Pope
Lv 7
Pope asked in Science & MathematicsMathematics · 10 years ago

One saddle, one minimum, and one maximum?

I received this from my nephew recently. I suppose we can forgo the sketching component.

Can a differentiable function f(x,y) of two variables have on the plane exactly three critical points, one saddle, one local minimum, and one local maximum? If no, explain why, if yes, give an example and sketch a graph.

Update:

This time the question is not actually my own. I quoted indirectly from an unknown source. To me it seemed clear that "on the plane" was a statement of the domain. The x-y plane is every real ordered pair (x, y).

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  • kb
    Lv 7
    10 years ago
    Favourite answer

    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!

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