
# 14.7: Maxima and minima

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: The Discriminant

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 $$(0,0)$$.

Solution

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:

$D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 2\cdot2-0=4>0,$

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 $$f(x,y)=x^2-y^2$$.

Solution

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.

Figure 14.7.1. A saddle point, neither a maximum nor a minimum.

Example 14.7.4

Find all local maxima and minima for $$f(x,y)=x^4+y^4$$.

Solution

The derivatives:

$f_x=4x^3 \qquad f_y=4y^3 \qquad f_{xx}=12x^2 \qquad f_{yy}=12y^2 \qquad f_{xy}=0.$

Again there is a single critical point, at $$(0,0)$$, and

$D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 0\cdot0-0=0,$

so we get no information. However, in this case it is easy to see that there is a minimum at $$(0,0)$$, because $$f(0,0)=0$$ and at all other points $$f(x,y)>0$$.

Example 14.7.5

Find all local maxima and minima for $$f(x,y)=x^3+y^3$$.

Solution

The derivatives:

$f_x=3x^2 \qquad f_y=3y^2 \qquad f_{xx}=6x^2 \qquad f_{yy}=6y^2 \qquad f_{xy}=0.$

Again there is a single critical point, at $$(0,0)$$, and

$D(0,0)=f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2= 0\cdot0-0=0,$

so we get no information. In this case, a little thought shows there is neither a maximum nor a minimum at $$(0,0)$$: when $$x$$ and $$y$$ are both positive, $$f(x,y)>0$$, and when $$x$$ and $$y$$ are both negative, $$f(x,y) < 0$$, and there are points of both kinds arbitrarily close to $$(0,0)$$. Alternately, if we look at the cross-section when $$y=0$$, we get $$f(x,0)=x^3$$, which does not have either a maximum or minimum at $$x=0$$.

Example 14.7.6

Suppose a box with no top is to hold a certain volume $$V$$. Find the dimensions for the box that result in the minimum surface area.

Solution

The area of the box is $$A=2hw+2hl+lw$$, and the volume is $$V=lwh$$, so we can write the area as a function of two variables,

$A(l,w)={2V\over l}+{2V\over w}+lw.$

Then

$A_l=-{2V\over l^2}+w \quad \text{and} A_w=-{2V\over w^2}+l.$

If we set these equal to zero and solve, we find $$w=(2V)^{1/3}$$ and $$l=(2V)^{1/3}$$, and the corresponding height is $$h=V/(2V)^{2/3}$$.

The second derivatives are

$A_{ll}={4V\over l^3} \qquad A_{ww}={4V\over w^3} \qquad A_{lw}=1,$

so the discriminant is

$D={4V\over l^3}{4V\over w^3}-1=4-1=3>0.$

Since $$A_{ll}$$ is 2, there is a local minimum at the critical point. Is this a global minimum? It is, but it is difficult to see this analytically; physically and graphically it is clear that there is a minimum, in which case it must be at the single critical point. Here is the graph as rendered by Sage, as an example.

Note that we must choose a value for $$V$$ in order to graph it.

### Contributors

• Integrated by Justin Marshall.