Supposef is a function of one variable which is twice differentiable in an intervalI. In this case a very simple sufficient condition for a stationary point inI to be a maximum point is thatf(x)≤0 for allxinI. (See Theorem 8.2.2.) The functionfis then called “concave”.
For functions of two variables there is a corresponding test for concavity based on the second-orderpartialderivatives. Provided the function has an interior stationary point, this test implies that its graph is a surface shaped like the one in Fig. 13.1.1.
Consider any curve parallel to thexz-plane which lies in the surface, likeQP Rin that figure. Any such curve is the graph of a concave function of one variable, implying that f11(x, y)≤0. A similar argument holds for any curve parallel to theyz-plane which lies in the surface, implying thatf22(x, y)≤0. In general, however, having these two second- order partial derivatives be nonpositive isnotsufficient on its own to ensure that the surface is shaped like the one in Fig. 13.1.1. This is clear from the next example.
E X A M P L E 1 The function
f (x, y)=3xy−x2−y2
hasf11(x, y)=f22(x, y)= −2. Each curve parallel to thexz-plane that lies in the surface defined by the graph has the equationz=3xy0−x2−y02for some fixedy0. It is therefore a concave parabola. So is each curve parallel to theyz-plane that lies in the surface. But along the liney =xthe function reduces tof (x, x)=x2, which is a convex rather than a concave parabola. It follows thatf has no maximum (or minimum) at(0,0), which is the only stationary point.
What Example 1 shows is that conditions ensuring that the graph off looks like the one in Fig. 13.1.1 cannot ignore the second-order mixed partial derivativef12(x, y). The following result will be discussed in FMEA. (See the end of Section 13.3 for a proof of the local version.) To formulate the theorem we need a new concept. A setS in thexy-plane is convexif, for each pair of pointsP andQinS, all the line segment betweenP andQlies inS.
T H E O R E M 1 3 . 2 . 1 ( S U F F I C I E N T C O N D I T I O N S F O R A M A X I M U M O R M I N I M U M ) Suppose that(x0, y0)is an interior stationary point for aC2 functionf (x, y) defined in a convex setSin⺢2.
(a) If for all(x, y)inS,
f11(x, y)≤0, f22(x, y)≤0, andf11(x, y)f22(x, y)−
f12(x, y)2
≥0 then(x0, y0)is a maximum point forf (x, y)inS.
(b) If for all(x, y)inS,
f11(x, y)≥0, f22(x, y)≥0, andf11(x, y)f22(x, y)−
f12(x, y)2
≥0 then(x0, y0)is a minimum point forf (x, y)inS.
S E C T I O N 1 3 . 2 / T W O V A R I A B L E S : S U F F I C I E N T C O N D I T I O N S 467 NOTE 1 The conditions in part (a) of Theorem 13.2.1 are sufficient for a stationary point to be a maximum point. They are far from being necessary. This is clear from the function whose graph is shown in Fig. 13.1.2, whichhasa maximum atP, but where the conditions in (a) are certainly not satisfied in the whole of its domain.
NOTE 2 If a twice differentiable function z = f (x, y) satisfies the inequalities in (a) throughout a convex setS, it is calledconcave, whereas it is calledconvexif it satisfies the inequalities in (b) throughoutS. It follows from these definitions thatf is concave if and only if−f is convex, just as in the one-variable case.
There are more general definitions of concave and convex functions which apply to functions that are not necessarily differentiable. These are presented in FMEA. (The one- variable case was briefly discussed in Section 8.7.)
E X A M P L E 2 Show that
f (x, y)= −2x2−2xy−2y2+36x+42y−158
has a maximum at the stationary point(x0, y0)=(5,8). (See Example 13.1.1.)
Solution: We found thatf1(x, y) = −4x−2y+36 andf2(x, y) = −2x−4y+42.
Furthermore,f11 = −4,f12 = −2, andf22 = −4. Thus f11(x, y)≤0, f22(x, y)≤0, f11(x, y)f22(x, y)−
f12(x, y)2
=16−4=12≥0 According to (a) in Theorem 13.2.1, these inequalities guarantee that the stationary point (5,8)is a maximum point.
E X A M P L E 3 Show that we have found the maximum in Example 13.1.4.
Solution: IfK >0 andL >0, we find that
πKK = −3K−3/2L1/4, πKL = 32K−1/2L−3/4, and πLL = −94K1/2L−7/4 Clearly,πKK <0,πLL <0, and moreover,
πKK πLL −(πKL )2 =274K−1L−3/2−94K−1L−3/2= 92K−1L−3/2>0 It follows that the stationary point(K, L)=(625,625)maximizes profit.
This section concludes with two examples that each involve a constraint. Nevertheless, a simple transformation can be used to convert the problem into the form we have been discussing, without any constraint.
E X A M P L E 4 Suppose that any production by the firm in Example 13.1.2 creates pollution, so it is legally restricted to produce a total of 320 units of the two kinds of output. The firm’s problem is then:
max −0.04x2−0.01xy−0.01y2+11x+7y−500 subject to x+y =320 What now are the optimal quantities of the two kinds of output?
Solution: The firm still wants to maximize its profits. But because of the restrictiony = 320−x, the new profit function is
ˆ
π (x)= −0.04x2−0.01x(320−x)−0.01(320−x)2+11x+7(320−x)−500 We easily findπˆ(x) = −0.08x +7.2, soπˆ(x) = 0 forx = 7.2/0.08 = 90. Since
ˆ
π(x)= −0.08<0 for allx, the pointx =90 does maximizeπˆ. The corresponding value ofyisy =320−90=230. The maximum profit is 1040.
E X A M P L E 5 A firm has three factories each producing the same item. Letx, y, and zdenote the respective output quantities that the three factories produce in order to fulfil an order for 2000 units in total. Hence,x+y+z=2000. The cost functions for the three factories are
C1(x)=200+ 1
100x2, C2(y)=200+y+ 1
300y3, C3(z)=200+10z The total cost of fulfilling the order is thus
C(x, y, z)=C1(x)+C2(y)+C3(z)
Find the values ofx,y, andzthat minimizeC. (Hint:Reduce the problem to one with only two variables by solvingx+y+z=2000 forz.)
Solution: Solving the equationx+y+z=2000 forzyieldsz=2000−x−y. Substituting this expression forzin the expression forCyields, after simplifying,
ˆ
C(x, y)=C(x, y,2000−x−y)= 1
100x2−10x+ 1
300y3−9y+20 600 Any stationary points ofCˆ must satisfy the two equations
ˆ
C1(x, y)= 1
50x−10=0, Cˆ2(x, y)= 1
100y2−9=0
The only solution isx = 500 andy = 30, implying thatz = 1470. The corresponding value ofCis 17 920.
The second-order partial derivatives areCˆ11(x, y)=501,Cˆ12(x, y)=0, andCˆ22(x, y)=
1
50y. It follows that for allx≥0,y≥0, one has ˆ
C11(x, y)≥0, Cˆ22(x, y)≥0, and Cˆ11(x, y)Cˆ22(x, y)−Cˆ12(x, y)2
= y 2500 ≥0 Part (b) of Theorem 13.2.1 implies that (500,30) is a minimum point of Cˆ within the convex domain of points(x, y)satisfyingx ≥0,y ≥0. It follows that(500,30,1470)is a minimum point ofC within the domain of(x, y, z)satisfyingx ≥0,y ≥0,z≥0, and x+y+z=2000.
S E C T I O N 1 3 . 2 / T W O V A R I A B L E S : S U F F I C I E N T C O N D I T I O N S 469
P R O B L E M S F O R S E C T I O N 1 3 . 2
1. Prove that the optimum has been found in: (a) Example 13.1.2; (b) Problem 13.1.1;
(c) Problem 13.1.3.
2. (a) A firm produces two different kindsAandBof a commodity. The daily cost of producing xunits ofAandyunits ofBis
C(x, y)=2x2−4xy+4y2−40x−20y+514
Suppose that the firm sells all its output at a price per unit of $24 forAand $12 forB. Find the daily production levelsxandythat maximize profit.
(b) The firm is required to produce exactly 54 units per day of the two kinds combined. What now is the optimal production plan?
⊂SM⊃3. Solve the utility-maximizing problem maxU=xyzsubject tox+3y+4z=108, by making Ua function ofyandzby eliminating the variablex.
4. The demands for a monopolist’s two products are determined by the equations p=25−x, q=24−2y
wherepandqare prices per unit of the two goods, andxandyare the corresponding quantities.
The costs of producing and sellingxunits of the first good andyunits of the other are C(x, y)=3x2+3xy+y2
(a) Find the monopolist’s profitπ(x, y)from producing and sellingxunits of the first good andyunits of the other.
(b) Find the values ofxandythat maximizeπ(x, y). Verify that you have found the maximum profit.
5. A firm produces two goods. The cost of producingxunits of good 1 andyunits of good 2 is C(x, y)=x2+xy+y2+x+y+14
Suppose that the firm sells all its output of each good at prices per unit ofpandqrespectively.
Find the values ofxandythat maximize profits. (Assume12p+12 < q <2p−1 andp >1.) 6. The profit function of a firm isπ(x, y)=px+qy−αx2−βy2, wherepandqare the prices per unit andαx2+βy2are the costs of producing and sellingxunits of the first good andy units of the other. The constants are all positive.
(a) Find the values ofxandythat maximize profits. Denote them byx∗andy∗. Verify that the second-order conditions are satisfied.
(b) Defineπ∗(p, q) =π(x∗, y∗). Verify that∂π∗(p, q)/∂p =x∗and∂π∗(p, q)/∂q =y∗. Give these results economic interpretations.
7. Find the smallest value ofx2+y2+z2when we require that 4x+2y−z=5. (Geometrically, the problem is to find the point in the plane 4x+2y−z=5 which is closest to the origin.) 8. Show thatf (x, y)=Axayb−px−qy−r(whereA,a,bare positive constants, andp,q, and
rare arbitrary constants) is concave forx >0,y >0 provided thata+b≤1. (See Note 2.)