Extrema of function, two variables

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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