Lec8[1]Multiplicadores de Lagrange

8/13/2019 Lec8[1]Multiplicadores de Lagrange

1/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

MATH 209Calculus, III

Volker Runde

University of Alberta

Edmonton, Fall 2011


2/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Maxima and minima with constraints, I

Problem

Find the maximum or minimum off(x, y)under the constraintg(x, y) =C.

Geometric intuition

A maximum or minimum occurs if the tangent lines to the levelcurves ofg and f at (x0, y0) are parallel, i.e.,

f(x0, y0) =g(x0, y0)

for some .


3/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Maxima and minima with constraints, II

More formally. . .Suppose that f(x, y, z) has a maximum or minimum value atP(x0, y0, z0) on the surface Sgiven by g(x, y, z) =C.Let r(t) = x(t), y(t), z(t) be a curve passing through P, andlet t0 be such that r(t0) =

x0, y0, z0

. Set

h(t) :=f(x(t), y(t), z(t)); then h(t) has a maximum orminimum at t0. Hence:

0 =h(t0)

= fx(x0, y0, z0)x(t0) + fy(x0, y0, z0)y

(t0)

+f

z(x0, y0, z0)z

(t0)

=

f(x0, y0, z0)

r(t0).


4/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Maxima and minima with constraints, III

More formally. . . (continued)

This meansf(x0, y0, z0) r(t0)

and thus

f(x0, y0, z0) tangent plane to S at P.

If

g(x0, y0, z0)

= 0, this means that

f(x0, y0, z0) =g(x0, y0, z0)

for some , aLagrange multiplier.


5/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, I

Example

Find the minimum and maximum of

f(x, y) =x2 + 2y2

on the circle linex2 +y2 = 1.

Set g(x, y) =x

2

+y

2

, so that

f = 2x, 4y and g= 2x, 2y.


6/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, II

Example (continued)

Need to find x, y, and , such thatf(x, y) =g(x, y), i.e.,

2x=2x(1)4y=2y(2)

and

x2 +y2 = 1.(3)


7/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, III

Example (continued)

Case 1: x= 0.Then (1) implies = 1, so that 4y= 2yby (2), and thusy= 0. Then (3) yields x=

1.

Thus check (1, 0).Case 2: x= 0.By (3), y= 1, and = 2 by (2).Thus check (0,1).We have

f(1, 0) = 1 and f(0,1) = 2,

i.e., the minimum is 1 attained at (1, 0), and the maximum is2 attained at (0,

1).


8/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, IV

Example (continued)

What about the maximum and minimum of f on the disc

x

2

+y2

1?To check on x2 +y2


9/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, V

Example

Find the minimum and maximum distance of (0, 0, 1) from theellipsoid x2 + 2y2 +z2 = 4.Set

f(x, y, z) =

x2 +y2 + (z 1)2x2 +y2 + (z 1)2

= (distance of (0, 0, 1) and (x, y, z))2

andg(x, y, z) =x2 + 2y2 +z2.

Need to find x, y, z, and such thatf(x, y, z) =g(x, y, z) and g(x, y, z) = 4.


10/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, VI

Example (continued)

As

f =

2x, 2y, 2z

2

and

g=2x, 4y, 2z

,

this amounts to solving:

2x=2x,(1)

2y=4y,(2)2z 2 =2z,(3)

x2 + 2y2 +z2 = 4.(4)


11/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, VII

Example (continued)Case 1: x= 0.By (1), we have = 1, so that z 1 =zby (3), which isimpossible.Case 2: x= 0.Case 2.1: y= 0. By (2), we have = 1

2. By (3), we have

2z 2 =z, i.e., z= 2. By (4),

0 + 2y2 + 4 = 4

and thus y= 0, which is a contradiction.Case 2.2: y= 0. In this case, (4) yields z2 = 4, i.e., z= 2.Check at (0, 0,2):

f(0, 0, 2) = 1 and f(0, 0,

2) = 9.


12/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, VIII

Example

Find the maximum and minimum of

f(x, y) =x2 2xy

under the constraint

g(x, y) :=1

2x2 +

2

3y2 = 2.

We havef = 2x 2y,2x

and

g=x,

4

3y

.


13/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, IX

Example (continued)

We need x, y, and satisfying

2x 2y=x,(1)2x=4

3y,(2)

1

2x2 +

2

3y2 = 2.(3)

Case 1: = 0.Then (2) yields x= 0, so that y= 0 by (1). This contradicts(3).


14/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, X

Example (continued)

Case 2: = 0.Then (2) yields 2y= 3

. Plugging into (1) gives

2 + 3

x=x.

Case 2.1: x= 0. Then y= 0 by (1), which contradicts (3).Case 2.2: x= 0. Then = 2 + 3

, i.e.,

0 =2 2 3 = ( 1)2 4.

This means= 3 or = 1.


15/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XI

Example (continued)

Case 2.2.1: = 3. Then (2) yields2x= 4y, i.e., x= 2y.Plugging into (3) leads to 2y2 + 2

3y2 = 2, i.e., y2 = 3

4, so that

y= 32 .Ify=

3

2 , then x= 3. Ify=

3

2 , then x=

3.

Test3,

3

2

and

3,

3

2

, and obtain

f

3, 3

2

=f

3,3

2

= 6.


16/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XII

Example (continued)

Case 2.2.2: = 1. Then (2) yields2x= 43y, i.e., x= 2

3y.

Plugging into (3) leads to 29y2 + 2

3y2 = 2, i.e., y2 = 9

4, so that

y= 32

.Ify= 3

2, then x= 1. Ify= 3

2, then x= 1.

Test

1, 32

and

1,32

, and obtain

f

1,32

=f1,32 = 2.


17/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XIII

Example (continued)

What is the maximum and minimum offon the ellipse1

2x2

+

2

3y2

2?We know it on 12x2 + 2

3y2 = 2.

For 12x2 + 2

3y2


18/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XIV

Example (continued)

Second derivative test:

fxx= 2, fyy = 0, and fxy = 2.Then

D= 2 0 (2)2 = 4,

i.e., fhas a saddle at (0, 0).Hence, all maxima and minima off on 1

2x2 + 2

3y2 2 occur

on the boundary of the ellipse.


19/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Maxima and minima with two constraints

The case of two constraints

Suppose that f(x, y, z) has a maximum or minimum at(x0, y0, z0) under the constraints g(x, y, z) =C1 andh(x, y, z) =C2.

Ifg(x0, y0, z0) andh(x0, y0, z0) are not zero and notparallel, there are and with

f(x0, y0, z0) =g(x0, y0, z0) +h(x0, y0, z0).

Hence, find x, y, z, , and with

f(x, y, z) =g(x, y, z) +h(x, y, z),

and check the values off at (x, y, z).


20/23

MATH 209

Calculus,III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XV

Example

Find the maximum value of

f(x, y, z) =x+ 2y+ 3z

on the intersection of the plane x y+z= 1 and the cylinderx2 +y2 = 1.Set

g(x, y, z) =x2 +y2 and h(x, y, z) :=x y+z,

so that

f := 1, 2, 3, g= 2x, 2y, 0,and

h=

1,

1, 1

.


21/23

MATH 209

Calculus,

III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XVI

Example (continued)

Find x, y, z, , and such that:

1 =2x+,(1)

2 =2y 3,(2)3 =,(3)

x

y+z= 1,(4)

x2 +y2 = 1.(5)


22/23

MATH 209

Calculus,

III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XVII

Example (continued)

By (3), = 3, and by (1) = 0 and x= 1

.By (2), y= 5

2.

Plugging into (5) yields 12

+ 2542

= 1, i.e., 2 = 294

and thus

= 292

.It follows that

x= 229

, y= 529

and

z= 1 x+y= 1 729

.


23/23

MATH 209

Calculus,

III

Volker Runde

Lagrange

multipliers

Two

constraints

Examples, XVIII

Example (continued)

Test f at 2

29,

529

, 1 + 7

29

and

229

, 529

, 1 729

.

The corresponding values off are

229

+ 2

5

29

+ 3

1 7

29

= 3 29.

The maximum is thus 3 +

29.

Lec8[1]Multiplicadores de Lagrange

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