impose the constraint.
     
    Last edited: Nov 9, 2003
  4. Nov 9, 2003 #3
    In my problem I have the constraint equal to 1, should I adjust it and have it equal to 0?
     
  5. Nov 9, 2003 #4
    It's traditional to write the constant as zero, but it really doesn't matter; it will work with any constraint g(x,y) = constant, because the constant goes away after you take the derivatives.
     
  6. Nov 9, 2003 #5
    So then for my problem I would get something like

    the partial of X: 2X+(lag. symbol)2X=0
    thus getting (Lag. Symbol)= -1
     
  7. Nov 9, 2003 #6
    Either that, or x=0.
     
  8. Nov 9, 2003 #7
    Then take that 1 and 0 and plug them into the orig. equation of F(X)
    and get (+-1,0) or (0,+-1) or (0,0). Then I have the conditions to make this either a max or min. Value. Is this correct?
     
  9. Nov 9, 2003 #8
    Almost; (0,0) is not a solution of the constraint. Otherwise, you're right.
     
  10. Nov 9, 2003 #9
    Alright, then for my next problem I have
    F(X);x^2y and G(X);x^2+2y^2=6

    The partials of x I get
    2xy+(Lang.)2x=0, giving me (Lang.)=y, can this be true?
     
  11. Nov 9, 2003 #10
    That's one solution (again, another is x=0), but you'll have to impose the other derivative constraints too, and you'll find that, in that case, they restrict what y can be.
     
  12. Nov 9, 2003 #11
    I have thought about the three equations
    Partial x
    2xy=(Lang.)2x
    Partial y
    x^2=(Lang.)4y
    and
    x^2+2y^2=6
    and am looking for numbers that satisfy all equations. So I would get something like...
     
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