Consider again the problem
max(min) f (x, y) subject to g(x, y)=c
Supposex∗andy∗are the values ofxandythat solve this problem. In general,x∗andy∗ depend onc. Weassumethatx∗=x∗(c)andy∗=y∗(c)are differentiable functions ofc.
The associated value off (x, y)is then also a function ofc, with
f∗(c)=f (x∗(c), y∗(c)) (1)
Heref∗(c)is called the(optimal) value functionfor the problem. Of course, the associ- ated value of the Lagrange multiplier also depends onc, in general. Provided that certain regularity conditions are satisfied, we have the remarkable result that
df∗(c)
dc =λ(c) (2)
Thus,the Lagrange multiplierλ=λ(c)is the rate at which the optimal value of the objective function changes with respect to changes in the constraint constantc.
In particular, ifdcis a small change inc, then
f∗(c+dc)−f∗(c)≈λ(c) dc (3)
In economic applications,coften denotes the available stock of some resource, andf (x, y) denotes utility or profit. Thenλ(c) dcmeasures the approximate change in utility or profit that can be obtained fromdcunits more (or−dcless, whendc <0). Economists callλa shadow priceof the resource. Iff∗(c)is the maximum profit when the resource input is c, then (3) says thatλindicates the approximate increase in profit per unit increase in the resource.
Proof of (2)(assuming thatf∗(c)is differentiable): Taking the differential of (1) gives df∗(c)=df (x∗, y∗)=f1(x∗, y∗) dx∗+f2(x∗, y∗) dy∗ (∗) But from the first-order conditions we havef1(x∗, y∗) = λg1(x∗, y∗)andf2(x∗, y∗) = λg2(x∗, y∗), so(∗)can be written as
df∗(c)=λg1(x∗, y∗) dx∗+λg2(x∗, y∗) dy∗=λ[g1(x∗, y∗) dx∗+g2(x∗, y∗) dy∗] (∗∗) Moreover, taking the differential of the identityg(x∗(c), y∗(c))=cyields
dg(x∗, y∗)=g1(x∗, y∗) dx∗+g2(x∗, y∗) dy∗=dc Substituting the last equality in(∗∗)implies thatdf∗(c)=λ dc
S E C T I O N 1 4 . 2 / I N T E R P R E T I N G T H E L A G R A N G E M U L T I P L I E R 505
E X A M P L E 1 Consider the following generalization of Example 14.1.1:
max xy subject to 2x+y=m
The first-order conditions again give y = 2x withλ = x. The constraint now becomes 2x+2x =m, sox=m/4. In the notation introduced above, the solution is
x∗(m)=m/4, y∗(m)=m/2, λ(m)=m/4
The value function is thereforef∗(m)=(m/4)(m/2)=m2/8. It follows thatdf∗(m)/dm
= m/4 = λ(m). Hence (2) is confirmed. Suppose in particular that m = 100. Then f∗(100)=1002/8. What happens to the value function ifm=100 increases by 1? The new value isf∗(101)=1012/8, sof∗(101)−f∗(100)=1012/8−1002/8 = 25.125.
Note that formula (3) withdc=1 givesf∗(101)−f∗(100)≈λ(100)ã1=25ã1=25, which is quite close to the exact value 25.125.
E X A M P L E 2 SupposeQ = F (K, L) denotes the output of a state-owned firm when the input of capital isKand that of labour isL. Suppose the prices of capital and labour arerandw, respectively, and that the firm is given a total budget ofmto spend on the two input factors.
The firm wishes to find the choice of inputs it can afford that maximizes output. So it faces the problem
maxF (K, L) subject to rK+wL=m
Solving this problem by using Lagrange’s method, the value of the Lagrange multiplier will tells us approximately the increase in output ifmis increased by 1 dollar.
Consider, for example, the specific problem
max 120KL subject to 2K+5L=m
Note that this is mathematically a special case of the problem in Example 14.1.3. Only the notation is different, along with the fact that the consumer has been replaced with a firm.
From(∗∗)in Example 14.1.3 we find the solution
K∗= 14m, L∗=101m, with λ=6m
The optimal output isQ∗(m)=120K∗L∗=12014m101m=3m2, sodQ∗/dm=6m=λ, and (2) is confirmed.
P R O B L E M S F O R S E C T I O N 1 4 . 2
1. Verify that equation (2) holds for the problem: maxx3ysubject to 2x+3y=m.
2. (a) With reference to Example 14.1.2, solve the problem minimizerK+wL subject to √
K+L=Q assuming thatQ > w/2r, wherer,w, andQare positive constants.
(b) Verify (2) in this case.
3. (a) Consider the problem of minimizingx2+y2subject tox+2y=a(whereais a constant).
Solve the problem by transforming it into an unconstrained optimization problem with one variable.
(b) Show that the Lagrange method leads to the same solution. Verify (2) in this case.
(c) Explain the solution by studying the level curves off (x, y)=x2+y2and the graph of the straight linex+2y =a. Can you give a geometric interpretation of the problem? Does the corresponding maximization problem have a solution?
⊂SM⊃4. (a) Solve the utility maximization problem maxU (x, y)=√
x+y subject to x+4y=100
using the Lagrange method, i.e. find the quantities demanded of the two goods.
(b) Suppose income increases from 100 to 101. What is the exact increase in the optimal value ofU (x, y)? Compare with the value found in (a) for the Lagrange multiplier.
(c) Suppose we change the budget constraint topx+qy = m, but keep the same utility function. Derive the quantities demanded of the two goods ifm > q2/4p.
⊂SM⊃5. (a) Consider the consumer demand problem maxx, y
U (x, y)=αln(x−a)+βln(y−b)
subject to px+qy=m (∗) whereα,β,a,b,p,q, andmare positive constants withα+β=1, and moreover, with m > ap+bq. Show that ifx∗,y∗solve problem(∗), then expenditure on the two goods is given by the two linear functions
px∗=αm+pa−α(pa+qb), qy∗=βm+qb−β(pa+qb) (∗∗) of the variables(m, p, q). (This is a special case of thelinear expenditure systemthat the British economist R. Stone fitted to UK consumption data in theEconomic Journal, 1954.) (b) LetU∗(p, q, m)=U (x∗, y∗)denote the indirect utility function. Show that∂U∗/∂m >0
and verify the so-called Roy’s identities, ∂U∗
∂p = −∂U∗
∂mx∗and∂U∗
∂q = −∂U∗
∂my∗. HARDER PROBLEM
⊂SM⊃6. An oil producer starts extracting oil from a well at time t = 0, and ends at a timet = T that the producer chooses. Suppose that the output flow at any timetin the interval [0, T] is xt (T −t )barrels per unit of time, where the intensityxcan also be chosen. The total amount of oil extracted in the given time span is thus given by the functiong(x, T )=T
0 xt (T−t ) dt ofxandT. Assume further that the sales price per barrel at timetisp=1+t, and that the cost per barrel extracted is equal toαT2, whereαis a positive constant. The profit per unit of time is then(1+t−αT2)xt (T −t ), so that the total profit earned during the time interval [0, T] is a function ofxandT given byf (x, T )=T
0(1+t−αT2) xt (T−t ) dt. If the total amount of extractable oil in the field isMbarrels, the producer can choose values ofxandT such that g(x, T )=M. The producer’s problem is thus
maxf (x, T ) subject to g(x, T )=M (∗) Find explicit expressions forf (x, T )andg(x, T )by calculating the given integrals, and then solve problem(∗). Verify Equation (2) in this case.
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