Use partial derivatives to locate critical points for a function of two variables.
Apply a second derivative test to identify a critical point as a local maximum, local minimum, or saddle point for a function of two variables.
Examine critical points and boundary points to find absolute maximum and minimum values for a function of two variables.
One of the most useful applications for derivatives of a function of one variable is the determination of maximum and/or minimum values. This application is also important for functions of two or more variables, but as we have seen in earlier sections of this chapter, the introduction of more independent variables leads to more possible outcomes for the calculations. The main ideas of finding critical points and using derivative tests are still valid, but new wrinkles appear when assessing the results.
Critical points
For functions of a single variable, we defined critical points as the values of the function when the derivative equals zero or does not exist. For functions of two or more variables, the concept is essentially the same, except for the fact that we are now working with partial derivatives.
Definition
Let
be a function of two variables that is differentiable on an open set containing the point
The point
is called a
critical point of a function of two variables
if one of the two following conditions holds:
Either
does not exist.
Finding critical points
Find the critical points of each of the following functions:
First, we calculate
Next, we set each of these expressions equal to zero:
Then, multiply each equation by its common denominator:
Therefore,
and
so
is a critical point of
We must also check for the possibility that the denominator of each partial derivative can equal zero, thus causing the partial derivative not to exist. Since the denominator is the same in each partial derivative, we need only do this once:
This equation represents a hyperbola. We should also note that the domain of
consists of points satisfying the inequality
Therefore, any points on the hyperbola are not only critical points, they are also on the boundary of the domain. To put the hyperbola in standard form, we use the method of completing the square:
Dividing both sides by
puts the equation in standard form:
Notice that point
is the center of the hyperbola.
First, we calculate
Next, we set each of these expressions equal to zero, which gives a system of equations in
Subtracting the second equation from the first gives
Substituting this into the first equation gives
so
Therefore
is a critical point of
(
[link] ). There are no points in
that make either partial derivative not exist.
The function
has a critical point at
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