1 / 18

Chapter 8 Multivariable Calculus

Chapter 8 Multivariable Calculus. Section 4 Maxima and Minima Using Lagrange Multipliers. Learning Objectives for Section 8.4: Max/Min with Lagrange Multipliers. The student will be able to solve problems involving Lagrange multipliers for functions of two independent variables

belden
Télécharger la présentation

Chapter 8 Multivariable Calculus

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 8Multivariable Calculus Section 4 Maxima and Minima Using Lagrange Multipliers

  2. Learning Objectives for Section 8.4: Max/Min with Lagrange Multipliers • The student will be able to solve problems involving • Lagrange multipliers for functions of two independent variables • Lagrange multipliers for functions of three independent variables.

  3. Theorem 1 - Lagrange Multipliers 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.

  4. 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:

  5. 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:

  6. 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:

  7. 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.

  8. Example 2 The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6y0.4. Maximize N under the constraint that 30x + 60y = 300,000. Step 1. Formulate: Maximize N(x, y) = 10 x0.6y0.4subject to g(x, y) = 30x + 60y – 300,000 = 0 Step 2. Form the function F:

  9. Example 2 The Cobb-Douglas production function for a product is given by N(x, y) = 10 x0.6y0.4. Maximize N under the constraint that 30x + 60y = 300,000. Step 1. Formulate: Maximize N(x, y) = 10 x0.6y0.4subject to g(x, y) = 30x + 60y – 300,000 = 0 Step 2. Form the function F:

  10. Example 2(continued) Step 3. Find the critical points for F:

  11. Example 2(continued) Step 3. Find the critical points for F: Solving yields one critical point (6000, 2000, – 0.1289) Step 4. Conclusion:

  12. Example 2(continued) 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.

  13. 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.

  14. Example 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:

  15. Example 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:

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

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

  18. Summary • We learned how to use Lagrange multipliers in the equationto find local extrema for two independent variable problems. • We learned how to use Lagrange multipliers in the equationto find local extrema for three independent variable problems.

More Related