Consider a differentiable functionz=f (x, y)defined on a setSin thexy-plane. Suppose thatf attains its largest value (its maximum) at an interior point(x0, y0)ofS, as indicated in Fig. 1. If we keepyfixed aty0, then the functiong(x)=f (x, y0)depends only onxand has its maximum atx =x0. (Geometrically, ifPis the highest point on the surface in Fig. 1, thenP is certainly also the highest point on the curve throughP that hasy =y0—i.e. on
the curve which is the intersection of the surface with the planey = y0.) From Theorem 8.1.1 we know thatg(x0)=0. But for allx, the derivativeg(x)is exactly the same as the partial derivativef1(x, y0). Atx=x0, therefore, one hasf1(x0, y0)=0. In the same way, we see that(x0, y0)must satisfyf2(x0, y0)=0, because the functionh(y)=f (x0, y)has its maximum aty=y0. A point(x0, y0)where both the partial derivatives are 0 is called a stationary(orcritical)pointoff.
Iff attains its smallest value (its minimum) at an interior point(x0, y0)ofS, a similar argument shows that the point again must be a stationary point. So we have the following important result:1
T H E O R E M 1 3 . 1 . 1 ( N E C E S S A R Y C O N D I T I O N S F O R I N T E R I O R E X T R E M A ) A differentiable functionz =f (x, y)can have a maximum or minimum at an interior point(x0, y0)ofSonly if it is astationary point—that is, if the point (x, y)=(x0, y0)satisfies the two equations
f1(x, y)=0, f2(x, y)=0 (first-order conditions, or FOCs)
z ⫽ f(x, y)
x0 y0
P ⫽ (x0, y0, f(x0, y0)) z
y
x
S Q
R
P
Q R z
x y
Figure 1 f (x, y)has maximum atP, the highest point on the surfacez =f (x, y), wheref1(x0, y0)=f2(x0, y0)=0.
Figure 2 Pis a maximum,Qis a local maximum, andRis a saddle point.
In Fig. 2, the three pointsP,Q, andRare all stationary points, but onlyP is a maximum.
(Later, we shall callQalocal maximum, whereasRis asaddle point.)
In the following examples and problems only the first-order conditions are considered.
The next section explains how to verify that we have found the optimum.
E X A M P L E 1 The functionf is defined for all(x, y)by
f (x, y)= −2x2−2xy−2y2+36x+42y−158 Assume thatf has a maximum point. Find it.
1 Interior point is defined precisely in Section 13.5.
S E C T I O N 1 3 . 1 / T W O V A R I A B L E S : N E C E S S A R Y C O N D I T I O N S 463 Solution: Theorem 13.1.1 applies. So a maximum point(x, y)must be a stationary point satisfying the first-order conditions
f1(x, y)= −4x−2y+36=0 f2(x, y)= −2x−4y+42=0
These are two linear equations which determinexandy. We find that(x, y)=(5,8)is the only pair of numbers which satisfies both equations. Assuming there is a maximum point, these must be its coordinates. The maximum value isf (5,8)=100. (In Example 13.2.2 we prove that(5,8)isa maximum point.)
E X A M P L E 2 A firm produces two different kindsAandBof a commodity. The daily cost of producing xunits ofAandyunits ofBis
C(x, y)=0.04x2+0.01xy+0.01y2+4x+2y+500
Suppose that the firm sells all its output at a price per unit of 15 forAand 9 forB. Find the daily production levelsxandythat maximize profit per day.
Solution: Profit per day isπ(x, y)=15x+9y−C(x, y), so
π(x, y)=15x+9y−0.04x2−0.01xy−0.01y2−4x−2y−500
= −0.04x2−0.01xy−0.01y2+11x+7y−500 Ifx >0 andy >0 maximize profit, then(x, y)must satisfy
∂π
∂x = −0.08x−0.01y+11=0, ∂π
∂y = −0.01x−0.02y+7=0
These two linear equations inx andy have the unique solutionx = 100,y =300, with π(100,300) = 1100. (We have not proved that this actually is a maximum. See Prob- lem 13.2.1(a).)
E X A M P L E 3 (Profit maximization) Suppose thatQ=F (K, L)is a production function withK as the capital input andL as the labour input. The price per unit of output isp, the cost (or rental) per unit of capital isr, and the wage rate isw. The constantsp,r, andware all positive. The profitπ from producing and sellingF (K, L)units is then given by the function
π(K, L)=pF (K, L)−rK−wL (1)
IfF is differentiable and π has a maximum withK > 0, L > 0, then the first-order conditions (FOCs) are
πK (K, L)=pFK(K, L)−r=0
πL(K, L)=pFL(K, L)−w=0 (∗)
Thus, a necessary condition for profit to be a maximum whenK=K∗andL=L∗is that pFK(K∗, L∗)=r, pFL(K∗, L∗)=w (∗∗) The first equation says thatr, the price of capital, must equal the sales value at the pricep per unit of the marginal product of capital. The second equation has a similar interpretation.
Suppose we think of increasing capital input from the levelK∗by 1 unit. How much would be gained? Production would increase by approximatelyFK(K∗, L∗)units. Because each extra unit is priced atp, the revenue gain is approximatelypFK(K∗, L∗). How much is lost? The answer isr, because this is the price of one unit of capital. These two must be equal.
The second equation in(∗∗)has a similar interpretation: Increasing labour input by one unit from levelL∗will lead to the approximate gainpFL(K∗, L∗)in revenue, whereas the extra labour cost isw. The profit-maximizing pair(K∗, L∗)thus has the property that the extra revenue from increasing either input by one unit is just offset by the extra cost.
Economists often divide the first-order conditions (∗∗) by the positive pricepto reach the alternative formFK(K, L)=r/pandFL(K, L) =w/p. So, to obtain maximum profit, the firm must chooseKandLto equate the marginal productivity of capital to its relative pricer/p, and also to equate the marginal productivity of labour to its relative pricew/p.
Note that the conditions in(∗∗)are necessary, but generally not sufficient for an interior maximum. Sufficient conditions for an optimum are given in Example 13.3.3.
E X A M P L E 4 Find the only possible solution to the following special case of Example 3:
max π(K, L)=12K1/2L1/4−1.2K−0.6L Solution: The first-order conditions are
πK (K, L)=6K−1/2L1/4−1.2=0, πL(K, L)=3K1/2L−3/4−0.6=0 These equations imply thatK−1/2L1/4 =K1/2L−3/4 =0.2=1/5. Multiplying each side of the first equation here byK1/2L3/4reduces it toL=K. HenceK−1/4 =L−1/4 =1/5.
It follows thatK=L=54 =625 is the only possible solution. (See Example 13.2.3 for a proof that this is indeed a maximum point.)
E X A M P L E 5 A firm is a monopolist in the domestic market but takes as given the fixed pricepw of its product in the world market. The quantities sold in the two markets are denoted byxd andxw, respectively. The price obtained in the domestic market, as a function of its sales, is given by the inverse demand functionpd =f (xd). The cost function isc(xd+xw). (Note thatcis here a function of one variable, not a constant.)
(a) Find the profit functionπ(xd, xw)and write down the first-order conditions for profit to be maximized atxd >0,xw >0. Give economic interpretations of these conditions.
(b) Suppose that the firm in the domestic market is faced with a demand curve whose price elasticity is a constant equal to−2. What is the relationship between the prices in the domestic and world markets?
S E C T I O N 1 3 . 1 / T W O V A R I A B L E S : N E C E S S A R Y C O N D I T I O N S 465 Solution: (a) The revenue from sellingxd units in the domestic market at the pricepd = f (xd)ispdxd =f (xd)xd. In the world market the revenue ispwxw. The profit function is π=π(xd, xw)=pdxd +pwxw−c(xd+xw). Thus the first-order conditions are
(i) π1 =pd +(dpd/dxd)xd−c(xd+xw)=0, (ii) π2 =pw−c(xd +xw)=0 According to (ii), the marginal cost in the world market must equal the price, which is the marginal revenue in this case. In the domestic market the marginal cost must also equal the marginal revenue. Here is an interpretation of (i): Suppose the firm contemplates producing and selling a little extra in its domestic market. The extra revenue per unit increase in output equalspdminus the loss that arises because of the induced price reduction for all domestic sales. The latter loss is approximatelyf(xd)xd =(dpd/dxd)xd. Since the cost of an extra unit of output is approximately the marginal costc(xd +xw), condition (i) expresses the requirement that, per unit of extra output, the domestic revenue gain is just offset by the cost increase.
(b) The price elasticity of demand is−2, meaning that Elpdxd =(pd/xd)(dxd/dpd)= −2.
By the rule for differentiating inverse functions one hasdpd/dxd =1/(dxd/dpd). It follows that(dpd/dxd)xd = −12pd. Then (i) and (ii) imply that12pd =c(xd+xw)=pw, so the world market price is half the domestic market price.
P R O B L E M S F O R S E C T I O N 1 3 . 1
1. The functionf defined for all(x, y)byf (x, y)= −2x2−y2+4x+4y−3 has a maximum.
Find the corresponding values ofxandy.
2. (a) The functionfdefined for all(x, y)byf (x, y)=x2+y2−6x+8y+35 has a minimum point. Find it.
(b) Show thatf (x, y)can be written in the formf (x, y)=(x−3)2+(y+4)2+10. Explain why this shows that you have really found the minimum in part (a).
3. In the profit-maximizing problem of Example 3, letp=1,r=0.65,w=1.2, and F (K, L)=80−(K−3)2−2(L−6)2−(K−3)(L−6) Find the only possible values ofKandLthat maximize profits.
4. Yearly profits (in millions of dollars) for a firm are given by P (x, y)= −x2−y2+22x+18y−102
wherexis the amount spent on research (in millions of dollars), andyis the amount spent on advertising (in millions of dollars).
(a) Find the profits whenx=10,y=8 and whenx=12,y=10.
(b) Find the only possible values ofxandythat can maximize profits, and the corresponding profit.