Section 14.7 – Minimum and Maximum Values
In single-variable calculus, we saw that one of the primary uses of the derivative was to find the
relative maxima (the peaks) and relative minima (the valleys) of a function, y  f ( x). We now
consider the analogous problem for a function of two variables, z  f ( x, y ). The partial
derivative will be the critical tool for our finding such extrema.
How did we find the possible relative extrema of a function of one variable, y  f ( x),
assuming that f '( x) existed at the point in question?
Theorem: If f has a local maximum or a local minimum at (a, b) and the first-order partial
derivatives of f exist there, then f x ( a, b)  0 and f y (a, b)  0.
What does this imply about the tangent plane to a surface at a relative extremum? How does this
relate to the tangent line to a curve at a relative extremum?
Definition: A point (a, b) is called a critical (or stationary) point of f if f x ( a, b)  0 and
f y (a, b)  0 or if one of these partial derivatives does not exist.
Example: Find all critical points of the function f ( x, y)  x  y  2 x  6 y  14.
2
2
Can we tell whether we have a max or min here?
Example: (#6) Find all critical points of the function f ( x, y)  x y  12 x  8 y.
3
2
Remember, though, that not all critical points in single variable calculus gave us a max or min.
At a critical point, a function could have a max, a min, or neither.
The analogous situation occurs in the two-variable case with a function such as z  y  x ,
which is pictured below.
2
2
The point (0, 0) is a critical point of the function, but it does not correspond to either a maximum
or a minimum. Instead, this point is called a saddle point of the function.
How could we tell whether a critical point corresponded to a relative maximum or a relative
minimum, aside from using the First Derivative Test?
Second Derivative Test for Functions of Two Variables: Suppose the second partial
derivatives of f are continuous on a disk with center (a, b) and suppose that f x ( a, b)  0 and
f y (a, b)  0 [that is, (a, b) is a critical point of f ]. Then, let
D  D(a, b)  f xx (a, b) f yy (a, b)  [ f xy (a, b)]2




If D  0 and f xx ( a, b)  0 , then f (a, b) is a local minimum.
If D  0 and f xx ( a, b)  0 , then f (a, b) is a local maximum.
If D  0, then f (a, b) is neither a local maximum nor a local minimum, and there is a
saddle point at the critical point in question. (i.e., the graph of f crosses its tangent plane
at this point)
If D  0, the test is inconclusive.
Why is this function called D ?
Example: Find all critical points of the function f ( x, y)  x  y  4 xy  1, and, if possible,
classify them as local maxima, local minima, or saddle points.
Can you make any predictions from the graph of f and its contour map?
4
4
Example: Find all critical points of f ( x, y)  10 x y  5 x  4 y  x  2 y . Classify them
as local maxima, local minima, or saddle points. Find the absolute maximum on the graph of f.
2
2
2
4
4
Here are 2 views of the graph of f and the contour map of f. The 5 critical points are shown in the
map.
Example: Find the shortest distance from the point 1,0, 2  to the plane x  2 y  z  4.
Example: A rectangular box with no top is to be constructed from 12 square meters of
cardboard. Find the maximum volume of such a box
To this point, we have been concerned with finding the local extrema of a function of two
variables. However, in the one-variable case, there was another situation that we considered.
We could ask: Given a function y  f ( x) on a closed interval, what are the absolute maximum
and minimum values of the function on that interval? How did we solve that problem?
Before we can phrase an analogous question for a function of two variables, we need a bit of
2
terminology. A closed set in
is one that contains all of its boundary points. For example,
the set D 
 x, y  | x
2
 y 2  1 is a closed set, because all of the boundary points of the set,
those on the unit circle, are in the set. However, the set D ' 
 x, y  | x
2
 y 2  1 is not a
closed set, because it does not contain all of its boundary points. If even one boundary point is
missing from the set, it is not closed.
2
We also defined a bounded set in
to be
one that is contained in some disk. In other
words, it is finite in extent, and does not extend
out infinitely far in any direction.
Now that we have a notion of closed and bounded sets, we can formulate the following theorem:
Extreme Value Theorem for Functions of Two Variables: If f is continuous on a closed,
bounded region of the plane, then f has an absolute maximum value f ( x1 , y1 ) and an absolute
minimum value f ( x2 , y2 ) at some points  x1 , y1  and  x2 , y2  in the region.
So how do we find the extreme values of a function of two variables on a closed, bounded region
in the plane?
1. Find the values of f at the critical points of f in the region.
2. Find the extreme values of f on the boundary of the region.
3. The largest of the values from the first two steps is the absolute maximum value; the
smallest of these values is the absolute minimum value.
Example: Find the absolute maximum and minimum values of the function
f ( x, y)  x 2  2 xy  2 y on the rectangle D   x, y  | 0  x  3, 0  y  2 .


f ( x, y)  x 2  2 xy  2 y
Example: #30- Find the absolute maximum and minimum values of the function
f ( x, y)  3  xy  x  2 y on the triangular region R which has vertices at (1,0), (5,0), and
(1,4).
Example: (#38)- If a function of 1 variable is continuous on an interval and has only 1 critical
point, then a local max has to be an absolute max. This is not true for functions of 2 variables.
Show that the function f ( x, y)  3xe y  x3  e3 y has exactly one critical point and that f has a
local max there that is not an absolute max.
We’ll look at a computer generated graph to show how this is possible.