One saddle, one minimum, and one maximum?

Updated version:

Such a function does exist (see below)!

We want a function whose cross section looks like this:

http://www.wolframalpha.com/input/?i=plot+z+%3D+3x...

So, f(x,y) from this perspective ought to have a relative max to the left, a relative min to the right, and a saddle point at the origin. All that needs to be done is to explicitly create a "smooth" version of this.

Here's one function that does the job (obtained via trial and error over a few hours):

f(x, y) = 3y^5 - 5y^3 + x^2 y

http://www.wolframalpha.com/input/?i=stationary+po...

---------------------

Calculus formalities:

f_x = 2xy

f_y = 15y^4 - 15y^2 + x^2 = 15y^2(y^2 - 1) + x^2.

Setting these equal to 0, the first equation yields x = 0 or y = 0.

(i) x = 0 ==> 15y^2 (y^2 - 1) = 0 ==> y = -1, 0, 1.

(ii) y = 0 ==> x^2 = 0 ==> x = 0.

Hence, we have exactly three critical points (x, y) = (0, -1), (0, 0), (0, 1).

Now, we try to test these with the second derivative test.

f_xx = 2y, f_xy = 2x, f_yy = 60y^3 - 30y = 30y(2y^2 - 1).

==> D = (f_xx)(f_yy) - (f_xy)^2 = 60y^2 (2y^2 - 1) - 4x^2.

(i) At (0, 1), we have D(0, 1) > 0 and f_xx(0, 1) > 0.

==> Local minimum at (0, 1).

(ii) At (0, -1), we have D(0, -1) > 0 and f_xx(0, -1) < 0.

==> Local maximum at (0, -1).

(iii) At (0, 0), D = 0; so the second derivative test does not apply.

Rewrite f as follows:

f(x, y) = (3y^4 - 5y^2 + x^2) y.

Even keeping x = 0, letting y be a small positive or negative number, we see that f(0, y) can be made positive or negative appropriately. Since f(0, 0) = 0, this means that f has neither a local max nor a local min at (0, 0); hence it's a saddle point.

-------------------

I hope this helps!