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Finding the critical points of a multivariable function

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data\
    Find the critical points of the function. Then use the Second Derivative Test to classify each critical point (or state that the test fails).

    f(x, y) = x3 + y4 - 21x - 18y2




    3. The attempt at a solution

    partial x derivative=3x^2-21=0

    partial y= 4y^3-36y=0

    i try to solve those for numbers, but the answer space in the question has space for 6 different critical point answers, so I am not sure where to go from here.
     
  2. jcsd
  3. Nov 4, 2011 #2

    SammyS

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    What did you get when you solved these for numbers?

    You should get two solutions for x and three solutions for y.
     
  4. Nov 4, 2011 #3
    partial x = 3x^2-21=0
    3x^2=21
    x=sqrt(21)/3

    partial y = 4y^3-36y=0
    4y^3=36y
    4y^2=36
    = 6
    y=3/2

    must be missing something, because I don't really see another way to solve it.


    Thanks,
    Matt
     
    Last edited: Nov 4, 2011
  5. Nov 4, 2011 #4
    Assignment due soon, anyone help please?
     
  6. Nov 4, 2011 #5

    SammyS

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    That should be [itex]x=\pm\sqrt{\frac{21}{3}}=\pm\sqrt{7}\,.[/itex]

    Similarly, use ± for the y solution.

    Also, when you divided by y, you got rid of the y=0 solution.
     
  7. Nov 4, 2011 #6

    Mark44

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    In addition to what SammyS said, there's another error here.
    Setting fy = 0 yields
    4y3-36y=0
    ==> 4y(y2 - 9) = 0
    I hope you can see that this equation has three solutions, none of which is 3/2.
     
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