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Math Help - stationary points for functions of two variables

  1. #1
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    stationary points for functions of two variables

    hi, i have a question concerning stationary points of functions of two variables, i have a problem sheet question - i have done the problem and got the answer but my class mate disagrees with my answer, im trying to find out who is right, its annoying me,

    the problem is:

    Find and classify the stationary points of the following function:

    f(x,y) = 1/3x^3 + 1/3y^3 - x^2 - y^2


    I have worked out that there are four stationary points at the following coordinates:

    (0,0) Maximum
    (0,2) Saddle point
    (2,0) Saddle point
    (2,2) Minimum

    My friend thinks there are only two stationary points -the maximum and minimum points from above,

    Could anybody help with this problem and settle the score for us as i hate admitting im wrong

    Any help is appreciated

    Thank you

    missmath
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  2. #2
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    Quote Originally Posted by missmath View Post
    hi, i have a question concerning stationary points of functions of two variables, i have a problem sheet question - i have done the problem and got the answer but my class mate disagrees with my answer, im trying to find out who is right, its annoying me,

    the problem is:

    Find and classify the stationary points of the following function:

    f(x,y) = 1/3x^3 + 1/3y^3 - x^2 - y^2


    I have worked out that there are four stationary points at the following coordinates:

    (0,0) Maximum Mr F says:
    (0,2) Saddle point Mr F says:
    (2,0) Saddle point Mr F says:
    (2,2) Minimum Mr F says:

    My friend thinks there are only two stationary points -the maximum and minimum points from above,

    Could anybody help with this problem and settle the score for us as i hate admitting im wrong

    Any help is appreciated

    Thank you

    missmath
    Tell your friend that Mr Fantastic said s/he is wrong.
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  3. #3
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    Yes! I love being right he he

    thankyou for taking time to go through it - i really do appreciate it

    Missmath
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