slide 6.5 - 1 copyright © 2008 pearson education, inc. publishing as pearson addison-wesley
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Slide 6.5 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Constrained Maximum and Minimum Values: Lagrange Multipliers
OBJECTIVES Find maximum and minimum values
using Lagrange multipliers. Solve applied problems involving
Lagrange multipliers.
6.5
Slide 6.5 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Method of Lagrange Multipliers To find a maximum or minimum value of a function f (x, y) subject to the constraint g(x, y) = 0:
1. Form a new function:F(x, y, λ) = f (x, y) – λg(x, y).
The variable λ (lambda) is called a Lagrange multiplier.
6.5 Constrained Maximum and
Minimum Values: Lagrange Multipliers
Slide 6.5 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
The Method of Lagrange Multipliers (continued)2. Find the first partial derivatives Fx, Fy, and Fλ.
3. Solve the systemFx = 0, Fy = 0, and Fλ = 0,
Let (a, b, λ) represent a solution of this system. We normally must determine whether (a, b, λ) yields a maximum or minimum of the function f. For the problems in this text, we will specify that a maximum or minimum exists.
The method of Lagrange multipliers can be extended to functions of three (or more) variables.
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 1: Find the maximum value ofA(x, y) = xy
subject to the constraint x + y = 20.
First note that x + y = 20 is equivalent to x + y – 20 = 0.
1. We form the new function, F, given byF(x, y, λ) = xy – λ·(x + y – 20).
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 1 (continued): 2. We find the first partial derivatives: Fx = y – λ
Fy = x – λ
Fλ = – ( x + y – 20)
3. We set each derivative equal to 0 and solve the resulting system:
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
y 0
x 0
(x y 20) 0
Slide 6.5 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 1 (concluded): From the first two equations, we can see that x = λ = y. Substituting x for y in the last equation, we get
Thus, y = x = 10. The maximum value of A subject to the constraint occurs at (10, 10) and is
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
10
202
020
x
x
xx
100
1010)10,10(
A
Slide 6.5 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 2: Find the maximum value off (x, y) = 3xy
subject to the constraint 2x + y = 8.
Note that we rewrite 2x + y = 8 as 2x + y – 8 = 0.
1. We form the new function, F, given byF(x, y, λ) = 3xy – λ(2x + y – 8).
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 2 (continued): 2. We find the first partial derivatives: Fx = 3y – 2λ
Fy = 3x – λ
Fλ = – (2x + y – 8)
3. We set each derivative equal to 0 and solve the resulting system:
0)82(
03
023
yx
x
y
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Example 2 (continued): Solving the second equation for λ , we get
λ = 3x.
Substituting this into the first equation gives
.2
63
0323
xy
xy
xy
Slide 6.5 - 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Example 2 (concluded): Substituting y = 2x into the third equation, we get
Then y = 2·2 = 4, and the maximum value of f subject to the constraint occurs at (2, 4) and is
.2
84
0822
x
x
xx
.24
423)4,2(
f
Slide 6.5 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3: The standard beverage can has a volume of 12 fl. oz, or 21.66 in3. What dimensions yield the minimum surface area? Find the minimum surface area. (Assume the shape of the can is a right circular cylinder.)
We want to minimize the function s, given bys(h, r) = 2πrh + 2πr2
subject to the volume constraint πr2h = 21.66 or πr2h – 21.66 = 0.
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3 (continued): 1. We form the new function, S, given by
S(h, r, λ) = 2πrh + 2πr2 – λ·(πr2h – 21.66).
2. We find the first partial derivatives: Sh = 2πr – λπr2
Sr = 2πh – 4πr – 2λπrh
Sλ = – (πr2h – 21.66)
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
Slide 6.5 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3 (continued): 3. We set each derivative equal to 0 and solve the
resulting system:
Since π is a constant, solve the first equation for r.
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
2r r2 0
2h 4r 2rh 0
(r2h 21.66) 0
2or 0
02or 0
0)2(
rr
rr
rr
Slide 6.5 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3 (continued): Since r = 0 cannot be a solution to the problem, we will continue by substituting 2/λ into the second equation.
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
2h 4 2
2
2
h 0
2h 8
4h 0
8
2h
4
h
Slide 6.5 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3 (continued): Since r = 2/λ and h = 4/λ, it follows that h = 2r. Substituting 2r for h in the third equation yields
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
r2 (2r) 21.66 0
2r3 21.66
r3 10.83
r 10.83
3 1.51 in
Slide 6.5 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example 3 (concluded): Thus, when r = 1.51 in., we have h = 3.02 in, and the surface area is then a minimum and is approximately
6.5 Constrained Maximum and Minimum Values: Lagrange Multipliers
2 (1.51)(3.02) 2 (1.51)2 42.98 in2