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