# Extremal value theorem.

1. Apr 2, 2005

### ArnfinnS

hi. ive got another problem.

i have the function :
f(x,y) = x^2 + y^2 +(xy)^-1

iam supposed to use the "extremalvalue theorem" to show that this function have a global minimum on the area x>= 0.1 ,y>= 0.1

and i also need to argue if the same thing is satisfied for x>0 and y>0.

for this one , there is given a hint : to set u=1/x and v=1/y..and then iam supposed to look at the function f(x,y) = g(u,v)

First of all , I tried to find the partials which i think is :
f_x = 2x - (1/(x^2*y))
f_y = 2y -(x/(y^2*x))

but i cant see which point those equals 0.
how can i do this? Can anyone help me?

2. Apr 2, 2005

### dextercioby

Redo the partial derivatives (the second) and set them to zero...

Daniel.

3. Apr 2, 2005

### ArnfinnS

yes , but how can i find out which points these partial derivatives is zero?
i really cant see which point it is...

4. Apr 2, 2005

### dextercioby

You'll get a system of 2 equations with 2 unknowns...That system needs to be solved.

Daniel.

5. Apr 2, 2005

### ArnfinnS

but how can i prove that it has a global minimum for X>0 and Y>0 ?
can someone help me?

6. Apr 2, 2005

### dextercioby

It's pretty weird,the system has real solutions only for (0,0),where the first derivatives are both 0.On the other hand,the origin is not in the function's domain...

There's something weird.

Daniel.

7. Apr 2, 2005

### ArnfinnS

i found that the system has solution at x=y=1/( 2^(0.25))
and x=y=-1/(2^(0.25))
but i also proved that this is a local mininum.
and those are the only critical points.
how can i prove that its global minimum?

8. Apr 2, 2005

### Data

It's differentiable whenever $x, y \neq 0$ so the only local maxima and minima are at critical points. You just need to check the boundary....

As either $x$ or $y$ approaches $0$ in the positive sense, the function goes to infinity.

Last edited: Apr 2, 2005