Your SlideShare is downloading. ×
0
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Lesson 22: Optimization (Section 021 slides)
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Lesson 22: Optimization (Section 021 slides)

603

Published on

Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.

Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.

Published in: Technology, Education
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
603
On Slideshare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
14
Comments
0
Likes
1
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. Section 4.5 Optimization Problems V63.0121.021, Calculus I New York University November 23, 2010 Announcements Turn in HW anytime between now and November 24, 2pm No Thursday recitation this week Quiz 5 on §§4.1–4.4 next week in recitation . . . . . .
  • 2. . . . . . . Announcements Turn in HW anytime between now and November 24, 2pm No Thursday recitation this week Quiz 5 on §§4.1–4.4 next week in recitation V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 2 / 31
  • 3. . . . . . . Objectives Given a problem requiring optimization, identify the objective functions, variables, and constraints. Solve optimization problems with calculus. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 3 / 31
  • 4. . . . . . . Outline Leading by Example The Text in the Box More Examples V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 4 / 31
  • 5. . . . . . . Leading by Example Example What is the rectangle of fixed perimeter with maximum area? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 5 / 31
  • 6. . . . . . . Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. . V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 5 / 31
  • 7. . . . . . . Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. .. ℓ V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 5 / 31
  • 8. . . . . . . Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solution Draw a rectangle. .. ℓ . w V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 5 / 31
  • 9. . . . . . . Solution Continued Let its length be ℓ and its width be w. The objective function is area A = ℓw. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 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. Since p = 2ℓ + 2w, we have ℓ = p − 2w 2 , V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 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. Since p = 2ℓ + 2w, we have ℓ = p − 2w 2 , so A = ℓw = p − 2w 2 · w = 1 2 (p − 2w)(w) = 1 2 pw − w2 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 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. Since p = 2ℓ + 2w, we have ℓ = p − 2w 2 , so A = ℓw = p − 2w 2 · w = 1 2 (p − 2w)(w) = 1 2 pw − w2 Now we have A as a function of w alone (p is constant). V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 14. . . . . . . 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. Since p = 2ℓ + 2w, we have ℓ = p − 2w 2 , so A = ℓw = p − 2w 2 · w = 1 2 (p − 2w)(w) = 1 2 pw − w2 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). V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 6 / 31
  • 15. . . . . . . Solution Concluded We use the Closed Interval Method for A(w) = 1 2 pw − w2 on [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 7 / 31
  • 16. . . . . . . Solution Concluded We use the Closed Interval Method for A(w) = 1 2 pw − w2 on [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. To find the critical points, we find dA dw = 1 2 p − 2w. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 7 / 31
  • 17. . . . . . . Solution Concluded We use the Closed Interval Method for A(w) = 1 2 pw − w2 on [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. To find the critical points, we find dA dw = 1 2 p − 2w. The critical points are when 0 = 1 2 p − 2w =⇒ w = p 4 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 7 / 31
  • 18. . . . . . . Solution Concluded We use the Closed Interval Method for A(w) = 1 2 pw − w2 on [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. To find the critical points, we find dA dw = 1 2 p − 2w. The critical points are when 0 = 1 2 p − 2w =⇒ w = p 4 Since this is the only critical point, it must be the maximum. In this case ℓ = p 4 as well. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 7 / 31
  • 19. . . . . . . Solution Concluded We use the Closed Interval Method for A(w) = 1 2 pw − w2 on [0, p/2]. At the endpoints, A(0) = A(p/2) = 0. To find the critical points, we find dA dw = 1 2 p − 2w. The critical points are when 0 = 1 2 p − 2w =⇒ w = p 4 Since this is the only critical point, it must be the maximum. In this case ℓ = p 4 as well. We have a square! The maximal area is A(p/4) = p2 /16. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 7 / 31
  • 20. . . . . . . Outline Leading by Example The Text in the Box More Examples V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 8 / 31
  • 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) V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 9 / 31
  • 22. . . . . . . The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 23. . . . . . . The Text in the Box 1. Understand the Problem. What is known? What is unknown? What are the conditions? 2. Draw a diagram. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 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 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 26. . . . . . . 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 27. . . . . . . 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 10 / 31
  • 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. crayons What number do I add to 5 to get 8? 8 - = 5 5 + 3 = 8 CHECK Does your answer make sense? Explain. Draw a picture to solve the problem. Write how many were given away. I. I had10 pencils. I gave some away. I have 3left. How many pencils did I give away? ~7 What number do I add to 3 to make 10? 13 i ft ill :i i ? 11 ft I '•' « I I ft A H 11 M i l U U U U> U U I I V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 11 / 31
  • 29. . . . . . . Recall: The Closed Interval Method See Section 4.1 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/most negative function value are the global minimum points. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 12 / 31
  • 30. . . . . . . Recall: The First Derivative Test See Section 4.3 Theorem (The First Derivative Test) Let f be continuous on (a, b) and c a critical point of f in (a, b). If f′ changes from negative to positive at c, then c is a local minimum. If f′ changes from positive to negative at c, then c is a local maximum. If f′ does not change sign at c, then c is not a local extremum. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 13 / 31
  • 31. . . . . . . Recall: The First Derivative Test See Section 4.3 Theorem (The First Derivative Test) Let f be continuous on (a, b) and c a critical point of f in (a, b). If f′ changes from negative to positive at c, then c is a local minimum. If f′ changes from positive to negative at c, then c is a local maximum. If f′ does not change sign at c, then c is not a local extremum. 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). V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 13 / 31
  • 32. . . . . . . Recall: The Second Derivative Test See Section 4.3 Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 14 / 31
  • 33. . . . . . . Recall: The Second Derivative Test See Section 4.3 Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous 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 derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 14 / 31
  • 34. . . . . . . Recall: The Second Derivative Test See Section 4.3 Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous 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 derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). 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 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 14 / 31
  • 35. . . . . . . Which to use when? CIM 1DT 2DT Pro – no need for inequalities – gets global extrema automatically – works on non-closed, non-bounded intervals – only one derivative – works on non-closed, non-bounded intervals – no need for inequalities Con – only for closed bounded intervals – Uses inequalities – More work at boundary than CIM – More derivatives – less conclusive than 1DT – more work at boundary than CIM V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 15 / 31
  • 36. . . . . . . Which to use when? CIM 1DT 2DT Pro – no need for inequalities – gets global extrema automatically – works on non-closed, non-bounded intervals – only one derivative – works on non-closed, non-bounded intervals – no need for inequalities Con – only for closed bounded intervals – Uses inequalities – More work at boundary than CIM – More derivatives – less conclusive than 1DT – more work at boundary than CIM 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. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 15 / 31
  • 37. . . . . . . Outline Leading by Example The Text in the Box More Examples V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 16 / 31
  • 38. . . . . . . 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? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 17 / 31
  • 39. . . . . . . Solution 1. Everybody understand? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 18 / 31
  • 40. . . . . . . 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? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 19 / 31
  • 41. . . . . . . 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 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 19 / 31
  • 42. . . . . . . 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 Objective: maximize area Constraint: fixed fence length V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 19 / 31
  • 43. . . . . . . Solution 1. Everybody understand? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 20 / 31
  • 44. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 20 / 31
  • 45. . . . . . . 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? . . .. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 21 / 31
  • 46. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 22 / 31
  • 47. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. Length and width are ℓ and w. Length of wire used is p. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 22 / 31
  • 48. . . . . . . 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? . . .. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 23 / 31
  • 49. . . . . . . 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? . . ... w . ℓ V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 23 / 31
  • 50. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. Length and width are ℓ and w. Length of wire used is p. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31
  • 51. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. Length and width are ℓ and w. Length of wire used is p. 4. Q = area = ℓw. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31
  • 52. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. 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 The domain of Q is [0, p/2] V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31
  • 53. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. 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 The domain of Q is [0, p/2] 6. dQ dw = p − 4w, which is zero when w = p 4 . V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31
  • 54. . . . . . . Solution 1. Everybody understand? 2. Draw a diagram. 3. 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 The domain of Q is [0, p/2] 6. dQ dw = p − 4w, which is zero when w = p 4 . Q(0) = Q(p/2) = 0, but Q (p 4 ) = p · p 4 − 2 · p2 16 = p2 8 = 80, 000m2 so the critical point is the absolute maximum. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31
  • 55. . . . . . . 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? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 25 / 31
  • 56. . . . . . . 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 f(w) = 2 A w + 3w. The domain is all positive numbers. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 25 / 31
  • 57. . . . . . . Diagram .... ℓ . w f = 2ℓ + 3w A = ℓw ≡ 216 V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 26 / 31
  • 58. . . . . . . Solution (Continued) We need to find the minimum value of f(w) = 2A w + 3w on (0, ∞). V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 27 / 31
  • 59. . . . . . . Solution (Continued) We need to find the minimum value of f(w) = 2A w + 3w on (0, ∞). We have df dw = − 2A w2 + 3 which is zero when w = √ 2A 3 . V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 27 / 31
  • 60. . . . . . . Solution (Continued) We need to find the minimum value of f(w) = 2A w + 3w on (0, ∞). We have df dw = − 2A w2 + 3 which is zero when w = √ 2A 3 . Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the critical point is a minimum, in fact the global minimum. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 27 / 31
  • 61. . . . . . . Solution (Continued) We need to find the minimum value of f(w) = 2A w + 3w on (0, ∞). We have df dw = − 2A w2 + 3 which is zero when w = √ 2A 3 . Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the critical point is a minimum, in fact the global minimum. So the area is minimized when w = √ 2A 3 = 12 and ℓ = A w = √ 3A 2 = 18. The amount of fence needed is f (√ 2A 3 ) = 2 · √ 3A 2 + 3 √ 2A 3 = 2 √ 6A = 2 √ 6 · 216 = 72m V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 27 / 31
  • 62. . . . . . . Try this one Example An advertisement consists of a rectangular printed region plus 1 in margins on the sides and 1.5 in margins on the top and bottom. If the total area of the advertisement is to be 120 in2 , what dimensions should the advertisement be to maximize the area of the printed region? V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 28 / 31
  • 63. . . . . . . Try this one Example An advertisement consists of a rectangular printed region plus 1 in margins on the sides and 1.5 in margins on the top and bottom. If the total area of the advertisement is to be 120 in2 , what dimensions should the advertisement be to maximize the area of the printed region? Answer The optimal paper dimensions are 4 √ 5 in by 6 √ 5 in. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 28 / 31
  • 64. . . . . . . Solution Let the dimensions of the printed region be x and y, P the printed area, and A the paper area. We wish to maximize P = xy subject to the constraint that A = (x + 2)(y + 3) ≡ 120 Isolating y in A ≡ 120 gives y = 120 x + 2 − 3 which yields P = x ( 120 x + 2 − 3 ) = 120x x + 2 −3x The domain of P is (0, ∞). .. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nam dapibus vehicula mollis. Proin nec tristique mi. Pellentesque quis placerat dolor. Praesent a nisl diam. Phasellus ut elit eu ligula accumsan euismod. Nunc condimentum lacinia risus a sodales. Morbi nunc risus, tincidunt in tristique sit amet, ultrices eu eros. Proin pellentesque aliquam nibh ut lobortis. Ut et sollicitudin ipsum. Proin gravida ligula eget odio molestie rhoncus sed nec massa. In ante lorem, imperdiet eget tincidunt at, pharetra sit amet felis. Nunc nisi velit, tempus ac suscipit quis, blandit vitae mauris. Vestibulum ante ipsum primis in faucibus orci luctus et ultrices posuere cubilia Curae; . 1.5 cm . 1.5 cm .1cm. 1cm . x . y V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 29 / 31
  • 65. . . . . . . Solution (Concluded) We want to find the absolute maximum value of P. dP dx = (x + 2)(120) − (120x)(1) (x + 2)2 − 3 = 240 − 3(x + 2)2 (x + 2)2 There is a single (positive) critical point when (x + 2)2 = 80 =⇒ x = 4 √ 5 − 2. The second derivative is d2 P dx2 = −480 (x + 2)3 , which is negative all along the domain of P. Hence the unique critical point x = ( 4 √ 5 − 2 ) cm is the absolute maximum of P. This means the paper width is 4 √ 5 cm, and the paper length is 120 4 √ 5 = 6 √ 5 cm. V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 30 / 31
  • 66. . . . . . . Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions 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. crayons What number do I add to 5 to get 8? 8 - = 5 5 + 3 = 8 CHECK Does your answer make sense? Explain. Draw a picture to solve the problem. Write how many were given away. I. I had10 pencils. I gave some away. I have 3left. How many pencils did I give away? ~7 What number do I add to 3 to make 10? 13 i ft ill :i i ? 11 ft I '•' « I I ft A H 11 M i l U U U U> U U I I V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 31 / 31

×