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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.

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) Lesson 22: Optimization II (Section 041 slides) Presentation Transcript

    • Section 4.5 Optimization II V63.0121.041, Calculus I New York University November 24, 2010Announcements No recitation this week Quiz 5 on §§4.1–4.4 next week in recitation Happy Thanksgiving! . . . . . .
    • 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
    • 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
    • OutlineRecallMore 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
    • 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
    • Recall: The Closed Interval MethodSee Section 4.1The Closed Interval MethodTo 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
    • Recall: The First Derivative TestSee Section 4.3Theorem (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
    • Recall: The First Derivative TestSee Section 4.3Theorem (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
    • Recall: The Second Derivative TestSee Section 4.3Theorem (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
    • Recall: The Second Derivative TestSee Section 4.3Theorem (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.WarningIf f′′ (c) = 0, the second derivative test is inconclusive (this does notmean 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
    • Recall: The Second Derivative TestSee Section 4.3Theorem (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.WarningIf f′′ (c) = 0, the second derivative test is inconclusive (this does notmean 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
    • Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequalities non-closed, non-closed, – gets global non-bounded non-bounded extrema intervals intervals automatically – only one derivative – no need for inequalities Con – only for closed – Uses inequalities – More derivatives bounded intervals – More work at – less conclusive boundary than CIM than 1DT – more work at boundary than CIM . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 9 / 25
    • Which to use when? CIM 1DT 2DT Pro – no need for – works on – works on inequalities non-closed, non-closed, – gets global non-bounded non-bounded extrema intervals intervals automatically – only one derivative – no need for inequalities Con – only for closed – Uses inequalities – More derivatives bounded intervals – More work at – less conclusive boundary than CIM 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
    • OutlineRecallMore 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
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible. . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 11 / 25
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible.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
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible.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
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible.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
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible.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
    • Addition with a constraintExampleFind two positive numbers x and y with xy = 16 and x + y as small aspossible.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
    • DistanceExampleFind 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
    • DistanceExampleFind the point P on the parabola y = x2 closest to the point (3, 0).Solution y (x, x2 ) . x . . . . 3 . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
    • DistanceExampleFind the point P on the parabola y = x2 closest to the point (3, 0).Solution yThe distance between (x, x2 )and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 (x, x2 ) . x . . . . 3 . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
    • DistanceExampleFind the point P on the parabola y = x2 closest to the point (3, 0).Solution yThe distance between (x, x2 )and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 We may instead minimize thesquare of f: g(x) = f(x)2 = (x − 3)2 + x4 (x, x2 ) . x . . . . 3 . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
    • DistanceExampleFind the point P on the parabola y = x2 closest to the point (3, 0).Solution yThe distance between (x, x2 )and (3, 0) is given by √ f(x) = (x − 3)2 + (x2 − 0)2 We may instead minimize thesquare of f: g(x) = f(x)2 = (x − 3)2 + x4 (x, x2 ) . xThe domain is (−∞, ∞). . . . . 3 . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 12 / 25
    • Distance problemminimization 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
    • Distance problemminimization 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
    • Distance problemminimization 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
    • Distance problemminimization 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
    • Distance problemminimization 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
    • Distance problemminimization 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
    • 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
    • A problem with a triangle.ExampleFind the rectangle of maximal area inscribed in a 3-4-5 right triangle with twosides on legs of the triangle and one vertex on the hypotenuse.Solution 5 4 . 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
    • A problem with a triangle.ExampleFind the rectangle of maximal area inscribed in a 3-4-5 right triangle with twosides on legs of the triangle and one vertex on the hypotenuse.Solution Let the dimensions of the rectangle be x and y. 5 x 4 y . 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
    • A problem with a triangle.ExampleFind the rectangle of maximal area inscribed in a 3-4-5 right triangle with twosides 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 4 = =⇒ 3y = 4(3 − x) 3−x 3 5 x 4 y . 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
    • A problem with a triangle.ExampleFind the rectangle of maximal area inscribed in a 3-4-5 right triangle with twosides 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 4 = =⇒ 3y = 4(3 − x) 3−x 3 5 x 4 4 So y = 4 − x and y 3 ( ) 4 4 . A(x) = x 4 − x = 4x − x2 3 3 3 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 15 / 25
    • Triangle Problemmaximization step 4 We want to find the absolute maximum of A(x) = 4x − x2 on the 3 interval [0, 3]. . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
    • Triangle Problemmaximization step 4 We want to find the absolute maximum of A(x) = 4x − x2 on the 3 interval [0, 3]. A(0) = A(3) = 0 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
    • Triangle Problemmaximization step 4 We want to find the absolute maximum of A(x) = 4x − x2 on the 3 interval [0, 3]. A(0) = A(3) = 0 8 12 A′ (x) = 4 − x, which is zero when x = = 1.5. 3 8 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
    • Triangle Problemmaximization step 4 We want to find the absolute maximum of A(x) = 4x − x2 on the 3 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
    • Triangle Problemmaximization step 4 We want to find the absolute maximum of A(x) = 4x − x2 on the 3 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 16 / 25
    • An Economics problemExampleLet r be the monthly rent per unit in an apartment building with 100units. A survey reveals that all units can be rented when r = 900 andthat one unit becomes vacant with each 10 increase in rent. Supposethe 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
    • An Economics problemExampleLet r be the monthly rent per unit in an apartment building with 100units. A survey reveals that all units can be rented when r = 900 andthat one unit becomes vacant with each 10 increase in rent. Supposethe 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 17 / 25
    • Economics ProblemFinishing 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
    • Economics ProblemFinishing 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. (Why?) . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
    • Economics ProblemFinishing 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. (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
    • Economics ProblemFinishing 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. (Why?) A(900) = $800 × 100 = $80, 000, A(1900) = 0 1 A′ (x) = − r + 200, which is zero when r = 1000. 5 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
    • Economics ProblemFinishing 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. (Why?) 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. . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 18 / 25
    • The Statue of LibertyExampleThe Statue of Liberty stands on top of a pedestal which is on top of onold fort. The top of the pedestal is 47 m above ground level. The statueitself 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 tomaximize the view of the statue? That is, what distance will maximizethe 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
    • The Statue of LibertySeting up the modelSolution The angle subtended by the a statue in the viewer’s eye can be expressed as ( ) ( ) b a+b b θ θ = arctan −arctan . x x xThe 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
    • The Statue of LibertyFinding 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 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 21 / 25
    • The Statue of LibertyFinding the critical points [ 2 ] [ ] dθ x + (a + b)2 b − (a + b) x2 + b2 = [ ] dx (x2 + b2 ) x2 + (a + b)2 . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
    • The Statue of LibertyFinding the critical points [ 2 ] [ ] dθ x + (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 ) . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
    • The Statue of LibertyFinding the critical points [ 2 ] [ ] dθ x + (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). . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
    • The Statue of LibertyFinding the critical points [ 2 ] [ ] dθ x + (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, ∞). . . . . . . V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 22 / 25
    • The Statue of LibertyFinal 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
    • The Statue of LibertyDiscussion √ 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
    • Summary 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? Remember the checklist UNDERSTAND What do you want to find out? • Draw a line under the question. Ask yourself: what is the objective? You can draw a picture to solve the problem. Remember your geometry: What number do I add to 5 to get 8? 8 - = 5 similar triangles crayons 5 + 3 = 8 right triangles CHECK Does your answer make sense? trigonometric functions Explain. What number Draw a picture to solve the problem. do I add to 3 Write how many were given away. to make 10? I. I had 10 pencils. ft ft ft A I gave some away. 13 ill i :i I • I I I have 3 left. How many i? « 11 I pencils did I give away? I H 11 M i l ~7 U U U U> U U . . . . . .V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 25 / 25