The problem to find the minimum or maximum value of a function is encountered very frequently in geometry, mechanics, physics, and other fields. It was one of the principal incentives for the development of the calculus in the seventeenth century. Our object for this page is to give a means of locating the extrema of two variable functions.

Recall the same problem in the case of single veriable function. We locate all critical points by solving f'(x) = 0. It is the same as to find all points with horizontal tangent line since at the points where a differentiable function attains relative maximum or minimum the tangent line is horizontal.                   Similarly, for the case of a two variable differentiable function f(x, y), we look for those points (also called critical or stationary points) whose tangent planes are horizontal.        Since the normal vector of tangent plane at (a, b) for the surface z = f(x, y) is

we want to solve and for critical points because horizontal planes have normal vector parallel to z-axis, i.e. (0, 0, 1) or (0, 0, -1).

• Example 1
  Find all critical points of the surface .
• Solution
  First of all, we compute

  

  Set both partial derivatives zero, we have x = 1, or -1 and y = 0. Therefore (1, 0) and (-1, 0) are the critical points of f.

In the single variable problem, we have the rules

to categorize the nature of critical points. There is an analogous theory, second-derivative test for extrema, for functions of two variables:

• Theorem Suppose (a, b) is a stationary point of a function f(x, y) with continuous second order partial derivatives in a domain containing (a, b). Let

  

  and let

  

  Then we have:

  

With this theorem we can summarize the procedure to find the extrema of a twice diffenentiable function f(x, y) of two variables as follows:

• Step 1
  Locate the critical points by solving and .
• Step 2
  Compute second partial derivatives and and evaluate the for each critical point found in step 1.
• Step 3
  Use the Theorem above to categorize the nature of critical points. Then compare their function values to find the absolute maximum or minimum.
• Step 4
  If the function is restricted to a domain (finite) with boundary, we should also search for the extrema along the boundary. It is then reduced to a single variable problem theoretically.

• Example 2
  Find all relative maximums and minimums for the function

  

• Solution
  • Step 1.  Compute the first order partial derivatives and set to zero.

    

    we have x = 0 or 2 and y = 1 or -1. Thus (0, 1), (0, -1), (2, 1) and (2, -1) are the critical points of f.

  • Step 2

    

    After some computations, we have

    

    Note that f is an exponential function, always positive.

  • Step 3
    From the Theorem, we know that f(0, 1) = exp(2/3) is a relative maximum and f(2,-1) = exp(-14/3) is a relative minimum. At both (0, -1) and (2, 1), f has saddle points.

Along y = 0, x-axis, it is easy to see that

It shows that f can grow as big as possible when x tends to infinity and as close to zero as well when x tends negative infinity (note f is positive for all (x, y)), so we conclude that f has no absolute maximum nor absolute minimum over xy-plane.

Suppose the problem was further restricted to the region R bounded by y = -x + 4 and , then we have more (step 4) to search for the possible extrema along boundaries. In this case, only (2, 1) and (2, -1) are inside R, they are still saddle point and relative minimum respectively.

• Step 4
  The restriction of f on the boundary of R can be expressed as

  

  respectively. Both are reduced to single variable problems. Readers should be able to find the maximums and minimums on each restriction. To determine the absolute extrema on the region R, you need to compare the function values of all critical points obtained in step 3 and 4 and corner points, such as the intersection points of two boundary curves.