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I would like some help finding the critical points of $f(x,y)=x^2+xy+y^2+\frac{1}{x}+\frac{1}{y}$. I tried solving $f_x=0, f_y=0$ (where $f_x, f_y$ are the partial derivatives) but the resulting equation is very complex. The exercise has a hint: think of $f_x-f_y$ and $f_x+f_y$. However, I can't see where to use it.

Thanks!

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$f_x$ and $f_y$ are both zero if and only if $f_x+f_y$ and $f_x-f_y$ are both zero. –  Rahul Apr 11 '12 at 2:06
    
Yeah, I know this much. I haven't figured out how to get anything out of the resulting set of equations, however. Maybe I'm missing something really simple, but I just can't see it. –  ro44 Apr 11 '12 at 2:09
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3 Answers 3

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After doing some computations I found the following (lets hope I didn't make any mistakes). You need to solve the equations

$$f_x = 2x + y - \frac{1}{x^2} = 0 \quad f_y = 2y + x -\frac{1}{y^2} = 0$$

therefore after subtracting and adding them as in the hint we get

$$\begin{align} f_x - f_y &= x - y - \frac{1}{x^2} + \frac{1}{y^2} = 0 \\ f_x + f_y &= 3x + 3y -\frac{1}{x^2} - \frac{1}{y^2} = 0 \end{align} $$

but you can factor them a little bit to get

$$ \begin{align} f_x - f_y &= x - y + \frac{x^2 - y^2}{x^2 y^2} = (x - y) \left ( 1 + \frac{x+ y}{x^2 y^2}\right ) = 0\\ f_x + f_y &= 3(x + y) -\frac{x^2 + y^2}{x^2 y^2} = 0 \end{align} $$

Now from the first equation you get two conditions, either $x = y$ or $x+y = -x^2 y^2$.

If $x = y$ you can go back to your first equation for $f_x$ and substitute to get

$$2x + x - \frac{1}{x^2} = 0 \implies 3x = \frac{1}{x^2} \implies x = \frac{1}{\sqrt[3]{3}}$$

and then you get the critical point $\left ( \dfrac{1}{\sqrt[3]{3}}, \dfrac{1}{\sqrt[3]{3}} \right )$

Now if instead $x + y = -x^2 y^2$ then if you substitute into the equation $f_x + f_y = 0$ we get the following

$$ 3(-x^2 y^2) - \frac{x^2 + y^2}{x^2 y^2} = 0 \implies 3x^4 y^4 + x^2 + y^2 = 0 \implies x = y = 0 $$

But this is actually one of the points where the partial derivatives or even your original function are not defined.

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Thank you! This is a good answer. –  ro44 Apr 11 '12 at 2:30
    
@ro44 I'm glad I could help. –  Adrián Barquero Apr 11 '12 at 2:38
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Here $f_{x}=2x+y-\frac{1}{x^{2}}$ and $f_{y}=2y+x-\frac{1}{y^{2}}$ For critical point $$f_{x}=0,\ f_{y}=0$$ $$2x+y-\frac{1}{x^{2}}=0\ ,\ 2y+x-\frac{1}{y^{2}}=0$$ $$2x^{3}+x^{2}y-1=0\ ,\ 2y^{3}+xy^{2}-1=0$$ Substracting this two equations we get, $$ 2(x^{3}-y^{3})+xy(x-y)=0$$ $$ 2(x-y)(x^{2}+xy+y^{2})+xy(x-y)=0$$ $$(x-y)(2x^{2}+2xy+2y^{2}+xy)=0$$ $$x-y=0\ \Rightarrow x=y$$ Put $x=y$ in $2x^{2}+2xy+2y^{2}+xy=0$,we get, $$ 2x^{2}+2x^{2}+2x^{2}+x^{2}=0\ \Rightarrow x=0 $$ Also $y=0$ \ $\therefore (0,0)$ is a critical point.

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However $(0,0)$ is not in the domain of the function. You don't need both factors to be $0$ for the product to be $0$, just one. –  Ross Millikan Apr 11 '12 at 2:04
    
You can see $(0,0)$ is a critical point much faster by showing you can't plug it in to either $f_x$ or $f_y$. Also, why are you plugging $x=y$ into $2x^2+2xy+2y^2+xy$? How do you know $2x^2 + 2xy + y^2 + xy (= 2x^2 + 3xy + y^2)$ doesn't have any other solutions? –  Jason DeVito Apr 11 '12 at 2:04
    
@ro44: When he subtracts the two equations he gets $f_x-f_y$ and the result factors as shown. You can show that the second factor is never $0$ except at the excluded $(0,0)$ Then you can take $x=y$ and follow Changwei Zhou to get to his answer. –  Ross Millikan Apr 11 '12 at 2:17
    
@RossMillikan: Yeah, but I get the feeling that's not how the hint is supposed to be used. I'm aware of how to show that $(0,0)$ is the only possible solution but in this course we haven't been taught (or are expected to use) stuff like that, so it's not very satisfying. –  ro44 Apr 11 '12 at 2:28
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You should solve $f_{x}=0,f_{y}=0$ and we have $2x+y-1/x^{2}=0,2y+x-1/y^{2}=0$. They are just $2x^{3}+x^{2}y-1=0, 2y^{3}+y^{2}x-1=0$. Consider symmetric solutions $(a,a)$ we have $$2a^{3}+a^{3}-1=0\leftrightarrow a^{3}=\frac{1}{3}\leftrightarrow x=y=3^{-\frac{1}{3}}$$

For the non-symmetric case, Solve $2x^{2}+2y^{2}+3xy=0$ implies no real solutions other than $x=y=0$, as the determinant is $9y^{2}-4*2*2y^{2}=-7y^{2}<0$ for $y\not=0$.

This ignores the original point $(0,0)$, which is not defined.

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You haven't justified the claim that all solutions will be symmetric. –  Ross Millikan Apr 11 '12 at 2:07
    
@RossMillikan: thanks, now justified. –  Kerry Apr 11 '12 at 2:15
    
@ChangweiZhou: Thanks for your answer. It's satisfying, but I'd still like to know how the hint about $f_x +/- f_y$ is supposed to be used. –  ro44 Apr 11 '12 at 2:22
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