Find Max/Min for Function: 2nd Derivative Test

[asi123]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[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.

 

[HallsofIvy]

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".

 

[asi123]

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".

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

 

[Redbelly98]

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

 

[HallsofIvy]

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

 

[Redbelly98]

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.