Extrema of function, two variables

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Homework Help Overview

The problem involves finding and classifying the extrema of the function f = x³ - 3xy² + y³, which is a multivariable calculus topic focusing on critical points and the second derivative test.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss finding partial derivatives and setting them to zero to identify critical points, with some questioning the validity of the critical point (0,0). There are attempts to apply the second derivative test and inquiries about its inconclusiveness. Participants also explore evaluating the function along specific paths to understand the nature of the critical point.

Discussion Status

The discussion is ongoing, with participants providing insights into the implications of the second derivative test being inconclusive and suggesting alternative methods to analyze the critical point. There is a recognition of the need to evaluate the function along various paths to draw conclusions about the nature of the critical point.

Contextual Notes

Participants express uncertainty about the definitions of local minima and maxima, and there is a focus on the behavior of the function around the critical point (0,0) to determine its classification.

philnow
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Homework Statement



Find and classify the extrema of f = x3 - 3xy2 + y3

The Attempt at a Solution



I find partial deriv. with respect to x is 3x2 - 3y2
and the partial deriv. with respect to y is -6xy + 3y2

I set these to zero, and for my critical point I get (0,0). This can't be right... where am I going wrong?
 
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fx = 3(x2 - y2)
fy = 3y(-2x + y)

fx = 0 ==> x = +/-y
fy = 0 ==> y = 0 or y = 2x

For both equations to be satisfied, x = y = 0

Now you need to figure out whether this is a local minimum, local maximum, or saddle point.
 
Ok, so I need to find D. Fxx = 6x = 0 at (0,0). Same for Fyy. Fxy is also 0 so D is zero. What does this tell me about the critical point (0,0)?
 
Anyone?
 
It doesn't tell you anything, since the test is inconclusive for D = 0. You're sure you have posted the problem exactly the way it is in your textbook?
 
It doesn't tell you anything about the critical point. The second derivative test failed. At this point the easiest thing to do is check some paths around the origin. Look at f(x,0). Is that a min, a max or saddle as a function of x?
 
Dick said:
It doesn't tell you anything about the critical point. The second derivative test failed. At this point the easiest thing to do is check some paths around the origin. Look at f(x,0). Is that a min, a max or saddle as a function of x?

f(x,0) = x3 = 0
f(0,y) = y3 = 0

along y=x, approaching (0,0) the function is = x3 - 3x3 + x3 = -1x3 = 0.

So it is approaching 0 from these paths, but what does that mean? :S
 
Approaching zero isn't the point. The point is that f(0,0)=0 but f(x,0)<0 if x<0 and f(x,0)>0 for x>0. Max, min or saddle?
 
I guess it would be saddle, but I'm iffy on this.
 
  • #10
philnow said:
I guess it would be saddle, but I'm iffy on this.

If f(0,0)=0 and there are negative values of f and positive values of f in any region around (0,0), it can't be a min or a max, can it? What are the basic definitions of min and max?
 
  • #11
That makes it more clear, thanks!
 

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