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# Lesson 20: (More) Optimization Problems

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What's the "best" design for a two-liter bottle? How do you get the best view of the Statue of Liberty?

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### Lesson 20: (More) Optimization Problems

1. 1. Section 4.6 More Optimization Problems Math 1a Introduction to Calculus March 21, 2008 Announcements ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 (not next week) ◮ Ofﬁce hours Tues, Weds, 2–4pm SC 323 (not next week) . . Image: Flickr user glassbeat . . . . . .
2. 2. Announcements ◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 (not next week) ◮ Ofﬁce hours Tues, Weds, 2–4pm SC 323 (not next week) . . . . . .
3. 3. Outline Modeling The Text in the Box More Examples Shortest Fence Norman Windows Worksheet Two-liter bottles The Statue of Liberty . . . . . .
4. 4. The Modeling Process . . Real-World . . Mathematical . f .ormulate Problems Model t .est s . olve . . Real-World . p . redict Mathematical . Predictions Conclusions . . . . . .
5. 5. Outline Modeling The Text in the Box More Examples Shortest Fence Norman Windows Worksheet Two-liter bottles The Statue of Liberty . . . . . .
6. 6. 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. . . . . . .
7. 7. Outline Modeling The Text in the Box More Examples Shortest Fence Norman Windows Worksheet Two-liter bottles The Statue of Liberty . . . . . .
8. 8. 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. . . . . . .
9. 9. . . w . . . ℓ f = 2ℓ + 3w A = ℓw ≡ 216 . . . . . .
10. 10. 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 ℓ= = = 18. The amount of fence needed is w 2 (√ ) √ √ 2A 2A 2A √ √ f =2· +3 = 2 6A = 2 6 · 216 = 72m 3 2 3 . . . . . .
11. 11. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + π feet of wood trim available. Discuss why a window designer might want to maximize the area of the window. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . . . . . . .
12. 12. Example A Norman window has the outline of a semicircle on top of a rectangle. Suppose there is 8 + π feet of wood trim available. Discuss why a window designer might want to maximize the area of the window. Find the dimensions of the rectangle and semicircle that will maximize the area of the window. . Answer The dimensions of the rectangular part are 2ft by 4ft. . . . . . .
13. 13. Solution We have to maximize A = ℓw + (w/2)2 π subject to the constraint that 2ℓ + w + π w = p. Solving for ℓ in terms of w gives ℓ = 1 (p − w − π w) 2 So A = 1 w(p − w − π w) + 1 π w2 . Differentiating gives 2 4 πw 1 1 A′ (w) = + (−1 − π)w + (p − π w − w) 2 2 2 p which is zero when w = . If p = 8 + 4π , w = 4. It follows 2+π that ℓ = 2. . . . . . .
14. 14. Outline Modeling The Text in the Box More Examples Shortest Fence Norman Windows Worksheet Two-liter bottles The Statue of Liberty . . . . . .
15. 15. Worksheet . . Image: Erick Cifuentes . . . . . .
16. 16. Two-liter bottles A two-liter soda bottle is roughly shaped like a cylinder with a spherical cap, and is made from a plastic called polyethylene terephthalate (PET). Its volume is ﬁxed at two liters. What dimensions of the bottle will minimize the cost of production? . . . . . .
17. 17. Solution I The volume of such a bottle is V = π r 2 h + 2 π r3 , 3 which is ﬁxed. Thus 2 V − 3 π r3 V 2 h= 2 = 2 − 3 r. πr πr The objective function is the surface area (since the material is all the same, cost is proportional to materials used) ( ) V − 2 π r2 2V 5 2 A = 2π rh + 3π r2 = 2π r 3 2 + 3 π r2 = + πr . πr r 3 . . . . . .
18. 18. Solution II The domain of this function is (0, ∞). To ﬁnd the critical points we need to ﬁnd dA 2V 10 −2V + 10 π r3 3 =− 2 + πr = . dr r 3 r2 dA The critical points are when = 0, or dr 10 3 0 = −2V + πr 3 ( )1 / 3 3V =⇒ r = . 5π . . . . . .
19. 19. Solution III Substituting into our expression for h tells us (after a lot of algebra) that h = r. That’s deﬁnitely a much squatter bottle that we see in the stores. So it’s not material costs that they’re minimizing (unless our shape is too off). . . . . . .
20. 20. The Statue of Liberty The Statue of Liberty stands on top of a pedestal which is on top of on old fort. The top of the pedestal is 47 m above ground level. The statue itself measures 46 m from the top of the pedestal to the tip of the torch. What distance should one stand away from the statue in order to maximize the view of the statue? That is, what distance will maximize the portion of the viewer’s vision taken up by the statue? . . . . . .
21. 21. Model I The angle subtended by the statue in the viewer’s eye can a be expressed as ( ) ( ) b a+b b θ θ = arctan −arctan . x x x . . . . . .
22. 22. Solution I Maximizing θ with respect to x is a simple matter of differentiation: dθ 1 −(a + b) 1 −b = ( )2 · − ( )2 · 2 dx a+b x2 b x 1+ x 1+ x b a+b = 2 − x2 + b x2 + (a + b)2 [ 2 ] [ ] x + (a + b)2 b − (a + b) x2 + b2 = (x2 + b2 ) [x2 + (a + b)2 ] . . . . . .
23. 23. Solution II This derivative is zero if and only if the numerator is zero, so we seek x such that [ ] [ ] 0 = x2 + (a + b)2 b − (a + b) x2 + b2 = a(ab + b2 − x2 ), or √ x= b(a + b). Using the ﬁrst derivative test, we see that dθ/dx > 0 if √ √ 0 < x < b(a + b) and dθ/dx < 0 if x > b(a + b). So this is deﬁnitely the absolute maximum on (0, ∞). . . . . . .
24. 24. Analysis and Discussion If we substitute in the numerical dimensions given, we have √ x = (46)(93) ≈ 66.1 meters This distance would put you pretty close to the front of the old fort which lies at the base of the island. Unfortunately, you’re not allowed to walk on this part of the lawn. √ The length b(a + b) is the geometric mean of the two distances measure from the ground—to the top of the pedestal (a) and the top of the statue (a + b). The geometric mean is of two numbers is always between them and greater than or equal to their average. . . . . . .