1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [tex]
    \begin{array}{rcl}
    x & = & \sqrt{2} \; \epsilon \cosh{t}, \; \epsilon = \pm 1 \\

    y & = & \sqrt{2} \sinh{t}, \; -\infty < t < \infty \\
    \end{array}
    [/tex]

    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

    cronxeh

    User Avatar
    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

    Dick

    User Avatar
    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:

    http://www.wolframalpha.com/input/?i=maximize+x-y+on+x^2-y^2-2%3D0


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

    Dick

    User Avatar
    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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Find the extrema of f subject to the stated constraint
  1. Find extremas (Replies: 2)

Loading...