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Homework Help: Find the extrema of f subject to the stated constraint

  1. May 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the extrema of f subject to the stated constraint:

    f(x,y) = x-y subject to x2-y2=2

    2. Relevant equations

    Apply the Lagrange Multiplier!

    3. The attempt at a solution

    This question was rather odd... I just did a problem similar to this one, and I got the answer right.

    Let g(x,y) = x2-y2-2

    Now let L(x,y) = f-[tex]\lambda[/tex]g (where [tex]\lambda[/tex] = the Lagrange Multiplier)

    Lx = 1 - 2[tex]\lambda[/tex]x = 0

    Ly = -1 + 2[tex]\lambda[/tex]y = 0

    I then solve for x and y.

    I get x = [tex]\frac{1}{2\lambda}[/tex] = y

    I plugged both of them into the constraint g(x,y) = x2-y2-2 = 0

    Both x and y cancels out!!! and so

    -2 = 0

    I am sure I am doing something wrong because there is an answer! I've checked using an online calculator.

    Can anyone please show me what I am doing wrong?
    Last edited: May 22, 2010
  2. jcsd
  3. May 22, 2010 #2
    If you parametrize x and y using hyperbolic substitution:

    x & = & \sqrt{2} \; \epsilon \cosh{t}, \; \epsilon = \pm 1 \\

    y & = & \sqrt{2} \sinh{t}, \; -\infty < t < \infty \\

    and substitute this into the function [itex]f(x, y)[/itex], you will get a function of a single variable [itex]\tilde{f}(t)[/itex]. Can you find its analytic expression? Does this function have extrema?
  4. May 22, 2010 #3
    you made an error in the last line. The first and second term cancel.
  5. May 22, 2010 #4


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    Gold Member

    Yes you right I made a mistake, the substitution is the way to go
  6. May 22, 2010 #5
    Sorry guys, but I still do not know what to do if the x's and y's cancels out...
  7. May 22, 2010 #6


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    Science Advisor
    Homework Helper

    x^2-y^2=2 is the same as (x+y)(x-y)=2 is the same as (x-y)=2/(x+y). Have you considered the possibility it may not have any extrema?
  8. May 22, 2010 #7
    Yes... but, I said I used the online calculator and it said the values for the extrema.

    Here is the link:


    Note*** I may be misreading the values on the online calculator
  9. May 22, 2010 #8


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    Science Advisor
    Homework Helper

    Don't trust wolfram alpha for everything, I don't even know what it's trying to say. I think it's simply confused. You've correctly concluded a local extremum exists if 0=(-2). It doesn't. There aren't any local extrema.
  10. May 22, 2010 #9
    Thank you!
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