As with functions of one variable, it is easy to find examples of functions of several variables that do not have any maximum or minimum points. For providing sufficient conditions to ensure that extreme points do exist, however, the extreme value theorem (Theorem 8.4.1) was very useful for functions of one variable. It can be directly generalized to functions of several variables. In order to formulate the theorem, however, we need a few new concepts.
For many of the results concerning functions of one variable, it was important to distin- guish between different kinds of domain for the functions. For functions of several variables, the distinction between different kinds of domain is no less important. In the one-variable case, most functions were defined over intervals, and there are not many different kinds of interval. For functions of several variables, however, there are many different kinds of domain. Fortunately, the distinctions that are relevant to the extreme value theorem can be made using only the concepts of open, closed, and bounded sets.
A point(a, b)is called aninterior pointof a setS in the plane ifthere existsa circle centred at(a, b)such that all points strictly inside the circle lie inS. (See Fig. 1.) A set is calledopenif it consists only of interior points. (See the second set illustrated in Fig. 1, where we indicate boundary points that belong to the set by a solid curve, and those that do not by a dashed curve.) The point(a, b)is called aboundary pointof a setSifevery circle centred at(a, b)contains points ofSas well as points in its complement, as illustrated in the first figure. A boundary point ofSdoes not necessarily lie inS. IfScontains all its boundary points, thenSis calledclosed. (See the third set in Fig. 1.) Note that a set that
S E C T I O N 1 3 . 5 / T H E E X T R E M E V A L U E T H E O R E M 483 contains some but not all of its boundary points, like the last of those illustrated in Fig. 1, is neither open nor closed. In fact, a set is closed if and only if its complement is open.4
Open Closed Neither open
nor closed Interior
point
Boundary point
S
Figure 1
These illustrations give only very loose indications of what it means for a set to be either open or closed. Of course, if a set is not even precisely defined, it is impossible to decide conclusively whether it is open or closed.
In many of the optimization problems considered in economics, sets are defined by one or more inequalities, and boundary points occur where one or more of these inequalities are satisfied with equality. For instance, provided thatp,q, andmare positive parameters, the (budget) set of points(x, y)that satisfy the inequalities
px+qy ≤m, x ≥0, y ≥0 (∗)
is closed. This set is a triangle, as shown in Fig. 4.4.12. Its boundary consists of the three sides of the triangle. Each of the three sides corresponds to having one of the inequalities in(∗)be satisfied with equality. On the other hand, the set that results from replacing≤by
<and≥by>in(∗)is open.
In general, ifg(x, y)is a continuous function andcis a real number, then the sets {(x, y):g(x, y)≥c}, {(x, y):g(x, y)≤c}, {(x, y):g(x, y)=c} are all closed. If≥is replaced by>, or≤is replaced by<, or=by=, then the corresponding set becomes open.
A set in the plane is boundedif the whole set is contained within a sufficiently large circle. The sets in Fig. 1 and the budget triangle in Fig. 4.4.12 are all bounded. On the other hand, the set of all(x, y)satisfying
x≥1 and y ≥0
is a closed, but unbounded set. (See Fig. 11.1.1.) The set is closed because it contains all its boundary points. How would you characterize the set in Fig. 11.1.2? (In fact, it is neither open nor closed, but it is bounded.) A set in the plane that is both closed and bounded is often calledcompact.
We are now ready to formulate the main result in this section.
4 In every day usage the words “open” and “closed” are antonyms: a shop is either open or closed.
In topology, however, a set that contains some but not all its boundary points is neither open nor closed. See the last set in Fig. 1.
T H E O R E M 1 3 . 5 . 1 ( E X T R E M E V A L U E T H E O R E M )
Suppose the functionf (x, y)is continuous throughout a nonempty, closed and bounded setSin the plane. Then there exists both a point(a, b)inSwheref has a minimum and a point(c, d)inSwheref has a maximum—that is,
f (a, b)≤f (x, y)≤f (c, d) for all(x, y)inS
Theorem 13.5.1 is a pure existence theorem. It tells us nothing abouthow to findthe extreme points. Its proof is found in most advanced calculus books and in FMEA. Also, even though the conditions of the theorem aresufficientto ensure the existence of extreme points, they are far from necessary. (See Note 8.4.1.)
Finding Maxima and Minima
Sections 13.1 and 13.2 presented some simple cases where we could find the maximum and minimum points of a function of two variables by finding its stationary points. The procedure set out in the following frame covers many additional optimization problems.
F I N D I N G M A X I M A A N D M I N I M A
Find the maximum and minimum values of a differentiable function f (x, y) defined on a closed, bounded setSin the plane.
Solution:
(I) Find all stationary points off in the interior ofS.
(II) Find the largest value and the smallest value offon the boundary ofS, along with the associated points. (If it is convenient to subdivide the boundary into several pieces, find the largest and smallest value on each piece of the boundary.)
(III) Compute the values of the function at all the points found in (I) and (II). The largest function value is the maximum value offinS. The smallest function value is the minimum value off inS.
(1)
We try out this procedure on the function whose graph is depicted in Fig. 2 below. (Because the function is not specified analytically, we can only give a rough geometric argument.) The function has a rectangular domainSof points(x, y)in thexy-plane. The only stationary point offit includes is(x0, y0), which corresponds to the pointPof the graph. The boundary ofSconsists of four straight-line segments. The pointRvertically above one corner point ofS represents the maximum value off along the boundary; similarly,Qrepresents the minimum value off along the boundary. The only candidates for a maximum/minimum
S E C T I O N 1 3 . 5 / T H E E X T R E M E V A L U E T H E O R E M 485 are, therefore, the three pointsP,Q, andR. By comparing the values off at these points, we see thatP represents the minimum value, whereasRrepresents the maximum value of f inS.
z ⫽ f(x, y)
(x0, y0) Q P
S z R
y
x Figure 2
As an aspiring economist you will be glad to hear that most optimization problems in economics, especially those appearing in textbooks, rarely create enough difficulties to call for the full recipe. Usually there is an interior optimum that can be found by equating all the first-order partial derivatives to zero. Conditions that are sufficient for this easier approach to work were already discussed in Section 13.2. Nevertheless, we consider an example of a harder problem which illustrates how the whole recipe is sometimes needed. This recipe is also needed in several of the problems for this section. In particular, Problem 3 gives an economic example.
E X A M P L E 1 Find the extreme values forf (x, y)defined overSwhen
f (x, y)=x2+y2+y−1, S= {(x, y):x2+y2 ≤1}
Solution: The setSconsists of all the points on or inside the circle of radius 1 centred at the origin, as shown in Fig. 3. The continuous functionf will attain both a maximum and a minimum overS, by the extreme value theorem.
According to the preceding recipe, we start by finding all the stationary points in the interior ofS. These stationary points satisfy the two equations
f1(x, y)=2x =0, f2(x, y)=2y+1=0
So(x, y) = (0,−1/2)is the only stationary point, and it is in the interior of S, with f (0,−1/2)= −5/4.
The boundary ofS consists of the circlex2 +y2 =1. Note that if(x, y)lies on this circle, then in particular bothxandylie in the interval [−1,1]. Insertingx2+y2 =1 into the expression forf (x, y)shows that,along the boundaryofS, the value off is determined by the following function of one variable:
g(y)=1+y−1=y, y ∈[−1,1]
The maximum value ofgis 1 fory =1, and thenx =0. The minimum value is−1 when y = −1, and then againx=0.
We have now found the only three possible candidates for extreme points, namely, (0,−1/2),(0,1), and(0,−1). Butf (0,−1/2)= −5/4,f (0,1)=1, andf (0,−1)= −1.
We conclude that themaximum valueoff inSis 1, which is attained at(0,1), whereas the minimum valueis−5/4, attained at(0,−1/2).
1 1
y
x S
Figure 3 The domain in Example 1
P R O B L E M S F O R S E C T I O N 1 3 . 5
1. Letf (x, y)=4x−2x2−2y2,S= {(x, y):x2+y2≤25}.
(a) Computef1(x, y)andf2(x, y), then find the only stationary point forf. (b) Find the extreme points forf overS.
⊂SM⊃2. Find the maximum and minimum points for the following:
(a) f (x, y)=x3+y3−9xy+27 subject to 0≤x≤4 and 0≤y≤4.
(b) f (x, y)=x2+2y2−xsubject tox2+y2≤1.
⊂SM⊃3. In one study of the quantitiesxandyof natural gas that Western Europe should import from Norway and Siberia, respectively, it was assumed that the benefits were given by the function f (x, y) =9x+8y−6(x+y)2. (The term−6(x+y)2occurs because the world price of natural gas rises as total imports increase.) Because of capacity constraints,xandymust satisfy 0≤x≤5 and 0≤y≤3. Finally, for political reasons, it was felt that imports from Norway should not provide too small a fraction of total imports at the margin, so thatx≥2(y−1), or equivalently−x+2y≤2. Thus, the optimization problem was cast as
max f (x, y)=9x+8y−6(x+y)2 subject to 0≤x≤5,0≤y≤3,−x+2y≤2 In thexy-plane, draw the setSof all points satisfying the three constraints, and then solve the problem.
4. (a) Determine values of the constantsa,b, andcsuch thatf (x, y)=ax2y+bxy+2xy2+c has a local minimum at the point(2/3,1/3)with local minimum value−1/9.
(b) With the values ofa,b, andcfound in part (a), find the maximum and minimum values of f overS= {(x, y):x≥0, y≥0,2x+y≤4}.
S E C T I O N 1 3 . 6 / T H R E E O R M O R E V A R I A B L E S 487
⊂SM⊃5. (a) Find all stationary points off (x, y) = xe−x(y2−4y)and classify them by using the second-derivative test.
(b) Show thatf has neither a global maximum nor a global minimum.
(c) LetS = {(x, y) : 0 ≤ x ≤ 5, 0 ≤ y ≤ 4}. Prove thatf has global maximum and minimum points inSand find them.
(d) Find the slope of the tangent to the level curvexe−x(y2−4y)=e−4 at the point where x=1 andy=4−e.
6. Which of the following sets are open, closed, bounded, or compact?
(a) {(x, y): 5x2+5y2≤9} (b) {(x, y):x2+y2>9} (c) {(x, y):x2+y2≤9} (d) {(x, y): 2x+5y≥6} (e) {(x, y): 5x+8y=8} (f) {(x, y): 5x+8y >8} HARDER PROBLEM
7. Give an example of a discontinuous function g of one variable such that the set {x:g(x)≤1}is not closed.