chapter 8 multivariable calculus

Chapter 8

Multivariable Calculus

Section 4

Maxima and Minima Using Lagrange

Multipliers

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Learning Objectives for Section 8.4: Max/Min with Lagrange Multipliers

The student will be able to solve problems involving

1. Lagrange multipliers for functions of two independent variables

2. Lagrange multipliers for functions of three independent variables.

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Step 1. Formulate the problem: Maximize (or minimize) z = f (x, y) subject to g (x, y) = 0.

Step 2. Form the function F:

Step 3. Find the critical points for F. That is, solve the system

Step 4. Conclusion: If (x0, y0, λ0) is the only critical point of F, we assume that (x0, y0) is the solution. If F has more than one critical point, we evaluate z = f (x, y) at each of them, to determine the maximum or minimum.

Theorem 1 - Lagrange Multipliers

( , , ) ( , ) ( , )F x y f x y g x y

Fx(x, y,) 0

Fy(x, y,) 0

F (x, y,) 0

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Example

Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10.

Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0

Step 2. Form the function F:

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Example

Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10.

Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0

Step 2. Form the function F:

Step 3. Find the critical points for F:

10225),( 22 yxyxyxF

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Example

Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10.

Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0

Step 2. Form the function F:

Step 3. Find the critical points for F:

Solving simultaneously yields one critical point at (4, 2, 4).

Step 4. Conclusion:

2 2 0, 2 0, 2 10 0x yF x F y F x y

10225),( 22 yxyxyxF

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Example

Maximize f (x, y) = 25 – x2 – y2, subject to 2x + y = 10.

Step 1. Formulate: Maximize z = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0

Step 2. Form the function F:

Step 3. Find the critical points for F:

Solving simultaneously yields one critical point at (4, 2, 4).

Step 4. Conclusion: Since (4, 2, 4) is the only critical point for F:

Max of f (x, y) with constraints = f (4, 2) = 25 – 42 – 22 = 5.

2 2 0, 2 0, 2 10 0x yF x F y F x y

10225),( 22 yxyxyxF

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Example 2

The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6 y0.4. Maximize N under the constraint that 30x + 60y = 300,000.

Step 1. Formulate: Maximize N(x, y) = 10 x0.6 y0.4 subject to g(x, y) = 30x + 60y – 300,000 = 0

Step 2. Form the function F:

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Example 2

The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6 y0.4. Maximize N under the constraint that 30x + 60y = 300,000.

Step 1. Formulate: Maximize N(x, y) = 10 x0.6 y0.4 subject to g(x, y) = 30x + 60y – 300,000 = 0

Step 2. Form the function F:

0.6 0.4( , , ) 10 (30 60 300,000)F x y x y x y

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Step 3. Find the critical points for F:

Example 2(continued)

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Step 3. Find the critical points for F:

Solving yields one critical point (6000, 2000, – 0.1289)

Step 4. Conclusion:

Example 2(continued)

0.4 0.4

0.6 0.6

6 30 0

4 60 0

30 60 300,000 0

x

y

F x y

F x y

F x y

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Step 3. Find the critical points for F:

Solving yields one critical point (6000, 2000, – 0.1289)

Step 4. Conclusion: Since (6000, 2000, – 0.1289) is the only critical point for F, we conclude that Max of N(x, y) under the given constraints = N(6000, 2000) = 10 · 6,0000.6 2,0000.4 = 38,664.

Example 2(continued)

0.4 0.4

0.6 0.6

6 30 0

4 60 0

30 60 300,000 0

x

y

F x y

F x y

F x y

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Functions of Three Independent Variables

Any local extremum of w = f (x, y, z) subject to the constraint g(x, y, z) = 0 will be among the set of points (x0, y0, z0, λ0) which are a solution to the system

Whereprovided that all the partial derivatives exist.This is an extension of the two-variable case.

( , , , ) ( , , ) ( , , )F x y z f x y z g x y z

Fx(x, y, z,) 0

Fy(x, y, z,) 0

Fz(x, y, z,) 0

F (x, y, z,) 0

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Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9.

Step 1. Formulate: Maximize w = f (x, y) = 25 – x2 – y2 subject to g(x, y) = 2x + y – 10 = 0

Step 2. Form the function F:

Example

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Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9.

Step 1. Formulate: Maximize w = f (x, y) = 2x + 4y + 4z subject to g(x, y) = x2 + y2 – z2 = 9

Step 2. Form the function F:

Step 3. Find the critical points for F:

Example

2 2 2( , , , ) 2 4 4 ( 9)F x y z x y z x y z

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Find the extrema of f (x, y, z) = 2x + 4y + 4z,subject to x2 + y2 + z2 = 9.

Step 1. Formulate: Maximize w = f (x, y) = 2x + 4y + 4z subject to g(x, y) = x2 + y2 – z2 = 9

Step 2. Form the function F:

Step 3. Find the critical points for F:

Example

2 2 2( , , , ) 2 4 4 ( 9)F x y z x y z x y z

2 2 2

2 2 0

4 2 0

4 2 0

9 0

x

y

z

F x

F y

F z

F x y z

Solving yields two critical points: (– 1, – 2, – 2, 1) and ( 1, 2, 2, –1)

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Example(continued)

Step 4. Conclusion.Since there are two candidates, we need to evaluate the function values:f (1, 2, 2) = 2 + 8 + 8 = 18f (–1, –2, –2) = – 2 – 8 – 8 = –18We conclude that the maximum of f occurs at (1, 2, 2), and the minimum of occurs at (–1, –2, –2).

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Summary

■ We learned how to use Lagrange multipliers in the equation

to find local extrema for two independent variable problems.

■ We learned how to use Lagrange multipliers in the equation

to find local extrema for three independent variable problems.

( , , ) ( , ) ( , )F x y f x y g x y

( , , , ) ( , , ) ( , , )F x y z f x y z g x y z