Classifying a stationary point of a function of two variables when the discriminant of that function is zero

Question

The given problem is:

Find any critical points of the function $f(x, y) = xy^2$ and classify them as either relative minima, relative maxima, or saddle points. Also, determine whether $f$ has an absolute maximum or absolute minimum.

By immediate observation, $f$ is differentiable everywhere. Also, the range of $f$ is $f(x, y) \in (-\infty, \infty)$, so there is no absolute maximum nor minimum.

$$ f_x(x, y) = 0 \quad f_y(x, y) = 0 $$ $$ y^2 = 0 \quad 2xy = 0 $$

From the two equations, $y$ must be $0$ but $x$ can be any value. Therefore, the critical (stationary) points are of the form $(x, 0)$.

Using the second derivative test, I will define the discriminant.

$$\begin{align} D(x, y) & = f_{xx}(x, y)f_{yy}(x, y) - \left[f_{xy}(x, y)\right]^2 \\ & = (0)(2x) - (2y)^2 \\ & = -4y^2 \end{align}$$

At $(x, 0)$, $D(x, 0) = -4(0)^2 = 0$, so the test yields no conclusion.

Is there a way around this?

Answer 1

This is more of a far-too-long hint than an answer, but it won't fit in a comment.

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In single-variable calculus, you had a fall back option when the second derivative test didn't tell you whether a point was a relative maximum or minimum: the first derivative test. This is where we look at points just to the left or right of our critical point and investigate the sign of the derivative, and then make some conclusion about whether our point is a minimum, a maximum, or neither.

In the two-variable case, we can do something similar. Again we want to investigate what happens to points near our critical point. This time we can't look just to the left or right of our point, but instead must look at a disk's worth of points near our critical point.

Suppose our critical point is $(x_0,y_0)$. If $(x_0,y_0)$ gives a relative minimum for our function, then we should have \begin{equation} f(x_0,y_0)\leq f(x,y) \end{equation} for points $(x,y)$ near $(x_0,y_0)$. That is, perturbing $(x_0,y_0)$ just a little should cause $f$ to increase (or at least not decrease). Similarly, these perturbations should have a non-increasing effect on $f$ when applied to a relative maximum. If $(x_0,y_0)$ is neither a maximum nor a minimum, then we should be able to find some points near $(x_0,y_0)$ with the property that \begin{equation} f(x,y)\leq f(x_0,y_0) \end{equation} and some so that \begin{equation} f(x,y)\geq f(x_0,y_0). \end{equation} At this point I would encourage you to take your critical point $(x,0)$ and perturb it. Does $f$ always increase (or decrease)? Is your point a minimum in one direction but a maximum in the other (and thus a saddle point)? Your answers to these questions will depend on $x$. In any case, your critical point is degenerate (since the Hessian is zero), so saying things like "relative minimum" or "relative maximum" becomes a bit weirder. (In particular, perturbing $x$ without changing $y$ does nothing, since $y=0$.)