We have explained the Lagrange multiplier method for solving the problem
max(min) f (x, y) subject to g(x, y)=c (1) In this section we give a geometric as well as an analytic argument for the method.
A Geometric Argument
The maximization problem in (1) can be given a geometric interpretation, as shown in Fig. 1.
The graph off is like the surface of an inverted bowl, whereas the equationg(x, y)=c represents a curve in thexy-plane. The curveK on the bowl is the one that lies directly above the curveg(x, y)=c. (The latter curve is the projection ofK onto thexy-plane.) Maximizingf (x, y)without taking the constraint into account gets us to the peakAin Fig. 1. The solution to problem (1), however, is atB, which is the highest point on the curve K. If we think of the graph off as representing a mountain, andK as a mountain path, then we seek the highest point on the path, which isB. Analytically, the problem is to find the coordinates ofB.
z ⫽ f(x, y)
g(x, y) ⫽ c A
B K
z
y
x
g(x, y) ⫽ c y
x P
A⬘ B⬘ Q
Figure 1 Illustrating a Lagrange problem
The right half of Fig. 1 shows some of the level curves forf, and also indicates the constraint curveg(x, y)=c. NowArepresents the point at whichf (x, y)has its unconstrained (free) maximum. The closer a level curve off is to pointA, the higher is the value off along that level curve. We are seeking that point on the constraint curveg(x, y)=cwheref has its highest value. If we start at pointP on the constraint curve and move along that curve towardA, we encounter level curves with higher and higher values off.
Obviously, the pointQindicated in Fig. 1 is not the point ong(x, y)=cat whichf has its highest value, because the constraint curve passesthroughthe level curve off at that point. Therefore, we can cross a level curve to higher values off by proceeding further along the constraint curve. However, when we reach pointB, we cannot go any higher. It is intuitively clear thatBis the point where the constraint curve touches (without intersecting) a level curve forf.
This observation implies that the slope of the tangent to the curveg(x, y)=cat(x, y) is equal to the slope of the tangent to the level curve off at that point.
Recall from Section 12.3 that the slope of the level curveF (x, y) = cis given by dy/dx = −F1(x, y)/F2(x, y). Thus, the condition that the slope of the tangent tog(x, y)= cis equal to the slope of a level curve forf (x, y)can be expressed analytically as:3
−g1(x, y)
g2(x, y) = −f1(x, y)
f2(x, y) or f1(x, y)
g1(x, y) = f2(x, y)
g2(x, y) (2) It follows that a necessary condition for(x, y)to solve problem (1) is that the left- and right-hand sides of the last equation in (2) are equal at(x, y). Letλdenote the common value of these fractions. This is the Lagrange multiplier introduced in Section 14.1. With this definition,
f1(x, y)−λg1(x, y)=0, f2(x, y)−λg2(x, y)=0 (3) Using the Lagrangian from Section 14.1, we see that (3) just tells us that the Lagrangian has a stationary point. An analogous argument for the problem of minimizingf (x, y)subject tog(x, y)=cgives the same condition.
An Analytic Argument
The geometric argument above is quite convincing. But the analytic argument we are about to offer is easier to generalize to more than two variables.
So far we have studied the problem of finding the absolute largest or smallest value of f (x, y)subject to the constraintg(x, y)=c. Sometimes we are interested in studying the corresponding local extrema. Briefly formulated, the problem is
local max(min) f (x, y) subject to g(x, y)=c (4) Possible solutions are illustrated in Fig. 2.
f(x, y) ⫽ 4 f(x, y) ⫽ 3 f(x, y) ⫽ 2 f(x, y) ⫽ 1 g(x, y) ⫽ c
y
x P
Q
R
Figure 2 Q,R, andPall satisfy the first-order conditions
The pointR is a local minimum point forf (x, y)subject tog(x, y)=c, whereasQand P are local maximum points. The global maximum off (x, y)subject tog(x, y) =cis
3 Disregard for the moment cases where any denominator is 0.
S E C T I O N 1 4 . 4 / W H Y T H E L A G R A N G E M E T H O D W O R K S 511 attained only atP. Each of the pointsP,Q, andRin Fig. 2 satisfies condition (2), so the first-order conditions are exactly as before. Let us derive them in a way that does not rely on geometric intuition. Except in some special cases, the equationg(x, y)=cdefinesy implicitly as a differentiable function ofxnear any local extreme point. Denote this function byy =h(x). According to formula (12.3.1), provided thatg2(x, y)=0, one has
y=h(x)= −g1(x, y)/g2(x, y)
Now, the objective functionz=f (x, y)=f (x, h(x))is, in effect, a function ofx alone.
By calculatingdz/dxwhile taking into account howydepends onx, we obtain a necessary condition for local extreme points by equatingdz/dxto 0. But
dz
dx =f1(x, y)+f2(x, y)y (withy=h(x))
So substituting the previous expression forh(x)gives the following necessary condition for(x, y)to solve problem (1):
dz
dx =f1(x, y)−f2(x, y)g1(x, y)
g2(x, y) =0 (5)
Assuming thatg2(x, y) =0, and definingλ =f2(x, y)/g2(x, y), we deduce that the two equationsf1(x, y)−λg1(x, y) =0 andf2(x, y)−λg2(x, y)=0 must both be satisfied.
Hence, the Lagrangian must be stationary at(x, y). The same result holds (by an analogous argument) providedg1(x, y)=0. To summarize, one can prove the following precise result:
T H E O R E M 1 4 . 4 . 1 ( L A G R A N G E ’ S T H E O R E M )
Suppose thatf (x, y)andg(x, y)have continuous partial derivatives in a domain Aof thexy-plane, and that(x0, y0)is both an interior point ofAand a local extreme point forf (x, y)subject to the constraintg(x, y)=c. Suppose further thatg1(x0, y0)andg2(x0, y0)are not both 0. Then there exists a unique number λsuch that the Lagrangian
L(x, y)=f (x, y)−λ (g(x, y)−c) has a stationary point at(x0, y0).
Problem 3 asks you to show how trouble can result from uncritical use of the Lagrange multiplier method, disregarding the assumptions in Theorem 14.4.1. Problem 4 asks you to show what can go wrong ifg1(x0, y0)andg2(x0, y0)are both 0.
In constrained optimization problems in economics, it is often implicitly assumed that the variables are nonnegative. This was certainly the case for the specific utility maximization problem in Example 14.1.3. Because the optimal solutions were positive, nothing was lost
by disregarding the nonnegativity constraints. Here is an example showing that sometimes we must take greater care.
E X A M P L E 1 Consider the utility maximization problem
maxxy+x+2y subject to 2x+y =m, x≥0, y≥0
where we have required that the amount of each good is nonnegative. The Lagrangian isL= xy+x+2y−λ(2x+y−m). So the first-order conditions (disregarding the nonnegativity constraints for the moment) are L1 = y +1−2λ = 0, L2(x, y) = x +2−λ = 0.
By eliminatingλ, we find thaty =2x+3. Inserting this into the budget constraint gives 2x+2x+3=m, sox =14(m−3). We easily find the corresponding value ofy, and the suggested solution that emerges isx∗ = 14(m−3),y∗= 12(m+3). Note that in the case whenm <3, thenx∗<0, so that the expressions we have found forx∗andy∗do not solve the given problem. The solution in this case is, as shown below,x∗=0,y∗=m. (So when income is low, the consumer should spend everything on just one commodity.)
Let us analyse the problem by converting it to one that is unconstrained. To do this, note how the constraint implies thaty =m−2x. In order for bothx andyto be nonnegative, one must require 0≤x ≤m/2 and 0≤y ≤m. Substitutingy =m−2x into the utility function, we obtain utility as functionU (x)ofx alone, where
U (x)=x(m−2x)+x+2(m−2x)= −2x2+(m−3)x+2m, x ∈[0, m/2]
This is a quadratic function withx = 14(m−3)as the stationary point. Ifm > 3, it is an interior stationary point for the concave functionU, so it is a maximum point. Ifm≤ 3, thenU(x)= −4x+(m−3)≤0 for allx≥0. Because of the constraintx ≥0, it follows thatU (x)must have its largest value forx =0.
Optimization problems with inequality constraints are generally known asnonlinear pro- gramming problems. Some relatively simple cases are discussed in Sections 14.8 and 14.9.
A much more systematic treatment of nonlinear programming is included in FMEA.
WARNING:One of the most frequently occurring errors in the economics literature (even in some leading textbooks) concerning the Lagrange multiplier method is the claim that it transforms a constrained optimization problem into one of finding an unconstrained optimum of the Lagrangian. Problem 1 shows that this is wrong. What the method does instead is to transform a constrained optimization problem into one of finding the appropriate stationary points of the Lagrangian. Sometimes these are maximum points, but often they are not.
P R O B L E M S F O R S E C T I O N 1 4 . 4
1. Consider the problem maxxysubject tox+y=2. Show that(x, y)=(1,1), withλ=1, is the only solution of the first-order conditions. (That this is indeed the solution of the problem is easily seen by reducing it to the one-variable problem of maximizingxy = x(2−x).) But(1,1)does not maximize the LagrangianL(x, y)=xy−1ã(x+y−2). Why not?
S E C T I O N 1 4 . 5 / S U F F I C I E N T C O N D I T I O N S 513 2. The following text taken from a book on mathematics for management containsgraveerrors. Sort them out. “Consider the general problem of finding the extreme points ofz=f (x, y)subject to the constraintg(x, y) =0. Clearly the extreme points must satisfy the pair of equations fx(x, y) =0,fy(x, y) =0 in addition to the constraintg(x, y) =0. Thus, there are three equations that must be satisfied by the pair of unknownsx, y. Because there are more equations than unknowns, the system is said to be overdetermined and, in general, is difficult to solve. In order to facilitate computation. . .” (A description of the Lagrange method follows.)
HARDER PROBLEMS
3. Consider the problem maxf (x, y)=2x+3ysubject tog(x, y)=√
x+ √y=5.
(a) Show that the Lagrange multiplier method suggests the wrong solution(x, y) = (9,4).
(Note thatf (9,4)=30, and yetf (25,0)=50.)
(b) Solve the problem by studying the level curves off (x, y)=2x+3ytogether with the graph of the constraint equation. (See Problem 5.4.2.)
(c) Which assumption of Theorem 14.4.1 is violated?
⊂SM⊃4. The functionsf andgare defined by
f (x, y)=(x+2)2+y2 and g(x, y)=y2−x(x+1)2
Find the minimum value off (x, y)subject tog(x, y)=0. Show that the Lagrange multiplier method cannot locate this minimum. (Hint:Draw a graph ofg(x, y)=0. Note in particular thatg(−1,0)=0.)