4.4. Concavity and Curve Sketching. Concavity

Transcription

1 4.4 Concavit and Curve Sketching Concavit and Curve Sketching f' decreases CONCAVE DOWN 3 f' increases CONCAVE UP FIGURE 4.25 The graph of ƒsd = 3 is concave down on s - q, d and concave up on s, q d (Eample 1a). In Section 4.3 we saw how the first derivative tells us where a function is increasing and where it is decreasing. At a critical point of a differentiable function, the First Derivative Test tells us whether there is a local maimum or a local minimum, or whether the graph just continues to rise or fall there. In this section we see how the second derivative gives information about the wa the graph of a differentiable function bends or turns. This additional information enables us to capture ke aspects of the behavior of a function and its graph, and then present these features in a sketch of the graph. Concavit As ou can see in Figure 4.25, the curve = 3 rises as increases, but the portions defined on the intervals s - q, d and s, q d turn in different was. As we approach the origin from the left along the curve, the curve turns to our right and falls below its tangents. The slopes of the tangents are decreasing on the interval s - q, d. As we move awa from the origin along the curve to the right, the curve turns to our left and rises above its tangents. The slopes of the tangents are increasing on the interval s, q d. This turning or bending behavior defines the concavit of the curve.

2 268 Chapter 4: Applications of Derivatives DEFINITION Concave Up, Concave Down The graph of a differentiable function = ƒsd is (a) concave up on an open interval I if ƒ is increasing on I (b) concave down on an open interval I if ƒ is decreasing on I. If = ƒsd has a second derivative, we can appl Corollar 3 of the Mean Value Theorem to conclude that ƒ increases if ƒ 7 on I, and decreases if ƒ 6. The Second Derivative Test for Concavit Let = ƒsd be twice-differentiable on an interval I. 1. If ƒ 7 on I, the graph of ƒ over I is concave up. 2. If ƒ 6 on I, the graph of ƒ over I is concave down. CONCAVE UP '' 1 '' CONCAVE UP If = ƒsd is twice-differentiable, we will use the notations ƒ and interchangeabl when denoting the second derivative. EXAMPLE 1 Appling the Concavit Test (a) The curve = 3 (Figure 4.25) is concave down on s - q, d where =6 6 and concave up on s, q d where =6 7. (b) The curve = 2 (Figure 4.26) is concave up on s - q, q d because its second derivative =2 is alwas positive. EXAMPLE 2 Determining Concavit Determine the concavit of = 3 + sin on [, 2p]. FIGURE 4.26 The graph of ƒsd = 2 is concave up on ever interval (Eample 1b) sin '' sin 2 FIGURE 4.27 Using the graph of to determine the concavit of (Eample 2). Solution The graph of = 3 + sin is concave down on s, pd, where =-sin is negative. It is concave up on sp, 2pd, where =-sin is positive (Figure 4.27). Points of Inflection The curve = 3 + sin in Eample 2 changes concavit at the point sp, 3d. We call sp, 3d a point of inflection of the curve. DEFINITION Point of Inflection A point where the graph of a function has a tangent line and where the concavit changes is a point of inflection. A point on a curve where is positive on one side and negative on the other is a point of inflection. At such a point, is either zero (because derivatives have the Intermediate Value Propert) or undefined. If is a twice-differentiable function, = at a point of inflection and has a local maimum or minimum.

3 4.4 Concavit and Curve Sketching 269 '' FIGURE 4.28 The graph of = 4 has no inflection point at the origin, even though = there (Eample 3). '' does not eist. 1/3 FIGURE 4.29 A point where fails to eist can be a point of inflection (Eample 4). EXAMPLE 3 An Inflection Point Ma Not Eist Where = The curve = 4 has no inflection point at = (Figure 4.28). Even though =12 2 is zero there, it does not change sign. EXAMPLE 4 An Inflection Point Ma Occur Where Does Not Eist The curve = 1>3 has a point of inflection at = (Figure 4.29), but does not eist there. We see from Eample 3 that a zero second derivative does not alwas produce a point of inflection. From Eample 4, we see that inflection points can also occur where there is no second derivative. To stud the motion of a bod moving along a line as a function of time, we often are interested in knowing when the bod s acceleration, given b the second derivative, is positive or negative. The points of inflection on the graph of the bod s position function reveal where the acceleration changes sign. EXAMPLE 5 = d 2 d 2 a1>3 b = d d a1 3-2>3 b = >3. Studing Motion Along a Line A particle is moving along a horizontal line with position function sstd = 2t 3-14t t - 5, t Ú. Find the velocit and acceleration, and describe the motion of the particle. Solution The velocit is and the acceleration is std = s std = 6t 2-28t + 22 = 2st - 1ds3t - 11d, astd = std = s std = 12t - 28 = 4s3t - 7d. When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left. Notice that the first derivative s = s d is zero when t = 1 and t = 11>3. Intervals Sign of Y s œ Behavior of s 6 t increasing 1 6 t 6 11>3 - decreasing 11>3 6 t + increasing Particle motion right left right The particle is moving to the right in the time intervals [, 1) and s11>3, q d, and moving to the left in (1, 11>3). It is momentaril stationar (at rest), at t = 1 and t = 11>3. The acceleration astd = s std = 4s3t - 7d is zero when t = 7>3. Intervals 6 t 6 7>3 7>3 6 t Sign of a s fl - + Graph of s concave down concave up

4 27 Chapter 4: Applications of Derivatives The accelerating force is directed toward the left during the time interval [, 7>3], is momentaril zero at t = 7>3, and is directed toward the right thereafter. Second Derivative Test for Local Etrema Instead of looking for sign changes in ƒ at critical points, we can sometimes use the following test to determine the presence and character of local etrema. THEOREM 5 Second Derivative Test for Local Etrema Suppose ƒ is continuous on an open interval that contains = c. 1. If ƒ scd = and ƒ scd 6, then ƒ has a local maimum at = c. 2. If ƒ scd = and ƒ scd 7, then ƒ has a local minimum at = c. 3. If ƒ scd = and ƒ scd =, then the test fails. The function ƒ ma have a local maimum, a local minimum, or neither. f', f'' local ma f', f'' local min Proof Part (1). If ƒ scd 6, then ƒ sd 6 on some open interval I containing the point c, since ƒ is continuous. Therefore, ƒ is decreasing on I. Since ƒ scd =, the sign of ƒ changes from positive to negative at c so ƒ has a local maimum at c b the First Derivative Test. The proof of Part (2) is similar. For Part (3), consider the three functions = 4, = - 4, and = 3. For each function, the first and second derivatives are zero at =. Yet the function = 4 has a local minimum there, = - 4 has a local maimum, and = 3 is increasing in an open interval containing = (having neither a maimum nor a minimum there). Thus the test fails. This test requires us to know ƒ onl at c itself and not in an interval about c. This makes the test eas to appl. That s the good news. The bad news is that the test is inconclusive if ƒ = or if ƒ does not eist at = c. When this happens, use the First Derivative Test for local etreme values. Together ƒ and ƒ tell us the shape of the function s graph, that is, where the critical points are located and what happens at a critical point, where the function is increasing and where it is decreasing, and how the curve is turning or bending as defined b its concavit. We use this information to sketch a graph of the function that captures its ke features. EXAMPLE 6 Using ƒ and ƒ to Graph ƒ Sketch a graph of the function ƒsd = using the following steps. (a) Identif where the etrema of ƒ occur. (b) Find the intervals on which ƒ is increasing and the intervals on which ƒ is decreasing. (c) Find where the graph of ƒ is concave up and where it is concave down. (d) Sketch the general shape of the graph for ƒ.

5 4.4 Concavit and Curve Sketching 271 (e) Plot some specific points, such as local maimum and minimum points, points of inflection, and intercepts. Then sketch the curve. Solution ƒ is continuous since ƒ sd = eists. The domain of ƒ is s - q, q d, and the domain of ƒ is also s - q, q d. Thus, the critical points of ƒ occur onl at the zeros of ƒ. Since ƒ sd = = 4 2 s - 3d the first derivative is zero at = and = 3. Intervals Sign of ƒ œ Behavior of ƒ decreasing decreasing increasing (a) Using the First Derivative Test for local etrema and the table above, we see that there is no etremum at = and a local minimum at = 3. (b) Using the table above, we see that ƒ is decreasing on s - q, ] and [, 3], and increasing on [3, q d. (c) ƒ sd = = 12s - 2d is zero at = and = 2. Intervals Sign of ƒ œ Behavior of ƒ concave up concave down concave up We see that ƒ is concave up on the intervals s - q, d and s2, q d, and concave down on (, 2). (d) Summarizing the information in the two tables above, we obtain < <<2 2<<3 3< decreasing decreasing decreasing increasing concave up concave down concave up concave up The general shape of the curve is decr decr decr incr General shape. conc up conc down conc up conc up 2 3 infl point infl point local min

6 272 Chapter 4: Applications of Derivatives 2 15 Inflection 1 point 5 (, 1) Inflection (2, 6) 1 point (3, 17) Local minimum FIGURE 4.3 The graph of ƒsd = (Eample 6). (e) Plot the curve s intercepts (if possible) and the points where and are zero. Indicate an local etreme values and inflection points. Use the general shape as a guide to sketch the curve. (Plot additional points as needed.) Figure 4.3 shows the graph of ƒ. The steps in Eample 6 help in giving a procedure for graphing to capture the ke features of a function and its graph. Strateg for Graphing ƒ() 1. Identif the domain of ƒ and an smmetries the curve ma have. 2. Find and. 3. Find the critical points of ƒ, and identif the function s behavior at each one. 4. Find where the curve is increasing and where it is decreasing. 5. Find the points of inflection, if an occur, and determine the concavit of the curve. 6. Identif an asmptotes. 7. Plot ke points, such as the intercepts and the points found in Steps 3 5, and sketch the curve. EXAMPLE 7 Using the Graphing Strateg Sketch the graph of ƒsd = s + 1d Solution 1. The domain of ƒ is s - q, q d and there are no smmetries about either ais or the origin (Section 1.4). 2. Find ƒ and ƒ. ƒsd = s + 1d ƒ sd = s1 + 2 d # 2s + 1d - s + 1d 2 # 2 s1 + 2 d 2 = 2s1-2 d s1 + 2 d 2 ƒ sd = s1 + 2 d 2 # 2s -2d - 2s1-2 d[2s1 + 2 d # 2] s1 + 2 d 4 = 4s2-3d s1 + 2 d 3 -intercept at = -1, -intercept s = 1d at = Critical points: = -1, = 1 After some algebra 3. Behavior at critical points. The critical points occur onl at = ;1 where ƒ sd = (Step 2) since ƒ eists everwhere over the domain of ƒ. At = -1, ƒ (-1) = 1 7 ielding a relative minimum b the Second Derivative Test. At = 1, ƒ s1d = -1 6 ielding a relative maimum b the Second Derivative Test. We will see in Step 6 that both are absolute etrema as well.

7 4.4 Concavit and Curve Sketching Increasing and decreasing. We see that on the interval s - q, -1d the derivative ƒ sd 6, and the curve is decreasing. On the interval s -1, 1d, ƒ sd 7 and the curve is increasing; it is decreasing on s1, q d where ƒ sd 6 again. 5. Inflection points. Notice that the denominator of the second derivative (Step 2) is alwas positive. The second derivative ƒ is zero when = -23,, and 23. The second derivative changes sign at each of these points: negative on A - q, - 23B, positive on A - 23, B, negative on A, 23B, and positive again on A 23, q B. Thus each point is a point of inflection. The curve is concave down on the interval A - q, - 23B, concave up on A - 23, B, concave down on A, 23B, and concave up again on A 23, q B. 6. Asmptotes. Epanding the numerator of ƒ() and then dividing both numerator and denominator b gives 2 2 (1, 2) Point of inflection where 3 ƒsd = s + 1d = = 1 + s2>d + s1>2 d s1> 2. d + 1 Epanding numerator Dividing b Point of inflection where 3 FIGURE 4.31 (Eample 7). 1 1 Horizontal asmptote s + 1d2 The graph of = We see that ƒsd : 1 + as : q and that ƒsd : 1 - as : - q. Thus, the line = 1 is a horizontal asmptote. Since ƒ decreases on s - q, -1d and then increases on s -1, 1d, we know that ƒs -1d = is a local minimum. Although ƒ decreases on s1, q d, it never crosses the horizontal asmptote = 1 on that interval (it approaches the asmptote from above). So the graph never becomes negative, and ƒs -1d = is an absolute minimum as well. Likewise, ƒs1d = 2 is an absolute maimum because the graph never crosses the asmptote = 1 on the interval s - q, -1d, approaching it from below. Therefore, there are no vertical asmptotes (the range of ƒ is 2). 7. The graph of ƒ is sketched in Figure Notice how the graph is concave down as it approaches the horizontal asmptote = 1 as : - q, and concave up in its approach to = 1 as : q. Learning About Functions from Derivatives As we saw in Eamples 6 and 7, we can learn almost everthing we need to know about a twice-differentiable function = ƒsd b eamining its first derivative. We can find where the function s graph rises and falls and where an local etrema are assumed. We can differentiate to learn how the graph bends as it passes over the intervals of rise and fall. We can determine the shape of the function s graph. Information we cannot get from the derivative is how to place the graph in the -plane. But, as we discovered in Section 4.2, the onl additional information we need to position the graph is the value of ƒ at one point. The derivative does not give us information about the asmptotes, which are found using limits (Sections 2.4 and 2.5).

8 274 Chapter 4: Applications of Derivatives f() f() f () Differentiable smooth, connected; graph ma rise and fall ' rises from left to right; ma be wav ' falls from left to right; ma be wav or or '' concave up throughout; no waves; graph ma rise or fall '' concave down throughout; no waves; graph ma rise or fall '' changes sign Inflection point or ' changes sign graph has local maimum or local minimum ' and '' at a point; graph has local maimum ' and '' at a point; graph has local minimum

9 274 Chapter 4: Applications of Derivatives EXERCISES 4.4 Analzing Graphed Functions Identif the inflection points and local maima and minima of the functions graphed in Eercises 1 8. Identif the intervals on which the functions are concave up and concave down ( 2 1) 4 2/3 9 1/3 ( 2 7) 14

10 4.4 Concavit and Curve Sketching tan 4, sin 2, sin, cos 2, 3 2 Sketching the General Shape Knowing Each of Eercises gives the first derivative of a continuous function = ƒsd. Find and then use steps 2 4 of the graphing procedure on page 272 to sketch the general shape of the graph of ƒ. 41. = = =s - 3d = 2 s2 - d 45. =s 2-12d 46. =s-1d 2 s2 + 3d 47. =s8-5 2 ds4 - d =s 2-2ds - 5d =sec 2, - p p 2 NOT TO SCALE =tan, - p p =cot u 2, 6 u 6 2p 52. =csc 2 u 2, 6 u 6 2p Graphing Equations Use the steps of the graphing procedure on page 272 to graph the equations in Eercises 9 4. Include the coordinates of an local etreme points and inflection points. 9. = = = = s6-2d = = = s - 2d = 1 - s + 1d = = 2 s 2-2d = = 2 s6-2 d - 4 = = 3 s4 - d = = 3 s + 2d = = 4 s - 5d = a 4 2-5b = + sin, 2p = - sin, 2p 25. = 1>5 26. = 3>5 27. = 2>5 28. = 4>5 29. = 2-3 2>3 3. = 5 2> = 2>3 a = 2>3 s - 5d 2 - b =tan 2 u - 1, - p 2 6 u 6 p 2 =1 - cot 2 u, 6 u 6 p =cos t, t 2p =sin t, t 2p 57. =s + 1d -2>3 58. =s - 2d -1>3 59. = -2>3 s - 1d 6. = -4>5 s + 1d =2 ƒ ƒ = e -2, 2, 7 =e -2, 2, 7 Sketching from Graphs of and Each of Eercises shows the graphs of the first and second derivatives of a function = ƒsd. Cop the picture and add to it a sketch of the approimate graph of ƒ, given that the graph passes through the point P f''() f'() P P f'() f''() ƒ ƒ ƒ ƒ ƒ ƒ 33. = = , Z = = 2 = e 2-, 2, 7 4. = 2-4 = s2-2 d 3>2 = = ƒ 2-2 ƒ 65. P f''() f'()

11 276 Chapter 4: Applications of Derivatives 66. Motion Along a Line The graphs in Eercises 71 and 72 show the position s = ƒstd of a bod moving back and forth on a coordinate f'() line. (a) When is the bod moving awa from the origin? Toward the origin? At approimatel what times is the (b) velocit equal to zero? (c) Acceleration equal to zero? (d) When is the acceleration positive? f''() Negative? 71. s P Theor and Eamples 67. The accompaning figure shows a portion of the graph of a twicedifferentiable function = ƒsd. At each of the five labeled points, classif and as positive, negative, or zero. Displacement s f(t) Time (sec) t 68. Sketch a smooth connected curve = ƒsd with ƒs -2d = 8, ƒsd = 4, ƒs2d =, ƒ sd 7 for ƒ ƒ 7 2, P 69. Sketch the graph of a twice-differentiable function = ƒsd with the following properties. Label coordinates where possible. Derivatives f() Q ƒ s2d = ƒ s -2d =, ƒ sd 6 for ƒ ƒ 6 2, ƒ sd 6 for 6, ƒ sd 7 for 7. 6, 7 =, 7 7, 7 7, = 7, 6 =, 6 6, 6 7. Sketch the graph of a twice-differentiable function = ƒsd that passes through the points s -2, 2d, s -1, 1d, s, d, s1, 1d and (2, 2) and whose first two derivatives have the following sign patterns: R S T 72. Displacement s 73. Marginal cost The accompaning graph shows the hpothetical cost c = ƒsd of manufacturing items. At approimatel what production level does the marginal cost change from decreasing to increasing? Cost 74. The accompaning graph shows the monthl revenue of the Widget Corporation for the last 12 ears. During approimatel what time intervals was the marginal revenue increasing? decreasing? c s f(t) Time (sec) c f() Thousands of units produced r(t) t : : t

12 4.4 Concavit and Curve Sketching Suppose the derivative of the function = ƒsd is =s - 1d 2 s - 2d. At what points, if an, does the graph of ƒ have a local minimum, local maimum, or point of inflection? (Hint: Draw the sign pattern for.) 76. Suppose the derivative of the function = ƒsd is =s - 1d 2 s - 2ds - 4d. At what points, if an, does the graph of ƒ have a local minimum, local maimum, or point of inflection? 77. For 7, sketch a curve = ƒsd that has ƒs1d = and ƒ sd = 1>. Can anthing be said about the concavit of such a curve? Give reasons for our answer. 78. Can anthing be said about the graph of a function = ƒsd that has a continuous second derivative that is never zero? Give reasons for our answer. 79. If b, c, and d are constants, for what value of b will the curve = 3 + b 2 + c + d have a point of inflection at = 1? Give reasons for our answer. 8. Horizontal tangents True, or false? Eplain. a. The graph of ever polnomial of even degree (largest eponent even) has at least one horizontal tangent. b. The graph of ever polnomial of odd degree (largest eponent odd) has at least one horizontal tangent. 81. Parabolas a. Find the coordinates of the verte of the parabola = a 2 + b + c, a Z. b. When is the parabola concave up? Concave down? Give reasons for our answers. 82. Is it true that the concavit of the graph of a twice-differentiable function = ƒsd changes ever time ƒ sd =? Give reasons for our answer. 83. Quadratic curves What can ou sa about the inflection points of a quadratic curve = a 2 + b + c, a Z? Give reasons for our answer. 84. Cubic curves What can ou sa about the inflection points of a cubic curve = a 3 + b 2 + c + d, a Z? Give reasons for our answer. COMPUTER EXPLORATIONS In Eercises 85 88, find the inflection points (if an) on the graph of the function and the coordinates of the points on the graph where the function has a local maimum or local minimum value. Then graph the function in a region large enough to show all these points simultaneousl. Add to our picture the graphs of the function s first and second derivatives. How are the values at which these graphs intersect the -ais related to the graph of the function? In what other was are the graphs of the derivatives related to the graph of the function? 85. = = = = Graph ƒsd = and its first two derivatives together. Comment on the behavior of ƒ in relation to the signs and values of ƒ and ƒ. 9. Graph ƒsd = cos and its second derivative together for 2p. Comment on the behavior of the graph of ƒ in relation to the signs and values of ƒ. 91. a. On a common screen, graph ƒsd = 3 + k for k = and nearb positive and negative values of k. How does the value of k seem to affect the shape of the graph? b. Find ƒ sd. As ou will see, ƒ sd is a quadratic function of. Find the discriminant of the quadratic (the discriminant of a 2 + b + c is b 2-4ac). For what values of k is the discriminant positive? Zero? Negative? For what values of k does ƒ have two zeros? One or no zeros? Now eplain what the value of k has to do with the shape of the graph of ƒ. c. Eperiment with other values of k. What appears to happen as k : - q? as k : q? 92. a. On a common screen, graph ƒsd = 4 + k , -2 2 for k = -4, and some nearb integer values of k. How does the value of k seem to affect the shape of the graph? b. Find ƒ sd. As ou will see, ƒ sd is a quadratic function of. What is the discriminant of this quadratic (see Eercise 91(b))? For what values of k is the discriminant positive? Zero? Negative? For what values of k does ƒ sd have two zeros? One or no zeros? Now eplain what the value of k has to do with the shape of the graph of ƒ. 93. a. Graph = 2>3 s 2-2d for Then use calculus to confirm what the screen shows about concavit, rise, and fall. (Depending on our grapher, ou ma have to enter 2>3 as s 2 d 1>3 to obtain a plot for negative values of.) b. Does the curve have a cusp at =, or does it just have a corner with different right-hand and left-hand derivatives? 94. a. Graph = 9 2>3 s - 1d for Then use calculus to confirm what the screen shows about concavit, rise, and fall. What concavit does the curve have to the left of the origin? (Depending on our grapher, ou ma have to enter 2>3 as s 2 d 1>3 to obtain a plot for negative values of.) b. Does the curve have a cusp at =, or does it just have a corner with different right-hand and left-hand derivatives? 95. Does the curve = sin 2 have a horizontal tangent near = -3? Give reasons for our answer.

13 278 Chapter 4: Applications of Derivatives 4.5 Applied Optimization Problems To optimize something means to maimize or minimize some aspect of it. What are the dimensions of a rectangle with fied perimeter having maimum area? What is the least epensive shape for a clindrical can? What is the size of the most profitable production run? The differential calculus is a powerful tool for solving problems that call for maimizing or minimizing a function. In this section we solve a variet of optimization problems from business, mathematics, phsics, and economics. 12 Eamples from Business and Industr 12 (a) EXAMPLE 1 Fabricating a Bo An open-top bo is to be made b cutting small congruent squares from the corners of a 12-in.-b-12-in. sheet of tin and bending up the sides. How large should the squares cut from the corners be to make the bo hold as much as possible? Solution We start with a picture (Figure 4.32). In the figure, the corner squares are in. on a side. The volume of the bo is a function of this variable: Vsd = s12-2d 2 = V = hlw Since the sides of the sheet of tin are onl 12 in. long, 6 and the domain of V is the interval 6. A graph of V (Figure 4.33) suggests a minimum value of at = and = 6 and a maimum near = 2. To learn more, we eamine the first derivative of V with respect to : (b) FIGURE 4.32 An open bo made b cutting the corners from a square sheet of tin. What size corners maimize the bo s volume (Eample 1)? Volume min Maimum 2 6 NOT TO SCALE (12 2) 2, 6 min FIGURE 4.33 The volume of the bo in Figure 4.32 graphed as a function of. Of the two zeros, = 2 and = 6, onl = 2 lies in the interior of the function s domain and makes the critical-point list. The values of V at this one critical point and two endpoints are The maimum volume is 128 in. 3. The cutout squares should be 2 in. on a side. EXAMPLE 2 dv d = = 12s d = 12s2 - ds6 - d. Designing an Efficient Clindrical Can You have been asked to design a 1-liter can shaped like a right circular clinder (Figure 4.34). What dimensions will use the least material? Solution Volume of can: If r and h are measured in centimeters, then the volume of the can in cubic centimeters is Surface area of can: Critical-point value: Vs2d = 128 Endpoint values: Vsd =, Vs6d =. A = 2pr 2 + 2prh ()* ()* circular circular ends wall pr 2 h = 1. 1 liter = 1 cm 3

14 4.5 Applied Optimization Problems 279 2r h FIGURE 4.34 This 1-L can uses the least material when h = 2r (Eample 2). How can we interpret the phrase least material? First, it is customar to ignore the thickness of the material and the waste in manufacturing.then we ask for dimensions r and h that make the total surface area as small as possible while satisfing the constraint pr 2 h = 1. To epress the surface area as a function of one variable, we solve for one of the variables in pr 2 h = 1 and substitute that epression into the surface area formula. Solving for h is easier: Thus, h = 1 pr 2. A = 2pr 2 + 2prh = 2pr 2 + 2pr a 1 pr 2 b = 2pr r. Our goal is to find a value of r 7 that minimizes the value of A. Figure 4.35 suggests that such a value eists. A Tall and thin can Short and wide can Tall and thin A 2 r 2 2, r r min Short and wide 3 5 r FIGURE 4.35 The graph of A = 2pr 2 + 2>r is concave up. Notice from the graph that for small r (a tall thin container, like a piece of pipe), the term 2>r dominates and A is large. For large r (a short wide container, like a pizza pan), the term 2pr 2 dominates and A again is large. Since A is differentiable on r 7, an interval with no endpoints, it can have a minimum value onl where its first derivative is zero. What happens at r = 2 3 5>p? da dr = 4pr - 2 r 2 = 4pr - 2 r 2 4pr 3 = 2 r = 3 A 5 p L 5.42 Set da>dr =. Multipl b r 2. Solve for r.

15 28 Chapter 4: Applications of Derivatives The second derivative d 2 A 4 = 4p + 2 dr r 3 is positive throughout the domain of A. The graph is therefore everwhere concave up and the value of A at r = 2 3 5>p an absolute minimum. The corresponding value of h (after a little algebra) is h = 1 pr 2 = 2 A 3 5 p = 2r. The 1-L can that uses the least material has height equal to the diameter, here with r L 5.42 cm and h L 1.84 cm. Solving Applied Optimization Problems 1. Read the problem. Read the problem until ou understand it. What is given? What is the unknown quantit to be optimized? 2. Draw a picture. Label an part that ma be important to the problem. 3. Introduce variables. List ever relation in the picture and in the problem as an equation or algebraic epression, and identif the unknown variable. 4. Write an equation for the unknown quantit. If ou can, epress the unknown as a function of a single variable or in two equations in two unknowns. This ma require considerable manipulation. 5. Test the critical points and endpoints in the domain of the unknown. Use what ou know about the shape of the function s graph. Use the first and second derivatives to identif and classif the function s critical points , 4 2 FIGURE 4.36 The rectangle inscribed in the semicircle in Eample 3. Eamples from Mathematics and Phsics EXAMPLE 3 Inscribing Rectangles A rectangle is to be inscribed in a semicircle of radius 2. What is the largest area the rectangle can have, and what are its dimensions? Solution Let s, 24-2 d be the coordinates of the corner of the rectangle obtained b placing the circle and rectangle in the coordinate plane (Figure 4.36). The length, height, and area of the rectangle can then be epressed in terms of the position of the lower right-hand corner: Notice that the values of are to be found in the interval 2, where the selected corner of the rectangle lies. Our goal is to find the absolute maimum value of the function on the domain [, 2]. Length: 2, Height: 24-2, Area: 2 # Asd = 224-2

16 4.5 Applied Optimization Problems 281 The derivative da d = is not defined when = 2 and is equal to zero when = s4-2 d = = 2 = 2 or = ;22. Of the two zeros, = 22 and = -22, onl = 22 lies in the interior of A s domain and makes the critical-point list. The values of A at the endpoints and at this one critical point are Critical-point value: AA 22B = = 4 Endpoint values: Asd =, As2d =. The area has a maimum value of 4 when the rectangle is 24-2 = 22 units high and 2 = 222 unit long. HISTORICAL BIOGRAPHY Willebrord Snell van Roen ( ) EXAMPLE 4 Fermat s Principle and Snell s Law The speed of light depends on the medium through which it travels, and is generall slower in denser media. Fermat s principle in optics states that light travels from one point to another along a path for which the time of travel is a minimum. Find the path that a ra of light will follow in going from a point A in a medium where the speed of light is c 1 to a point B in a second medium where its speed is c 2. A Angle of incidence Medium 1 1 a 1 P d Angle of b 2 refraction Medium 2 d FIGURE 4.37 A light ra refracted (deflected from its path) as it passes from one medium to a denser medium (Eample 4). B Solution Since light traveling from A to B follows the quickest route, we look for a path that will minimize the travel time. We assume that A and B lie in the -plane and that the line separating the two media is the -ais (Figure 4.37). In a uniform medium, where the speed of light remains constant, shortest time means shortest path, and the ra of light will follow a straight line. Thus the path from A to B will consist of a line segment from A to a boundar point P, followed b another line segment from P to B. Distance equals rate times time, so The time required for light to travel from A to P is From P to B, the time is Time = distance rate. t 1 = AP c 1 = 2a2 + 2 c 1. t 2 = PB c 2 = 2b 2 + sd - d 2 c 2.

17 282 Chapter 4: Applications of Derivatives The time from A to B is the sum of these: t = t 1 + t 2 = 2a2 + 2 c 1 + 2b 2 + sd - d 2 c 2. This equation epresses t as a differentiable function of whose domain is [, d]. We want to find the absolute minimum value of t on this closed interval. We find the derivative dt d = c 1 2a In terms of the angles u 1 and u 2 in Figure 4.37, dt d = sin u 1 c 1 - sin u 2 c 2. d - c 2 2b 2 + sd - d 2. dt/d negative dt/d zero dt/d positive If we restrict to the interval d, then t has a negative derivative at = and a positive derivative at = d. B the Intermediate Value Theorem for Derivatives (Section 3.1), there is a point H [, d] where dt>d = (Figure 4.38). There is onl one such point because dt>d is an increasing function of (Eercise 54). At this point d FIGURE 4.38 The sign pattern of dt>d in Eample 4. sin u 1 c 1 = sin u 2 c 2. This equation is Snell s Law or the Law of Refraction, and is an important principle in the theor of optics. It describes the path the ra of light follows. Eamples from Economics In these eamples we point out two was that calculus makes a contribution to economics. The first has to do with maimizing profit. The second has to do with minimizing average cost. Suppose that rsd = the revenue from selling items csd = the cost of producing the items psd = rsd - csd = the profit from producing and selling items. The marginal revenue, marginal cost, and marginal profit when producing and selling items are dr d dc d dp d = marginal revenue, = marginal cost, = marginal profit. The first observation is about the relationship of p to these derivatives. If r() and c() are differentiable for all 7, and if psd = rsd - csd has a maimum value, it occurs at a production level at which p sd =. Since p sd = r sd - c sd, p sd = implies that r sd - c sd = or r sd = c sd.

18 4.5 Applied Optimization Problems 283 Therefore At a production level ielding maimum profit, marginal revenue equals marginal cost (Figure 4.39). Cost c() Dollars Revenue r() Break-even point B Maimum profit, c'() r'() Local maimum for loss (minimum profit), c'() r'() Items produced FIGURE 4.39 The graph of a tpical cost function starts concave down and later turns concave up. It crosses the revenue curve at the break-even point B. To the left of B, the compan operates at a loss. To the right, the compan operates at a profit, with the maimum profit occurring where c sd = r sd. Farther to the right, cost eceeds revenue (perhaps because of a combination of rising labor and material costs and market saturation) and production levels become unprofitable again. c() r() 9 Maimum for profit EXAMPLE 5 Maimizing Profit Suppose that rsd = 9 and csd = , where represents thousands of units. Is there a production level that maimizes profit? If so, what is it? Solution Notice that r sd = 9 and c sd = The two solutions of the quadratic equation are 1 = 2 = = = = L Set c sd = r sd. = 2-22 L.586 and Local maimum for loss NOT TO SCALE FIGURE 4.4 The cost and revenue curves for Eample 5. The possible production levels for maimum profit are L.586 thousand units or L thousand units. The second derivative of psd = rsd - csd is p sd = -c sd since r sd is everwhere zero. Thus, p sd = 6s2 - d which is negative at = and positive at = B the Second Derivative Test, a maimum profit occurs at about = (where revenue eceeds costs) and maimum loss occurs at about =.586. The graph of r() is shown in Figure 4.4.

19 284 Chapter 4: Applications of Derivatives Cost c() 5 25 min Ccle length 25 5 FIGURE 4.41 The average dail cost c() is the sum of a hperbola and a linear function (Eample 6). EXAMPLE 6 Minimizing Costs A cabinetmaker uses plantation-farmed mahogan to produce 5 furnishings each da. Each deliver of one container of wood is $5, whereas the storage of that material is $1 per da per unit stored, where a unit is the amount of material needed b her to produce 1 furnishing. How much material should be ordered each time and how often should the material be delivered to minimize her average dail cost in the production ccle between deliveries? Solution If she asks for a deliver ever das, then she must order 5 units to have enough material for that deliver ccle. The average amount in storage is approimatel one-half of the deliver amount, or 5 >2. Thus, the cost of deliver and storage for each ccle is approimatel Cost per ccle = deliver costs + storage costs Cost per ccle = 5 ()* + a 5 deliver cost 2 b # # ()* 1 ()* average amount stored number of das stored ()* storage cost per da We compute the average dail cost c() b dividing the cost per ccle b the number of das in the ccle (see Figure 4.41). As : and as : q, the average dail cost becomes large. So we epect a minimum to eist, but where? Our goal is to determine the number of das between deliveries that provides the absolute minimum cost. We find the critical points b determining where the derivative is equal to zero: Of the two critical points, onl 22 lies in the domain of c(). The critical-point value of the average dail cost is We note that c() is defined over the open interval s, q d with c sd = 1> 3 7. Thus, an absolute minimum eists at = 22 L das. The cabinetmaker should schedule a deliver of 5s14d = 7 units of the eotic wood ever 14 das. In Eamples 5 and 6 we allowed the number of items to be an positive real number. In realit it usuall onl makes sense for to be a positive integer (or zero). If we must round our answers, should we round up or down? EXAMPLE 7 c A 22B = 5 22 csd = , 7. c sd = = = ;22 L ; Sensitivit of the Minimum Cost = 522 L $ Should we round the number of das between deliveries up or down for the best solution in Eample 6?

20 4.5 Applied Optimization Problems 285 Solution The average dail cost will increase b about $.3 if we round down from to 14 das: and cs14d = s14d = $77.14 cs14d - cs14.14d = $ $77.11 = $.3. On the other hand, cs15d = $78.33, and our cost would increase b $ $77.11 = $1.22 if we round up. Thus, it is better that we round down to 14 das.

21 4.5 Applied Optimization Problems 285 EXERCISES 4.5 Whenever ou are maimizing or minimizing a function of a single variable, we urge ou to graph it over the domain that is appropriate to the problem ou are solving. The graph will provide insight before ou calculate and will furnish a visual contet for understanding our answer. Applications in Geometr 1. Minimizing perimeter What is the smallest perimeter possible for a rectangle whose area is 16 in. 2, and what are its dimensions? 2. Show that among all rectangles with an 8-m perimeter, the one with largest area is a square. 3. The figure shows a rectangle inscribed in an isosceles right triangle whose hpotenuse is 2 units long. a. Epress the -coordinate of P in terms of. (Hint: Write an equation for the line AB.) b. Epress the area of the rectangle in terms of. c. What is the largest area the rectangle can have, and what are its dimensions? 1 4. A rectangle has its base on the -ais and its upper two vertices on the parabola = What is the largest area the rectangle can have, and what are its dimensions? 5. You are planning to make an open rectangular bo from an 8-in.- b-15-in. piece of cardboard b cutting congruent squares from the corners and folding up the sides. What are the dimensions of B P(,?) A 1 the bo of largest volume ou can make this wa, and what is its volume? 6. You are planning to close off a corner of the first quadrant with a line segment 2 units long running from (a, ) to (, b). Show that the area of the triangle enclosed b the segment is largest when a = b. 7. The best fencing plan A rectangular plot of farmland will be bounded on one side b a river and on the other three sides b a single-strand electric fence. With 8m of wire at our disposal, what is the largest area ou can enclose, and what are its dimensions? 8. The shortest fence A 216 m 2 rectangular pea patch is to be enclosed b a fence and divided into two equal parts b another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fence will be needed? 9. Designing a tank Your iron works has contracted to design and build a 5 ft 3, square-based, open-top, rectangular steel holding tank for a paper compan. The tank is to be made b welding thin stainless steel plates together along their edges. As the production engineer, our job is to find dimensions for the base and height that will make the tank weigh as little as possible. a. What dimensions do ou tell the shop to use? b. Briefl describe how ou took weight into account. 1. Catching rainwater A 1125 ft 3 open-top rectangular tank with a square base ft on a side and ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not onl the material from which the tank is made but also an ecavation charge proportional to the product. a. If the total cost is c = 5s 2 + 4d + 1, what values of and will minimize it? b. Give a possible scenario for the cost function in part (a).

22 286 Chapter 4: Applications of Derivatives T c. Use a graphical method to find the maimum volume and the 11. Designing a poster You are designing a rectangular poster to contain 5 in. 2 of printing with a 4-in. margin at the top and bottom value of that gives it. and a 2-in. margin at each side. What overall dimensions will d. Confirm our result in part (c) analticall. minimize the amount of paper used? T 17. Designing a suitcase A 24-in.-b-36-in. sheet of cardboard is 12. Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 3. accompaning figure. Then four congruent squares of side length folded in half to form a 24-in.-b-18-in. rectangle as shown in the are cut from the corners of the folded rectangle. The sheet is unfolded, and the si tabs are folded up to form a bo with sides and a lid. 3 a. Write a formula V() for the volume of the bo. b. Find the domain of V for the problem situation and graph V 3 over this domain. c. Use a graphical method to find the maimum volume and the value of that gives it. d. Confirm our result in part (c) analticall. e. Find a value of that ields a volume of 112 in Two sides of a triangle have lengths a and b, and the angle between them is u. What value of u will maimize the triangle s f. Write a paragraph describing the issues that arise in part (b). area? (Hint: A = s1>2dab sin u. ) 14. Designing a can What are the dimensions of the lightest opentop right circular clindrical can that will hold a volume of 1 cm 3? Compare the result here with the result in Eample Designing a can You are designing a 1 cm 3 right circular 24" 24" clindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius r will be cut from squares that measure 2r units on a side. The total amount of aluminum used up b the can will therefore be 36" 18" rather than the A = 2pr 2 + 2prh in Eample 2. In Eample 2, the ratio of h to r for the most economical can was 2 to 1. What is the ratio now? 16. Designing a bo with a lid A piece of cardboard measures 1 in. b 15 in. Two equal squares are removed from the corners of a 1-in. side as shown in the figure. Two equal rectangles are removed from the other corners so that the tabs can be folded to form a rectangular bo with lid. NOT TO SCALE 1" A = 8r 2 + 2prh Base 15" a. Write a formula V() for the volume of the bo. b. Find the domain of V for the problem situation and graph V over this domain. Lid The sheet is then unfolded. 24" Base 18. A rectangle is to be inscribed under the arch of the curve = 4 cos s.5d from = -p to = p. What are the dimensions of the rectangle with largest area, and what is the largest area? 19. Find the dimensions of a right circular clinder of maimum volume that can be inscribed in a sphere of radius 1 cm. What is the maimum volume? 2. a. The U.S. Postal Service will accept a bo for domestic shipment onl if the sum of its length and girth (distance around) does not eceed 18 in. What dimensions will give a bo with a square end the largest possible volume? 36"

23 4.5 Applied Optimization Problems 287 Girth distance around here 24. The trough in the figure is to be made to the dimensions shown. Onl the angle u can be varied. What value of u will maimize the trough s volume? 1' 1' Length Square end 1' 2' T b. Graph the volume of a 18-in. bo (length plus girth equals 18 in.) as a function of its length and compare what ou see with our answer in part (a). 21. (Continuation of Eercise 2.) a. Suppose that instead of having a bo with square ends ou have a bo with square sides so that its dimensions are h b h b w and the girth is 2h + 2w. What dimensions will give the bo its largest volume now? Girth 25. Paper folding A rectangular sheet of 8.5-in.-b-11-in. paper is placed on a flat surface. One of the corners is placed on the opposite longer edge, as shown in the figure, and held there as the paper is smoothed flat. The problem is to make the length of the crease as small as possible. Call the length L. Tr it with paper. a. Show that L 2 = 2 3 >s2-8.5d. b. What value of minimizes L 2? c. What is the minimum value of L? D R C h T b. Graph the volume as a function of h and compare what ou see with our answer in part (a). 22. A window is in the form of a rectangle surmounted b a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits onl half as much light per unit area as clear glass does. The total perimeter is fied. Find the proportions of the window that will admit the most light. Neglect the thickness of the frame. h w L 2 2 A L Crease 26. Constructing clinders Compare the answers to the following two construction problems. a. A rectangular sheet of perimeter 36 cm and dimensions cm b cm to be rolled into a clinder as shown in part (a) of the figure. What values of and give the largest volume? b. The same sheet is to be revolved about one of the sides of length to sweep out the clinder as shown in part (b) of the figure. What values of and give the largest volume? P B Q (originall at A) 23. A silo (base not included) is to be constructed in the form of a clinder surmounted b a hemisphere. The cost of construction per square unit of surface area is twice as great for the hemisphere as it is for the clindrical sidewall. Determine the dimensions to be used if the volume is fied and the cost of construction is to be kept to a minimum. Neglect the thickness of the silo and waste in construction. Circumference (a) (b)

24 288 Chapter 4: Applications of Derivatives 27. Constructing cones A right triangle whose hpotenuse is 23 m long is revolved about one of its legs to generate a right circular cone. Find the radius, height, and volume of the cone of greatest volume that can be made this wa. h r 3 a. Find the dimensions of the strongest beam that can be cut from a 12-in.-diameter clindrical log. b. Graph S as a function of the beam s width w, assuming the proportionalit constant to be k = 1. Reconcile what ou see with our answer in part (a). c. On the same screen, graph S as a function of the beam s depth d, again taking k = 1. Compare the graphs with one another and with our answer in part (a). What would be the effect of changing to some other value of k? Tr it. 28. What value of a makes ƒsd = 2 + sa>d have a. a local minimum at = 2? b. a point of inflection at = 1? 29. Show that ƒsd = 2 + sa>d cannot have a local maimum for an value of a. 3. What values of a and b make ƒsd = 3 + a 2 + b have a. a local maimum at = -1 and a local minimum at = 3? b. a local minimum at = 4 and a point of inflection at = 1? Phsical Applications 31. Vertical motion The height of an object moving verticall is given b with s in feet and t in seconds. Find a. the object s velocit when t = b. its maimum height and when it occurs c. its velocit when s =. s = -16t t + 112, 32. Quickest route Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can row 2 mph and can walk 5 mph. Where should she land her boat to reach the village in the least amount of time? 33. Shortest beam The 8-ft wall shown here stands 27 ft from the building. Find the length of the shortest straight beam that will reach to the side of the building from the ground outside the wall. Beam 8' wall 27' Building T 35. Stiffness of a beam The stiffness S of a rectangular beam is proportional to its width times the cube of its depth. a. Find the dimensions of the stiffest beam that can be cut from a 12-in.-diameter clindrical log. b. Graph S as a function of the beam s width w, assuming the proportionalit constant to be k = 1. Reconcile what ou see with our answer in part (a). c. On the same screen, graph S as a function of the beam s depth d, again taking k = 1. Compare the graphs with one another and with our answer in part (a). What would be the effect of changing to some other value of k? Tr it. 36. Motion on a line The positions of two particles on the s-ais are s 1 = sin t and s 2 = sin st + p>3d, with s 1 and s 2 in meters and t in seconds. a. At what time(s) in the interval t 2p do the particles meet? b. What is the farthest apart that the particles ever get? c. When in the interval t 2p is the distance between the particles changing the fastest? 37. Frictionless cart A small frictionless cart, attached to the wall b a spring, is pulled 1 cm from its rest position and released at time t = to roll back and forth for 4 sec. Its position at time t is s = 1 cos pt. a. What is the cart s maimum speed? When is the cart moving that fast? Where is it then? What is the magnitude of the acceleration then? b. Where is the cart when the magnitude of the acceleration is greatest? What is the cart s speed then? 12" w d T 34. Strength of a beam The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. (See accompaning figure.) 1 s

25 4.5 Applied Optimization Problems Two masses hanging side b side from springs have positions s 1 = 2 sin t and s 2 = sin 2t, respectivel. a. At what times in the interval 6 t do the masses pass each other? (Hint: sin 2t = 2 sin t cos t. ) b. When in the interval t 2p is the vertical distance between the masses the greatest? What is this distance? (Hint: cos 2t = 2 cos 2 t - 1. ) Light source A Angle of incidence Normal 1 2 Plane mirror Light receiver Angle of reflection B m Distance between two ships At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a nautical mile is 2 d) and continued to do so all da. Ship B was sailing east at 8 knots and continued to do so all da. a. Start counting time with t = at noon and epress the distance s between the ships as a function of t. b. How rapidl was the distance between the ships changing at noon? One hour later? c. The visibilit that da was 5 nautical miles. Did the ships ever sight each other? T d. Graph s and ds>dt together as functions of t for -1 t 3, using different colors if possible. Compare the graphs and reconcile what ou see with our answers in parts (b) and (c). e. The graph of ds>dt looks as if it might have a horizontal asmptote in the first quadrant. This in turn suggests that ds>dt approaches a limiting value as t : q. What is this value? What is its relation to the ships individual speeds? 4. Fermat s principle in optics Fermat s principle in optics states that light alwas travels from one point to another along a path that minimizes the travel time. Light from a source A is reflected b a plane mirror to a receiver at point B, as shown in the figure. Show that for the light to obe Fermat s principle, the angle of incidence must equal the angle of reflection, both measured from the line normal to the reflecting surface. (This result can also be derived without calculus. There is a purel geometric argument, which ou ma prefer.) s 1 s 2 s m Tin pest When metallic tin is kept below 13.2 C, it slowl becomes brittle and crumbles to a gra powder. Tin objects eventuall crumble to this gra powder spontaneousl if kept in a cold climate for ears. The Europeans who saw tin organ pipes in their churches crumble awa ears ago called the change tin pest because it seemed to be contagious, and indeed it was, for the gra powder is a catalst for its own formation. A catalst for a chemical reaction is a substance that controls the rate of reaction without undergoing an permanent change in itself. An autocataltic reaction is one whose product is a catalst for its own formation. Such a reaction ma proceed slowl at first if the amount of catalst present is small and slowl again at the end, when most of the original substance is used up. But in between, when both the substance and its catalst product are abundant, the reaction proceeds at a faster pace. In some cases, it is reasonable to assume that the rate = d>dt of the reaction is proportional both to the amount of the original substance present and to the amount of product. That is, ma be considered to be a function of alone, and where = the amount of product a = the amount of substance at the beginning k = a positive constant. = ksa - d = ka - k 2, At what value of does the rate have a maimum? What is the maimum value of? 42. Airplane landing path An airplane is fling at altitude H when it begins its descent to an airport runwa that is at horizontal ground distance L from the airplane, as shown in the figure. Assume that the landing path of the airplane is the graph of a cubic polnomial function = a 3 + b 2 + c + d, where s -Ld = H and sd =. a. What is d>d at =? b. What is d>d at = -L? c. Use the values for d>d at = and = -L together with sd = and s -Ld = H to show that sd = H c2 a L b a L b 2 d.