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Math Help - [SOLVED] Finding stationary points.

  1. #1
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    [SOLVED] Finding stationary points.

    so i have a function

     f(x,y) = x^2+3y^2-y^3+5

    I want to find the stationary point

     f_x = 2 , f_y = 6y-3y^2

    so stationary point is at (0,2) ??

    Then to find if its a max or min

     f_{xx} = 2 , f_{yy}=6-6y , f_{xy}=0

    So  f_{xx}f_{yy}-f_{xy}^2 = 2 \cdot (6-6y)-(0)^2= 12-12y

    So do i then sub in my value for y e.g y = 2 so ifnd out what that answer is? Or have i done something wrong?
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by el123 View Post
    so i have a function

     f(x,y) = x^2+3y^2-y^3+5

    I want to find the stationary point

     f_x = 2 , f_y = 6y-3y^2

    so stationary point is at (0,2) ??

    Then to find if its a max or min

     f_{xx} = 2 , f_{yy}=6-6y , f_{xy}=0

    So  f_{xx}f_{yy}-f_{xy}^2 = 2 \cdot (6-6y)-(0)^2= 12-12y

    So do i then sub in my value for y e.g y = 2 so ifnd out what that answer is? Or have i done something wrong?
    Given f_x=2x and f_y=3y(2-y), we have that the stationary points are (0,0) and (0,2).

    Other than that, what you are doing looks fine.
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  3. #3
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    So how do i define if it is a max or min?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by el123 View Post
    So how do i define if it is a max or min?
    Define M=f_{xx}f_{yy}-f_{xy}^2

    If M>0 and f_{xx}>0, then you have a local minimum.
    If M>0 and f_{xx}<0, then you have a local maximum.
    If M<0, then you have a saddle point.
    If M=0, then the test is inconclusive.

    Check out the Wikipedia page on the second derivative test.
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  5. #5
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    So since M = 12-12y

    How do i figure what M equals? Do i use both values for y from each stationary point and put them in equation? So...12-12(0) and 12-12(2)?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Yes
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