trench lagrange multipliers
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William F. TrenchProfessor Emeritus
Department of Mathematics
San Antonio, Texas, USA
Copyright November 2012, WilliamF. Trench
This is a supplement to the authors
Introduction to Real Analysis
Reproduction and online posting are permitted for any valid noncommercial edu-
cational, mathematical, or scientific purpose. However, charges for profit beyond
reasonable printing costs are strictly prohibited. A complete instructors solutions
manual is available on request by email to the author, subject to verification of the
requestors faculty status.
THE METHOD OF LAGRANGE MULTIPLIERS
This is a revised and extended version of Section 6.5 of my Advanced Calculus (Harper
& Row, 1978). It is a supplement to my textbook Introduction to Real Analysis, which
is referenced via hypertext links.
To avoid repetition, it is to be understood throughout that f and g1, g2,. . . , gm are
continuously differentiable on an open setD in Rn.
Suppose thatm < n and
g1.X/ D g2.X/ D D gm.X/ D 0 (1)
on a nonempty subsetD1 ofD. If X0 2 D1 and there is a neighborhoodN of X0 suchthat
f .X/ f .X0/ (2)for every X inN\D1, thenX0 is a local maximum point of f subject to the constraints(1). However, we will usually say subject to rather than subject to the constraint(s).
If (2) is replaced by
f .X/ f .X0/; (3)then maximum is replaced by minimum. A local maximum or minimum of f
subject to (1) is also called a local extreme point of f subject to (1). More briefly, we
also speak of constrained local maximum, minimum, or extreme points. If (2) or (3)
holds for all X inD1, we omit local.
Recall that X0 D .x10; x20; : : : ; xn0/ is a critical point of a differentiable functionL D L.x1; x2; : : : ; xn/ if
Lxi .x10; x20; : : : ; xn0/ D 0; 1 i n:
Therefore, every local extreme point of L is a critical point of L; however, a critical
point of L is not necessarily a local extreme point of L (pp. 334-5).
Suppose that the system (1) of simultaneous equations can be solved for x1, . . . ,
xm in terms of the xmC1, . . . , xn; thus,
xj D hj .xmC1; : : : ; xn/; 1 j m: (4)
Then a constrained extreme value of f is an unconstrained extreme value of
f .h1.xmC1; : : : ; xn/; : : : ; hm.xmC1; : : : ; xn/; xmC1; : : : ; xn/: (5)
However, it may be difficult or impossible to find explicit formulas for h1, h2, . . . , hm,
and, even if it is possible, the composite function (5) is almost always complicated.
Fortunately, there is a better way to to find constrained extrema, which also requires
the solvability assumption, but does not require an explicit formulas as indicated in (4).
It is based on the following theorem. Since the proof is complicated, we consider two
special cases first.
Theorem 1 Suppose that n > m: If X0 is a local extreme point of f subject to
g1.X/ D g2.X/ D D gm.X/ D 0and
:::: : :
for at least one choice of r1 < r2 < < rm in f1; 2; : : : ; ng; then there are constants1; 2; . . . ; m such that X0 is a critical point of
f 1g1 2g2 mgmIthat is;
1 i n.The following implementation of this theorem is the method of Lagrange
(a) Find the critical points of
f 1g1 2g2 mgm;treating 1, 2, . . .m as unspecified constants.
(b) Find 1, 2, . . . , m so that the critical points obtained in (a) satisfy the con-
(c) Determine which of the critical points are constrained extreme points of f . This
can usually be done by physical or intuitive arguments.
If a and b1, b2, . . . , bm are nonzero constants and c is an arbitrary constant, then the
local extreme points of f subject to g1 D g2 D D gm D 0 are the same as the localextreme points of af c subject to b1g1 D b2g2 D D bmgm D 0. Therefore, wecan replace f 1g12g2 mgm by af 1b1g12b2g2 mbmgmcto simplify computations. (Usually, the c indicates dropping additive constants.)We will denote the final form by L (for Lagrangian).
3 Extrema subject to one constraint
Here is Theorem 1 withm D 1.
Theorem 2 Suppose that n > 1: If X0 is a local extreme point of f subject to g.X/ D0 and gxr .X0/ 0 for some r 2 f1; 2; : : : ; ng; then there is a constant such that
fxi .X0/ gxi .X0/ D 0; (7)
1 i nI thus; X0 is a critical point of f g:
Proof For notational convenience, let r D 1 and denote
U D .x2; x3; : : : xn/ and U0 D .x20; x30; : : : xn0/:
Since gx1.X0/ 0, the Implicit Function Theorem (Corollary 6.4.2, p. 423) impliesthat there is a unique continuously differentiable function h D h.U/; defined on aneighborhoodN Rn1 ofU0; such that .h.U/;U/ 2 D for allU 2 N , h.U0/ D x10,and
g.h.U/;U/ D 0; U 2 N: (8)Now define
which is permissible, since gx1 .X0/ 0. This implies (7) with i D 1. If i > 1,differentiating (8) with respect to xi yields
@xiD 0; U 2 N: (10)
@xiD @f .h.U/;U/
@xiC @f .h.U/;U/
@xi; U 2 N: (11)
Since .h.U0/;U0/ D X0, (10) implies that@g.X0/
@xiD 0: (12)
If X0 is a local extreme point of f subject to g.X/ D 0, then U0 is an unconstrainedlocal extreme point of f .h.U/;U/; therefore, (11) implies that
@xiC @f .X0/
@xiD 0: (13)
Since a linear homogeneous systema b
has a nontrivial solution if and only if a bc d
(Theorem 6.1.15, p. 376), (12) and (13) imply that@f .X0/
D 0; so
since the determinants of a matrix and its transpose are equal. Therefore, the system
has a nontrivial solution (Theorem 6.1.15, p. 376). Since gx1.X0/ 0, u must benonzero in a nontrivial solution. Hence, we may assume that u D 1, so
In particular,@f .X0/
@x1C v @g.X0/
@x1D 0; so v D fx1 .X0/
Now (9) implies that v D , and (14) becomes26664
Computing the topmost entry of the vector on the left yields (7).
Example 1 Find the point .x0; y0/ on the line
ax C by D d
closest to a given point .x1; y1/.
SolutionWe must minimizep.x x1/2 C .y y1/2 subject to the constraint. This
is equivalent to minimizing .x x1/2 C .y y1/2 subject to the constraint, which issimpler. For, this we could let
L D .x x1/2 C .y y1/2 .ax C by d/Ihowever,
L D .x x1/2 C .y y1/22
.ax C by/is better. Since
Lx D x x1 a and Ly D y y1 b;.x0; y0/ D .x1 C a; y1 C b/, where we must choose so that ax0 C by0 D d .Therefore,
ax0 C by0 D ax1 C by1 C .a2 C b2/ D d;so
D d ax1 by1a2 C b2 ;
x0 D x1 C.d ax1 by1/a
a2 C b2 ; and y0 D y1 C.d ax1 by1/b
a2 C b2 :
The distance from .x1; y1/ to the line is
p.x0 x1/2 C .y0 y1/2 D
jd ax1 by1jpa2 C b2
Example 2 Find the extreme values of f .x; y/ D 2x C y subject tox2 C y2 D 4:
L D 2xC y 2.x2 C y2/I
Lx D 2 x and Ly D 1 y;so .x0; y0/ D .2=; 1=/. Since x20 C y20 D 4, D
p5=2. Hence, the constrained
maximum is 2p5, attained at .4=
p5/, and the constrained minimum is 2
attained at .4=p5;2=
Example 3 Find the point in the plane
3xC 4y C D 1 (15)closest to .1; 1; 1/.
SolutionWe must minimize
f .x; y; / D .x C 1/2 C .y 1/2 C . 1/2
subject to (15). Let
L D .x C 1/2 C .y 1/2 C . 1/2
2 .3x C 4y C /I
Lx D x C 1 3; Ly D y 1 4; and L D 1 ;so
x0 D 1C 3; y0 D 1C 4; 0 D 1C :From (15),
3.1C 3/C 4.1C 4/C .1C / 1 D 1C 26 D 0;
so D 1=26 and.x0; y0; 0/ D
The distance from .x0; y0; 0/ to .1; 1; 1/ isp.x0 C 1/2 C .y0 1/2 C .0 1/2 D
Example 4 Assume that n 2 and xi 0, 1 i n.(a) Find the extreme values of
xi subject to
x2i D 1.
(b) Find the minimum value of
x2i subject to
xi D 1.
Solution (a) Let
Lxi D 1 xi ; so xi0 D1
; 1 i n:
x2i0 D n=2, so D pn and
.x10; x20; : : : ; xn0/ D
1pn;1pn; : : : ;
Therefore, the constrained maximum ispn and the constrain