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Math Help - Critical points of a function of 2 variables

  1. #1
    MHF Contributor arbolis's Avatar
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    Critical points of a function of 2 variables

    f(x,y)=x^4+y^4-4xy+1.
    Taking the partial derivatives of f and setting them to 0 gives y^9=y and x^9=x.
    Then my professor states that the critical points of f are (0,0), (1,1) and (-1,-1). Fine, I understand why, but I don't understand why the points (0,1), (1,0), (-1,0), (0,-1), (-1,1) and (1,-1) are not critical points.
    How do you know that when you take y=0, then you have to chose x=0 and not x=1 and x=-1?

    Because I know of an example ( f(x,y)= x^2+y^2+\frac{1}{x^2y^2}) which gives x=\pm \frac{1}{\sqrt{y\sqrt 2}} and y=\pm \frac{1}{\sqrt{x\sqrt 2}}. It still remains to solve for x and y (I see that x=\frac{1}{\sqrt {\sqrt 2}} and y=1 works for example. I've plotted the function and I saw 4 critical points (minimums), instead of 2 as my professor "would" find here.

    So what am I overlooking?
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  2. #2
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    Quote Originally Posted by arbolis View Post
    f(x,y)=x^4+y^4-4xy+1.
    Taking the partial derivatives of f and setting them to 0 gives y^9=y and x^9=x.
    Then my professor states that the critical points of f are (0,0), (1,1) and (-1,-1). Fine, I understand why, but I don't understand why the points (0,1), (1,0), (-1,0), (0,-1), (-1,1) and (1,-1) are not critical points.
    How do you know that when you take y=0, then you have to chose x=0 and not x=1 and x=-1?

    Because I know of an example ( f(x,y)= x^2+y^2+\frac{1}{x^2y^2}) which gives x=\pm \frac{1}{\sqrt{y\sqrt 2}} and y=\pm \frac{1}{\sqrt{x\sqrt 2}}. It still remains to solve for x and y (I see that x=\frac{1}{\sqrt {\sqrt 2}} and y=1 works for example. I've plotted the function and I saw 4 critical points (minimums), instead of 2 as my professor "would" find here.

    So what am I overlooking?
    When you calculate the first partials you obtain

     <br />
f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).<br />

    Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example (0,1) ).
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Danny View Post
    When you calculate the first partials you obtain

     <br />
f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).<br />

    Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example (0,1) ).
    Thank you very much for the quick response!
    Ok, so I have to test each point? It does not bothers me by the way.
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