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# Lesson 22: Optimization II (Section 041 slides)

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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 more good examples.

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### Lesson 22: Optimization II (Section 041 slides)

1. 1. Section 4.5 Optimization II V63.0121.041, Calculus I New York University November 24, 2010 Announcements No recitation this week Quiz 5 on §§4.1–4.4 next week in recitation Happy Thanksgiving! . . . . . .
2. 2. . . . . . . Announcements No recitation this week Quiz 5 on §§4.1–4.4 next week in recitation Happy Thanksgiving! V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 2 / 25
3. 3. . . . . . . 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 4.5 Optimization II November 24, 2010 3 / 25
4. 4. . . . . . . Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 4 / 25
5. 5. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 5 / 25
6. 6. . . . . . . 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 4.5 Optimization II November 24, 2010 6 / 25
7. 7. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 7 / 25
8. 8. . . . . . . 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 4.5 Optimization II November 24, 2010 7 / 25
9. 9. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 8 / 25
10. 10. . . . . . . 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). V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 8 / 25
11. 11. . . . . . . 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 f V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 8 / 25
12. 12. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 9 / 25
13. 13. . . . . . . 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 4.5 Optimization II November 24, 2010 9 / 25
14. 14. . . . . . . Outline Recall More examples Addition Distance Triangles Economics The Statue of Liberty V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 10 / 25
15. 15. . . . . . . Addition with a constraint Example Find two positive numbers x and y with xy = 16 and x + y as small as possible. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
16. 16. . . . . . . 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 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
17. 17. . . . . . . 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, ∞). V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
18. 18. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
19. 19. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
20. 20. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
21. 21. . . . . . . Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
22. 22. . . . . . . Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution .. x. y .. (x, x2 ) .. 3 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
23. 23. . . . . . . Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution The distance between (x, x2 ) and (3, 0) is given by f(x) = √ (x − 3)2 + (x2 − 0)2 .. x. y .. (x, x2 ) .. 3 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
24. 24. . . . . . . Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution 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. y .. (x, x2 ) .. 3 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
25. 25. . . . . . . Distance Example Find the point P on the parabola y = x2 closest to the point (3, 0). Solution 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 The domain is (−∞, ∞). .. x. y .. (x, x2 ) .. 3 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
26. 26. . . . . . . Distance problem minimization step We want to find the global minimum of g(x) = (x − 3)2 + x4 . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
27. 27. . . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
28. 28. . . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
29. 29. . . . . . . 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) V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
30. 30. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
31. 31. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 13 / 25
32. 32. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 14 / 25
33. 33. . . . . . . A problem with a triangle .. Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with two sides on legs of the triangle and one vertex on the hypotenuse. Solution .. 3 . 4 . 5 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
34. 34. . . . . . . A problem with a triangle .. Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with two sides on legs of the triangle and one vertex on the hypotenuse. Solution Let the dimensions of the rectangle be x and y. .. 3 . 4 . 5 . y . x V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
35. 35. . . . . . . A problem with a triangle .. Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with two sides on legs of the triangle and one vertex on the hypotenuse. Solution Let the dimensions of the rectangle be x and y. Similar triangles give y 3 − x = 4 3 =⇒ 3y = 4(3 − x) .. 3 . 4 . 5 . y . x V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
36. 36. . . . . . . A problem with a triangle .. Example Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with two sides on legs of the triangle and one vertex on the hypotenuse. Solution Let the dimensions of the rectangle be x and y. Similar triangles give y 3 − x = 4 3 =⇒ 3y = 4(3 − x) So y = 4 − 4 3 x and A(x) = x ( 4 − 4 3 x ) = 4x− 4 3 x2 .. 3 . 4 . 5 . y . x V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
37. 37. . . . . . . Triangle Problem maximization step We want to find the absolute maximum of A(x) = 4x − 4 3 x2 on the interval [0, 3]. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
38. 38. . . . . . . Triangle Problem maximization step We want to find the absolute maximum of A(x) = 4x − 4 3 x2 on the interval [0, 3]. A(0) = A(3) = 0 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
39. 39. . . . . . . Triangle Problem maximization step We want to find the absolute maximum of A(x) = 4x − 4 3 x2 on the interval [0, 3]. A(0) = A(3) = 0 A′ (x) = 4 − 8 3 x, which is zero when x = 12 8 = 1.5. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
40. 40. . . . . . . Triangle Problem maximization step We want to find the absolute maximum of A(x) = 4x − 4 3 x2 on the interval [0, 3]. A(0) = A(3) = 0 A′ (x) = 4 − 8 3 x, which is zero when x = 12 8 = 1.5. Since A(1.5) = 3, this is the absolute maximum. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
41. 41. . . . . . . Triangle Problem maximization step We want to find the absolute maximum of A(x) = 4x − 4 3 x2 on the interval [0, 3]. A(0) = A(3) = 0 A′ (x) = 4 − 8 3 x, which is zero when x = 12 8 = 1.5. Since A(1.5) = 3, this is the absolute maximum. So the dimensions of the rectangle of maximal area are 1.5 × 2. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
42. 42. . . . . . . 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? V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 17 / 25
43. 43. . . . . . . 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). We have n(900) = 100 and ∆n ∆r = − 1 10 . So n − 100 = − 1 10 (r − 900) =⇒ n(r) = − 1 10 r + 190 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 17 / 25
44. 44. . . . . . . Economics Problem Finishing the model and maximizing The profit per unit rented is r − 100, so P(r) = (r − 100)n(r) = (r − 100) ( − 1 10 r + 190 ) = − 1 10 r2 + 200r − 19000 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
45. 45. . . . . . . Economics Problem Finishing the model and maximizing The profit per unit rented is r − 100, so P(r) = (r − 100)n(r) = (r − 100) ( − 1 10 r + 190 ) = − 1 10 r2 + 200r − 19000 We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?) V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
46. 46. . . . . . . Economics Problem Finishing the model and maximizing The profit per unit rented is r − 100, so P(r) = (r − 100)n(r) = (r − 100) ( − 1 10 r + 190 ) = − 1 10 r2 + 200r − 19000 We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?) A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
47. 47. . . . . . . Economics Problem Finishing the model and maximizing The profit per unit rented is r − 100, so P(r) = (r − 100)n(r) = (r − 100) ( − 1 10 r + 190 ) = − 1 10 r2 + 200r − 19000 We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?) A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 A′ (x) = − 1 5 r + 200, which is zero when r = 1000. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
48. 48. . . . . . . Economics Problem Finishing the model and maximizing The profit per unit rented is r − 100, so P(r) = (r − 100)n(r) = (r − 100) ( − 1 10 r + 190 ) = − 1 10 r2 + 200r − 19000 We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?) A(900) = \$800 × 100 = \$80, 000, A(1900) = 0 A′ (x) = − 1 5 r + 200, which is zero when r = 1000. n(1000) = 90, so P(r) = \$900 × 90 = \$81, 000. This is the maximum intake. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
49. 49. . . . . . . The Statue of Liberty Example 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? V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 19 / 25
50. 50. . . . . . . The Statue of Liberty Seting up the model Solution The angle subtended by the statue in the viewer’s eye can be expressed as θ = arctan ( a + b x ) −arctan ( b x ) . a b θ x The domain of θ is all positive real numbers x. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 20 / 25
51. 51. . . . . . . The Statue of Liberty Finding the derivative θ = arctan ( a + b x ) − arctan ( b x ) So dθ dx = 1 1 + ( a+b x )2 · −(a + b) x2 − 1 1 + ( b x )2 · −b x2 = b x2 + b2 − a + b x2 + (a + b)2 = [ x2 + (a + b)2 ] b − (a + b) [ x2 + b2 ] (x2 + b2 ) [ x2 + (a + b)2 ] V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 21 / 25
52. 52. . . . . . . The Statue of Liberty Finding the critical points dθ dx = [ x2 + (a + b)2 ] b − (a + b) [ x2 + b2 ] (x2 + b2 ) [ x2 + (a + b)2 ] V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
53. 53. . . . . . . The Statue of Liberty Finding the critical points dθ dx = [ x2 + (a + b)2 ] b − (a + b) [ x2 + b2 ] (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 ) V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
54. 54. . . . . . . The Statue of Liberty Finding the critical points dθ dx = [ x2 + (a + b)2 ] b − (a + b) [ x2 + b2 ] (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). V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
55. 55. . . . . . . The Statue of Liberty Finding the critical points dθ dx = [ x2 + (a + b)2 ] b − (a + b) [ x2 + b2 ] (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, ∞). V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
56. 56. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 23 / 25
57. 57. . . . . . . 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. V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 24 / 25
58. 58. . . . . . . Summary Remember the checklist Ask yourself: what is the objective? Remember your geometry: similar triangles right triangles trigonometric functions Name [_ Problem Solving Strategy Draw a Picture Kathy had a box of 8 crayons. She gave some crayons away. She has 5 left. How many crayons did Kathy give away? UNDERSTAND • What do you want to find out? Draw a line under the question. You can draw a picture to solve the problem. crayons What number do I add to 5 to get 8? 8 - = 5 5 + 3 = 8 CHECK Does your answer make sense? Explain. Draw a picture to solve the problem. Write how many were given away. I. I had10 pencils. I gave some away. I have 3left. How many pencils did I give away? ~7 What number do I add to 3 to make 10? 13 i ft ill :i i ? 11 ft I '•' « I I ft A H 11 M i l U U U U> U U I I V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 25 / 25