# Thread: Stationary Points

1. ## Stationary Points

I am trying to find the stationary points for:

$f(x,y)=(x^2+y^2)e^{-(x+y)}$

So I have:

$f_x=0 \rightarrow -e^{-(x+y)}(y^2+x(x-2))=0$

with solutions $(0,0),(1,-1),(1,1),(2,0)$

$f_y=0 \rightarrow -e^{-(x+y)}(x^2+y(y-2))=0$

with solutions $(0,0),(1,1),(-1,1),(0,2)$

Do I just take the solutions that satisfy both $f_x$ and $f_y$ , i.e. $(0,0),(1,1)$ as the stationary points?

2. Originally Posted by billym
I am trying to find the stationary points for:

$f(x,y)=(x^2+y^2)e^{-(x+y)}$

So I have:

$f_x=0 \rightarrow -e^{-(x+y)}(y^2+x(x-2))=0$

with solutions $(0,0),(1,-1),(1,1),(2,0)$
HOW did you get solutions for both x and y from a single equation"
$-e^{-(x+y)}(y^2+ x^2- 2x)= 0$ is equivalent to $y^2+ x^2- 2x= 0$ and there are an infinite number of (x, y) pairs that satisfy this equation. (Surely, you are not assuming that the stationary points must have integer cooridinates?)

$f_y=0 \rightarrow -e^{-(x+y)}(x^2+y(y-2))=0$

with solutions $(0,0),(1,1),(-1,1),(0,2)$
Same comment as before.

Do I just take the solutions that satisfy both $f_x$ and $f_y$ , i.e. $(0,0),(1,1)$ as the stationary points?
If you subtract the two equations, the square terms cancel leaving 2x- 2y= 0 so y= x. Put y= x into either of the equations: $2x- x^2- y^2= 0$ becomes $2x- x^2- x^2= 2x- 2x^2= 2x(1- x)= 0$.
That givex x= 0 and x= 1 as roots. (Which are integer but I hope you understand you cannot assume that. That is an artifact of teachers and text writers focusing on concepts and not arithmetic.)

Now, use the other condition, y= x, to find the corresponding y. The stationary points are (0,0) and (1,1).

You really need to work on this. Your technique appears to be "try easy numbers and see if they work".