# Maximum and Minimum

1. Aug 8, 2008

### asi123

1. The problem statement, all variables and given/known data

Hey.
I need to find maximum and minimum for this function, as you can see, I found (0,0) as a "suspicious point" (I don't know how it's called in English ).
My question is, how can I determine if it's Max or Min (if it's possible)?
I know the test of the second derivative, but how should I get it? I mean should I use the definition of the derivative? because I think there is a problem around the point (0,0).
Any idea guys?

2. Relevant equations

3. The attempt at a solution

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2. Aug 8, 2008

### Focus

It might help if you wrote out the question. You should differentiate the function and then plug in your value, unless the value is at a boundary, it should make the derivative zero. That shows its a candidate for an extrema. Then check the double derivative to see if it is min or max, then compare the values of the function at the point (0,0) with your boundary values.

3. Aug 8, 2008

### HallsofIvy

Staff Emeritus
The function is $z= 1- \sqrt{x^2+ y^2}$

asi123 gives the derivatives, correctly, as $-\frac{x}{\sqrt{x^2+y^2}}$ and $-\frac{y}{\sqrt{x^2+y^2}}$ and notes that they are 0 only at (0,0),

As for whether that is a max or min, just note that $\sqrt{x^2+y^2}$ is never negative.

By the way, I think the English phrase you are looking for is "critical point". Here "critical" is in the sense of "important".

4. Aug 8, 2008

### asi123

Yeah, I only notice later on that this is actually an upside cone.
10x, and 10x for the new word

5. Aug 8, 2008

### Redbelly98

Staff Emeritus
The derivitives are not zero at (0,0). Try plugging that point into the derivitive expressions and see what you get.

6. Aug 8, 2008

### HallsofIvy

Staff Emeritus
Well, okay. Fortunately what that means is that the derivatives do not exist which is also a criterion for a critical point!

7. Aug 8, 2008

### Redbelly98

Staff Emeritus
True enough. And the op figured out the shape of the function, which makes it pretty clear that (0,0) is the maximum after all.