1. 4  Applications of the First Derivative  Applications of the Second Derivative  Curve Sketching  Optimization I  Optimization II Applications of the Derivative

2. 4.1 Applications of the First Derivative

3. Increasing and Decreasing Functions  A function f is increasing on an interval (a, b) if for any two numbers x 1 and x 2 in (a, b), f(x 1 ) < f(x 2 ) wherever x 1 < x 2. x y a f(x2)f(x2)f(x2)f(x2) f(x1)f(x1)f(x1)f(x1) b x2x2x2x2 x1x1x1x1

4. Increasing and Decreasing Functions  A function f is decreasing on an interval (a, b) if for any two numbers x 1 and x 2 in (a, b), f(x 1 ) > f(x 2 ) wherever x 1 f(x 2 ) wherever x 1 < x 2. x y a f(x1)f(x1)f(x1)f(x1) f(x2)f(x2)f(x2)f(x2) b x2x2x2x2 x1x1x1x1

5. Theorem 1  If f ′ (x) > 0 for each value of x in an interval (a, b), then f is increasing on (a, b).  If f ′ (x) < 0 for each value of x in an interval (a, b), then f is decreasing on (a, b).  If f ′ (x) = 0 for each value of x in an interval (a, b), then f is constant on (a, b).

6. Example  Find the interval where the function f(x) = x 2 is increasing and the interval where it is decreasing. Solution  The derivative of f(x) = x 2 is f ′ (x) = 2x.  f ′ (x) = 2x > 0 if x > 0 and f ′ (x) = 2x < 0 if x < 0.  Thus, f is increasing on the interval (0,  ) and decreasing on the interval (– , 0). y f(x) = x 2 Example 1, page 245

7. Determining the Intervals Where a Function is Increasing or Decreasing 1.Find all the values of x for which f ′ (x) = 0 or f ′ is discontinuous and identify the open intervals determined by these numbers. 2.Select a test number c in each interval found in step 1 and determine the sign of f ′ (c) in that interval. a.If f ′ (c) > 0, f is increasing on that interval. b.If f ′ (c) < 0, f is decreasing on that interval.

8. Examples  Determine the intervals where the function f(x) = x 3 – 3x 2 – 24x + 32 is increasing and where it is decreasing. Solution 1.Find f ′ and solve for f ′ (x) = 0: f ′ (x) = 3x 2 – 6x – 24 = 3(x + 2)(x – 4) = 0 f ′ (x) = 3x 2 – 6x – 24 = 3(x + 2)(x – 4) = 0 ✦ Thus, the zeros of f ′ are x = –2 and x = 4. ✦ These numbers divide the real line into the intervals (– , –2), (– 2, 4), and (4,  ). Example 2, page 246

9. Examples  Determine the intervals where the function f(x) = x 3 – 3x 2 – 24x + 32 is increasing and where it is decreasing. Solution 2.To determine the sign of f ′ (x) in the intervals we found (– , –2), (– 2, 4), and (4,  ), we compute f ′ (c) at a convenient test point in each interval. ✦ Lets consider the values –3, 0, and 5: f ′ (–3)= 3(–3) 2 – 6(–3) – 24 = 27 +18 – 24 = 21 > 0 f ′ (–3)= 3(–3) 2 – 6(–3) – 24 = 27 +18 – 24 = 21 > 0 f ′ (0)= 3(0) 2 – 6(0) – 24 = 0 +0 – 24 = –24 < 0 f ′ (0)= 3(0) 2 – 6(0) – 24 = 0 +0 – 24 = –24 < 0 f ′ (5)= 3(5) 2 – 6(5) – 24 = 75 – 30 – 24 = 21 > 0 f ′ (5)= 3(5) 2 – 6(5) – 24 = 75 – 30 – 24 = 21 > 0 ✦ Thus, we conclude that f is increasing on the intervals (– , –2), (4,  ), and is decreasing on the interval (– 2, 4). Example 2, page 246

10. –7 –5 –3 –11357 Examples  Determine the intervals where the function f(x) = x 3 – 3x 2 – 24x + 32 is increasing and where it is decreasing. Solution  So, f increases on (– , –2), (4,  ), and decreases on (– 2, 4): 604020–20–40 – 60 x y y = x3 – 3x 2 – 24x + 32 Example 2, page 246

11. Examples  Determine the intervals where is increasing and where it is decreasing. Solution 1.Find f ′ and solve for f ′ (x) = 0: ✦ f ′ (x) = 0 when the numerator is equal to zero, so: ✦ Thus, the zeros of f ′ are x = –1 and x = 1. ✦ Also note that f ′ is not defined at x = 0, so we have four intervals to consider: (– , –1), (– 1, 0), (0, 1), and (1,  ). Example 4, page 247

12. Examples  Determine the intervals where is increasing and where it is decreasing. Solution 2.To determine the sign of f ′ (x) in the intervals we found (– , –1), (– 1, 0), (0, 1), and (1,  ), we compute f ′ (c) at a convenient test point in each interval. ✦ Lets consider the values –2, –1/2, 1/2, and 2: ✦ So f is increasing in the interval (– , –1). Example 4, page 247

13. Examples  Determine the intervals where is increasing and where it is decreasing. Solution 2.To determine the sign of f ′ (x) in the intervals we found (– , –1), (– 1, 0), (0, 1), and (1,  ), we compute f ′ (c) at a convenient test point in each interval. ✦ Lets consider the values –2, –1/2, 1/2, and 2: ✦ So f is decreasing in the interval (– 1, 0). Example 4, page 247

14. Examples  Determine the intervals where is increasing and where it is decreasing. Solution 2.To determine the sign of f ′ (x) in the intervals we found (– , –1), (– 1, 0), (0, 1), and (1,  ), we compute f ′ (c) at a convenient test point in each interval. ✦ Lets consider the values –2, –1/2, 1/2, and 2: ✦ So f is decreasing in the interval (0, 1). Example 4, page 247

15. Examples  Determine the intervals where is increasing and where it is decreasing. Solution 2.To determine the sign of f ′ (x) in the intervals we found (– , –1), (– 1, 0), (0, 1), and (1,  ), we compute f ′ (c) at a convenient test point in each interval. ✦ Lets consider the values –2, –1/2, 1/2, and 2: ✦ So f is increasing in the interval (1,  ). Example 4, page 247

16. Examples  Determine the intervals where is increasing and where it is decreasing. Solution  Thus, f is increasing on (– , –1) and (1,  ), and decreasing on (– 1, 0) and(0, 1): –4 –2 2 4 42–2–4 x y Example 4, page 247

17. Relative Extrema  The first derivative may be used to help us locate high points and low points on the graph of f: ✦ High points are called relative maxima ✦ Low points are called relative minima.  Both high and low points are called relative extrema. x y y = f(x) Relative Maxima Relative Minima

18. Relative Extrema Relative Maximum  A function f has a relative maximum at x = c if there exists an open interval (a, b) containing c such that f(x)  f(c) for all x in (a, b). x1x1x1x1 x2x2x2x2 x y y = f(x) Relative Maxima

19. Relative Extrema Relative Minimum  A function f has a relative minimum at x = c if there exists an open interval (a, b) containing c such that f(x)  f(c) for all x in (a, b). x3x3x3x3 x4x4x4x4 x y y = f(x) Relative Minima

20. Finding Relative Extrema  Suppose that f has a relative maximum at c.  The slope of the tangent line to the graph must change from positive to negative as x increases.  Therefore, the tangent line to the graph of f at point (c, f(c)) must be horizontal, so that f ′ (x) = 0 or f ′ (x) is undefined. x y c a b f ′(x) < 0 f ′ (x) > 0 f ′(x) = 0

21. Finding Relative Extrema  Suppose that f has a relative minimum at c.  The slope of the tangent line to the graph must change from negative to positive as x increases.  Therefore, the tangent line to the graph of f at point (c, f(c)) must be horizontal, so that f ′ (x) = 0 or f ′ (x) is undefined. x y c a b f ′(x) > 0 f ′(x) < 0 f ′(x) = 0

22. Finding Relative Extrema  In some cases a derivative does not exist for particular values of x.  Extrema may exist at such points, as the graph below shows: x y a b Relative Maximum Relative Minimum

23. Critical Numbers  We refer to a number in the domain of f that may give rise to a relative extremum as a critical number. Critical number of f ✦ A critical number of a function f is any number x in the domain of f such that f ′ (x) = 0 or f ′ (x) does not exist.

24. Critical Numbers  The graph below shows us several critical numbers.  At points a, b, and c, f ′ (x) = 0.  There is a corner at point d, so f ′ (x) does not exist there.  The tangent to the curve at point e is vertical, so f ′ (x) does not exist there either.  Note that points a, b, and d are relative extrema, while points c and e are not. x y a b Relative Extrema Not Relative Extrema cde

25. The First Derivative Test 1.Determine the critical numbers of f. 2.Determine the sign of f ′ (x) to the left and right of each critical point. a.If f ′ (x) changes sign from positive to negative as we move across a critical number c, then f(c) is a relative maximum. b.If f ′ (x) changes sign from negative to positive as we move across a critical number c, then f(c) is a relative minimum. c.If f ′ (x) does not change sign as we move across a critical number c, then f(c) is not a relative extremum.  Procedure for Finding Relative Extrema of a Continuous Function f

26. Examples  Find the relative maxima and minima of y f(x) = x 2 Solution  The derivative of f is f ′ (x) = 2x.  Setting f ′ (x) = 0 yields x = 0 as the only critical number of f.  Sincef ′ (x) < 0 if x < 0 andf ′ (x) > 0 if x > 0 we see that f ′ (x) changes sign from negative to positive as we move across 0.  Thus, f(0) = 0 is a relative minimum of f. f ′(x) < 0 f ′(x) > 0 Relative Minimum Example 5, page 252

27. Examples  Find the relative maxima and minima of Solution  The derivative of f is f ′ (x) = 2/3x –1/3.  f ′ (x) is not defined at x = 0, is continuous everywhere else, and is never equal to zero in its domain.  Thus x = 0 is the only critical number of f.  Sincef ′ (x) < 0 if x < 0 andf ′ (x) > 0 if x > 0 we see that f ′ (x) changes sign from negative to positive as we move across 0.  Thus, f(0) = 0 is a relative minimum of f. f´(x) < 0 f´(x) > 0 – 4 – 2 24 – 4 – 2 24 42 x y f(x) = x 2/3 Relative Minimum Example 6, page 252

28. Examples  Find the relative maxima and minima of Solution  The derivative of f and equate to zero:  The zeros of f ′ (x) are x = –2 and x = 4.  f ′ (x) is defined everywhere, so x = –2 and x = 4 are the only critical numbers of f. Example 7, page 253

29. Examples  Find the relative maxima and minima of Solution  Since f ′ (x) > 0 if x 0 if x < –2 and f ′ (x) < 0 if 0 < x < 4, we see that f ′ (x) changes sign from positive to negative as we move across –2.  Thus, f(–2) = 60 is a relative maximum. –7 –5 –3 –11357 604020–20–40 – 60 x y f(x)f(x)f(x)f(x) Relative Maximum Example 7, page 253

30. Examples  Find the relative maxima and minima of Solution  Since f ′ (x) 0 if, x > 4 we see that f ′ (x) changes sign from negative to positive as we move across 4.  Thus, f(4) = – 48 is a relative minimum. –7 –5 –3 –11357 604020–20–40 – 60 x y f(x)f(x)f(x)f(x) Relative Minimum Relative Maximum Example 7, page 253

31. 4.2 Applications of the Second Derivative

32. Concavity  A curve is said to be concave upwards when the slope of tangent line to the curve is increasing:  Thus, if f is differentiable on an interval (a, b), then f is concave upwards on (a, b) if f ′ is increasing on (a, b). f is concave upwards on (a, b) if f ′ is increasing on (a, b). x y

33. Concavity  A curve is said to be concave downwards when the slope of tangent line to the curve is decreasing:  Thus, if f is differentiable on an interval (a, b), then f is concave downwards on (a, b) if f ′ is decreasing on (a, b). f is concave downwards on (a, b) if f ′ is decreasing on (a, b). x y

34. Theorem 2  Recall that f ″ (x) measures the rate of change of the slope f ′ (x) of the tangent line to the graph of f at the point (x, f(x)).  Thus, we can use f ″ (x) to determine the concavity of f. a.If f ″ (x) > 0 for each value of x in (a, b), then f is concave upward on (a, b). b.If f ″ (x) < 0 for each value of x in (a, b), then f is concave downward on (a, b).

35. Steps in Determining the Concavity of f 1.Determine the values of x for which f ″ is zero or where f ″ is not defined, and identify the open intervals determined by these numbers. 2.Determine the sign of f ″ in each interval found in step 1. To do this compute f ″ (c), where c is any conveniently chosen test number in the interval. a.If f ″ (c) > 0, f is concave upward on that interval. b.If f ″ (c) < 0, f is concave downward on that interval.

36. Examples  Determine where the function is concave upward and where it is concave downward. Solution  Here,and  Setting f ″ (c) = 0 we find which gives x = 1.  So we consider the intervals (– , 1) and (1,  ): ✦ f ″ (x) < 0 when x < 1, so f is concave downward on (– , 1). ✦ f ″ (x) > 0 when x > 1, so f is concave upward on (1,  ). Example 1, page 265

37. Examples  Determine where the function is concave upward and where it is concave downward. Solution  The graph confirms that f is concave downward on (– , 1) and concave upward on (1,  ): –7 –5 –3 –11357 604020–20–40 – 60 x y f(x)f(x)f(x)f(x) Example 1, page 265

38. Examples  Determine the intervals where the function is concave upward and concave downward. Solution  Here,and  So, f ″ cannot be zero and is not defined at x = 0.  So we consider the intervals (– , 0) and (0,  ): ✦ f ″ (x) < 0 when x < 0, so f is concave downward on (– , 0). ✦ f ″ (x) > 0 when x > 0, so f is concave upward on (0,  ). Example 2, page 266

39. Examples  Determine the intervals where the function is concave upward and concave downward. Solution  The graph confirms that f is concave downward on (– , 0) and concave upward on (0,  ): Example 2, page 266

40. Inflection Point  A point on the graph of a continuous function where the tangent line exists and where the concavity changes is called an inflection point.  Examples: x y Inflection Point

41.  A point on the graph of a continuous function where the tangent line exists and where the concavity changes is called an inflection point.  Examples: x y Inflection Point

42.  A point on the graph of a continuous function where the tangent line exists and where the concavity changes is called an inflection point.  Examples: x y Inflection Point

43. Finding Inflection Points To find inflection points: 1.Compute f ″ (x). 2.Determine the numbers in the domain of f for which f ″ (x) = 0 or f ″ (x) does not exist. 3.Determine the sign of f ″ (x) to the left and right of each number c found in step 2. If there is a change in the sign of f ″ (x) as we move across x = c, then (c, f(c)) is an inflection point of f.

44. Examples  Find the points of inflection of the function Solution  We haveand  So, f ″ is continuous everywhere and is zero for x = 0.  We see that f ″ (x) 0 when x > 0.  Thus, we find that the graph of f : ✦ Has one and only inflection point at f(0) = 0. ✦ Is concave downward on the interval (– , 0). ✦ Is concave upward on the interval (0,  ). –2 2 –2 2 42–2–4 x y f(x)f(x)f(x)f(x) Inflection Point Example 3, page 268

45. Examples  Find the points of inflection of the function Solution  We haveand  So, f ″ is not defined at x = 1, and f ″ is never equal to zero.  We see that f ″ (x) 0 when x > 1.  Thus, we find that the graph of f : ✦ Has one and only inflection point at f(1) = 0. ✦ Is concave downward on the interval (– , 1). ✦ Is concave upward on the interval (1,  ). –2 –1 12 –2 –1 12 321–1–2–3 x y f(x)f(x)f(x)f(x) Inflection Point Example 4, page 268

46. Applied Example: Effect of Advertising on Sales  The total sales S (in thousands of dollars) of Arctic Air Co., which makes automobile air conditioners, is related to the amount of money x (in thousands of dollars) the company spends on advertising its product by the formula  Find the inflection point of the function S.  Discuss the meaning of this point. Applied Example 7, page 270

47. Applied Example: Effect of Advertising on Sales Solution  The first two derivatives of S are given by  Setting S ″ (x) = 0 gives x = 50.  So (50, S(50)) is the only candidate for an inflection point.  Since S ″ (x) > 0 for x 50, the point (50, 2700) is, in fact, an inflection point of S.  This means that the firm experiences diminishing returns on advertising beyond $50,000: ✦ Every additional dollar spent on advertising increases sales by less than previously spent dollars. Applied Example 7, page 270

48. The Second Derivative Test  Maxima occur when a curve is concave downwards, while minima occur when a curve is concave upwards.  This is the basis of the second derivative test: 1.Compute f ′ (x) and f ″ (x). 2.Find all the critical numbers of f at which f ′ (x) = 0. 3.Compute f ″ (c), if it exists, for each critical number c. a.If f ″ (c) < 0, then f has a relative maximum at c. b.If f ″ (c) > 0, then f has a relative minimum at c. c.If f ″ (c) = 0, then the test fails (it is inconclusive).

49. Example  Determine the relative extrema of the function Solution  We have so f ′ (x) = 0 gives the critical numbers x = –2 and x = 4.  Next, we have which we use to test the critical numbers: so, f(–2) = 60 is a relative maximum of f. so, f(4) = –48 is a relative minimum of f. Example 9, page 272

50. 4.3 Curve Sketching

51. Vertical Asymptotes  The line x = a is a vertical asymptote of the graph of a function f if either or

52. Finding Vertical Asymptotes of Rational Functions  Suppose f is a rational function where P and Q are polynomial functions.  Then, the line x = a is a vertical asymptote of the graph of f if Q(a) = 0 but P(a) ≠ 0.

53. Example  Find the vertical asymptotes of the graph of the function Solution  f is a rational function with P(x) = x 2 and Q(x) = 4 – x 2.  The zeros of Q are found by solving giving x = –2 and x = 2.  Examine x = –2: P(–2) = (–2) 2 = 4 ≠ 0, so x = –2 is a vertical asymptote. P(–2) = (–2) 2 = 4 ≠ 0, so x = –2 is a vertical asymptote.  Examine x = 2: P(2) = (2) 2 = 4 ≠ 0, so x = 2 is also a vertical asymptote. P(2) = (2) 2 = 4 ≠ 0, so x = 2 is also a vertical asymptote. Example 1, page 285

54. Horizontal Asymptotes  The line y = b is a horizontal asymptote of the graph of a function f if either

55. Example  Find the horizontal asymptotes of the graph of the function Solution  We compute and so y = –1 is a horizontal asymptote.  We compute also yielding y = –1 as a horizontal asymptote. Example 2, page 287

56. Example  Find the horizontal asymptotes of the graph of the function Solution  So, the graph of f has two vertical asymptotes x = –2 and x = 2, and one horizontal asymptote y = –1: x y y = –1 x = –2 x = 2 f(x)f(x)f(x)f(x) Example 2, page 287

57. Asymptotes and Polynomials  A polynomial function has no vertical asymptotes.  To see this, note that a polynomial function P(x) can be written as a rational function with a denominator equal to 1.  Thus,  Since the denominator is never zero, P has no vertical asymptotes.

58. Asymptotes and Polynomials  A polynomial function has no horizontal asymptotes.  If P(x) is a polynomial of degree greater or equal to 1, then are either infinity or minus infinity; that is, they do not exist.  Therefore, P has no horizontal asymptotes.

59. A Guide to Sketching a Curve 1.Determine the domain of f. 2.Find the x- and y-intercepts of f. 3.Determine the behavior of f for large absolute values of x. 4.Find all horizontal and vertical asymptotes of f. 5.Determine the intervals where f is increasing and where f is decreasing. 6.Find the relative extrema of f. 7.Determine the concavity of f. 8.Find the inflection points of f. 9.Plot a few additional points to help further identify the shape of the graph of f and sketch the graph.

60. Examples  Sketch the graph of the function Solution 1.The domain of f is the interval (– ,  ). 2.By setting x = 0, we find that the y-intercept is 2. (The x-intercept is not readily obtainable) 3.Since we see that f decreases without bound as x decreases without bound and that f increases without bound when x increases without bound. 4.Since f is a polynomial function, there are no asymptotes. Example 3, page 288

61. Examples  Sketch the graph of the function Solution 5. 5. Setting f ′ (x) = 0 gives x = 1 and x = 3 as critical points. Setting f ′ (x) = 0 gives x = 1 and x = 3 as critical points. Testing with different values of x we find that f ′ (x) > 0 when x 0 when x > 3. Testing with different values of x we find that f ′ (x) > 0 when x 0 when x > 3. Thus, f is increasing in the intervals (– , 1) and (3,  ), and f is decreasing in the interval (1, 3). Example 3, page 288

62. Examples  Sketch the graph of the function Solution 6. f ′ changes from positive to negative as we go across x = 1, so a relative maximum of f occurs at (1, f(1)) = (1, 6). f ′ changes from negative to positive as we go across x = 3, so a relative minimum of f occurs at (3, f(3)) = (1, 2). f ′ changes from negative to positive as we go across x = 3, so a relative minimum of f occurs at (3, f(3)) = (1, 2). Example 3, page 288

63. Examples  Sketch the graph of the function Solution 7. 7. which is equal to zero when x = 2. which is equal to zero when x = 2. Testing with different values of x we find that f ″ (x) 0 when 2 0 when 2 < x. Thus, f is concave downward in the interval (– , 2) and concave upward in the interval (2,  ). 8.Since f ″ (2) = 0, we have an inflection point at (2, f(2)) = (2, 4). Example 3, page 288

64. Examples  Sketch the graph of the function Solution  Summarizing, we’ve found the following: ✦ Domain: (– ,  ). ✦ Intercept: (0, 2). ✦ ✦ Asymptotes: None. ✦ f is increasing in the intervals (– , 1) and (3,  ), and f is decreasing in the interval (1, 3). ✦ A relative maximum of f occurs at (1, 6). ✦ A relative minimum of f occurs at (1, 2). ✦ f is concave downward in the interval (– , 2) and f is concave upward in the interval (2,  ). ✦ f has an inflection point at (2, 4). Example 3, page 288

65. 7654321 x y 123456123456123456123456Examples  Sketch the graph of the function Solution  Sketch the graph: (1, 6) Relative maximum (2, 4) Inflection point (3, 2) Relative minimum Intercept Example 3, page 288

66. 7654321 x y 123456123456123456123456Examples  Sketch the graph of the function Solution  Sketch the graph: Example 3, page 288

67. Examples  Sketch the graph of the function Solution 1. f is undefined when x = 1, so the domain of f is the set of all real numbers other than x = 1. 2.Setting y = 0, gives an x-intercept of –1. Setting x = 0, gives an y-intercept of –1. Example 4, page 290

68. Examples  Sketch the graph of the function Solution 3.Since we see that f(x) approaches the line y = 1 as |x| becomes arbitrarily large. ✦ For x > 1, f(x) > 1, so f approaches the line y = 1 from above. ✦ For x < 1, f(x) < 1, so f approaches the line y = 1 from below. 4.From step three we conclude that y = 1 is a horizontal asymptote of f. Also, the straight line x = 1 is a vertical asymptote of f. Example 4, page 290

69. Examples  Sketch the graph of the function Solution 5. 5. So, f ′ (x) is discontinuous at x = 1 and is never equal to zero. Testing we find that f ′ (x) < 0 wherever it is defined. 6.From step 5 we see that there are no critical numbers of f, since f ′ (x) is never equal to zero. Example 4, page 290

70. Examples  Sketch the graph of the function Solution 7. 7. Testing with different values of x we find that f ″ (x) 0 when 1 0 when 1 < x. Thus, f is concave downward in the interval (– , 1) and concave upward in the interval (1,  ). 8.From point 7 we see there are no inflection points of f, since f ″ (x) is never equal to zero. Example 4, page 290

71. Examples  Sketch the graph of the function Solution  Summarizing, we’ve found the following: ✦ Domain: (– , 1)  (1,  ). ✦ Intercept: (0, –1); (–1, 0). ✦ ✦ Asymptotes:x = 1 is a vertical asymptote. y = 1 is a horizontal asymptote. ✦ f is decreasing everywhere in the domain of f. ✦ Relative extrema: None. ✦ f is concave downward in the interval (– , 1) and f is concave upward in the interval (1,  ). ✦ f has no inflection points. Example 4, page 290

72. Examples  Sketch the graph of the function Solution  Sketch the graph: 4 2 –2 x y –2 2 4 –2 2 4 y = 1 x = 1 Example 4, page 290

73. 4.4 Optimization I

74. Absolute Extrema  The absolute extrema of a function f ✦ If f(x)  f(c) for all x in the domain of f, then f(c) is called the absolute maximum value of f. ✦ If f(x)  f(c) for all x in the domain of f, then f(c) is called the absolute minimum value of f.

75. Examples  f has an absolute minimum at (0, 0): Absolute minimum 4 2 x y –2–1 12

76. Examples  f has an absolute maximum at (0, 4): Absolute maximum 4 2 x y –2–1 12

77. Examples  f has an absolute minimum at and an absolute maximum at 1 1/2 –1/2–1 x y –1 1 Absolute minimum Absolute maximum

78. Examples  f has no absolute extrema: 7654321 x y 123456123456123456123456

79. Theorem 3 Absolute Extrema in a Closed Interval  If a function f is continuous on a closed interval [a, b], then f has both an absolute maximum value and an absolute minimum value on [a, b].

80. Example  The relative minimum of f at x 3 is also the absolute minimum of f.  The right endpoint b of the interval [a, b] gives rise to the absolute maximum value f(b) of f. x y Relative minimum Relative maximum Absolute minimum Absolute maximum a x1x1x1x1 x2x2x2x2 x3x3x3x3b Relative maximum

81. Finding Absolute Extrema To find the absolute extrema of f on a closed interval [a, b]. 1.Find the critical numbers of f that lie on (a, b). 2.Compute the value of f at each critical number found in step 1 and compute f(a) and f(b). 3.The absolute maximum value and absolute minimum value of f will correspond to the largest and smallest numbers, respectively, found in step 2.

82. Examples  Find the absolute extrema of the function F(x) = x 2 defined on the interval [–1, 2]. Solution  The function F is continuous on the closed interval [–1, 2] and differentiable on the open interval (–1, 2).  Setting F ′ = 0, we get F ′ (x) = 2x = 0, so there is only one critical point at x = 0.  So, F(–1) = (–1) 2 = 1, F(0) = (0) 2 = 0, and F(2) = (2) 2 = 4.  It follows that 0 is the absolute minimum of F, and 4 is the absolute maximum of F. Example 1, page 300

83. –2–1 12 Examples  Find the absolute extrema of the function F(x) = x 2 defined on the interval [–1, 2]. Solution Absolute minimum 4321 x y Absolute maximum Example 1, page 300

84. Examples  Find the absolute extrema of the function defined on the interval [0, 3]. Solution  The function f is continuous on the closed interval [0, 3] and differentiable on the open interval (0, 3).  Setting f ′ = 0, we get which gives two critical points at x = – 2/3 and x = 2.  We drop x = – 2/3 since it lies outside the interval [0, 3].  So, f(0) = 4, f(2) = – 4, and f(3) = 1.  It follows that – 4 is the absolute minimum of f, and 4 is the absolute maximum of f. Example 2, page 300

85. Examples  Find the absolute extrema of the function defined on the interval [0, 3]. Solution –2 24 Absolute minimum 42–2–4 x y Absolute maximum Example 2, page 300

86. Applied Example: Maximizing Profits  Acrosonic’s total profit (in dollars) from manufacturing and selling x units of their model F speakers is given by  How many units of the loudspeaker system must Acrosonic produce to maximize profits? Solution  To find the absolute maximum of P on [0, 20,000], first find the stationary points of P on the interval (0, 20,000).  Setting f ′ = 0, we get which gives us only one stationary point at x = 7500. Applied Example 4, page 301

87. Applied Example: Maximizing Profits  Acrosonic’s total profit (in dollars) from manufacturing and selling x units of their model F speakers is given by  How many units of the loudspeaker system must Acrosonic produce to maximize profits? Solution  Evaluating the only stationary point we get P(7500)= 925,000  Evaluating the endpoints we get P(0)= –200,000 P(20,000)= –2,200,000  Thus, Acrosonic will realize the maximum profit of $925,000 by producing 7500 speakers. Applied Example 4, page 301

88. Applied Example: Maximizing Profits  Acrosonic’s total profit (in dollars) from manufacturing and selling x units of their model F speakers is given by  How many units of the loudspeaker system must Acrosonic produce to maximize profits? Solution 1000800600400200–200 x y 246810121416 (Thousands of speakers) (Thousands of dollars) Maximum Profit Applied Example 4, page 301

89. 4.5 Optimization II

90. Guidelines for Solving Optimization Problems 1.Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure. 2.Find an expression for the quantity to be optimized. 3.Use the conditions given in the problem to write the quantity to be optimized as a function f of one variable. Note any restrictions to be placed on the domain of f from physical considerations of the problem. 4.Optimize the function f over its domain using the methods of Section 4.4.

91. Applied Maximization Problem: Packaging  By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an open box.  If the cardboard is 16 inches long and 10 inches wide, find the dimensions of the box that will yield the maximum volume. 16 10 Applied Example 2, page 314

92. Applied Maximization Problem: Packaging Solution 1.Let x denote the length in inches of one side of each of the identical squares to be cut out of the cardboard. The dimensions of the box are (16 – 2x) by (10 – 2x) by x in. x 16 – 2x 10 16 x xx 10 – 2x Applied Example 2, page 314

93. Applied Maximization Problem: Packaging Solution 2.Let V denote the volume (in cubic inches) of the resulting box. The volume, is the quantity to be maximized. x 16 – 2x 10 – 2x Applied Example 2, page 314

94. Applied Maximization Problem: Packaging Solution 3.Each side of the box must be nonnegative, so x must satisfy the inequalities x  0, 16 – 2x  0, and 10 – 2x  0. ✦ All these inequalities are satisfied if 0  x  5. ✦ Therefore, the problem at hand is equivalent to finding the absolute maximum of on the closed interval [0, 5]. Applied Example 2, page 314

95. Applied Maximization Problem: Packaging Solution 4. f is continuous on [0, 5]. Setting f ′ (x) = 0 we get which yields the critical numbers x = 20/3 and x = 2. We discard x = 20/3 for being outside the interval [0, 5]. We evaluate f at the critical point and at the endpoints: f(0) = 0 f(2) = 144 f(5) = 0 Thus, the volume of the box is maximized by taking x = 2. The resulting dimensions of the box are 12 ″ ☓ 6 ″ ☓ 2 ″. Applied Example 2, page 314

96. Applied Minimization Problem: Inventory Control  Dixie’s Import-Export is the sole seller of the Excalibur 250 cc motorcycle.  Management estimates that the demand for these motorcycles will be 10,000 for the coming year and that they will sell at a uniform rate throughout the year.  The cost incurred in ordering each shipment of motorcycles is $10,000, and the cost per year of storing each motorcycle is $200.  Dixie’s management faces the following problem: ✦ Ordering too many motorcycles at one time increases storage cost. ✦ On the other hand, ordering too frequently increases the ordering costs.  How large should each order be, and how often should orders be placed, to minimize ordering and storage costs? Applied Example 5, page 317

97. Applied Minimization Problem: Inventory Control Solution  Let x denote the number of motorcycles in each order.  Assuming each shipment arrives just as the previous shipment is sold out, the average number of motorcycles in storage during the year is x/2, as shown below:  Thus, Dixie’s storage cost for the year is given by 200(x/2), or 100x dollars. x Inventory Level Time Average inventory Applied Example 5, page 317

98. Applied Minimization Problem: Inventory Control Solution  Next, since the company requires 10,000 motorcycles for the year and since each order is for x motorcycles, the number of orders required is  This gives an ordering cost of dollars for the year.  Thus, the total yearly cost incurred by Dixie’s, including ordering and storage costs, is given by Applied Example 5, page 317

99. Applied Minimization Problem: Inventory Control Solution  The problem is reduced to finding the absolute minimum of the function C in the interval (0,10,000].  To accomplish this, we compute  Setting C ′ (x) =0 and solving we obtain x = + 1000.  We reject the negative for being outside the domain.  So we have x = 1000 as the only critical number. Applied Example 5, page 317

100. Applied Minimization Problem: Inventory Control Solution  So we have x = 1000 as the only critical number.  Now we find  Since C ″ (1000) > 0, the second derivative test implies that x = 1000 is a relative minimum of C.  Also, since C ″ (1000) > 0 for all x in (0,10,000], the function C is concave upward everywhere so that also gives the absolute minimum of C.  Thus, to minimize the ordering and storage costs, Dixie’s should place 10,000/1000, or 10, orders per year, each for a shipment of 1000 motorcycles. Applied Example 5, page 317

101. End of Chapter