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# Finding extreme point in a multivariable function when the gradient is not defined

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1. Finding extreme point in a multivariable function when the gradient is not defined
Hello,
As far as I know, critical points are the point in which the gradient equals zero or not defined. In the following function, the gradient is not defined for all or . How do I prove that (-1,-1) is an extreme point (maximum)?
Note: There is another extreme point, but I am having trouble only with this one.

Michael
2. Re: Finding extreme point in a multivariable function when the gradient is not define
What you could try is changing coordinates: Let u=1+X and v=1+Y. This gives: and you're interested in what happens when u and v get close to (0,0).

You can show that for u and v sufficiently close to (0,0), Z is negative and hence less than the value of Z for u=v=0.
3. Re: Finding extreme point in a multivariable function when the gradient is not define
(Original post by ttoby)
What you could try is changing coordinates: Let u=1+X and v=1+Y. This gives: and you're interested in what happens when u and v get close to (0,0).

You can show that for u and v sufficiently close to (0,0), Z is negative and hence less than the value of Z for u=v=0.
I understand what your are saying, however the 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?

Last edited by msokol; 06-05-2012 at 12:41.
4. Re: Finding extreme point in a multivariable function when the gradient is not define
god!! first define the gradient... blates

jk, i do media studies
5. Re: Finding extreme point in a multivariable function when the gradient is not define
(Original post by msokol)
I understand what your are saying, however the 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?
That's another question. Because the function is differentiable on , you can take partial derivatives there to check for maximum/minimum points.

Now let's take points of the form (X,-1) for X non-zero. here . From there, you could do a contradiction arguement. Suppose there was a neighbourhood around a (X,-1) point such that Z of (that point) is greater than Z of (any other point in that neighbourhood). Then there exists a ball of radius epsilon around (X,-1) that is still in that neighbourhood, however Z of is greater than Z of (X,-1) giving a contradiction (this is easier to explain using function notation but hopefully you understand where I'm going here).

Similarly, there are no minimum points along that line, and also similarly when we look in the Y direction instead.

There might also be another approach you can use when only some of the partial derivatives are defined, but I don't know it so I'd probably use this method.
6. Re: Finding extreme point in a multivariable function when the gradient is not define
Got it, thanks a lot!

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Last updated: May 6, 2012
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