Chapter 12: Multivariable Functions and partial Derivatives

|.'i'

Free maximum

Lagrange MultipliersAs we saw in section 12.8, we sometimes need to find the extreme values ofa function whose domain is constrained to lie within some particular subset ofthe plane-a disk, for example, or a closed triangular region. But, as Fig. 12.5gsuggests, a function may be subject to other kinds of constraints as well.

In this section, we explore a powerful method for finding extreme values ofconstrained functions: the method of ktgrange multipliers. Lagrange developedthe method in 1755 to solve max-min problems in geometry. Today the method isimportant in economics, in engineering (where it is used in designing multistagerockets, for example), and in mathematics.

Constrained Maxima and MinimaEXAMPLE 1 Find the point p (x, y, z) closest to the origin on rhe plane 2x ay-z-5:0.

solution The problem asks us to find the minimum value of the function

loFl :W:Jx2+v2+it

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 valueof f (x,y, z) subject to the constraint 2x * y - z -5 : 0. If we regard x and y asthe independent variables in this equation and write z as

z:2x*y-5,

our problem reduces to 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, thefirst derivative test of Section 12.8 tells us that any minima that ft might have mustoccur at points where

h, :2x *2(2x * y - 5)(2) : O, hy:2y *2(2x f y - 5) :0.

This leads to

lOx * 4y:20, 4x *4y: lQ,

and the solution

- t -

we may apply a geometric argurnent together with the second derivative test toshow that these values minimize h. The z-coordinate of the corresponding point on

z=49-*2-y2 Constrainedmaximum

x+3y-10=constraint onx andy

x

12.58 The function f (x, y) : 49 - x2 - y2,subject to the constraint g(x,y) :x+3y-10:0.

56J

12.59 The hyperbolic cylinderx2-22-1:0inExample2.

The hJperbolic cylinderx2 - z2 = |

12.60 The region in the xy-plane fromwhich the first two coordinates of thepoints (x,y,z) on the hyperbolic cylinderx2 - z2: 1 are selected excludes the band-1 <x<l inthexy-plane.

12.9 Lagrange Mult ip l iers 981

theplane z:2x *y-5is

Therefore, the point we seek is

. ^ /ss s\Closest point : r ( r ,a.-O/

The distance from P to the origin is 5lJ6 x 2'04. D

Attempts to solve a constrained maximum or minimum problem by substitution,

as we might call the method of Example l, do not always go smoothly. This is one

of the reasons for learning the new method of this section.

EXAMPLE 2 Find the points closest to the origin on the hyperbolic cylinder

x2-22-1:0.

Solution 1 The cylinder is shown inFig. 12.59. We seek the points on the cylinder

closest to the origin. These are the points whose coordinates minimize the value of

the function

f(x, y, z) : xz + yz + z2 Square of the distance

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

andthevaluesoff(x,y,Z):x2+y'+z2onthecyl inderaregivenbythefunct ion

h(x,y) : x2 * y2 + @2 - 1) : 2x2 + Y2 - l .

To find the points on the cylinder whose coordinates minimize /, we look for the

points in the xy-plane whose coordinates minimize h.The only extreme value of

ft occurs where

h, :4x - 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 happened was 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

of thepoints (x,y,z) onthecyl inder isrestr ictedtothe"shadow"of thecyl inder

on the ry-plane; it does not include the band between the lines x : -l and x : I

(Fig. 12.60).We can avoid this problem if we treat y and z as independent variables (instead

of x and y) and express .x in terms of y and z as

x2:22*1.

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

/5\ s 5z:2l l l+;-5---- \31 6 6

982 Chapter 12: Mult ivar iable Funct ions and Part ia l Der ivat ives

k in the yz-plane now matches the domain from which we select the y- andz-coordinates of the points (x, y, z) on the cylinder. Hence, the points that minimizeft in the plane will have corresponding points on the cylinder. The smallest valuesof k occur where

kY:2Y-0 and

or where | : z :0. This leads to

x2:22*l :1,

kr:42-0,

- - -Lt

"2_22_l :0

12.61 A sphere expanding l ike a soapbubble centered at the origin unti l i t justtouches the hyperbolic cylinder

x2-22-1:0.

See Solution 2 of Example 2.

The conesponding points on the cylinder are (* 1,0,0). We can see from theinequality

k(Y.d: l+y2+222>l

that the points (* 1,0,0) give a minimum value for k. we can also see that theminimum distance from the origin to a point on the cylinder is 1 unit.

solution 2 Another way to find the points on the cylinder closest to the origin isto imagine a small sphere centered at the origin expanding like a soap bubble untilit just touches the cylinder (Fig. 12.61). At each point of contact, the cylinder andsphere have the same tangent plane and normal line. Therefore, if the sphere andcylinder are represented as the level surfaces obtained by setting

f(x,y,Z):x2+y' ,+22-a2 and g(x,y,Z):x2-z ' - l

equal to 0, then the gradients V/ and vg will be parallel where the surfaces touch.At any point of contact we should therefore be able to find a scalar l. ("lambda")such that

V/: lVg'

or

2x i | 2y j * Zzk : ).(2x i - 2zk).

Thus, the coordinates x, y, and z of any point of tangency will have to satisfy thethree scalar equations

2x : 2) .x, 2y :0, 2z : -2) ,2. ( l )

For what values of ), will a point (x, y, z) whose coordinates satisfy the equa-tions in (1) also lie on the surface x2 - z2 - I : 0? To answer this question, weuse the fact that no point on the surface has a zero x-coordinate to conclude thatx l0 in the first equation in (l). This means that 2x :2)"x only if

2:2) ' , or ,1" : 1.

For )" : l, the equation 27 : -2)"2 becomes 2z : -Zz. If this equation is to besatisfied as well, z must be zero. Since y : 0 also (from the equation 2y :0), weconclude 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 - 1072 :1, x2 :1, or x : *1.

The points on the cylinder closest to the origin are the points (* 1,0,0). D

x2+y2+a2-o2=

The Method of Lagrange Mult ipl iersIn Solution 2 of Example 2, we solved the problem by the method of Lagrange

multipliers. In general terms, the method says that the extreme values of a function

f (*, y, z) whose variables are subject to a constraint ,g(-r, y, z) : 0 are to be found

on the surf'ace 8 : 0 at the points where

Y1 : )"Vg

for some scalar ,tr (called a Lagrange multiplier).

To explore the method further and see why it works, we first make the following

observation, which we state as a theorem.

Theorem 9The Orthogonal Gradient TheoremSuppose that /(x, ), z) is differentiable in 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

values of f on C are given by the composite /(S(t), h(t),k(t)), whose derivative

with respect to t is

df :af lg*{4!"{4! :vr . r .dt 0x dt 3y dt 3z dt

12.9 Lagrange Mult ip l iers 983

has a local maximum or minimum relative to its values

SO

At any point P6 where /on the curve, df ldt :0,

V./ .v:0.

By dropping the z-terms in Theorem 9, we obtain a similar result for functions

of two variables.

Corollary of Theorem 9At the points on a smooth curve r: g(/) i+h(t)i where a differentiable

function f (x, y) takes on 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

f (x,y,z) and g?,y,z) are differentiable and that Pe is a point on the surface

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

C@,y,2):0. Therefore, V/ is orthogonal to the velocity vector of every such

differentiable curve through Ps. But so is Vg (because Vg 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.

D

984 Chapter '12: Multivariable Functions and partial Derivatives

12.62 Example 3 shows how to find thelargest and smallest values of the productxy on this ell ipse.

The Method of Lagrange Multipliers

Suppose that f (x,),2) and g(r,y,a) are differentiable. To find rhe localmaximum and minimum values of / subject to the constraint g(x, !, z) :0,find the values of x, y,z, and.i. that simultaneously satisfy the equations

yf : Lyg and g(x, y, z) : 0.

For functions of two independent variables, the appropriate equations areyf : ) ,Vg and Ek, i :0.

EXAMPLE 3 Find the greatest and smallest varues rhar the function

f(x ' Y) : xY

takes on the ellipse (Fig. 12.62)

Solution We want the extreme values of f (x,y):x), subject to the constraint

x2 r,2s\x,y) :T* i_r :0.

To do so, we first findthe values of x,y, and ), for which

Vf : ),YS and g(x, y) :0.

The gradient equation gives

Yi-fx i :1* i*ry i ,4

from which we find

-2 . ,2

^ , /T-t : t '

lJ :4*, x:) 'Y, and

r . tA.- L '

V: : ( , lV):-v-- A'

a+

s: l ' * :

12.53 When subjected to the constraintsU,9 : x2l8 + y2/2- I : o, the functiont(x,y): xy takes on extreme values atthe four points (+2, *1). These are thepoints on the ell ipse when Vf (red) is ascalar multiple of vg (blue) (Example 3).

so that ) : 0 or I : * 2. We now consider these two cases.

Case 1: If y :0, then x : | :0. But (0,0) is not on the ell ipse. Hence, y + 0.Case 2: If y * 0, then i_: * 2 and x : +2y.Substituting this in the equations(x,y) :0gives

The function f (x, y) : .ry therefore takes on its extreme values on the ellipse atthe four points (+ 2, l), (+2, -l). The extreme values are x! : 2 and xy _ _2.

The Geometry of the Solution The level curves of rhe funcrion .f (x, y) : xyare the hyperbolas xy -- c (Fig. 12.63). The farther the hyperboras lie from theorigin, the larger the absolute value of /. we want to find the extreme values off (x, y), given that the poinr (x, y) also lies on the ellipse xz + 4y2 : g. Whichhyperbolas intersecting the ellipse lie farthest from the origin? The hyperbolas that

and v-* l

just graze the ellipse, the ones that are tangent to it.normal to the hyperbola is normal to the ellipse, so(l : + 2) of Yg : (x 14) i + t ' j . At the point (2, 1),

Yf: i+2i , vs: j i+ j . and

At the point ( -2, l ) ,

vf: i -2 j , vg:- ] i+1, andz

12.9 Lagrange Mult ipl iers 985

At these points, any vectorYf :y i*x j is a mult ip lefor example,

Yf :2Yg.

Yf : -2Yg. D

24., - _ - -{- -'i"

)

and ,( j ) . - ( - i )

EXAMPLE 4 Findthemaximumandminimumvaluesofthefunction f(x, y) :

3x -l 4y on the circle x2 + y2 : 1.

Solution We model this as a Lagrange multiplier problem with

f (x,y) - - 3x *4y, SO,9: x2 + y2 - |

and look for the values of x, y, and I that satisfy the equations

V/:) ,Vg : 3 i+4j :2x), i*2y), j ,

8(x 'y) :o: x2+Y2-l :0.

The gradient equation implies that )" l0 and gives

-t

D.

These equations tell us, among other things, that r and y have the same sign. Withthese values for x and y, the equation g(x, )) :0 gives

94so

aiz +

u: I '

Thus.

zy:; .

z l r2 / . t \2

( ; , ) . ( ; ) - , :0

g + 16 - 4),2, 4),2 : 25, and ), : + :.2

JJ- - l -_

11 {vf . :3 i + 4i : zv8- and f(x,y) :3x l4y has extreme values at (x,y) : +(3/5,4/5).

* 3i By calculating the value of 3x * 4y at the points * (3 15 , 415) , we see that its

maxlmum and minimum values on the circle x2 + v2: 1 are

6.>l:)4l) )

x/3\ /4\ 2s

' ls /** ls/ : s:5' )<

3x+4y:J

3x+4y=-5

12.64 The function f(x,y) :3x + 4y takeson its largest value on the unit circleS(x,y) : x'+ y' - 1 : 0 at the point(315, 415) and its smallest value at thepoint (-3l5, -4l5) (Example 4). At each ofthese points, Vf is a scalar multiple ofVg. The figure shows the gradients at thefirst point but not the second.

The Geometry of the Solution (FiS. 12.6q The level curves of f (x,y):

3x * 4y are the lines 3r * 4y : c. The farther the lines lie from the origin, thelarger the absolute value of /. We want to find the extreme values of /(x, y) giventhat the point (r,y) also lies on the circle x2 +y2: l. Which lines intersectingthe circle lie farthest from the origin? The lines tangent to the circle. At the pointsof tangency, any vector normal to the line is normal to the circle, so the gradientYf :3 i f 4 j is a mult ip le () , : * 512) of the gradient Yg:2" i+2y:. At thepoint (3 /5 , 4 I 5) , for example,

vf:3i+4i , vs: : t+; j , and vf : t tvs E

986 Chapter 12: Mult ivar iable Funct ions and part ia l Der ivat ives

12.65 The vectors Vg1 and Vg2 lie in aplane perpendicular to the curve Cbecause Vg1 is normal to the surface9r : 0 and Vg2 is normal to the surface9z:0'

Lagrange Multipliers with Two ConstraintsMany problems require us to find the extreme values of a differentiable functionf (x, y, z) whose variables are subject to two constraints. If the constraints are

gt@, y, z) : 0 and gz/ , y, z) :0

and g1 and g2 are differentiable, with Vg1 not parallel to yg2, we find the constrainedlocal maxima and minima of / by introducing two Lagrange multipliers I and pr(mu, pronounced "mew"). That is, we locate the points p (x, y, z) where / takeson its constrained extreme values by finding the values of x,y,z,)", and, p 111a1simultaneously satisfy the equations

V/ : lVgr * t tYgz, gr?, y, z) : 0, 8z@, y, z) : o. (2)

The equations in (2) have a nice geometric interpretation. The surfaces gr :0and g2: 0 (usually) intersect in a smooth curve, say c (Fig. 12.65), and alonsthis curve we seek the points where / has local maximum and minimum valueirelative to its other values on the curve. These are the points where V/ is normalto c, as we saw in Theorem 9. But Vg1 and yg2 are also normal to c at thesepoints because C lies in the surfaces gr : 0 and 92:0. Therefore V/ lies in theplane 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 mustsatisfy the equations g{x, y, z) : 0 and g2(x, !, z) :0, which are the remaininerequirements in Eqs. (2).

EXAMPLE 5 The plane xly lz: l cuts the cyl inder , r+yr: l in anellipse (Fig- 12.66). Find the points on the ellipse that lie closest to and farthestfrom 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

8z@,y,2):x*y*z- l :0.

The gradient equation in (2) then gives

V/ : lVSr * ltYgz Eq (2)

2xi*2y j*2zk: L(2xi+2y j ) +rr . ( i+ j+k)

2xi*2y j - t2zk - (Z).x * tDi* (2)"y * t t ) j+ pk

2x : 2)"x * l.t, 2y :2Ly * 1t, 2z : F.The scalar equations in (5) yield

(3)

(4)

2x : 2),x l2z

2y :2Ly *22

+ ( l - X)x :7,

+ ( l - L)y :7.

(s)

(6)12.66 On the ell ipse where the planeand cylinder meet, what are the pointsclosest to and farthest from the origin(Example 5)?

Equations (6) are satisfied simultaneously if either ). : I and z : 0 or ), I I and.tr : . I : z/ ( l - )") .

x- l !*7: l

Exercises 12.9 987

If z : 0, then solving Eqs. (3) and (4) simultaneously to find the corresponding

poinrs on the ell ipse gives the two points (1,0,0) and (0, 1,0). This makes sense

when you look at Fig. 12.66.If x : ), then Eqs. (3) and (4) give

x * x 11- I : 0

6, \L

^--2

The corresponding points on the ellipse are

,,:(+,+,'-t)

z : | -2x

z: l+J2.

/ -IJ ' J '

and P>: | - - - : - , - -- t 1 2 '\ -

t+Jr)

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 closest to the origin are (1, 0, 0) and (0, 1, 0). The

point on ihe ellipse farthest from the origin is P2. D

Exercises 12.9

Two Independent Variables with One Constraint1. Find the points on the el l ipse x2 +2y2: 1 where f (x,y): ay

has its extreme values.

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 *3y :10 (Fie. 12.58).

4. Find the local extreme values of f(x,y):r 'y on the l ine

x+Y:3.

5. Find the points on the curve x!2 : 54 nearest the origin.

6. pind the points on the curve ,ty:2 nearest the origin.

7. Use the method of Lagrange multipliers to find

a) the minimum value of x * y, subject to the constraints

xJ:16, ' r>0,Y>0;b) 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 smallest surface area whose volume is 16rr cm3-

10. Find the radius and height of the open right circular cylinder of

largest surface area that can be inscribed in a sphere of radius a.

What is the largest surface area?

11. Use the method of Lagrange multipliers to find the dimensions of

the rectangle of greatest area that can be inscribed in the ellipse

x2 116 + y2 19 :1 with sides parallel to the coordinate axes'

12. Find the dimensions of the rectangle of largest perimeter that can

be inscribed in the ellipse x2 1a2 + y2 lb2 :1 with sides parallel

to the coordinate axes. What is the largest perimeter?

13. Find the maximum and minimum values of x2 + y2 subject to

the constraint x2 - 2x I y2 - 4y :0.

14. Find the maximum and minimum values of 3x - y * 6 subject

to the constraint x2 * Y2 :4.

15. The temperature at 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

temperatures encountered by the ant?

16. Your firm has been asked to design a storage tank for liquid

petroleum gas. The customer's specifications call 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?

Three Independent Variables with One Constraint17. Find the point on the plane x * 2y * 3z : 13 closest to the point

(1,1,1).

18. Find the point on the sphere x2 + y2 * z2 :4 which is farthest

from the point (1, -1, l) .