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Math Help - Local Minimum and Maximum; Saddle Point

  1. #1
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    Local Minimum and Maximum; Saddle Point

    Hi everyone. I have a question on one of my homework problems.

    Q. Find the local max./min. values and the saddle point(s) of the function  f(x,y) = (x^2+y^2)e^{y^2-x^2}

    So I started doing this problem and it was looking really messy so I went on cramster and got this:

    Step 1


    Step 2





    Step 3





    so critical points are (0,0), (1,0), and (-1,0)

    Step 4




    Step 5


    at (0,0):
    at (1,0):
    at (-1,0):

    Therefore, f has a local minimum of 0 at (0,0) and saddle points at (1,0) and (-1,0).

    The answer is correct but the work makes no sense. When computing f_x, y^2e^{y^2-x^2} was completely neglected. f_x should have been:

    f_x = 2xe^{y^2-x^2}-2x^3e^{y^2-x^2} -2xy^2e^{y^2-x^2} = 0

    Same problem with f_y

    Is there something wrong with the solution or am I doing something wrong?

    Thanks in advance!
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  2. #2
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    Quote Originally Posted by lilaziz1 View Post
    Hi everyone. I have a question on one of my homework problems.

    Q. Find the local max./min. values and the saddle point(s) of the function  f(x,y) = (x^2+y^2)e^{y^2-x^2}

    So I started doing this problem and it was looking really messy so I went on cramster and got this:

    Step 1


    Step 2



    There's already a mistake here since you forgot to differentiate the second summand above. It should be:

    f_x=2xe^{y^2-x^2}-2x^3e^{y^2-x^2}-2xy^2e^{y^2-x^2}=0\Longrightarrow 2xe^{y^2-x^2}(1-x^2-y^2)=0 , and this

    alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative...

    Likewise f_y is wrong...you seem to have some problems with partial derivatives. Check this.

    Tonio







    Step 3





    so critical points are (0,0), (1,0), and (-1,0)

    Step 4




    Step 5


    at (0,0):
    at (1,0):
    at (-1,0):

    Therefore, f has a local minimum of 0 at (0,0) and saddle points at (1,0) and (-1,0).

    The answer is correct but the work makes no sense. When computing f_x, y^2e^{y^2-x^2} was completely neglected. f_x should have been:

    f_x = 2xe^{y^2-x^2}-2x^3e^{y^2-x^2} -2xy^2e^{y^2-x^2} = 0

    Same problem with f_y

    Is there something wrong with the solution or am I doing something wrong?

    Thanks in advance!
    .
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  3. #3
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    Ah okay. As I stated at the end of the thread, those mistakes are not mine. they are from cramster. I also corrected f_x in the first thread. Anyhow, my question is what do you mean by, "this alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative..." How does it tell you that?
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  4. #4
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    Quote Originally Posted by lilaziz1 View Post
    Ah okay. As I stated at the end of the thread, those mistakes are not mine. they are from cramster. I also corrected f_x in the first thread. Anyhow, my question is what do you mean by, "this alone already tells you that all the points on the unit circle around the origin are zeroes of this derivative..." How does it tell you that?

    Check the corrected f_x . This can be equal to zero if 1-x^2-y^2=0\Longleftrightarrow x^2+y^2=1 , and this last eq.

    describes the unit circle centered at the origin...

    Tonio
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  5. #5
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    Yes I got that but how does this tell you that all the derivative are zero around the cirlce?
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  6. #6
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    Quote Originally Posted by lilaziz1 View Post
    Yes I got that but how does this tell you that all the derivative are zero around the cirlce?

    I already answered: because if a point (x,y) belongs to THAT circle then inputting it in f_x we get zero...what else?!

    Tonio
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