Constrained optimization problems in economics usually involve more than just two vari- ables. The typical problem withnvariables can be written in the form
max(min) f (x1, . . . , xn) subject to g(x1, . . . , xn)=c (1) The Lagrange multiplier method presented in the previous sections can be easily generalized.
As before, associate a Lagrange multiplierλwith the constraint and form the Lagrangian L(x1, . . . , xn)=f (x1, . . . , xn)−λ
g(x1, . . . , xn)−c
(2) Next, find all the first-order partial derivatives ofLand equate them to zero, so that
L1 =f1(x1, . . . , xn)−λg1(x1, . . . , xn)=0 . . . . Ln=fn(x1, . . . , xn)−λgn(x1, . . . , xn)=0
(3)
Thesenequations, together with the constraint, formn+1 equations that should be solved simultaneously to determine then+1 unknownsx1, . . . , xn, andλ.
S E C T I O N 1 4 . 6 / A D D I T I O N A L V A R I A B L E S A N D C O N S T R A I N T S 517 NOTE 1 This method will (in general) fail to give correct necessary conditions if all the first- order partial derivatives ofg(x1, . . . , xn)vanish at the stationary point of the Lagrangian.
Otherwise, the proof is an easy generalization of the analytic argument in Section 14.4 for the first-order conditions. If, say,∂g/∂xn =0, we “solve”g(x1, . . . , xn)=cforxnnear the stationary point, and thus reduce the problem to an unconstrained extremum problem in the remainingn−1 variablesx1,. . .,xn−1.
E X A M P L E 1 Solve the consumer’s demand problem
max U (x, y, z)=x2y3z subject to x+y+z=12
Solution: WithL(x, y, z)=x2y3z−λ(x+y+z−12), the first-order conditions are L1 =2xy3z−λ=0, L2 =3x2y2z−λ=0, L3=x2y3−λ=0 (∗) Ifanyof the variablesx,y, andzis 0, thenx2y3z=0, which isnotthe maximum value.
So suppose thatx,y, andzare all positive. From the two first equations in(∗), we have 2xy3z = 3x2y2z, so y = 3x/2. The first and third equations in(∗)likewise imply that z=x/2. Insertingy=3x/2 andz=x/2 into the constraint yieldsx+3x/2+x/2=12, sox =4. Theny=6 andz=2. Thus, the only possible solution is(x, y, z)=(4,6,2).
E X A M P L E 2 Solve the problem
minimize f (x, y, z)=(x−4)2+(y−4)2+ z−122
subject to x2+y2 =z Can you supply a geometric interpretation of the problem?
Solution: The Lagrangian isL(x, y, z)=(x−4)2+(y−4)2+ z−122
−λ(x2+y2−z), and the first-order conditions are:
L1(x, y, z)=2(x−4)−2λx=0 (i) L2(x, y, z)=2(y−4)−2λy=0 (ii) L3(x, y, z)=2
z−12
+λ=0 (iii)
x2+y2=z (iv)
From (i) we see thatx =0 is impossible. Equation (i) thus givesλ=1−4/x. Inserting this into (ii) and (iii) givesy =xandz=2/x. Using these results, equation (iv) reduces to 2x2 =2/x, that is, x3 = 1, sox = 1. It follows that(x, y, z) =(1,1,2)is the only solution candidate to the problem.
The expression(x−4)2+(y−4)2+(z−1/2)2measures the square of the distance from the point(4,4,1/2)to the point(x, y, z). The set of points(x, y, z)that satisfyz=x2+y2 is a surface known as a paraboloid, part of which is shown in Fig. 1. The minimization problem is therefore to find that point on the paraboloid which has the smallest (square) distance from(4,4,1/2). It is “geometrically obvious” that this problem has a solution. On the other hand, the problem of finding the largest distance from(4,4,1/2)to a point on the paraboloid does not have a solution, because the distance can be made as large as we like.
z ⫽ x2⫹ y2
4
4 z
y x
Figure 1
E X A M P L E 3 The general consumer optimization problem withngoods is
xmax1,...,xn
U (x1, . . . , xn) subject to p1x1+ ã ã ã +pnxn=m (4) whereUis defined forx1 ≥0, . . . , xn≥0. The Lagrangian is
L(x1, . . . , xn)=U (x1, . . . , xn)−λ(p1x1+ ã ã ã +pnxn−m) The first-order conditions are
Li(x1, . . . , xn)=Ui(x1, . . . , xn)−λpi =0, i=1, . . . , n Writingx=(x1, . . . , xn), we have
U1(x)
p1 =U2(x)
p2 = ã ã ã = Un(x)
pn =λ (5)
Apart from the last equation, which serves only to determine the Lagrange multiplierλ, we haven−1 equations. (Forn=2, there is one equation; forn=3, there are two equations;
and so on.) In addition, the constraint must hold. Thus, we havenequations to determine the values ofx1, . . . , xn. From (5) it also follows that
Uj(x) Uk(x)= pj
pk for every pair of goodsj andk (6) The left-hand side is the marginal rate of substitution (MRS) of goodkfor goodj, whereas the right-hand side is their price ratio, or rate of exchange of goodkfor goodj. So condition (6) equates the MRS for each pair of goods to the corresponding price ratio.
Consider the equations in (5), together with the budget constraint. Assume that this system is solved for x1, . . . , xn and λ as functions of p1, . . . , pn and m, giving xi = Di(p1, . . . , pn, m), for i = 1, . . . , n. Then Di(p1, . . . , pn, m)gives the amount of the ith commodity demanded by the individual when facing pricesp1, . . . , pnand incomem.
For this reasonD1, . . . , Dnare called the(individual) demand functions. By the same argument as in Example 14.1.3, the demand functions are homogeneous of degree 0. As one check that you have correctly derived the demand functions, it is a good idea to verify that the functions you find are indeed homogeneous of degree 0, and satisfy the budget constraint.
S E C T I O N 1 4 . 6 / A D D I T I O N A L V A R I A B L E S A N D C O N S T R A I N T S 519 In the case when the consumer has a Cobb–Douglas utility function, the constrained maximization problem is
max Ax1a1ã ã ãxann subject to p1x1+ ã ã ã +pnxn=m (∗) where we assume that each “taste” parameterai >0. Then the demand functions are
Di(p1, . . . , pn, m)= ai a1+ ã ã ã +an
m
pi, i=1, . . . , n (∗∗) (see Problem 8(a)). We see how the pattern of the two-variable case in Example 14.1.3 is repeated, with a constant fraction of incomemspent on each good, independent of all prices. Note also that the demand for any goodiis completely unaffected by changes in the price of any other good. This is an argument against using Cobb–Douglas utility functions, because we expect realistic demand functions to depend on prices of other goods that are either complements or substitutes.
More Constraints
Occasionally economists need to consider optimization problems with more than one equal- ity constraint (although it is much more common to have many inequality constraints). The corresponding general Lagrange problem is
max(min) f (x1, . . . , xn) subject to
⎧⎨
⎩
g1(x1, . . . , xn)=c1 . . . . gm(x1, . . . , xn)=cm
(7) The Lagrange multiplier method can be extended to treat problem (7). To do so, associate a Lagrange multiplier with each constraint, then form the Lagrangian sum
L(x1, . . . , xn)=f (x1, . . . , xn)− m j=1
λj
gj(x1, . . . , xn)−cj
(8) Except in special cases, this Lagrangian must be stationary at any optimal point. That is, its partial derivative w.r.t. each variableximust vanish. Hence,
∂L
∂xi = ∂f (x1, . . . , xn)
∂xi −
m j=1
λj∂gj(x1, . . . , xn)
∂xi =0, i=1,2, . . . , n (9) Together with themequality constraints, thesenequations form a total ofn+mequations in then+munknownsx1, . . . , xn,λ1, . . . , λm.
E X A M P L E 4 Solve the problem
min x2+y2+z2 subject to
x+2y+ z=30 (i) 2x− y−3z=10 (ii)
Solution: The Lagrangian is
L(x, y, z)=x2+y2+z2−λ1(x+2y+z−30)−λ2(2x−y−3z−10) The first-order conditions (9) require that
∂L
∂x =2x−λ1−2λ2 =0 (iii)
∂L
∂y =2y−2λ1+λ2 =0 (iv)
∂L
∂z =2z−λ1+3λ2=0 (v)
So there are five equations, (i) to (v), to determine the five unknownsx,y,z,λ1, andλ2. Solving (iii) and (iv) simultaneously forλ1andλ2gives
λ1= 25x+45y , λ2 =45x−25y
Inserting these expressions forλ1andλ2into (v) and rearranging yields
x−y+z=0 (vi)
This equation together with (i) and (ii) constitutes a system of three linear equations in the unknownsx,y, andz. Solving this system by elimination gives(x, y, z)=(10,10,0). The corresponding values of the multipliers areλ1=12 andλ2 =4.
Here is a geometric argument to confirm that we have solved the minimization problem.
Each of the two constraints represents a plane in⺢3, and the points satisfying both constraints consequently lie on the straight line where the two planes intersect. Now x2 +y2 +z2 measures (the square of) the distance from the origin to a point on this straight line, which we want to make as small as possible by choosing the point on the line that is nearest to the origin. No maximum distance can possibly exist, but it is geometrically obvious that there is a minimum distance, and it must be attained at this nearest point.
An easier alternative method to solve this particular problem is to reduce it to a one- variable optimization problem by using (i) and (ii) to gety =20−xandz=x−10, the equations of the straight line where the planes intersect. Then the square of the distance from the origin isx2+y2+z2 =x2+(20−x)2+(x−10)2 =3(x−10)2+200, and this function is easily seen to have a minimum whenx=10. See also Problem 5.
P R O B L E M S F O R S E C T I O N 1 4 . 6
1. Consider the problem minx2+y2+z2subject tox+y+z=1.
(a) Write down the Lagrangian for this problem, and find the only point(x, y, z)that satisfies the necessary conditions.
(b) Give a geometric argument for the existence of a solution. Does the corresponding maxi- mization problem have any solution?
2. Use the result in(∗∗)in Example 3 to solve the utility maximizing problem max 10x1/2y1/3z1/4 subject to 4x+3y+6z=390
S E C T I O N 1 4 . 6 / A D D I T I O N A L V A R I A B L E S A N D C O N S T R A I N T S 521
3. A consumer’s demandsx, y, zfor three goods are chosen to maximize the utility function U (x, y, z)=x+√y−1/z (x≥0, y >0, z >0)
subject to the budget constraintpx+qy+rz=m, wherep,q,r >0 andm≥ √pr+p2/4q.
(a) Write out the first-order conditions for a constrained maximum.
(b) Find the utility-maximizing demands for all three goods as functions of the four variables (p, q, r, m).
(c) Show that the maximized utility is given by the indirect utility function U∗(p, q, r, m)=m
p + p 4q−2
r p
(d) Find∂U∗/∂mand comment on your answer.
4. Each week an individual consumes quantitiesx andyof two goods, and works forlhours.
These three quantities are chosen to maximize the utility function U (x, y, l)=αlnx+βlny+(1−α−β)ln(L−l)
which is defined for 0≤l < Land forx,y >0. Hereαandβare positive parameters satisfying α+β <1. The individual faces the budget constraintpx+qy =wl, wherewis the wage per hour. Defineγ =(α+β)/(1−α−β). Find the individual’s demandsx∗,y∗, and labour supplyl∗as functions ofp,q, andw.
5. Consider the problem in Example 4, and let(x, y, z)=(10+h,10+k, l). Show that if(x, y, z) satisfies both constraints, thenk= −handl=h. Then show thatx2+y2+z2=200+3h2. Conclusion?
6. A statistical problem requires solving
min a12x12+a22x22+ ã ã ã +a2nx2n subject to x1+x2+ ã ã ã +xn=1
Here all the constantsaiare nonzero. Solve the problem, taking it for granted that the minimum value exists. What is the solution if one of theai’s is zero?
⊂SM⊃7. Solve the problem:
max(min) x+y subject to
x2+2y2+z2=1 x + y +z =1
HARDER PROBLEM
⊂SM⊃8. Consider the consumer optimization problem in Example 3. Find the demand functions when:
(a) U (x1, . . . , xn)=Ax1a1ã ã ãxnan (A >0,a1>0,. . .,an>0) (b) U (x1, . . . , xn)=x1a+ ã ã ã +xna (0< a <1)