Find Max/Min for Function: 2nd Derivative Test

In summary, the conversation discusses finding the maximum and minimum of a given function and determining if a point is a maximum or minimum using the second derivative test. The function is also clarified to be an upside cone, and the English phrase for a critical point is mentioned. The conversation concludes with the determination that (0,0) is the maximum point of the function.
  • #1
asi123
258
0

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

 

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  • #2
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
The function is [itex]z= 1- \sqrt{x^2+ y^2}[/itex]

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

As for whether that is a max or min, just note that [itex]\sqrt{x^2+y^2}[/itex] 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
HallsofIvy said:
The function is [itex]z= 1- \sqrt{x^2+ y^2}[/itex]

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

As for whether that is a max or min, just note that [itex]\sqrt{x^2+y^2}[/itex] 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 :smile:
 
  • #5
The derivitives are not zero at (0,0). Try plugging that point into the derivitive expressions and see what you get.
 
  • #6
Well, okay. Fortunately what that means is that the derivatives do not exist which is also a criterion for a critical point!
 
  • #7
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.
 

1. What is the 2nd Derivative Test?

The 2nd Derivative Test is a method used in calculus to determine the maximum and minimum values of a function. It involves taking the second derivative of a function and analyzing its concavity to determine the nature of the critical points.

2. Why is the 2nd Derivative Test important?

The 2nd Derivative Test is important because it can help us determine whether a critical point is a maximum, minimum, or a point of inflection. This information is crucial in understanding the behavior of a function and making predictions about its graph.

3. How do you use the 2nd Derivative Test?

To use the 2nd Derivative Test, follow these steps:

  1. Find the critical points of the function by setting the first derivative equal to zero.
  2. Take the second derivative of the function.
  3. Plug in the critical points into the second derivative.
  4. If the result is positive, the critical point is a minimum. If the result is negative, the critical point is a maximum. If the result is zero, the test is inconclusive.

4. When is the 2nd Derivative Test not applicable?

The 2nd Derivative Test is not applicable when the second derivative is undefined or when the function is not continuous at the critical point.

5. Can the 2nd Derivative Test be used to find absolute maximum/minimum values?

No, the 2nd Derivative Test can only determine the maximum and minimum values of a function at its critical points. To find absolute maximum/minimum values, the endpoints of the interval must also be considered.

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