# Thread: Finding extreme point in a multivariable function

1. ## Finding extreme point in a multivariable function

Hello,
As far as I know, critical points are the point in which the gradient equals zero or not defined. In the following function$\displaystyle Z=X(1+Y)^{1/2}+Y(1+X)^{1/2}$, the gradient is not defined for all $\displaystyle X\leq-1$ or $\displaystyle Y\leq-1$. How do I prove that only (-1,-1) is an extreme point (maximum)?

Michael

2. ## Re: Finding extreme point in a multivariable function

Are you sure (-1, -1) is the maximum of z? For a function that is continuous on a closed interval we know that the function will achieve its extrema at some f(a) and f(b) with a and b contained in the closed interval, and f'(a)=f'(b)=0. How can you get the interval on which z is continuous to be closed?

If you are still struggling, a good first step would be to reconsider your statement that z(-1, -1) does not exist.

3. ## Re: Finding extreme point in a multivariable function

Originally Posted by letepsilonbenegative
Are you sure (-1, -1) is the maximum of z? For a function that is continuous on a closed interval we know that the function will achieve its extrema at some f(a) and f(b) with a and b contained in the closed interval, and f'(a)=f'(b)=0. How can you get the interval on which z is continuous to be closed?

If you are still struggling, a good first step would be to reconsider your statement that z(-1, -1) does not exist.
I did not say that z(-1,-1) does bot exist, it does. In addition, it is a maximum point because z(-1,-1)=0, and the points in the small neighborhood are either negative or not defined.

4. ## Re: Finding extreme point in a multivariable function

The function itself is not defined for x< -1 or y< -1 so there can't be a max or min there.