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Critical point multivariable

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data
    find the critical point of,

    x^(2)+y^(2)+2x^(-1)y^(-1)


    2. Relevant equations
    none



    3. The attempt at a solution
    Ok so first we differentiate the function such that fx = 0 and fy=0

    doing this yields,

    fx = 2x-2x^(-2)y^(-1)
    fy = 2y-2x^(-1)y^(-2)

    both set to equal this is where i lose it,
    so i can i get 1/x^(3) = y but subbing this into fy helps me little.

    any help would be appreciated thanks,
    dan
     
  2. jcsd
  3. Jan 23, 2012 #2
    OK here's the function, as I read it from your post:

    [itex]f = x^{2} + y^{2} + \frac{2}{xy}[/itex]

    You get two lines describing the stationary points for each partial derivative
    [itex]\frac{\partial f}{\partial y} = 0 ; 2y - \frac{2}{xy^{2}} = 0[/itex] therefore [itex]x = y^{-3}[/itex]

    and similarly [itex]y = x^{-3}[/itex] from [itex]\frac{\partial f}{\partial x}[/itex]

    So the place(s) where those lines cross is/are stationary.
     
    Last edited: Jan 23, 2012
  4. Jan 23, 2012 #3
    book gives the answer as (-1,-1) and (1,1)
     
  5. Jan 23, 2012 #4
    I'm totally lost on how the book arrives at the aforementioned answers.

    Dan.
     
  6. Jan 23, 2012 #5
    How can you find out where [itex]x=y^{-3}[/itex] crosses [itex]y=x^{-3}[/itex] ? Or are you lost earlier than this?
     
  7. Jan 23, 2012 #6
    lost at that point.
     
  8. Jan 23, 2012 #7
    Subbing either of those values into fx or fy doesn't seem to help me :(.
    Dan.
     
  9. Jan 23, 2012 #8

    ehild

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    Your equation are identical with xy3 and x3y=1. Divide them and solve for x/y. What do you get?

    ehild
     
  10. Jan 23, 2012 #9
    Substituting the two equations into each other, as you said, you get x=(x-3)-3 (and/or y=(y-3)-3). From there it shouldn't be too hard.
     
  11. Jan 24, 2012 #10
    ok i get how the book gets point (1,1) but how does it get (-1,-1).

    thanks for your assistance so far.
    Dan
     
  12. Jan 24, 2012 #11

    ehild

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    How did you get point (1,1)?

    ehild
     
  13. Jan 24, 2012 #12

    SammyS

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    You have x3y = 1 and xy3 = 1.

    You could graph these.

    Or use algebra to solve them simultaneously.
    Solving the way Joffan suggested is one way. Maybe the exponents involved confused you.

    Alternatively, you can notice that since x3y and xy3 each equal 1, they equal each other.

    So, x3y = xy3 → x2 = y2 → x = ±y.

    Substituting that back into either equation gives ±x4 = 1. (or ±y4 = 1)

    Only the + sign works for real numbers, so you have x = y.

    Finally you get x4 = 1.

    Solving this for x should give the two answers you're looking for. ​
     
  14. Jan 25, 2012 #13
    subbing y=1/x^(3) into x=1/y^(3), which leads to 1=x^(8).
    so x = 1 and y = 1
     
  15. Jan 25, 2012 #14

    ehild

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    But x=-1 is also solution, isn't it? (-1)^8=1.


    ehild
     
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