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Prove that f(sqrt(2),-sqrt(2))=-8<f(x,y)

  1. Aug 7, 2005 #1
    I have an exercise i would really appreciate if you could help me with:

    Given f:R^2->R, f(x,y)=x^4+y^4-2(x-y)^2
    1-Prove that (sqrt(2),-sqrt(2)) and (-sqrt(2),sqrt(2)) are absolute minimums
    2-are there any local maximums?

    1-I found out that the critical points lie on the line y=-x, and i suppose i should prove that f(sqrt(2),-sqrt(2))=-8<f(x,y) for every (x,y) but i dont know how to do this.

    2-I found that there arent any local maximums, but i would like you to correct me if i am wrong.

    Thank you very much, Paul.
  2. jcsd
  3. Aug 7, 2005 #2

    matt grime

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    2. what are the partial derivatves, where do they vanish and what is the discrominant there? post the working.
  4. Aug 7, 2005 #3
    my work

    The partial derivatives are: Fx=4x^3-4(x-y) and Fy=4y^3+4(x-y)
    they vanish at x=-y and the discriminant there is 144y^4-96y^2 if my calculations are correct.
    Fx and Fy are the partial derivatives with respect to x and y respectively.
    What i did next, was to say that if there is a relative maximum at (x,-x):

    (a)the discriminant has to be positive and Fxx>0 or
    (b)the discriminant has to be equal to 0.

    if (a) happens it would mean that Fxx=12x^2-4>0 and discriminant=144y^4-96y^2>0 which is not possible for any point over y=-x.

    (b) can only happen if (x,y)=(0,0) but i can prove that it is a saddle point approaching from different directions to (0,0).

    So i concluded that there are not relative maximums, is this correct?????

    If you ask me, i think i can explain it a little bit clearer.
    Thank you for your help, Paul.
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