THE METHOD OF LAGRANGE MULTIPLIERS

1 THE METHOD OFLAGRANGE MULTIPLIERSW illiam F. TrenchAndrew G. Cowles Distinguished Professor EmeritusDepartment of MathematicsTrinity UniversitySan Antonio, Texas, is a supplement to the author sIntroduction to Real Analysis. It has beenjudged to meet the evaluation criteria set by the Editorial Board of the AmericanInstitute of Mathematics in connection with the Institute sOpen Textbook may be copied, modified, redistributed, translated, and built upon subject to theCreative CommonsAttribution-NonCommercial-ShareAl ike Unported License. Acomplete instructor s solution manual is available by email to verification of the requestor s faculty METHOD OF LAGRANGE MULTIPLIERSW illiam F.

2 Trench1 ForewordThis is a revised and extended version of Section of myAdvanced Calculus(Harper& Row, 1978). It is a supplement to my textbookIntroduction to Real Analysis, whichis referenced via hypertext IntroductionTo avoid repetition, it is to be understood throughout thatfandg1,g2,..,gmarecontinuously differentiable on an open thatm < (1)on a nonempty subsetD1ofD. IfX02D1and there is a neighborhoodNofX0suchthatf .X/ f .X0/(2)for everyXinN\D1, thenX0isa local maximum point offsubject to the constraints(1). However, we will usually say subject to rather than subject to the constraint(s).

3 If (2) is replaced byf .X/ f .X0/;(3)then maximum is replaced by minimum. A local maximum or minimum offsubject to (1) is also called alocal extreme point offsubject to(1). More briefly, wealso speak ofconstrained local maximum, minimum, or extreme points. If (2) or (3)holds for allXinD1, we omit local. Recall ; x20; : : : ; xn0/is acritical pointof a differentiable ; x2; : : : ; ; x20; : : : ; xn0/D0; 1 i n:Therefore, every local extreme point ofLis a critical point ofL; however, a criticalpoint ofLis not necessarily a local extreme point ofL(pp.)

4 334-5).Suppose that the system (1) of simultaneous equations can be solved forx1, ..,xmin terms of thexmC1, ..,xn; thus, ; : : : ; xn/; 1 j m:(4)Then a constrained extreme value offis an unconstrained extreme value off . ; : : : ; xn/; : : : ; ; : : : ; xn/; xmC1; : : : ; xn/:(5)2 However, it may be difficult or impossible to find explicit formulas forh1,h2, ..,hm,and, even if it is possible, the composite function (5) is almost always , there is a better way to to find constrained extrema, which also requiresthe solvability assumption, but does not require an explicit formula as indicated in (4).

5 It is based on the following theorem. Since the proof is complicated, we consider twospecial cases 1 Suppose thatn > m:IfX0is a local extreme point offsubject @ @xr2 @ @ @xr2 @ @ @xr2 @xrm 0(6)for at least one choice ofr1< r2< < rminf1; 2; : : :; ng;then there are constants 1; 2;..; msuch thatX0is a critical point off 1g1 2g2 mgmIthat is;@f .X0/@xi @xi @xi @xiD0;1 i following implementation of this theorem is themethod ofLagrangemultipliers.(a)Find the critical points off 1g1 2g2 mgm;treating 1, 2.

6 Mas unspecified constants.(b)Find 1, 2, .., mso that the critical points obtained in (a) satisfy the con-straints.(c)Determine which of the critical points are constrained extreme points off. Thiscan usually be done by physical or intuitive ,b2, ..,bmare nonzero constants andcis an arbitrary constant, then thelocal extreme points offsubject tog1Dg2D DgmD0are the same as the localextreme points ofaf csubject tob1g1Db2g2D DbmgmD0. Therefore, wecan replacef 1g1 2g2 mgmbyaf 1b1g1 2b2g2 mbmgm cto simplify computations. (Usually, the c indicates dropping additive constants.)

7 We will denote the final form byL(forLagrangian).33 Extrema subject to one constraintHere is 2 Suppose thatn > 1:IfX0is a local extreme point offsubject 0for somer2 f1; 2; : : :; ng;then there is a constant such ;(7)1 i nIthus;X0is a critical point off g:ProofFor notational convenience, letrD1and ; x3; : : : ; x30; : : : xn0 0, the Implicit Function Theorem(Corollary , p. 423)impliesthat there is a unique continuously differentiable ;defined on aneighborhoodN Rn 1ofU0;such ;U/2 Dfor allU2N, , ;U/D0;U2N:(8)Now define ;(9)which is permissible, 0.

8 This implies (7) withiD1. Ifi > 1,differentiating (8) with respect @ @ @xiD0;U2N:(10)Also,@f . ;U//@xiD@f . ;U/@xiC@f . ;U/@ (10) implies @ @ @xiD0:(12)IfX0is a local extreme point offsubject , thenU0is an unconstrainedlocal extreme point off . ;U/; therefore, (11) implies that@f .X0/@xiC@f .X0/@ @xiD0:(13)Since a linear homogeneous system a bc d uv D 00 4has a nontrivial solution if and only if a bc d D0;(Theorem , p. 376), (12) and (13) imply that @f .X0/@xi@f .X0/@ @ @x1 D0;so @f .X0/@ @xi@f .X0/@ @x1 D0;since the determinants of a matrix and its transpose are equal.

9 Therefore, the system26664@f .X0/@ @xi@f .X0/@ @x137775 uv D 00 has a nontrivial solution(Theorem , p. 376). 0,umust benonzero in a nontrivial solution. Hence, we may assume thatuD1, so26664@f .X0/@ @xi@f .X0/@ @x137775 1v D 00 :(14)In particular,@f .X0/@ @x1D0;so :Now (9) implies that vD , and (14) becomes26664@f .X0/@ @xi@f .X0/@ @x137775 1 D 00 :Computing the topmost entry of the vector on the left yields (7).Example 1 Find the ; y0/on the lineaxCbyDdclosest to a given ; y1/.5 SolutionWe must x1 y1/2subject to the constraint. Thisis equivalent to x1 y1/2subject to the constraint, which issimpler.

10 For, this we could x1 y1/2 .axCby d/Ihowever, x1 y1/22 .axCby/is better. SinceLxDx x1 aandLyDy y1 b;.x0; y0 a; y1C b/, where we must choose so ,ax0 Cby0 Dax1 Cby1C .a2Cb2/Dd;so Dd ax1 by1a2Cb2; ax1 by1/aa2Cb2; ax1 by1/ba2Cb2:The distance ; y1/to the line x1 y1/2 Djd ax1 by1jpa2Cb2:Example 2 Find the extreme values off .x; y/D2xCysubject tox2Cy2D4:SolutionLetLD2xCy xandLyD1 y; ; y0 ; 1= /. Sincex20Cy20D4, D p5=2. Hence, the constrainedmaximum is2p5, attained ; 2=p5/, and the constrained minimum is 2p5,attained at. 4=p5; 2=p5/.