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- 1. Sec on 4.5 Op miza on Problems V63.0121.011: Calculus I Professor Ma hew Leingang New York University April 18, 2011.
- 2. Announcements Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm I am teaching Calc II MW 2:00pm and Calc III TR 2:00pm both Fall ’11 and Spring ’12
- 3. Objectives Given a problem requiring op miza on, iden fy the objec ve func ons, variables, and constraints. Solve op miza on problems with calculus.
- 4. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area?
- 5. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solu on
- 6. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solu on Draw a rectangle. .
- 7. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solu on Draw a rectangle. . ℓ
- 8. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solu on Draw a rectangle. w . ℓ
- 9. Solution Solu on (Con nued) Let its length be ℓ and its width be w. The objec ve func on is area A = ℓw.
- 10. Solution Solu on (Con nued) Let its length be ℓ and its width be w. The objec ve func on is area A = ℓw. This is a func on of two variables, not one. But the perimeter is ﬁxed.
- 11. Solution Solu on (Con nued) This is a func on of two variables, not one. But the perimeter is ﬁxed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2
- 12. Solution Solu on (Con nued) p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2 Now we have A as a func on of w alone (p is constant).
- 13. Solution Solu on (Con nued) p − 2w Since p = 2ℓ + 2w, we have ℓ = , so 2 p − 2w 1 1 A = ℓw = · w = (p − 2w)(w) = pw − w2 2 2 2 Now we have A as a func on of w alone (p is constant). The natural domain of this func on is [0, p/2] (we want to make sure A(w) ≥ 0).
- 14. Solution Solu on (Con nued) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2].
- 15. Solution Solu on (Con nued) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0.
- 16. Solution Solu on (Con nued) 1 We use the Closed Interval Method for A(w) = pw − w2 on 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 To ﬁnd the cri cal points, we ﬁnd = p − 2w. dw 2
- 17. Solution Solu on (Con nued) dA 1 To ﬁnd the cri cal points, we ﬁnd = p − 2w. dw 2 1 p The cri cal points are when p − 2w = 0, or w = . 2 4
- 18. Solution Solu on (Con nued) dA 1 To ﬁnd the cri cal points, we ﬁnd = p − 2w. dw 2 1 p The cri cal points are when p − 2w = 0, or w = . 2 4 Since this is the only cri cal point, it must be the maximum. In p this case ℓ = as well. 4
- 19. Solution Solu on (Con nued) dA 1 To ﬁnd the cri cal points, we ﬁnd = p − 2w. dw 2 1 p The cri cal points are when p − 2w = 0, or w = . 2 4 Since this is the only cri cal point, it must be the maximum. In p this case ℓ = as well. 4 We have a square! The maximal area is A(p/4) = p2 /16.
- 20. Outline The Text in the Box More Examples
- 21. Strategies for Problem Solving 1. Understand the problem 2. Devise a plan 3. Carry out the plan 4. Review and extend György Pólya (Hungarian, 1887–1985)
- 22. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons?
- 23. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons? 2. Draw a diagram.
- 24. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons? 2. Draw a diagram. 3. Introduce Nota on.
- 25. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons? 2. Draw a diagram. 3. Introduce Nota on. 4. Express the “objec ve func on” Q in terms of the other symbols
- 26. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons? 2. Draw a diagram. 3. Introduce Nota on. 4. Express the “objec ve func on” Q in terms of the other symbols 5. If Q is a func on of more than one “decision variable”, use the given informa on to eliminate all but one of them.
- 27. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the condi ons? 2. Draw a diagram. 3. Introduce Nota on. 4. Express the “objec ve func on” Q in terms of the other symbols 5. If Q is a func on of more than one “decision variable”, use the given informa on to eliminate all but one of them. 6. Find the absolute maximum (or minimum, depending on the problem) of the func on on its domain.
- 28. Polya’s Method in Kindergarten Name [_ Problem Solving Strategy Draw a Picture Kathy had a box of 8 crayons. She gave some crayons away. She has 5 left. How many crayons did Kathy give away? UNDERSTAND • What do you want to find out? Draw a line under the question. You can draw a picture to solve the problem. What number do I add to 5 to get 8? 8 - = 5 crayons 5 + 3 = 8 CHECK Does your answer make sense? Explain. What number Draw a picture to solve the problem. do I add to 3 Write how many were given away. to make 10? I. I had 10 pencils. ft ft ft A I gave some away. 13 ill i :i I • I I I have 3 left. How many i? « 11 I
- 29. The Closed Interval MethodSee Section 4.1 The Closed Interval Method To ﬁnd the extreme values of a func on f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the cri cal points x where either f′ (x) = 0 or f is not diﬀeren able at x. The points with the largest func on value are the global maximum points The points with the smallest/most nega ve func on value are the global minimum points.
- 30. The First Derivative TestSee Section 4.3 Theorem (The First Deriva ve Test) Let f be con nuous on (a, b) and c a cri cal point of f in (a, b). If f′ changes from nega ve to posi ve at c, then c is a local minimum. If f′ changes from posi ve to nega ve at c, then c is a local maximum. If f′ does not change sign at c, then c is not a local extremum.
- 31. The First Derivative TestSee Section 4.3 Corollary If f′ < 0 for all x < c and f′ (x) > 0 for all x > c, then c is the global minimum of f on (a, b). If f′ < 0 for all x > c and f′ (x) > 0 for all x < c, then c is the global maximum of f on (a, b).
- 32. Recall: The Second Derivative TestSee Section 4.3 Theorem (The Second Deriva ve Test) Let f, f′ , and f′′ be con nuous on [a, b]. Let c be in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum.
- 33. Recall: The Second Derivative TestSee Section 4.3 Theorem (The Second Deriva ve Test) Let f, f′ , and f′′ be con nuous on [a, b]. Let c be in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. Warning If f′′ (c) = 0, the second deriva ve test is inconclusive (this does not mean c is neither; we just don’t know yet).
- 34. Recall: The Second Derivative TestSee Section 4.3 Corollary If f′ (c) = 0 and f′′ (x) > 0 for all x, then c is the global minimum of f If f′ (c) = 0 and f′′ (x) < 0 for all x, then c is the global maximum of f
- 35. Which to use when? CIM 1DT 2DT Pro Con
- 36. Which to use when? CIM 1DT 2DT Pro – no need for inequali es Con
- 37. Which to use when? CIM 1DT 2DT Pro – no need for inequali es – gets global extrema automa cally Con
- 38. Which to use when? CIM 1DT 2DT Pro – no need for inequali es – gets global extrema automa cally Con – only for closed bounded intervals
- 39. Which to use when? CIM 1DT 2DT Pro – no need for – works on inequali es non-closed, – gets global extrema non-bounded automa cally intervals Con – only for closed bounded intervals
- 40. Which to use when? CIM 1DT 2DT Pro – no need for – works on inequali es non-closed, – gets global extrema non-bounded automa cally intervals – only one deriva ve Con – only for closed bounded intervals
- 41. Which to use when? CIM 1DT 2DT Pro – no need for – works on inequali es non-closed, – gets global extrema non-bounded automa cally intervals – only one deriva ve Con – only for closed – Uses inequali es bounded intervals
- 42. Which to use when? CIM 1DT 2DT Pro – no need for – works on inequali es non-closed, – gets global extrema non-bounded automa cally intervals – only one deriva ve Con – only for closed – Uses inequali es bounded intervals – More work at boundary than CIM
- 43. Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequali es non-closed, non-closed, – gets global extrema non-bounded non-bounded automa cally intervals intervals – only one deriva ve Con – only for closed – Uses inequali es bounded intervals – More work at boundary than CIM
- 44. Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequali es non-closed, non-closed, – gets global extrema non-bounded non-bounded automa cally intervals intervals – only one deriva ve – no need for inequali es Con – only for closed – Uses inequali es bounded intervals – More work at boundary than CIM
- 45. Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequali es non-closed, non-closed, – gets global extrema non-bounded non-bounded automa cally intervals intervals – only one deriva ve – no need for inequali es Con – only for closed – Uses inequali es – More deriva ves bounded intervals – More work at boundary than CIM
- 46. Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequali es non-closed, non-closed, – gets global extrema non-bounded non-bounded automa cally intervals intervals – only one deriva ve – no need for inequali es Con – only for closed – Uses inequali es – More deriva ves bounded intervals – More work at – less conclusive than boundary than CIM 1DT
- 47. Which to use when? If domain is closed and bounded, use CIM. If domain is not closed or not bounded, use 2DT if you like to take deriva ves, or 1DT if you like to compare signs.
- 48. Outline The Text in the Box More Examples
- 49. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
- 50. Solution Solu on 1. Everybody understand?
- 51. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions?
- 52. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Known: amount of fence used Unknown: area enclosed
- 53. Another Example Example (The Best Fencing Plan) A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? Known: amount of fence used Unknown: area enclosed Objec ve: maximize area Constraint: ﬁxed fence length
- 54. Solution Solu on 1. Everybody understand?
- 55. Solution Solu on 1. Everybody understand? 2. Draw a diagram.
- 56. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . .
- 57. Solution Solu on 1. Everybody understand? 2. Draw a diagram.
- 58. Solution Solu on 1. Everybody understand? 2. Draw a diagram. 3. Length and width are ℓ and w. Length of wire used is p.
- 59. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . .
- 60. Diagram A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? ℓ w . .
- 61. Solution Solu on 1. Everybody understand? 2. Draw a diagram. 3. Length and width are ℓ and w. Length of wire used is p.
- 62. Solution Solu on 4. Q = area = ℓw.
- 63. Solution Solu on 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2]
- 64. Solution Solu on 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 The domain of Q is [0, p/2] dQ p 6. = p − 4w, which is zero when w = . dw 4
- 65. Solution Solu on 7. Q(0) = Q(p/2) = 0, but (p) p p2 p2 Q =p· −2· = = 80, 000m2 4 4 16 8 so the cri cal point is the absolute maximum.
- 66. Solution Solu on 7. Q(0) = Q(p/2) = 0, but (p) p p2 p2 Q =p· −2· = = 80, 000m2 4 4 16 8 so the cri cal point is the absolute maximum. 8. The dimensions of the op mal rectangle are p p w= = 200 m2 and ℓ = = 400 m2 4 2
- 67. Your turn Example (The shortest fence) A 216m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of its sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed?
- 68. Your turn Example (The shortest fence) A 216m2 rectangular pea patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of its sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? Solu on Let the length and width of the pea patch be ℓ and w. The amount of fence needed is f = 2ℓ + 3w. Since ℓw = A, a constant, we have A f(w) = 2 + 3w. The domain is all posi ve numbers. w
- 69. Diagram w . ℓ f = 2ℓ + 3w A = ℓw ≡ 216
- 70. Solution (Continued) 2A We need to ﬁnd the minimum value of f(w) = + 3w on (0, ∞). w
- 71. Solution (Continued) 2A We need to ﬁnd the minimum value of f(w) = + 3w on (0, ∞). w √ df 2A 2A We have = − 2 + 3 which is zero when w = . dw w 3
- 72. Solution (Continued) 2A We need to ﬁnd the minimum value of f(w) = + 3w on (0, ∞). w √ df 2A 2A We have = − 2 + 3 which is zero when w = . dw w 3 Since f′′ (w) = 4Aw−3 , which is posi ve for all posi ve w, the cri cal point is a minimum, in fact the global minimum.
- 73. Solution (Continued) 2A We need to ﬁnd the minimum value of f(w) = + 3w on (0, ∞). w √ 2A So the area is minimized when w = = 12 and √ 3 A 3A ℓ= = = 18. w 2
- 74. Solution (Continued) 2A We need to ﬁnd the minimum value of f(w) = + 3w on (0, ∞). w √ 2A So the area is minimized when w = = 12 and √ 3 A 3A ℓ= = = 18. w 2 The amount of fence needed is (√ ) √ √ 2A 3A 2A √ √ f =2· +3 = 2 6A = 2 6 · 216 = 72m 3 2 3
- 75. Try this one Example An adver sement consists of a rectangular printed region plus 1 in margins on the sides and 1.5 in margins on the top and bo om. If the total area of the adver sement is to be 120 in2 , what dimensions should the adver sement be to maximize the area of the printed region?
- 76. Try this one Example An adver sement consists of a rectangular printed region plus 1 in margins on the sides and 1.5 in margins on the top and bo om. If the total area of the adver sement is to be 120 in2 , what dimensions should the adver sement be to maximize the area of the printed region? Answer √ √ The op mal paper dimensions are 4 5 in by 6 5 in.
- 77. Solution of the printed region be x and y, Let the dimensions P the printed area, and A the paper area. We wish to maximize P = xy subject to the constraint that 1.5 cm A = (x + 2)(y + 3) ≡ 120 Lorem ipsum dolor sit amet, consectetur adipiscing elit. 1 cm 1 cm 120 y Nam dapibus Isola ng y in A ≡ 120 gives y = − 3 which vehicula mollis. Proin nec tris que x+2 mi. Pellentesque quis placerat yields . dolor. Praesent ( ) 1.5 cm 120 120x x P=x −3 = − 3x x+2 x+2 The domain of P is (0, ∞).
- 78. Solution (Concluded) We want to ﬁnd the absolute maximum value of P. dP (x + 2)(120) − (120x)(1) 240 − 3(x + 2)2 = −3= dx (x + 2)2 (x + 2)2 There is a single (posi ve) √ cal point when cri (x + 2) = 80 =⇒ x = 4 5 − 2. 2 d2 P −480 The second deriva ve is 2 = , which is nega ve all dx (x + 2)3 along the domain of P. ( √ ) Hence the unique cri cal point x = 4 5 − 2 cm is the absolute maximum of P.
- 79. Solution (Concluded) ( √ ) Hence the unique cri cal point x = 4 5 − 2 cm is the absolute maximum of P. √ This means the paper width is 4 5 cm. 120 √ the paper length is √ = 6 5 cm. 4 5
- 80. Summary Name [_ Problem Solving Strategy Draw a Picture Kathy had a box of 8 crayons. She gave some crayons away. Remember the checklist She has 5 left. How many crayons did Kathy give away? UNDERSTAND Ask yourself: what is the • What do you want to find out? Draw a line under the question. objec ve? You can draw a picture to solve the problem. Remember your geometry: crayons What number do I add to 5 to get 8? 8 - 5 + 3 = 8 = 5 similar triangles CHECK Does your answer make sense? right triangles Explain. What number do I add to 3 Draw a picture to solve the problem. trigonometric func ons Write how many were given away. I. I had 10 pencils. to make 10? ft ft ft A I gave some away. 13 ill i :i I • I I I have 3 left. How many i? « 11 I pencils did I give away? I H 11 M i l ~7 U U U U> U U

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