Lesson 22: Optimization Problems (handout)

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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.

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Lesson 22: Optimization Problems (handout)

  1. 1. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes Sec on 4.5 Op miza on Problems V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 18, 2011 . . Notes 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 . . Notes 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. . . . 1.
  2. 2. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes Leading by Example Example What is the rectangle of fixed perimeter with maximum area? Solu on Draw a rectangle. w . ℓ . . Notes 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. 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 . . Notes Solution Solu on (Con nued) 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). 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. . . . 2.
  3. 3. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes 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. . . Notes Outline The Text in the Box More Examples . . Notes 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) . . . 3.
  4. 4. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes 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. . . The Closed Interval Method Notes See 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. . . The First Derivative Test Notes See 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. . . . 4.
  5. 5. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 The First Derivative Test Notes See 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). . . Recall: The Second Derivative Test Notes See 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). . . Recall: The Second Derivative Test Notes See 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 . . . 5.
  6. 6. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes 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 – more work at boundary than CIM . . Notes 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. . . Notes Outline The Text in the Box More Examples . . . 6.
  7. 7. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes 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 . . Notes Solution Solu on 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] . . Notes Solution Solu on dQ p 6. = p − 4w, which is zero when w = . dw 4 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. . . . 7.
  8. 8. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes 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 . . . . Notes 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 . . Notes Diagram . . . 8.
  9. 9. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes Solution (Continued) . . Notes 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. . . Notes Solution . . . 9.
  10. 10. . V63.0121.001: Calculus I . Sec on 4.5: Op miza on Problems . April 18, 2011 Notes Solution (Concluded) . . Notes Summary Remember the checklist Ask yourself: what is the objec ve? Remember your geometry: similar triangles right triangles trigonometric func ons . . Notes . . . 10.

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