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Classifying Critical Points

  1. Mar 8, 2009 #1
    1. The problem statement, all variables and given/known data
    f(x,y)=xy+(144/x)+(12/y)
    f has a relative minimum at ( , , )

    2. Relevant equations

    Partial derivatives

    3. The attempt at a solution
    fx=y-(144/x^2)
    fy=x-(12/y^2)

    I setted these to zero. Multiplies both sides so there is no fraction.

    Came out to be something like~~~ x2y-144=0
    xy2-12=0

    then did the factoring stuff and got

    x2-12xy2=0

    This tells me that the critical point is (0,0), but this is the wrong answer.

    PLEASE TELL ME WHAT I AM DOING WRONG.

    THANK YOU SO MUCH TO THOSE WHO HELP. THIS FORUM IS HELPING ME BECOME A BETTER MATHMATICIAN AND THEREFORE, MY TEST GRADES ARE GOING UP. THANK YOU.
     
  2. jcsd
  3. Mar 8, 2009 #2

    gabbagabbahey

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    Your partial derivatives and general procedure are correct:smile:; but your attempt to solve for x and y is very confusing...try solving the f_y=0 equation for x, and substitute that result into your f_x=0 equation and solve for y....
     
  4. Mar 9, 2009 #3

    lanedance

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    just to add - there can be (and is in this case) more than one stationary point

    as you mentioned (0,0) is a stationary point, but that doesn't automatically make it an extremum - need to look at the 2nd order partial dervatievs to determine that
    Edit - (0,0) is not a critical point
     
    Last edited: Mar 9, 2009
  5. Mar 9, 2009 #4

    HallsofIvy

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    (0, 0) most certainly is NOT a stationary point- it's not even a point on the graph. f is not defined at (0,0).
    And I get only one critical point.
     
  6. Mar 9, 2009 #5

    lanedance

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    yeah - looked at that a little too quickly, sorry for the mis-steer
     
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