Thread: Critical points of a two-variable function.

1. Critical points of a two-variable function.

Hi!
I Can't seem to find the critical points of the function f(x,y) = x*y*ln(x^2 + y^2)

I did the df/dx = y*[ (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) ]
and the df/dy = x*[ (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) ]

I equalled them both to 0, but now I'm stuck because I don't know how to solve the system:

y*[ (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) ] = 0
x*[ (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) ] = 0

I have the solutions. There are 4 critical points and none of them is x=0 or y=0.
But I don't know how to get to the solution.

Tony

2. Re: Critical points of a two-variable function.

if x and y don't equal zero, then surely you can divide through by y in equation 1, x in equation 2. Then you have two things equal to 0... if you equate them what do you get?

This is just a starting step...

3. Re: Critical points of a two-variable function.

I tried that but it got me nowhere. The solutions are:
[ 1/(sqrt(2)*e); 1/(sqrt(2)*e) ], [ -1/(sqrt(2)*e); -1/(sqrt(2)*e) ],
[-sqrt(e)/2 ; sqrt(e)/2] and [ sqrt(e)/2 ; -sqrt(e)/2 ].

I don't see how I could get this from (2 x^2)/(x^2 + y^2) + ln(x^2 + y^2) = 0
and (2 y^2)/(x^2 + y^2) + ln(x^2 + y^2) = 0

I tried adding them, substracting, moving the 'ln' to the other side and raising them as power of 'e', but I can't seem to solve it.

4. Re: Critical points of a two-variable function.

Are your first solution supposed to be:

$x=y= \pm\frac{1}{\sqrt{2e}}$

Hopefully it is....

divide through by x and y in the corresponding equation, you have two things that equal 0 then, so they equal each other...

5. Re: Critical points of a two-variable function.

Yes they are the two first solutions.

I tried equalling them too but as I said I can't seem to move on. Please could you show me how you get these results?

6. Re: Critical points of a two-variable function.

your last 2 critical points are wrong i'm afraid, I have spent the last 10 minutes puzzling over those solutions, but now I have graphed it, and I have attained the correct solution.

The set of solutions is...
$\left(\frac{1}{\sqrt{2e}},\frac{1}{\sqrt{2e}} \right),\left(-\frac{1}{\sqrt{2e}},-\frac{1}{\sqrt{2e}} \right),\left(\frac{1}{\sqrt{2e}},-\frac{1}{\sqrt{2e}} \right),\left(-\frac{1}{\sqrt{2e}},\frac{1}{\sqrt{2e}} \right),$

I will post working shortly

7. Re: Critical points of a two-variable function.

So you are right with your first 2 equations. Dividing through by y in the first, x in the second gives:

$\log(x^2+y^2)+\frac{2x^2}{x^2+y^2}=0=\log(x^2+y^2) + \frac{2y^2}{x^2+y^2}$

Can you see where to go from here? if $a+b=a+c$ then what do we know about b and c?

8. Re: Critical points of a two-variable function.

Yes, I've gotten to the point that x^2/(x^2 + y^2) = y^2/(x^2+y^2) if that's what you mean, but I can't go further.
I don't know how to get the solutions [ 1/(sqrt(2)*e); 1/(sqrt(2)*e) ], etc. from it.

9. Re: Critical points of a two-variable function.

Never mind. I solved it. Your results were correct.
Thanks for the help!