Lesson 17: The Method of Lagrange Multipliers

32,416 views
32,139 views

Published on

The Method of Lagrange multipliers allows us to find constrained extrema. It's more equations, more variables, but less algebra.

Published in: Technology, Economy & Finance
4 Comments
6 Likes
Statistics
Notes
No Downloads
Views
Total views
32,416
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
542
Comments
4
Likes
6
Embeds 0
No embeds

No notes for slide

Lesson 17: The Method of Lagrange Multipliers

  1. 1. Section 11.8 Lagrange Multipliers Math 21a March 14, 2008 Announcements ◮ Midterm is graded ◮ Office hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . Image: Flickr user Tashland . . . . . .
  2. 2. Announcements ◮ Midterm is graded ◮ Office hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . . . . .
  3. 3. Happy Pi Day! 3:14 PM Digit recitation contest! Recite all the digits you know of π (in order, please). Please let us know in advance if you’ll recite π in a base other than 10 (the usual choice), 2, or 16. Only positive integer bases allowed – no fair to memorize π in base π /(π − 2)... 4 PM — Pi(e) eating contest! Cornbread are square; pie are round. You have 3 minutes and 14 seconds to stuff yourself with as much pie as you can. The leftovers will be weighed to calculate how much pie you have eaten. Contests take place in the fourth floor lounge of the Math Department. . . Image: Flickr user Paul Adam Smith . . . . . .
  4. 4. Outline Introduction The Method of Lagrange Multipliers Examples For those who really must know all . . . . . .
  5. 5. The problem Last time we learned how to find the critical points of a function of two variables: look for where ∇f = 0. That is, ∂f ∂f = =0 ∂x ∂y Then the Hessian tells us what kind of critical point it is. . . . . . .
  6. 6. The problem Last time we learned how to find the critical points of a function of two variables: look for where ∇f = 0. That is, ∂f ∂f = =0 ∂x ∂y Then the Hessian tells us what kind of critical point it is. Sometimes, however, we have a constraint which restricts us from choosing variables freely: ◮ Maximize volume subject to limited material costs ◮ Minimize surface area subject to fixed volume ◮ Maximize utility subject to limited income . . . . . .
  7. 7. Example Maximize the function √ f( x , y ) = xy subject to the constraint g(x, y) = 20x + 10y = 200. . . . . . .
  8. 8. √ Maximize the function f(x, y) = xy subject to the constraint 20x + 10y = 200. . . . . . .
  9. 9. √ Maximize the function f(x, y) = xy subject to the constraint 20x + 10y = 200. Solution Solve the constraint for y and make f a single-variable function: 2x + y = 20, so y = 20 − 2x. Thus √ √ f(x) = x(20 − 2x) = 20x − 2x2 1 10 − 2x f ′ (x ) = √ (20 − 4x) = √ . 2 20x − 2x 2 20x − 2x2 Then f′ (x) = 0 when 10 − 2x = 0, or x = 5. Since y = 20 − 2x, y = 10. √ f(5, 10) = 50. . . . . . .
  10. 10. Checking maximality: Closed Interval Method Cf. Section 4.2 Once the function is restricted to the line 20x + 10y = 200, we can’t plug in negative numbers for f(x). Since √ f(x) = x(20 − 2x) we have a restricted domain of 0 ≤ x ≤ 10. We only need to check f on these two endpoints and its critical√ point to find the maximum value. But f(0) = f(10) = 0 so f(5) = 50 is the maximum value. . . . . . .
  11. 11. Checking maximality: First Derivative Test Cf. Section 4.3 We have 10 − 2x f ′ (x ) = √ 20x − 2x2 The denominator is always positive, so the fraction is positive exactly when the numerator is positive. So f′ (x) < 0 if x < 5 and f′ (x) > 0 if x > 5. This means f changes from increasing to decreasing at 5. So 5 is the global maximum point. . . . . . .
  12. 12. Checking maximality: Second Derivative Test Cf. Section 4.3 We have 100 f′′ (x) = − (20x − 2x2 )3/2 So f′′ (5) < 0, which means f has a local maximum at 5. Since there are no other critical points, this is the global maximum. . . . . . .
  13. 13. Example Find the maximum and minimum values of f (x, y) = x2 + y2 − 2x − 2y + 14. subject to the constraint g (x, y) = x2 + y2 − 16 ≡ 0. . . . . . .
  14. 14. Example Find the maximum and minimum values of f (x, y) = x2 + y2 − 2x − 2y + 14. subject to the constraint g (x, y) = x2 + y2 − 16 ≡ 0. This one’s harder. Solving for y in terms of x involves the square root, of which there’s two choices. . . . . . .
  15. 15. Example Find the maximum and minimum values of f (x, y) = x2 + y2 − 2x − 2y + 14. subject to the constraint g (x, y) = x2 + y2 − 16 ≡ 0. This one’s harder. Solving for y in terms of x involves the square root, of which there’s two choices. There’s a better way! . . . . . .
  16. 16. Outline Introduction The Method of Lagrange Multipliers Examples For those who really must know all . . . . . .
  17. 17. Consider a path that moves across a hilly terrain. Where are the critical points of elevation along your path? .. . . . . . .
  18. 18. Simplified map l .evel curves of f .evel curve g = 0 l - .9. . . . - - - . 10 876 5 4. 3. 2 . 1 -. - - - - - . . . . . . .
  19. 19. Simplified map l .evel curves of f .evel curve g = 0 l - .9. . . . - - - . 10 876 5 4. 3. 2 . 1 -. - - - - - . . . . . . .
  20. 20. Simplified map l .evel curves of f .evel curve g = 0 l . At the constrained critical point, the . tangents to the - .9. . . . - - - . 10 876 5 4. 3. 2 . 1 -. - - - - - level curves of f and g are in the same direction! . . . . . .
  21. 21. The slopes of the tangent lines to these level curves are ( ) ( ) dy f dy g = − x and =− x dx f fy dx g gy So they are equal when fx g f fy = x ⇐⇒ x = fy gy gx gy If λ is the common ratio on the right, we have fx g = x =λ fy gy So f x = λg x f y = λg y This principle works with any number of variables. . . . . . .
  22. 22. Theorem (The Method of Lagrange Multipliers) Let f(x1 , x2 , . . . , xn ) and g(x1 , x2 , . . . , xn ) be functions of several variables. The critical points of the function f restricted to the set g = 0 are solutions to the equations: ∂f ∂g (x1 , x2 , . . . , xn ) = λ (x1 , x2 , . . . , xn ) for each i = 1, . . . , n ∂ xi ∂ xi g (x 1 , x 2 , . . . , x n ) = 0 . Note that this is n + 1 equations in the n + 1 variables x1 , . . . , xn , λ. . . . . . .
  23. 23. Outline Introduction The Method of Lagrange Multipliers Examples For those who really must know all . . . . . .
  24. 24. Example √ Maximize the function f(x, y) = xy subject to the constraint 20x + 10y = 200. . . . . . .
  25. 25. Let’s set g(x, y) = 20x + 10y − 200. We have √ ∂f 1 y ∂g = = 20 ∂x 2 x ∂x √ ∂f 1 x ∂g = = 10 ∂y 2 y ∂y So the equations we need to solve are √ √ 1 y 1 x = 20λ = 10λ 2 x 2 y 20x + 10y = 200. . . . . . .
  26. 26. Solution (Continued) Dividing the first by the second gives us y = 2, x which means y = 2x. We plug this into the equation of constraint to get 20x + 10(2x) = 200 =⇒ x = 5 =⇒ y = 10. . . . . . .
  27. 27. Caution When dividing equations, one must take care that the equation we divide by is not equal to zero. So we should verify that there is no solution where √ 1 x = 10λ = 0 2 y If this were true, then λ = 0. Since y = 800λ2 x, we get y = 0. Since x = 200λ2 y, we get x = 0. But then the equation of constraint is not satisfied. So we’re safe. Make sure you account for these because you can lose solutions! . . . . . .
  28. 28. Example Find the maximum and minimum values of f (x, y) = x2 + y2 − 2x − 6y + 14. subject to the constraint g (x, y) = x2 + y2 − 16 ≡ 0. . . . . . .
  29. 29. Example Find the maximum and minimum values of f (x, y) = x2 + y2 − 2x − 6y + 14. subject to the constraint g (x, y) = x2 + y2 − 16 ≡ 0. Solution We have the two equations 2x − 2 = λ(2x) 2y − 6 = λ(2y). as well as the third x2 + y2 = 16. . . . . . .
  30. 30. Solution (Continued) Solving both of these for λ and equating them gives x−1 y−3 = . x y Cross multiplying, xy − y = xy − 3x =⇒ y = 3x. Plugging this in the equation of constraint gives x2 + (3x)2 = 16, √ √ which gives x = ± 8/5, and y = ±3 8/5. . . . . . .
  31. 31. Solution (Continued) Looking at the function f (x, y) = x2 + y2 − 2x − 2y + 14 We see that ( ) √ √ √ 94 + 10 5 f −2 2/5, −6 2/5 = 5 is the maximum and ( √ √ √ ) 94 − 10 5 f 2 2/5, 6 2/5 = 5 is the minimum value of the constrained function. . . . . . .
  32. 32. Contour Plot 4 2 The green curve is the 0 constraint, and the two green points are the constrained max and min. 2 4 4 2 0 2 4 . . . . . .
  33. 33. Compare and Contrast Elimination Lagrange Multipliers ◮ solve, then differentiate ◮ differentiate, then solve ◮ messier (usually) ◮ nicer (usually) equations equations ◮ more equations ◮ fewer equations ◮ adaptable to more than ◮ more complex with more one constraint constraints ◮ second derivative test ◮ second derivative test is (won’t do) is harder easier ◮ multipliers have contextual meaning . . . . . .
  34. 34. Example A rectangular box is to be constructed of materials such that the base of the box costs twice as much per unit area as does the sides and top. If there are D dollars allocated to spend on the box, how should these be allocated so that the box contains the maximum possible value? . . . . . .
  35. 35. Example A rectangular box is to be constructed of materials such that the base of the box costs twice as much per unit area as does the sides and top. If there are D dollars allocated to spend on the box, how should these be allocated so that the box contains the maximum possible value? Answer √ √ 1 D 1 D x=y= z= 3 c 2 c where c is the cost per unit area of the sides and top. . . . . . .
  36. 36. Solution Let the sides of the box be x, y, and z. Let the cost per unit area of the sides and top be c; so the cost per unit area of the bottom is 2c. If x and y are the dimensions of the bottom of the box, then we want to maximize V = xyz subject to the constraint that 2cyz + 2cxz + 3cxy − D = 0. Thus yz = λc(2z + 3y) xz = λc(3x + 2z) xy = λc(2x + 2y) . . . . . .
  37. 37. Before dividing, check that none of x, y, z, or λ can be zero. Each of those possibilities eventually leads to a contradiction to the constraint equation. Dividing the first two gives y 2z + 3y = =⇒ y(3x + 2z) = x(2z + 3y) =⇒ 2yz = 2xz x 3x + 2z Since z ̸= 0, we have x = y. . . . . . .
  38. 38. The last equation now becomes x2 = 4λcx. Dividing the second equation by this gives z 3x + 2z = =⇒ z = 3 x. 2 x 4x Putting these into the equation of constraint we have D = 3cxy + 2cyz + 2xz = 3cx2 + 3cx2 + 3cx2 = 9cx2 . So √ √ 1 D 1 D x=y= z= 3 c 2 c It also follows that √ x 1 D λ= = 4c 12 c3 . . . . . .
  39. 39. Interpretation of λ Let V∗ be the maximum volume found by solving the Lagrange multiplier equations. Then ( √ )( √ )( √ ) √ 1 D 1 D 1 D 1 D3 V∗ = = 3 c 3 c 2 c 18 c3 Now √ √ dV∗ 3 1 D 1 D = 3 = =λ dD 2 18 c 12 c3 This is true in general; the multiplier is the derivative of the extreme value with respect to the constraint. . . . . . .
  40. 40. Outline Introduction The Method of Lagrange Multipliers Examples For those who really must know all . . . . . .
  41. 41. The second derivative test for constrained optimization Constrained extrema of f subject to g = 0 are unconstrained critical points of the Lagrangian function L(x, y, λ) = f(x, y) − λg(x, y) The hessian at a critical point is   0 gx g y HL = gx fxx fxy  gy fxy fyy For (x, y, λ) to be minimal, we need det(HL) < 0, and for (x, y, λ) to be maximal, we need det(HL) > 0. . . . . . .

×