# Homework Help: Extrema of function, two variables

1. Oct 25, 2009

### philnow

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

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

3. 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?

2. Oct 25, 2009

### Staff: Mentor

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.

3. Oct 25, 2009

### philnow

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)?

4. Oct 25, 2009

### philnow

Anyone?

5. Oct 25, 2009

### Staff: Mentor

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?

6. Oct 25, 2009

### Dick

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?

7. Oct 25, 2009

### philnow

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

8. Oct 25, 2009

### Dick

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?

9. Oct 25, 2009

### philnow

I guess it would be saddle, but I'm iffy on this.

10. Oct 25, 2009

### Dick

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. Oct 25, 2009

### philnow

That makes it more clear, thanks!