extremal value theorem.

[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?

[dextercioby]

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

Daniel.

[ArnfinnS]

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

[dextercioby]

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

Daniel.

[ArnfinnS]

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

[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.

[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?

[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.