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# Lesson 22: Optimization (Section 041 handout)

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### Lesson 22: Optimization (Section 041 handout)

1. 1. Section 4.4 Optimization Problems V63.0121.041, Calculus I New York University November 22, 2010 Announcements There is class on Wednesday, November 24 Announcements There is class on Wednesday, November 24 V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 2 / 31 Objectives Given a problem requiring optimization, identify the objective functions, variables, and constraints. Solve optimization problems with calculus. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 3 / 31 Notes Notes Notes 1 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
2. 2. Outline Leading by Example The Text in the Box More Examples V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 4 / 31 Leading by Example Example What is the rectangle of ﬁxed perimeter with maximum area? Solution Draw a rectangle. w V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 5 / 31 Solution Continued Let its length be and its width be w. The objective function is area A = w. This is a function of two variables, not one. But the perimeter is ﬁxed. Since p = 2 + 2w, we have = p − 2w 2 , so A = w = 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). V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 6 / 31 Notes Notes Notes 2 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
3. 3. 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 = p 4 as well. We have a square! The maximal area is A(p/4) = p2 /16. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 7 / 31 Outline Leading by Example The Text in the Box More Examples V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 8 / 31 Strategies for Problem Solving 1. Understand the problem 2. Devise a plan 3. Carry out the plan 4. Review and extend Gy¨orgy P´olya (Hungarian, 1887–1985) V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 9 / 31 Notes Notes Notes 3 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
4. 4. 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. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 10 / 31 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. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 12 / 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). V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 13 / 31 Notes Notes Notes 4 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
5. 5. 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 fV63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 14 / 31 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. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 15 / 31 Outline Leading by Example The Text in the Box More Examples V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 16 / 31 Notes Notes Notes 5 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
6. 6. 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 V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 17 / 31 Solution 1. Everybody understand? 2. Draw a diagram. 3. Length and width are and w. Length of wire used is p. 4. Q = area = w. 5. Since p = + 2w, we have = 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. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 18 / 31 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 V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 21 / 31 Notes Notes Notes 6 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
7. 7. 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 and w. The amount of fence needed is f = 2 + 3w. Since w = A, a constant, we have f (w) = 2 A w + 3w. The domain is all positive numbers. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 25 / 31 Diagram w f = 2 + 3w A = w ≡ 216 V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 26 / 31 Solution (Continued) V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 27 / 31 Notes Notes Notes 7 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
8. 8. 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. V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 28 / 31 Solution V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 29 / 31 Solution (Concluded) V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 30 / 31 Notes Notes Notes 8 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010
9. 9. Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions V63.0121.041, Calculus I (NYU) Section 4.4 Optimization Problems November 22, 2010 31 / 31 Notes Notes Notes 9 Section 4.4 : Optimization ProblemsV63.0121.041, Calculus I November 22, 2010