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Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
Lesson 22: Optimization II (Section 10 version)
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Lesson 22: Optimization II (Section 10 version)

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More examples, including distance, triangles, economics, and the Statue of Liberty.

More examples, including distance, triangles, economics, and the Statue of Liberty.

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  • 1. Section 4.5 Optimization Problems, Part Deux V63.0121, Calculus I April 8, 2009 Announcements Quiz 5 is next week, covering Sections 4.1–4.4 I am moving to WWH 624 sometime next week (April 13th) Happy Passover/Easter . . . . . .
  • 2. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
  • 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. 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. 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. 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 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. 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. Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty . . . . . .
  • 9. Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. . . . . . .
  • 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. 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. 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. 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. 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. Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). . . . . . .
  • 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. 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. 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: . x, x2 ) ( g(x) = f(x)2 = (x − 3)2 + x4 . . . x . 3 . . . . . . .
  • 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: . x, x2 ) ( g(x) = f(x)2 = (x − 3)2 + x4 . . . x . The domain is (−∞, ∞). 3 . . . . . . .
  • 20. Distance problem minimization step We want to find the global minimum of g(x) = (x − 3)2 + x4 . . . . . . .
  • 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. 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. 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. 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) > 1 if x > 1. So 1 is the global minimum. . . . . . .
  • 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) > 1 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. 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 . . . . . . .
  • 27. 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 . . . . . . .
  • 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. Similar triangles give y 4 5 . x . =⇒ 3y = 4(3−x) = 4 . 3−x 3 y . . 3 . . . . . . .
  • 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 4 y . So y = 4 − x and 3 ( ) . 4 4 3 . A(x) = x 4 − x = 4x− x2 3 3 . . . . . .
  • 30. Triangle Problem maximization step 42 We want to find the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. . . . . . .
  • 31. Triangle Problem maximization step 42 We want to find the absolute maximum of A(x) = 4x − x on 3 the interval [0, 3]. A(0) = A(3) = 0 . . . . . .
  • 32. Triangle Problem maximization step 42 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 . . . . . .
  • 33. Triangle Problem maximization step 42 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. . . . . . .
  • 34. Triangle Problem maximization step 42 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. . . . . . .
  • 35. An Economics problem Example Let r be the monthly rent per unit in an apartment building with 100 units. A survey revelas 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? . . . . . .
  • 36. An Economics problem Example Let r be the monthly rent per unit in an apartment building with 100 units. A survey revelas 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). 1 ∆n = − . So We have n(900) = 100 and 10 ∆r 1 1 n − 100 = − (r − 900) =⇒ n(r) = − r + 190 10 10 . . . . . .
  • 37. 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 = − r2 + 200r − 19000 10 . . . . . .
  • 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 = − r2 + 200r − 19000 10 We want to maximize P on the interval 0 ≤ r ≤ 1900. . . . . . .
  • 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 = − r2 + 200r − 19000 10 We want to maximize P on the interval 0 ≤ r ≤ 1900. A(0) = −19000, A(1900) = 0, so these are probably not maximal . . . . . .
  • 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 = − r2 + 200r − 19000 10 We want to maximize P on the interval 0 ≤ r ≤ 1900. A(0) = −19000, A(1900) = 0, so these are probably not maximal 1 A′ (x) = − r + 200, which is zero when r = 1000. 5 . . . . . .
  • 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 = − r2 + 200r − 19000 10 We want to maximize P on the interval 0 ≤ r ≤ 1900. A(0) = −19000, A(1900) = 0, so these are probably not maximal 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. . . . . . .
  • 42. 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? . . . . . .
  • 43. The Statue of Liberty Seting up the model The angle subtended by the a statue in the viewer’s eye can be expressed as ( ) () b a+b b θ −arctan θ = arctan . x x x The domain of θ is all positive real numbers x. . . . . . .
  • 44. The Statue of Liberty Finding the derivative ( ) () a+b b − arctan θ = arctan x x So −(a + b) −b dθ 1 1 )2 · − ( )2 · 2 ( = x2 dx x a+b b 1+ 1+ x x b a+b − = 2 x2 + (a + b)2 x2+b [ ] [2 ] x + (a + b)2 b − (a + b) x2 + b2 = (x2 + b2 ) [x2 + (a + b)2 ] . . . . . .
  • 45. The Statue of Liberty Finding the critical points [ ] [ ] x2 + (a + b)2 b − (a + b) x2 + b2 dθ = (x2 + b2 ) [x2 + (a + b)2 ] dx . . . . . .
  • 46. The Statue of Liberty Finding the critical points [ ] [ ] x2 + (a + b)2 b − (a + b) x2 + b2 dθ = (x2 + b2 ) [x2 + (a + b)2 ] dx 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 ) . . . . . .
  • 47. The Statue of Liberty Finding the critical points [ ] [ ] x2 + (a + b)2 b − (a + b) x2 + b2 dθ = (x2 + b2 ) [x2 + (a + b)2 ] dx 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). . . . . . .
  • 48. The Statue of Liberty Finding the critical points [ ] [ ] x2 + (a + b)2 b − (a + b) x2 + b2 dθ = (x2 + b2 ) [x2 + (a + b)2 ] dx 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, ∞). . . . . . .
  • 49. 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. . . . . . .
  • 50. The Statue of Liberty Discussion √ 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. . . . . . .
  • 51. Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions . . . . . .

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