Chapter 12: MultivariableFunctionsand partial Derivatives
|.'
LagrangeMultipliers
i'
Free maximum
z=4 9-* 2 -y2
Constrained
maximum
x + 3 y -1 0 =
constraint on
x andy
As we saw in section 12.8, we sometimesneed to find the extreme values
of
a function whose domain is constrained to lie within some particular subset of
the plane-a disk, for example, or a closed triangular region. But, as Fig. 12.5g
suggests,a function may be subjectto other kinds of constraintsas well.
In this section, we explore a powerful method for finding extreme values of
constrained functions: the method of ktgrange multipliers. Lagrange developed
the method in 1755 to solve max-min problems in geometry. Today the method is
important in economics, in engineering (where it is used in designing multistage
rockets, for example), and in mathematics.
Constrained
Maximaand Minima
Find the point p (x, y, z) closest to the origin on rhe plane 2x a
EXAMPLE 1
y-z-5:0.
solution
The problem asks us to find the minimum value of the function
lo F l:W
: J x 2 + v 2 + it
x
12.58 The function f (x, y) : 49 - x2 - y2,
subject to the constraint g(x,y) :
x+3y-10 :0.
subject to the constraint that
2x* j -z-5:0.
Since lOFl has a minimum value wherever the tunction
f(x,y;Z):x2+ y' + z'
has a minimum value, we may solve the problem by finding the minimum value
of f (x,y, z) subjectto the constraint2x * y - z -5 : 0. If we regardx and y as
the independentvariables in this equation and write z as
z:2x* y-5,
our problem reducesto one of finding the points (x, y) at which the function
h(x,y) : f(x,y,2x
- 5) : x2 * y2 + (Zx + y _ 5) z
*)
has its minimum value or values. since the domain of ft is the entire ly-plane, the
first derivative test of Section 12.8 tells us that any minima that ft might have must
occur at points where
h, :2x
* 2(2x * y - 5)(2) : O,
hy:2y
* 2(2x f y - 5) : 0.
This leads to
l Ox * 4y:20,
4x * 4y:
l Q,
and the solution
J
-t-
5
6
we may apply a geometric argurnent together with the second derivative test to
show that these values minimize h. The z-coordinate of the corresponding point on
12.9 LagrangeMult iplier s 981
th e pl anez:2x
* y-5i s
/5\ s
z:2lll+
;- 5- - \31 6
5
6
Therefore, the point we seek is
.
Clo s eps o
t in t :
s\
/ss
r^ (r,
a . -O /
D
The distancefrom P to the origin is 5lJ6 x 2'04.
Attemptsto solvea constrainedmaximum or minimum problemby substitution,
as we might call the methodof Example l, do not always go smoothly.This is one
of the reasonsfor learning the new method of this section.
EXAMPLE 2
x 2 -22-1:0.
12.59 The hyperboliccylinder
x2- 22- 1: 0inE x a m p l e 2 .
Find the points closest to the origin on the hyperbolic cylinder
Solution 1 The cylinder is shown inFig. 12.59.We seekthe points on the cylinder
closest to the origin. These are the points whose coordinatesminimize the value of
the function
f(x, y, z) : xz + yz + z2
of thedistance
Square
subject to the constraintthat x2 - zz - 1 : 0. If we regard x and y as independent
variables in the constraint equation, then
z2:x2-l
a n d theval uesoff(x,y,Z):x2+ y' + z2onthecyl i nderaregi venbyt hef unct ion
h(x,y) : x2 * y2 + @ 2- 1) : 2x2 + Y 2 - l .
To find the points on the cylinder whose coordinates minimize /, we look for the
points in the xy-plane whose coordinatesminimize h.The only extremevalue of
ft occurs where
h, :4x
The hJperbolic cylinderx2 - z2 = |
- 0
and
hy:2Y
- 0,
that is, at the point (0, 0). But now we're in trouble-there are no points on the
cylinder where both x and y are zero. What went wrong?
What happenedwas that the first derivative test found (as it should have) the
point ln the domain of h whete ft has a minimum value. we, on the other hand, want
ihe points on the cylinder where h has a minimum value. While the domain of /r is
the entire xy-plane, the domain from which we can select the first two coordinates
o f t hepoi nts(x,y,z) onthecyl i nderi srestri ctedtothe" shadow"oft hecylinder
on the ry-plane; it doesnot include the band betweenthe lines x : -l and x : I
(Fig. 12.60).
We can avoid this problem if we treat y and z as independentvariables (instead
of x and y) and express.x in terms of y and z as
x2:22* 1.
12.60 The regionin the xy-planefrom
of the
which the first two coordinates
points(x,y,z) on the hyperboliccylinder
the band
x2 - z2: 1 are selectedexcludes
-1 < x < l int hex y -p l a n e .
With this substitution,f (x, y, z) : x2 + y' + z2 becomes
k(y, z) :
722* I) * yz * zz : | + y2 + 2zz
and we look for the points where ft takes on its smallest value. The domain of
982
Chapt er12: M u l ti v a ri a b l eF u n c ti o n sa n d Parti alD eri vati ves
k in the yz-plane now matches the domain from which we select the y- and
z-coordinatesof the points (x, y, z) on the cylinder.Hence,the points that minimize
ft in the plane will have correspondingpoints on the cylinder. The smallestvalues
of k occur where
kY :2Y -0
or where | : z :0.
and
kr:42-0,
This leadsto
x 2 : 2 2 *l : 1 ,
- - - Lt
The conesponding points on the cylinder are (* 1,0,0). We can see from the
inequality
k(Y .d:l + y2+ 222> l
that the points (* 1,0,0) give a minimum value for k. we can also see that the
minimum distance from the origin to a point on the cylinder is 1 unit.
" 2_ 2 2 _ l : 0
x 2+ y 2+ a2- o2=
solution 2 Another way to find the points on the cylinder closest to the origin is
to imagine a small spherecenteredat the origin expanding like a soap bubble until
it just touches the cylinder (Fig. 12.61). At each point of contact, the cylinder and
sphere have the same tangent plane and normal line. Therefore, if the sphere and
cylinder are representedas the level surfacesobtained by setting
g(x,y,Z):x2-z' -l
and
f(x,y,Z):x2+ y' ,+ 22-a2
equal to 0, then the gradients V/ and vg will be parallel where the surfacestouch.
At any point of contact we should therefore be able to find a scalar l. ("lambda")
such that
V /:
l V g'
or
2x i | 2y j * Zzk : ).(2x i - 2zk).
12.61 A sphereexpandinglike a soap
bubblecenteredat the origin until it just
touchesthe hyperboliccylinder
Thus, the coordinatesx, y, and z of any point of tangency will have to satisfy the
three scalarequations
x 2- 22- 1 :0 .
SeeSolution2 of Example2.
2x : 2).x,
2y :0,
2z : -2),2.
( l)
For what valuesof ), will a point (x, y, z) whose coordinatessatisfy the equations in (1) also lie on the surfacex2 - z2 - I : 0? To answer this question,we
use the fact that no point on the surface has a zero x-coordinate to conclude that
x l0 in the first equationin (l). This meansthat 2x :2)"x only if
2:2)' ,
or
,1": 1.
For )" : l, the equation 27 : -2)"2 becomes2z : -Zz. If this equation is to be
satisfiedas well, z must be zero. Since y : 0 also (from the equation2y :0), we
conclude that the points we seek all have coordinates of the form
(-r,0, 0).
What points on the surface x2 - zz : I have coordinates of this form? The points
(x, 0, 0) for which
x2 -
:1,
1072
x2 :1,
or
x : * 1.
The points on the cylinder closestto the origin are the points (* 1,0,0).
D
12.9 LagrangeM ult iplier s 983
Multipliers
The Methodof Lagrange
In Solution 2 of Example 2, we solved the problem by the method of Lagrange
multipliers. In generalterms,the methodsaysthat the extremevaluesof a function
: 0 are to be found
f (*, y, z) whosevariablesare subjectto a constraint,g(-r,y, z)
:
on the surf'ace8 0 at the points where
Y1 : )"Vg
for some scalar,tr(called a Lagrange multiplier).
To explorethe methodfurther and seewhy it works, we first make the following
observation,which we stateas a theorem.
Theorem9
Theorem
Gradient
TheOrthogonal
Supposethat /(x, ), z) is differentiablein a region whose interior contains
a smooth curve
C: r : s (t )i+ h (t )i+ f t (/ )k .
If Ps is a point on C where / has a local. maximum or minimuin relative
to its values on C, then V/ is orthogonal to C at Ps.
Proof We show that V.f is orthogonal to the curve's velocity vector at Ps. The
valuesof f on C are given by the composite/(S(t), h(t),k(t)), whosederivative
with respectto t is
d f : a f lg * { 4 ! " { 4 !
dt
0x dt
3y dt
3z dt
:vr.r.
At any point P6 where / has a local maximum or minimum relative to its values
SO
o n the curve,df l dt:0,
D
V ./.v:0.
By droppingthe z-termsin Theorem9, we obtain a similar result for functions
of two variables.
Corollaryof Theorem 9
At the pointson a smoothcurve r: g(/) i+h(t)i where a differentiable
function f (x, y) takeson its local maxima and minima relative to its values
on the curve,V / . v:0.
Theorem 9 is the key to the method of Lagrange multipliers. Suppose that
(x,y,z)
and g?,y,z) are differentiableand that Pe is a point on the surface
f
g@,y,2):0
where / has a local maximum or minimum value relative to its
other values on the surface. Then / takes on a local maximum or minimum at
Pe relative to its values on every differentiable curve through Ps on the surface
Therefore,V/ is orthogonalto the velocity vector of every such
C@,y,2):0.
differentiable curve through Ps. But so is Vg (becauseVg is orthogonal to the level
surface 8 : 0, as we saw in Section 12.7). Therefore, at Ps, V/ is some scalar
multiple I of V6.
984
Chapter '12:MultivariableFunctionsand partial Derivatives
The Method of Lagrange Multipliers
Supposethat f (x,),2) and g(r,y,a) are differentiable.To find
rhe local
maximum and minimum valuesof subjectto the constraintg(x,
/
!, z) :0,
find the valuesof x, y,z, and.i. that simultaneouslysatisfy the equations
yf :
Lyg
and
g(x, y, z) : 0.
For functions of two independentvariables,the appropriateequations
are
yf : ),V g
EXAMPLE3
and
:0.
E k, i
Find the greatestand smallestvaruesrharthe function
f(x' Y) : xY
takeson the ellipse(Fig. 12.62)
^-2 ,/
.,2
T- t:t'
12.62 Example3 showshow to find the
largestand smallestvaluesof the product
xy on this ellipse.
Solution We want the extremevaluesof (x,y):x),
f
x2
subjectto the constraint
r,2
s \ x , y ) : T * i_ r : 0 .
To do so, we first findthe valuesof x,y,
Vf : ),YS
and), for which
and
g(x, y) :0.
The gradient equation gives
Y i-f x i: 1 * i* ry i,
4
from which we find
l
J:4* ,
r
x:)' Y ,
and
A .V ::(,l V
):-vA'
.t
L'
a+
s: l'*:
so that ) : 0 or I : * 2. We now considerthesetwo cases.
Case1: If y :0,
then x : | :0. But (0,0) is not on the ellipse.Hence,y
+ 0.
Case2: If y * 0, then i_: * 2 and x : +2y.Substituting
this in the equation
s(x,y):0gi ves
and
v- * l
The function f (x, y) : .ry therefore takes on its extreme values
on the ellipse at
the four points (+ 2, l), (+2, -l). The extremevalues
are x! : 2 andxy _ _2.
12.53 When subjectedto the constraint
sU,9 : x2l8+ y2/2- I : o, the function
t(x,y): xy takeson extremevaluesat
the four points(+2,*1). Theseare the
pointson the ellipsewhen Vf (red)is a
scalarmultipleof vg (blue)(Example3).
The Geometry of the Solution
The level curvesof rhe funcrion (x, y) : xy
.f
are the hyperbolas xy -- c (Fig. 12.63). The farther the
hyperboras lie from the
origin, the larger the absolute value of
/. we want to find the extreme values of
f (x, y), given that the poinr (x, y) also lies on the ellipse xz + 4y2: g. Which
hyperbolas intersecting the ellipse lie farthest from the
origin? The hyperbolas that
Multipliers 985
12.9 Lagrange
just graze the ellipse, the ones that are tangentto it. At these points, any vector
is a m ult iple
normal to the hyperbola is normal to the ellipse, so Y f :yi * xj
(l : + 2) of Yg : (x 14)i + t' j. At the point (2, 1), for example,
vs: ji+ j.
Yf:i+ 2i,
and
Y f :2Yg.
and
Yf : -2Yg.
At the poi nt (-2,l ),
v f : i-2 j,
v g : -] i+ 1 ,
z
D
Findthemaximumandminimumvaluesofthefunction
EXAMPLE4
f(x, y) :
3x -l 4y on the circle x2 + y2 : 1.
Solution We model this as a Lagrangemultiplier problem with
f (x,y) -- 3x * 4y,
x2 + y2 - |
S O,9:
and look for the valuesof x, y, and I that satisfy the equations
V /:),V g
:
3i + 4j :2x),i * 2y),j ,
x2+ Y 2-l :0.
8(x' y):o:
and gives
The gradientequationimplies that )" l0
-t
z
y:;.
D.
These equationstell us, among other things, that r and y have the same sign. With
thesevaluesfor x and y, the equationg(x, )) :0 gives
z l r2
94
+
aiz u:
Thus.
so
g + 16 - 4),2,
I'
JJ
-
vf.:3i + 4i: zv8
6.
>l
:) * 3i
/.t\2
.(;)
(;,)
11
-l-_
{
-,:0
4),2 :
25,
and
), : + :.
2
24
., - _ - -{-'i"
)
and f(x,y) :3x l 4y has extremeval uesat (x,y) : + (3/5,4/ 5) .
By calculatingthe value of 3x * 4y at the points * (315, 415), we seethat its
maxlmum and minimum valueson the circle x2 + v2: 1 are
4l
))
/3\
x
3x + 4y : J
3x + 4y = - 5
12.64 The functionf(x,y) :3x + 4y takes
on its largestvalueon the unit circle
S(x,y) : x'+ y' - 1 : 0 at the point
(315,415)and its smallestvalueat the
point (-3l5, -4l5) (Example4). At each of
thesepoints,Vf is a scalarmultipleof
Vg. The figure showsthe gradientsat the
first point but not the second.
/4\
2s
' l s /* * l s /: s :5
and , ( j) . - ( - i)
')<
(FiS. 12.6q The level curvesof f (x,y):
The Geometry of the Solution
3x * 4y are the lines 3r * 4y : c. The farther the lines lie from the origin, the
larger the absolutevalue of /. We want to find the extreme values of /(x, y) given
l. Which lines intersecting
that the point (r,y) also lies on the circle x2 +y2:
the circle lie farthest from the origin? The lines tangent to the circle. At the points
of tangency, any vector normal to the line is normal to the circle, so the gradient
At t he
Y f :3i f
4j i s a mul ti pl e(),: * 512) of the gradi entY g:2"i+2y: .
point (3/5 , 4 I 5) , for example,
vf:3i+ 4i,
vs::t+;j,
and
vf:ttv s
E
986
Chapt e r1 2 : Mu l ti v a ri a b l eF u n c ti o n sa n d parti alD eri vati ves
LagrangeMultiplierswith Two Constraints
Many problemsrequire us to find the extreme values of a differentiablefunction
f (x, y, z) whose variablesare subjectto two constraints.If the constraintsare
gt@ , y, z) : 0
gz/, y, z) :0
and
and g1 and g2 are differentiable, with Vg1 not parallel to yg2, we find the constrained
local maxima and minima of / by introducing two Lagrangemultipliers I and pr
(mu, pronounced"mew"). That is, we locate the points p (x, y,
z) where / takes
on its constrainedextreme values by finding the values of x,y,z,)", and,p 111a1
simultaneouslysatisfy the equations
V/ : lVgr * ttYgz,
12.65 The vectorsVg1 and Vg2 lie in a
planeperpendicular
to the curveC
becauseVg1 is normalto the surface
9r : 0 and Vg2 is normalto the surface
9z : 0'
gr?, y, z) : 0,
8z@,y, z) : o.
(2)
The equationsin (2) have a nice geometricinterpretation.The surfacesgr :0
and g2: 0 (usually) intersectin a smooth curve, say c (Fig. 12.65),and alons
this curve we seek the points where / has local maximum and minimum valuei
relative to its other values on the curve. These are the points where V/ is normal
to c, as we saw in Theorem 9. But Vg1 and yg2 are also normal to c at these
points becauseC lies in the surfacesgr : 0 and
92:0. Therefore V/ lies in the
plane determined by Vg1 and Vg2, which means that V/ : lVgr *
ttygz for some
)' and p'. Since the points we seek also lie in both surfaces,their coordinates must
satisfy the equationsg{x, y, z) : 0 and g2(x, !, z) :0, which are the remainine
requirementsin Eqs. (2).
E XA MP LE 5
The pl ane xl yl z:l
cuts the cyl i nder,r +yr : l
in an
ellipse (Fig- 12.66). Find the points on the ellipse that lie closest to and farthest
from the origin.
Solution
We find the extreme values of
f(x,y,z):x2+ y2+ 22
(the square of the distance from (x, y, z) to the origin) subject to the constraints
gr?,y,Z):x2+ y2-l :O
(3)
8z@ ,y,2):x* y* z-l :0.
(4)
The gradient equation in (2) then gives
V/ : lVSr * ltYgz
2xi * 2y
j * 2zk:
2xi * 2y
j -t2zk
Eq (2)
L(2xi + 2y j ) + rr.(i + j + k)
- (Z).x* tD i * (2)" y* tt)j + pk
x- l!*7 :l
2x : 2)"x * l.t,
2y :2Ly
* 1t,
2z : F.
(s)
The scalarequationsin (5) yield
12.66 On the ellipsewherethe plane
and cylindermeet,what are the points
closestto and farthest from the origin
(Example5)?
2x : 2),xl2z
+
(l - X)x :7,
2y :2Ly *22
+
(l - L)y :7.
Equations (6) are satisfied simultaneously if either ). : I and z : 0 or ), I
.tr:. I : z/(l - )" ).
(6)
I and
12.9
Exercises
987
If z : 0, then solving Eqs.(3) and (4) simultaneouslyto find the corresponding
poinrson the ellipsegivesthe two points(1,0,0) and (0, 1,0). This makessense
when you look at Fig. 12.66.
If x : ), then Eqs. (3) and (4) give
x * x 11-
I :
0
z :
| -2x
z:
l+J2 .
6
,\L
^--2
pointson the ellipseare
The corresponding
,,:(+,+,'-t)
and
P->:
/-
IJ'
---:-,
|
t
\-
1
--
J' t+Jr)
2'
But here we need to be careful.While Pr and P2 both give local maxima of / on
the ellipse, P2 is farther from the origin than P1.
The points on the ellipse closestto the origin are (1, 0, 0) and (0, 1, 0). The
D
point on ihe ellipse farthest from the origin is P2.
12.9
Exercises
Two Independent Variableswith One Constraint
1. Find the pointson the ellipsex2 +2y2:
has its extreme values.
1 where f (x,y):
ay
2' Find the extreme values of f (r ' y) : ;ry subject to the constraint
s(x,y):x2 * Y2- lo: 0.
3. Find the maximum value of f (x, y) : 49 - x2 - yz on the line
x * 3 y :10 ( Fie. 12. 58) .
4. Find the local extreme values of f(x,y):r'y
on the line
x+Y:3.
5. Find the points on the curve x!2 : 54 nearestthe origin.
6. pind the points on the curve ,ty:2
nearestthe origin.
7. Use the method of Lagrange multipliers to find
a)
b)
the minimum value of x * y, subject to the constraints
xJ:16 ,'r > 0, Y> 0;
the maximum value of ;ry, subject to the constraint
x+Y:16 '
Comment on the geometry of each solution.
8. Find the points on the curve x2 + xy -f y2 :1 in the .ry-plane
that are nearest to and farthest from the origin.
9. Find the dimensions of the closed right circular cylindrical can
of smallestsurfacearea whose volume is 16rr cm310. Find the radius and height of the open right circular cylinder of
largest surface area that can be inscribed in a sphereof radius a.
What is the largest surface area?
11. Use the method of Lagrange multipliers to find the dimensions of
the rectangle of greatestarea that can be inscribed in the ellipse
x2 116+ y219 :1 with sidesparallel to the coordinateaxes'
12. Find the dimensions of the rectangle of largest perimeter that can
be inscribedin the ellipse x21a2 + y2lb2 :1 with sidesparallel
to the coordinate axes. What is the largest perimeter?
13. Find the maximum and minimum valuesof x2 + y2 subjectto
the constraintx2 - 2x I y2 - 4y :0.
14. Find the maximum and minimum values of 3x - y * 6 subject
to the constraintx2 * Y2 :4.
15. The temperatureat a point (x,)) on a metal plate is T(x'y):
4x2 - 4xy + y2. An ant on the plate walks around the circle of
radius 5 centered at the origin. What are the highest and lowest
temperaturesencounteredby the ant?
16. Your firm has been asked to design a storage tank for liquid
petroleum gas. The customer's specificationscall for a cylindrical
iank with hemispherical ends, and the tank is to hold 8000 m3
of gas. The customer also wants to use the smallest amount of
material possible in building the tank. What radius and height do
you recofirmend for the cylindrical portion of the tank?
with OneConstraint
Variables
ThreeIndependent
17. Find the point on the planex * 2y * 3z : 13 closestto the point
(1,1,1).
18. Find the point on the spherex2 + y2 * z2 :4
from the point (1, -1, l).
which is farthest