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Finding Stationary Points

  1. Jan 7, 2014 #1
    1. The problem statement, all variables and given/known data
    See attached for problem

    Attempt at solution

    ∂f/∂x = 3x2+12x
    ∂f/∂y = 3y2-12y

    Factorise

    x(3x+12)

    x=0 or,
    3x+12 = 0 → x= -4

    y(3y-12)

    y=0 or,
    3y-12=0 → y=4

    I find the next stage difficult, because there are no mixed terms in the polynomial any of the values i substitute into another gives me 0, because of this i need help finding the stationary points, I'm fine with classifying them

    Thanks
     

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  3. Jan 7, 2014 #2

    haruspex

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    I don't see the difficulty. You have two values for x and two for y. How many combinations does that give for (x, y)?
     
  4. Jan 7, 2014 #3
    This just gives me two stationary points doesn't it?
     
  5. Jan 7, 2014 #4
    I see, it is 4, i thought there would be other values i would need, so my points would be

    (0,0)
    (0,4)
    (-4,0)
    (-4,4)

    Is this correct?
     
  6. Jan 7, 2014 #5

    HallsofIvy

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    If you had had "mixed" terms, xy, then you would have equations with both x and y leading to the result that a specific value of x leads to a specific value of y. Here, that does not happen. The derivative with respect to x will be 0 if x= 0 or -4 no matter what y is. The derivative with respect to y will be 0 if y= 0 or 4 no matter what x is. In particular, both derivatives will be 0 at (0, 0), (0, 4), (-4, 0), and (-4, 4) as you say.
     
  7. Jan 7, 2014 #6
    Thanks, that explanation helps me understand that better
     
  8. Jan 7, 2014 #7

    Ray Vickson

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    Yes, those are the stationary points. As for classification: have you taken (multivariate) second-order tests yet?
     
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