Show Global Min. of f(x,y): x>= 0.1, y>= 0.1

[ArnfinnS]

hi. I've 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))

how can i do this? Can anyone help me?

i got that the function has local minimum in the point x=y= 1/(2^(1/4))
if this is a local minimum , can i then directly say that it has to be a global minimum for x>0 and y>0 ?

 

[anathema]

Hi there,

Yes, you are on the right track with finding the partial derivatives of the function. To use the extremal value theorem, we need to show that the function has a critical point (where both partial derivatives are equal to 0) and that it is a minimum point.

To find the critical point, we can set the partial derivatives equal to 0 and solve for x and y. We get x=y=1/(2^(1/4)) as you mentioned. However, we also need to check that this point satisfies the given conditions of x>= 0.1 and y>= 0.1. Since 1/(2^(1/4)) is approximately 0.84, it does satisfy the given conditions.

Now, to show that this point is a minimum, we can use the second derivative test. Taking the second derivatives, we get:

f_xx = 2 + 2/(x^3*y)
f_yy = 2 + 2/(y^3*x)
f_xy = -1/(x^2*y^2)

Plugging in x=y=1/(2^(1/4)), we get:

f_xx = 4 + 2/(1/8) = 4 + 16 = 20
f_yy = 4 + 2/(1/8) = 4 + 16 = 20
f_xy = -1/(1/16) = -16

Using the discriminant (D = f_xx*f_yy - f_xy^2), we get D = 20*20 - (-16)^2 = 400 - 256 = 144. Since D > 0 and f_xx > 0, we can conclude that this point is a local minimum.

Now, to show that it is a global minimum for x>0 and y>0, we can use the hint given and set u=1/x and v=1/y. This means that f(x,y) = g(u,v) = u^2 + v^2 + uv. We can find the partial derivatives of g(u,v) and set them equal to 0 to find the critical point. We get u=v=1/2.

Using the inverse of our substitution, we get x=y=2. This point satisfies the given conditions of x>0 and y>0, and since it is the only critical point found, it

 

[thepatientmental]

To show the global minimum of a function, we need to find the point where the function has the lowest value in the given domain. In this case, we are given the domain x>= 0.1 and y>= 0.1. To use the "extremal value theorem", we need to find the critical points of the function within this domain.

Using the partial derivatives you have found, we can set them equal to 0 to find the critical points:

f_x = 2x - (1/(x^2*y)) = 0
f_y = 2y -(x/(y^2*x)) = 0

Solving these equations, we get x = y = 1/√2, which is within the given domain. This means that the point (1/√2, 1/√2) is a critical point and could potentially be the global minimum.

To confirm this, we can use the second derivative test to check if this point is a minimum or maximum. Taking the second partial derivatives, we get:

f_xx = 2 + (2/(x^3*y))
f_xy = -(1/(x^2*y^2))
f_yy = 2 + (2/(y^3*x))

Evaluating these at x=y=1/√2, we get f_xx = f_yy = 6 and f_xy = -4. This means that the point (1/√2, 1/√2) is a local minimum.

Now, to show that this is also a global minimum, we need to show that there are no other points within the given domain that have a lower value than this point. Since the function is continuous and differentiable within the given domain, we can use the "extremal value theorem" to conclude that the global minimum must occur at either the critical point we found or at the boundaries of the domain.

To check the boundaries, we can use the hint given and set u=1/x and v=1/y. This will transform the function into g(u,v) = u^2 + v^2 + u*v. We can then use the same process as before to find the critical points of this new function.

Setting the partial derivatives of g(u,v) equal to 0, we get u = v = 1/√2, which is equivalent to x=y