Suppose a surface given by $f(x,y)$ has a local maximum at $(x_0,y_0,z_0)$; geometrically, this point on the surface looks like the top of a hill. If we look at the cross-section in the plane $y=y_0$, we will see a local maximum on the curve at $(x_0,z_0)$, and we know from single-variable calculus that ${\partial z\over\partial x}=0$ at this point. Likewise, in the plane $x=x_0$, ${\partial z\over\partial y}=0$. So if there is a local maximum at $(x_0,y_0,z_0)$, both partial derivatives at the point must be zero, and likewise for a local minimum. Thus, to find local maximum and minimum points, we need only consider those points at which both partial derivatives are 0. As in the single-variable case, it is possible for the derivatives to be 0 at a point that is neither a maximum or a minimum, so we need to test these points further.

You will recall that in the single variable case, we examined three methods to identify maximum and minimum points; the most useful is the second derivative test, though it does not always work. For functions of two variables there is also a second derivative test; again it is by far the most useful test, though it doesn't always work.

Theorem 14.7.1 Suppose that the second partial derivatives of $f(x,y)$ are continuous near $(x_0,y_0)$, and $f_x(x_0,y_0)=f_y(x_0,y_0)=0$. We denote by $D$ the discriminant $D(x_0,y_0)=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}(x_0,y_0)^2$. If $D>0$ and $f_{xx}(x_0,y_0)< 0$ there is a local maximum at $(x_0,y_0)$; if $D>0$ and $f_{xx}(x_0,y_0)>0$ there is a local minimum at $(x_0,y_0)$; if $D< 0$ there is neither a maximum nor a minimum at $(x_0,y_0)$; if $D=0$, the test fails.

Example 14.7.2 Verify that $f(x,y)=x^2+y^2$ has a minimum at $(0,0)$.

First, we compute all the needed derivatives: $$f_x=2x \qquad f_y=2y \qquad f_{xx}=2 \qquad f_{yy}=2 \qquad f_{xy}=0.$$ The derivatives $f_x$ and $f_y$ are zero only at $(0,0)$. Applying the second derivative test there: $$D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 2\cdot2-0=4>0$$ and $$f_{xx}(0,0)=2>0,$$ so there is a local minimum at $(0,0)$, and there are no other possibilities.

Example 14.7.3 Find all local maxima and minima for $f(x,y)=x^2-y^2$.

The derivatives: $$f_x=2x \qquad f_y=-2y \qquad f_{xx}=2 \qquad f_{yy}=-2 \qquad f_{xy}=0.$$ Again there is a single critical point, at $(0,0)$, and $$D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 2\cdot-2-0=-4< 0,$$ so there is neither a maximum nor minimum there, and so there are no local maxima or minima. The surface is shown in figure 14.7.1.