Exploring Extreme Point (-1,-1) in Z=X(1+Y)^{\frac{1}{2}}+Y(1+X)^{\frac{1}{2}}

[Mike s]

Hello,
As far as I know, critical points are the point in which the gradient equals zero or not defined. In the following functionZ=X(1+Y)^{1/2}+Y(1+X)^{1/2}, the partial derivatives are not defined for all the points: (-1,Y) or (X,-1) in which X,Y are bigger or equal to -1.
Why is the point (-1,-1) the only extreme point?

\nabla Z=(Z_{X},Z_{Y})=(\dfrac{1}{2} Y(1+X)^{-{\frac{1}{2}}}+(1+Y)^{\frac{1}{2}},\dfrac{1}{2} X(1+Y)^{-{\frac{1}{2}}}+(1+X)^{\frac{1}{2}})Note: There is another extreme point, but I am having trouble only with this one.Thanks in advance,
Michael

 

[DryRun]

This problem is not about finding the total derivative, but the total differential.
dZ=\dfrac{1}{2} Y(1+X)^{-{\frac{1}{2}}}+(1+Y)^{\frac{1}{2}}+\dfrac{1}{2} X(1+Y)^{-{\frac{1}{2}}}+(1+X)^{\frac{1}{2}}
Since you're after the critical points, let dZ=0.
You can easily see that the critical points are (-1,-1) and the origin.

 

[Ray Vickson]

This problem is not about finding the total derivative, but the total differential.
dZ=\dfrac{1}{2} Y(1+X)^{-{\frac{1}{2}}}+(1+Y)^{\frac{1}{2}}+\dfrac{1}{2} X(1+Y)^{-{\frac{1}{2}}}+(1+X)^{\frac{1}{2}}
Since you're after the critical points, let dZ=0.
You can easily see that the critical points are (-1,-1) and the origin.

This is false. The function F(X,Y) = X(1+Y)^{1/2} + Y(1+X)^{1/2} has gradient (-\infty,-\infty) at (X,Y) = (-1,-1) (in the sense of limits as X,Y → -1 from above). The point (-1,-1) is a max but not a critical point, because of the essential constraints X,Y ≥ -1 that are needed to have a well-defined F(X,Y). In fact, the easiest solution would be to change variables to u = (1+X)^{1/2}, \text{ and } v = (1+Y)^{1/2}, and to look at f(u,v) = F(X,Y) in the region u,v ≥ 0. The point (0,0) is still not a critical point of f, but is the maximum of f; just look at directional derivatives of f at (0,0).

RGV

 

[Mike s]

This is false. The function F(X,Y) = X(1+Y)^{1/2} + Y(1+X)^{1/2} has gradient (-\infty,-\infty) at (X,Y) = (-1,-1) (in the sense of limits as X,Y → -1 from above). The point (-1,-1) is a max but not a critical point, because of the essential constraints X,Y ≥ -1 that are needed to have a well-defined F(X,Y). In fact, the easiest solution would be to change variables to u = (1+X)^{1/2}, \text{ and } v = (1+Y)^{1/2}, and to look at f(u,v) = F(X,Y) in the region u,v ≥ 0. The point (0,0) is still not a critical point of f, but is the maximum of f; just look at directional derivatives of f at (0,0).

RGV

Thanks!
Since we haven't learned yet how to find extreme points under constraints, I've tried to solve this in a different way:
Z\left(x,-1\right)=-{\left(1+x\right)}^{\frac{1}{2}}
Z_x\left(x,-1\right)=-\frac{1}{2}{\left(1+x\right)}^{-\frac{1}{2}}
Since Z_x is negative for all x\ge -1 , (-1,-1) is maximum.

Then I do the same for (-1,y)
Is this correct?

 

[Ray Vickson]

Thanks!
Since we haven't learned yet how to find extreme points under constraints, I've tried to solve this in a different way:
Z\left(x,-1\right)=-{\left(1+x\right)}^{\frac{1}{2}}
Z_x\left(x,-1\right)=-\frac{1}{2}{\left(1+x\right)}^{-\frac{1}{2}}
Since Z_x is negative for all x\ge -1 , (-1,-1) is maximum.

Then I do the same for (-1,y)
Is this correct?

You really do have a problem of optimization under constraints, whether or not you have taken this material. I did not assume you knew methods for optimization under constraints, but I did assume you know about directional derivatives, and that is why I mentioned them.

It is easiest to work with _finite_ derivatives, and that is why I suggested you change variables to u = sqrt(1+X), v = sqrt(1+Y). Have you done that yet? By doing that you can look at changes in F (or f) for _any_ direction that leads to feasible points, not just the two directions (1,0) and (0,1) that you have chosen. But, aside from that, your argument is OK.

RGV

 

[Mike s]

You really do have a problem of optimization under constraints, whether or not you have taken this material. I did not assume you knew methods for optimization under constraints, but I did assume you know about directional derivatives, and that is why I mentioned them.

It is easiest to work with _finite_ derivatives, and that is why I suggested you change variables to u = sqrt(1+X), v = sqrt(1+Y). Have you done that yet? By doing that you can look at changes in F (or f) for _any_ direction that leads to feasible points, not just the two directions (1,0) and (0,1) that you have chosen. But, aside from that, your argument is OK.

RGV

For all the other directions except (-1,y) and (x,-1), I can find the other critical points simply by solving the two equations:
Zx=0
Zy=0

right?

I specifically chose to check (-1,y) and (x,-1) because the gradient is not defined in them.

 

[Ray Vickson]

For all the other directions except (-1,y) and (x,-1), I can find the other critical points simply by solving the two equations:
Zx=0
Zy=0

right?
I specifically chose to check (-1,y) and (x,-1) because the gradient is not defined in them.

Your remarks about setting Zx = 0 and Zy = 0 are true, but irrelevant, since we are trying to determine how F(X,Y) behaves near (X,Y) = (-1,-1).

You have the beginning of a solution. You have shown that the function goes down if you go away from the point (-1,-1) along the +X axis or the +Y axis; that is, your point is a max for small movements at angles of 0 or 90 degrees. But, what happens if you go away from (-1,-1) at some other angle? There ARE many examples of functions in which a point is a max when going away from it along the x or y axes, but is a min when you go away at some angles between 0 and 90 degrees. You need to know whether or not your F(X,Y) behaves like that.

RGV