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Thread: 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

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

    Because I know of an example ( $\displaystyle f(x,y)= x^2+y^2+\frac{1}{x^2y^2}$) which gives $\displaystyle x=\pm \frac{1}{\sqrt{y\sqrt 2}}$ and $\displaystyle y=\pm \frac{1}{\sqrt{x\sqrt 2}}$. It still remains to solve for $\displaystyle x$ and $\displaystyle y$ (I see that $\displaystyle x=\frac{1}{\sqrt {\sqrt 2}}$ and $\displaystyle y=1$ 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
    $\displaystyle f(x,y)=x^4+y^4-4xy+1$.
    Taking the partial derivatives of $\displaystyle f$ and setting them to $\displaystyle 0$ gives $\displaystyle y^9=y$ and $\displaystyle x^9=x$.
    Then my professor states that the critical points of $\displaystyle f$ are $\displaystyle (0,0)$, $\displaystyle (1,1)$ and $\displaystyle (-1,-1)$. Fine, I understand why, but I don't understand why the points $\displaystyle (0,1)$, $\displaystyle (1,0)$, $\displaystyle (-1,0)$, $\displaystyle (0,-1)$, $\displaystyle (-1,1)$ and $\displaystyle (1,-1)$ are not critical points.
    How do you know that when you take $\displaystyle y=0$, then you have to chose $\displaystyle x=0$ and not $\displaystyle x=1$ and $\displaystyle x=-1$?

    Because I know of an example ( $\displaystyle f(x,y)= x^2+y^2+\frac{1}{x^2y^2}$) which gives $\displaystyle x=\pm \frac{1}{\sqrt{y\sqrt 2}}$ and $\displaystyle y=\pm \frac{1}{\sqrt{x\sqrt 2}}$. It still remains to solve for $\displaystyle x$ and $\displaystyle y$ (I see that $\displaystyle x=\frac{1}{\sqrt {\sqrt 2}}$ and $\displaystyle y=1$ 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

    $\displaystyle
    f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).
    $

    Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example $\displaystyle (0,1)$ ).
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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

    $\displaystyle
    f_x = 4(x^3-y),\;\;\; f_y = 4(y^3-x).
    $

    Critical points are points where both vanish. As you can see many of the points you mention don't satisfy these (for example $\displaystyle (0,1)$ ).
    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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