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Local Max/Min and saddle points

  1. Oct 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the local max/min or saddle points of f(x,y) = (x-y)(1-xy)

    2. Relevant equations



    3. The attempt at a solution

    I expanded the equation to f(x,y) = x-y-(x^2)y+xy^2.

    Then I found the partial derivatives of the function.
    fx = 1-2xy +y^2
    fy = -x^2-2xy

    I'm stuck after this part. Usually I can set the function to 0 and solve for x or y, but I can't do that here.
     
  2. jcsd
  3. Oct 13, 2012 #2

    Ray Vickson

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    Your fy is incorrect.

    RGV
     
  4. Oct 13, 2012 #3
    Oh, I managed to type out the whole fy wrong.

    It is -x^2 -1 +2yx
     
  5. Oct 13, 2012 #4

    SammyS

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    So, can you solve the problem now?
     
  6. Oct 13, 2012 #5
    No. I don't know how to solve for 0. I can't set both y or x on either side of the equation.
     
  7. Oct 13, 2012 #6

    SammyS

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    Solve these equations simultaneously:

    1-2xy +y2 = 0

    -x2 -1 +2yx = 0

    Use elimination: add them together.
     
  8. Oct 13, 2012 #7
    Thanks man, figured it out
     
  9. Oct 13, 2012 #8

    Ray Vickson

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    So what solution or solutions do you get?

    RGV
     
  10. Oct 13, 2012 #9
    Oh, I got saddle points at (1,1) and (-1,-1)
     
  11. Oct 13, 2012 #10

    Ray Vickson

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    Correct.

    RGV
     
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