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# Lesson 22: Optimization I (Section 10 Version)

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We know the procedure for optimization. When applied to more real-world problems there is a wealth of questions to answer.

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### Lesson 22: Optimization I (Section 10 Version)

1. 1. Section 4.5 Optimization Problems V63.0121, Calculus I April 5, 2009 Announcements Quiz 5 is next week, covering Sections 4.1–4.4 I am moving to WWH 624 sometime next week (April 13th) Happy Opening Day! . . Image credit: wallyg . . . . . .
2. 2. Ofﬁce Hours and other help In addition to recitation Day Time Who/What Where in WWH M 1:00–2:00 Leingang OH 718/624 3:30–4:30 Katarina OH 707 5:00–7:00 Curto PS 517 T 1:00–2:00 Leingang OH 718/624 4:00–5:50 Curto PS 317 W 1:00–2:00 Katarina OH 707 2:00–3:00 Leingang OH 718/624 R 9:00–10:00am Leingang OH 718/624 5:00–7:00pm Maria OH 807 F 2:00–4:00 Curto OH 1310 I am moving to WWH 624 sometime next week (April 13th) . . . . . .
3. 3. Outline Leading by Example The Text in the Box More Examples . . . . . .
4. 4. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? . . . . . .
5. 5. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solution Draw a rectangle. . . . . . . . .
6. 6. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solution Draw a rectangle. . . . ℓ . . . . . .
7. 7. Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solution Draw a rectangle. . w . . . ℓ . . . . . .
8. 8. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. . . . . . .
9. 9. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is ﬁxed. . . . . . .
10. 10. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function of two variables, not one. But the perimeter is ﬁxed. p − 2w Since p = 2ℓ + 2w, we have ℓ = , 2 . . . . . .
11. 11. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function 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 · w = (p − 2w)(w) = pw − w2 A = ℓw = 2 2 2 . . . . . .
12. 12. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function 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 · w = (p − 2w)(w) = pw − w2 A = ℓw = 2 2 2 Now we have A as a function of w alone (p is constant). . . . . . .
13. 13. Solution (Continued) Let its length be ℓ and its width be w. The objective function is area A = ℓw. This is a function 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 · w = (p − 2w)(w) = pw − w2 A = ℓw = 2 2 2 Now we have A as a function of w alone (p is constant). The natural domain of this function is [0, p/2] (we want to make sure A(w) ≥ 0). . . . . . .
14. 14. Solution (Concluded) 1 pw − w2 on We use the Closed Interval Method for A(w) = 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. . . . . . .
15. 15. Solution (Concluded) 1 pw − w2 on We use the Closed Interval Method for A(w) = 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 = p − 2w. To ﬁnd the critical points, we ﬁnd dr 2 . . . . . .
16. 16. Solution (Concluded) 1 pw − w2 on We use the Closed Interval Method for A(w) = 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 = p − 2w. To ﬁnd the critical points, we ﬁnd dr 2 The critical points are when p 1 p − 2w =⇒ w = 0= 2 4 . . . . . .
17. 17. Solution (Concluded) 1 pw − w2 on We use the Closed Interval Method for A(w) = 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 = p − 2w. To ﬁnd the critical points, we ﬁnd dr 2 The critical points are when p 1 p − 2w =⇒ w = 0= 2 4 Since this is the only critical point, it must be the maximum. p In this case ℓ = as well. 4 . . . . . .
18. 18. Solution (Concluded) 1 pw − w2 on We use the Closed Interval Method for A(w) = 2 [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. dA 1 = p − 2w. To ﬁnd the critical points, we ﬁnd dr 2 The critical points are when p 1 p − 2w =⇒ w = 0= 2 4 Since this is the only critical point, it must be the maximum. p In this case ℓ = as well. 4 We have a square! The maximal area is A(p/4) = p2 /16. . . . . . .
19. 19. Outline Leading by Example The Text in the Box More Examples . . . . . .
20. 20. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? . . . . . .
21. 21. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. . . . . . .
22. 22. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. . . . . . .
23. 23. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols . . . . . .
24. 24. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols 5. If Q is a function of more than one “decision variable”, use the given information to eliminate all but one of them. . . . . . .
25. 25. The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. 3. Introduce Notation. 4. Express the “objective function” Q in terms of the other symbols 5. If Q is a function of more than one “decision variable”, use the given information to eliminate all but one of them. 6. Find the absolute maximum (or minimum, depending on the problem) of the function on its domain. . . . . . .
26. 26. The Closed Interval Method To ﬁnd the extreme values of a function f on [a, b], we need to: Evaluate f at the endpoints a and b Evaluate f at the critical points x where either f′ (x) = 0 or f is not differentiable at x. The points with the largest function value are the global maximum points The points with the smallest or most negative function value are the global minimum points. . . . . . .
27. 27. The First Derivative Test Theorem (The First Derivative Test) Let f be continuous on [a, b] and c a critical point of f in (a, b). If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local maximum. If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local minimum. If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. . . . . . .
28. 28. Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point 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. If f′′ (c) = 0, the second derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). . . . . . .
29. 29. Which to use when? CIM 1DT 2DT Pro no need for in- w o r k s on w o r k s on equalities non-closed, non-closed, gets global ex- non-bounded non-bounded trema automati- intervals intervals cally only one deriva- no need for in- tive equalities Con only for closed Uses inequalities More derivatives bounded inter- More work at less conclusive vals boundary than than 1DT CIM more work at boundary than CIM . . . . . .
30. 30. Which to use when? The bottom line Use CIM if it applies: the domain is a closed, bounded interval If domain is not closed or not bounded, use 2DT if you like to take derivatives, or 1DT if you like to compare signs. . . . . . .
31. 31. Outline Leading by Example The Text in the Box More Examples . . . . . .
32. 32. 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 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . .
33. 33. Solution 1. Everybody understand? . . . . . .
34. 34. 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 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? . . . . . .
35. 35. 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 800 m 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 . . . . . .
36. 36. 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 800 m 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 Objective: maximize area Constraint: ﬁxed fence length . . . . . .
37. 37. Solution 1. Everybody understand? . . . . . .
38. 38. Solution 1. Everybody understand? 2. Draw a diagram. . . . . . .
39. 39. Diagram . . . . . . . . . .
40. 40. Solution 1. Everybody understand? 2. Draw a diagram. . . . . . .
41. 41. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. . . . . . .
42. 42. Diagram . . . . . . . . . .
43. 43. Diagram . ℓ w . . . . . . . . . . .
44. 44. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. . . . . . .
45. 45. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. . . . . . .
46. 46. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 . . . . . .
47. 47. Solution 1. Everybody understand? 2. Draw a diagram. 3. Introduce notation: Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. 5. Since p = ℓ + 2w, we have ℓ = p − 2w and so Q(w) = (p − 2w)(w) = pw − 2w2 dQ p = p − 4w, which is zero when w = . 6. dw 4 (p) p2 p2 p = 80000m2 =p· −2· Q = 4 4 16 8 Since Q(0) = Q(p/2) = 0, this critical point is a maximum. . . . . . .
48. 48. 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? Solution 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. w The domain is all positive numbers. . . . . . .
49. 49. . . w . . . ℓ A = ℓw ≡ 216 f = 2ℓ + 3w . . . . . .
50. 50. Solution (Continued) So df 2A =− 2 +3 dw w √ 2A which is zero when w = . 3 Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the critical point is a minimum, in fact the global minimum. √ 2A So the area is minimized when w = = 12 and 3 √ A 3A = 18. The amount of fence needed is ℓ= = w 2 (√ ) √ √ √ √ 2A 2A 2A =2· = 2 6A = 2 6 · 216 = 72m f +3 3 2 3 . . . . . .
51. 51. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . . . . . . .
52. 52. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + 2π feet of wood trim available. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . Answer The dimensions are 4ft by 2ft. . . . . . .
53. 53. Solution Let h and w be the height and width of the window. We have π ( w )2 π L = 2h + w + w A = wh + 2 22 If L is ﬁxed to be 8 + 2π, we have 16 + 4π − 2w − πw h= , 4 so ( ) w 1π π A = (16 + 4π − 2w − πw) + w2 = (π + 4)w − w2 . + 4 8 28 ( ) π So A′ = (π + 4)w − 1 + , which is zero when 4 π+4 w= = 4 ft. The dimensions are 4ft by 2ft. 1+ π2 . . . . . .
54. 54. Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions . . . . . .