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# Lesson 24: Optimization II

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More examples of optimization problems (from the Wednesday class, but Thursday students are welcome to look at it!)

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### Lesson 24: Optimization II

1. 1. Section 4.5 Optimization II V63.0121.034, Calculus I November 25, 2009 Announcements Final Exam, December 18, 2:00–3:50pm . . . . . .
2. 2. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
3. 3. Checklist for optimization problems 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. . . . . . .
4. 4. Recall: The Closed Interval Method See Section 4.1 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. . . . . . .
5. 5. Recall: The First Derivative Test See Section 4.3 Theorem (The First Derivative Test) Let f be a continuous function and c a critical point of f in (a, b). If f′ (x) > 0 on (a, c) (i.e., “before c”) and f′ (x) < 0 on (c, b) (i.e., “after c”, then c is a local maximum for f. If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local minimum for f. If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. . . . . . .
6. 6. Recall: The Second Derivative Test See Section 4.3 Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous. 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). . . . . . .
7. 7. 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. . . . . . .
8. 8. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
9. 9. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. . . . . . .
10. 10. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. Solution Objective: minimize S = x + y subject to the constraint that xy = 16 . . . . . .
11. 11. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. Solution Objective: minimize S = x + y subject to the constraint that xy = 16 Eliminate y: y = 16/x so S = x + 16/x. The domain of consideration is (0, ∞). . . . . . .
12. 12. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. Solution Objective: minimize S = x + y subject to the constraint that xy = 16 Eliminate y: y = 16/x so S = x + 16/x. The domain of consideration is (0, ∞). Find the critical points: S′ (x) = 1 − 16/x2 , which is 0 when x = 4. . . . . . .
13. 13. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. Solution Objective: minimize S = x + y subject to the constraint that xy = 16 Eliminate y: y = 16/x so S = x + 16/x. The domain of consideration is (0, ∞). Find the critical points: S′ (x) = 1 − 16/x2 , which is 0 when x = 4. Classify the critical points: S′′ (x) = 32/x3 , which is always positive. So the graph is always concave up, 4 is a local min, and therefore the global min. . . . . . .
14. 14. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. Solution Objective: minimize S = x + y subject to the constraint that xy = 16 Eliminate y: y = 16/x so S = x + 16/x. The domain of consideration is (0, ∞). Find the critical points: S′ (x) = 1 − 16/x2 , which is 0 when x = 4. Classify the critical points: S′′ (x) = 32/x3 , which is always positive. So the graph is always concave up, 4 is a local min, and therefore the global min. So the numbers are x = y = 4, Smin = 8. . . . . . .
15. 15. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). . . . . . .
16. 16. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution y . . x, x2 ) ( . . . x . 3 . . . . . . .
17. 17. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution y . The distance between (x, x2 ) and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 . x, x2 ) ( . . . x . 3 . . . . . . .
18. 18. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution y . The distance between (x, x2 ) and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 We may instead minimize the square of f: g(x) = f(x)2 = (x − 3)2 + x4 . x, x2 ) ( . . . x . 3 . . . . . . .
19. 19. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution y . The distance between (x, x2 ) and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 We may instead minimize the square of f: g(x) = f(x)2 = (x − 3)2 + x4 . x, x2 ) ( . . . x . The domain is (−∞, ∞). 3 . . . . . . .
20. 20. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . . . . . . .
21. 21. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . g′ (x) = 2(x − 3) + 4x3 = 4x3 + 2x − 6 = 2(2x3 + x − 3) . . . . . .
22. 22. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . g′ (x) = 2(x − 3) + 4x3 = 4x3 + 2x − 6 = 2(2x3 + x − 3) If a polynomial has integer roots, they are factors of the constant term (Euler) . . . . . .
23. 23. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . g′ (x) = 2(x − 3) + 4x3 = 4x3 + 2x − 6 = 2(2x3 + x − 3) If a polynomial has integer roots, they are factors of the constant term (Euler) 1 is a root, so 2x3 + x − 3 is divisible by x − 1: f′ (x) = 2(2x3 + x − 3) = 2(x − 1)(2x2 + 2x + 3) The quadratic has no real roots (the discriminant b2 − 4ac < 0) . . . . . .
24. 24. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . g′ (x) = 2(x − 3) + 4x3 = 4x3 + 2x − 6 = 2(2x3 + x − 3) If a polynomial has integer roots, they are factors of the constant term (Euler) 1 is a root, so 2x3 + x − 3 is divisible by x − 1: f′ (x) = 2(2x3 + x − 3) = 2(x − 1)(2x2 + 2x + 3) The quadratic has no real roots (the discriminant b2 − 4ac < 0) We see f′ (1) = 0, f′ (x) < 0 if x < 1, and f′ (x) > 0 if x > 1. So 1 is the global minimum. . . . . . .
25. 25. Distance problem minimization step We want to ﬁnd the global minimum of g(x) = (x − 3)2 + x4 . g′ (x) = 2(x − 3) + 4x3 = 4x3 + 2x − 6 = 2(2x3 + x − 3) If a polynomial has integer roots, they are factors of the constant term (Euler) 1 is a root, so 2x3 + x − 3 is divisible by x − 1: f′ (x) = 2(2x3 + x − 3) = 2(x − 1)(2x2 + 2x + 3) The quadratic has no real roots (the discriminant b2 − 4ac < 0) We see f′ (1) = 0, f′ (x) < 0 if x < 1, and f′ (x) > 0 if x > 1. So 1 is the global minimum. The point on the parabola closest to (3, 0) is (1, 1). The √ minimum distance is 5. . . . . . .
26. 26. Remark We’ve used each of the methods (CIM, 1DT, 2DT) so far. Notice how we argued that the critical points were absolute extremes even though 1DT and 2DT only tell you relative/local extremes. . . . . . .
27. 27. A problem with a triangle Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle as shown. Solution 5 . 4 . . 3 . . . . . . .
28. 28. A problem with a triangle Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle as shown. Solution Let the dimensions of the rectangle be x and y. 5 . x . 4 . y . . 3 . . . . . . .
29. 29. A problem with a triangle Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle as shown. Solution Let the dimensions of the rectangle be x and y. Similar triangles give y 4 5 . x . = =⇒ 3y = 4(3−x) 4 . 3−x 3 y . . 3 . . . . . . .
30. 30. A problem with a triangle Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle as shown. Solution Let the dimensions of the rectangle be x and y. Similar triangles give y 4 5 . x . = =⇒ 3y = 4(3−x) 4 . 3−x 3 4 y . So y = 4 − x and 3 ( ) . 4 4 3 . A(x) = x 4 − x = 4x− x2 3 3 . . . . . .
31. 31. Triangle Problem maximization step 4 2 We want to ﬁnd the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. . . . . . .
32. 32. Triangle Problem maximization step 4 2 We want to ﬁnd the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. A(0) = A(3) = 0 . . . . . .
33. 33. Triangle Problem maximization step 4 2 We want to ﬁnd the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. A(0) = A(3) = 0 8 12 A′ (x) = 4 − x, which is zero when x = = 1.5. 3 8 . . . . . .
34. 34. Triangle Problem maximization step 4 2 We want to ﬁnd the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. A(0) = A(3) = 0 8 12 A′ (x) = 4 − x, which is zero when x = = 1.5. 3 8 Since A(1.5) = 3, this is the absolute maximum. . . . . . .
35. 35. Triangle Problem maximization step 4 2 We want to ﬁnd the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. A(0) = A(3) = 0 8 12 A′ (x) = 4 − x, which is zero when x = = 1.5. 3 8 Since A(1.5) = 3, this is the absolute maximum. So the dimensions of the rectangle of maximal area are 1.5 × 2. . . . . . .
36. 36. An Economics problem Example Let r be the monthly rent per unit in an apartment building with 100 units. A survey reveals that all units can be rented when r = 900 and that one unit becomes vacant with each 10 increase in rent. Suppose the average monthly maintenance costs per occupied unit is \$100/month. What rent should be charged to maximize proﬁt? . . . . . .
37. 37. An Economics problem Example Let r be the monthly rent per unit in an apartment building with 100 units. A survey reveals that all units can be rented when r = 900 and that one unit becomes vacant with each 10 increase in rent. Suppose the average monthly maintenance costs per occupied unit is \$100/month. What rent should be charged to maximize proﬁt? Solution Let n be the number of units rented, depending on price (the demand function). ∆n 1 We have n(900) = 100 and = − . So ∆r 10 1 1 n − 100 = − (r − 900) =⇒ n(r) = − r + 190 10 10 . . . . . .
38. 38. Economics Problem Finishing the model and maximizing The proﬁt per unit rented is r − 100, so ( ) 1 P(r) = (r − 100)n(r) = (r − 100) − r + 190 10 1 2 = − r + 200r − 19000 10 . . . . . .
39. 39. Economics Problem Finishing the model and maximizing The proﬁt per unit rented is r − 100, so ( ) 1 P(r) = (r − 100)n(r) = (r − 100) − r + 190 10 1 2 = − r + 200r − 19000 10 We want to maximize P on the interval 900 ≤ r ≤ 1900. . . . . . .
40. 40. Economics Problem Finishing the model and maximizing The proﬁt per unit rented is r − 100, so ( ) 1 P(r) = (r − 100)n(r) = (r − 100) − r + 190 10 1 2 = − r + 200r − 19000 10 We want to maximize P on the interval 900 ≤ r ≤ 1900. A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 . . . . . .
41. 41. Economics Problem Finishing the model and maximizing The proﬁt per unit rented is r − 100, so ( ) 1 P(r) = (r − 100)n(r) = (r − 100) − r + 190 10 1 2 = − r + 200r − 19000 10 We want to maximize P on the interval 900 ≤ r ≤ 1900. A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 1 A′ (x) = − r + 200, which is zero when r = 1000. 5 . . . . . .
42. 42. Economics Problem Finishing the model and maximizing The proﬁt per unit rented is r − 100, so ( ) 1 P(r) = (r − 100)n(r) = (r − 100) − r + 190 10 1 2 = − r + 200r − 19000 10 We want to maximize P on the interval 900 ≤ r ≤ 1900. A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 1 A′ (x) = − r + 200, which is zero when r = 1000. 5 n(1000) = 90, so P(r) = \$900 × 90 = \$81, 000. This is the maximum intake. . . . . . .
43. 43. 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? . . . . . .
44. 44. The Statue of Liberty Seting up the model 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 The domain of θ is all positive real numbers x. . . . . . .
45. 45. The Statue of Liberty Finding the derivative ( ) ( ) a+b b θ = arctan − arctan x x So 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 ] . . . . . .
46. 46. The Statue of Liberty Finding the critical points [ ] [ ] dθ x2 + (a + b)2 b − (a + b) x2 + b2 = dx (x2 + b2 ) [x2 + (a + b)2 ] . . . . . .
47. 47. The Statue of Liberty Finding the critical points [ ] [ ] dθ x2 + (a + b)2 b − (a + b) x2 + b2 = dx (x2 + b2 ) [x2 + (a + b)2 ] 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 ) . . . . . .
48. 48. The Statue of Liberty Finding the critical points [ ] [ ] dθ x2 + (a + b)2 b − (a + b) x2 + b2 = dx (x2 + b2 ) [x2 + (a + b)2 ] 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 ) √ The only positive solution is x = b(a + b). . . . . . .
49. 49. The Statue of Liberty Finding the critical points [ ] [ ] dθ x2 + (a + b)2 b − (a + b) x2 + b2 = dx (x2 + b2 ) [x2 + (a + b)2 ] 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 ) √ The only positive solution is 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, ∞). . . . . . .
50. 50. The Statue of Liberty Final answer 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. . . . . . .
51. 51. The Statue of Liberty Discussion √ The length b(a + b) is the geometric mean of the two distances measured 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. . . . . . .
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