# Extrema test in Multivariable calculus

I'm trying to find extrema for a two variable problem:

$$g(x,y) = x^3+(x+y)^2.$$

After finding the critical point $(0, 0)$ using the gradient function, I found the Hessian matrix which has determinant zero and hence the test is inconclusive.

How shall I find whether the critical point is local or isolated or global extrema or a saddle point?

If we look at the 3D-plot, we have:

You are correct, using the typical tests (Find all critical points of $f(x,y) = x^3 - 12xy + 8y^3$ and state maximum, minimum, or saddle points.), the results are inconclusive and you do not have a local or global minimum or maximum at the single critical point $(0,0)$.

• without plotting the function graphically, how can I conclude that (0,0) is a saddle point by mathematical analysis? – gopi_rm Nov 15 '13 at 10:03

It should be mentioned just how the critical point is determined, as it will be relevant to what follows. Setting the first partial derivatives of $\ g(x,y) \ = \ x^3 \ + \ (x+y)^2 \$ equal to zero yields

$$g_x \ = \ 3x^2 \ + \ 2 \ (x+y) \cdot 1 \ = \ 0 \quad , \quad g_y \ = \ 2 \ (x+y) \cdot 1 \ = \ 0$$

$$\Rightarrow \ \ x \ + \ y \ = \ 0 \ \ \Rightarrow \ \ 3x^2 \ = \ 0 \ \ \Rightarrow \ \ x \ = \ 0 \ , \ y \ = \ 0 \ \ .$$

So $\ (0, \ 0) \$ is the only critical point on the surface $\ z \ = \ g(x,y) \$ , but there is something special about the line $\ y \ = \ -x \$ as well.

One way to evaluate the behavior of the function without graphing it is to take "cross-sections" on various planes containing the $\ z-$ axis. In the $\ yz-$ plane $\ (x = 0) \$ , our function becomes $\ z \ = \ f(0, \ y) \ = \ y^2 \$ , which we know is "concave upward" about the origin. However, in the plane $\ y \ = \ -x \$ , the "cross-sectional" function is $\ z \ = \ f(x, \ -x) \ = \ x^3 \$ , which has a change in concavity about the origin (point of inflection).

A point on a "surface" can only be a local extremum if the direction of concavity about that point is the same in every direction ("concave upward" in all directions for a local minimum and "downward" for a local maximum). This is not the case at $\ (0, \ 0, \ 0) \$ for our function, so it is a saddle point.

[Side note -- If we take other cross-sections through the surface on the planes $\ y \ = \ 0 \$ (the $\ xz-$ plane) and $\ y \ = \ x \$ , we obtain the functions $\ x^3 \ + \ x^2 \$ and $\ x^3 \ + \ 4x^2 \$ , respectively. These have zero slope at the origin, but positive second derivatives. In these planes, $\ (0, \ 0) \$ is a local minimum, and we find that there is a nearby local maximum for small negative values of $\ x \$ . This appears to be exploring the developing "ridge" that appears toward the front of Amzoti's graph-cube. (Note that the "vertical" scale is highly compressed there.)]