10.3 Principles in Action 361 10.3 PRINCIPLES IN ACTION In this section we put the principles discussed in Sections 10.1 and 10.2 into action. ◆ EXAMPLE 10.6 Below is a graph of f , the derivative of f . The questions that follow refer to f . The domain of f is all real numbers. f′ (not f ) x –2–5 2 2 1 –1 –2 –3 57 Figure 10.21 (a) Identify all critical points of f . Which of these critical points are also stationary points of f ? (b) On a number line plot all the critical points of f . Indicate the sign of f and indicate where f is increasing and where f is decreasing. Is this enough information to classify all the extrema? Does f have a global maximum? If so, can we determine where it is attained? Does f have a global minimum? If so, can we determine where it is attained? (c) Where is the graph of f increasing and concave down? Where is the graph of f decreasing and concave up? (d) Determine all points of inflection of f . SOLUTION (a) Critical points of f are points at which either f = 0, f is undefined, or endpoints of the domain. The critical points in this example are x = 0 and x = 5. Both are points at which f = 0, so they are both stationary points. (b) 0+ ––5 graph of f sign of f′ We can see that f has a local maximum at x = 0 because f changes from increasing to decreasing at x = 0. We can tell that x = 0 is also the global maximum point. (f is increasing for all negative x and decreasing for all positive x.) x = 5 is not an extreme point, because f is decreasing both before and after x = 5. f =−1 for x>7; consequently, f is decreasing with a slope of −1on(7, ∞). Therefore there is no global minimum. (c) f is increasing and concave down on (−2, 0), where f is positive and decreasing. (Check the graph of f .) f is decreasing and concave up where f is negative and increasing. That is, on (2, 5). 362 CHAPTER 10 Optimization (d) Points of inflection of f are points at which the concavity of f changes. That is, points of inflection of f are points at which f changes sign. f is the slope of f ,sof changes sign at x =−2, x = 2, and x = 5. Below is a sketch 6 of f . (It is drawn without the x-axis because we don’t know where the x-axis is located.) pt. of inflection x = –2 x = 0 x = 2 x = 5 x = 0 local max global max pt. of inflection pt. of inflection Figure 10.22 ◆ EXERCISE 10.2 Notice that in Example 10.6 one of the x-intercepts of f corresponded to a local extrema of f and the other did not. Assuming f is continuous, come up with a graphical criteria for determining which x-intercepts of f correspond to local extrema of f . EXERCISE 10.3 Explain why local extrema of f correspond to points of inflection of f . (Assume f is continuous.) ◆ EXAMPLE 10.7 f is a continuous function defined on [−3, 3]. Its derivative f is given below. f is an odd function, that is, f (−x) =−f (x). Classify the critical points of f . f ′ x 3 –3 Figure 10.23 SOLUTION The critical points of f are x = 0, x =−3, and x = 3. x = 0 is a critical point because f is undefined there. x =−3and x = 3 are endpoints of the domain. –30 3 + – graph of f sign of f ′ 6 Note that because f is differentiable we know f is continuous. We proved this in Chapter 7. 10.3 Principles in Action 363 x = 0 is both a local maximum and the absolute maximum point. x =−3and x = 3 are both absolute minimum points, because f is an odd function. x = 0 absolute max local max x = –3 absolute min x = 3 absolute min Figure 10.24 ◆ ◆ EXAMPLE 10.8 Consider the function f(x)=x 3 +3x 2 −9x+3. Identify the x-coordinates of all local maxima and minima. Is there a global maximum? A global minimum? SOLUTION Begin by identifying the critical points. f (x) = 3x 2 + 6x − 9 = 3(x 2 + 2x − 3) = 3(x + 3)(x − 1) f is always defined, so the critical points are simply the stationary points. f (x) = 0atx=−3and x = 1 We test the sign of f in each of the intervals (−∞, −3), (−3, 1), and (1, ∞). This is most efficiently done when f is factored. –3 1 ++– graph of f sign of f ′ f has a relative maximum at x =−3and a relative minimum at x = 1. Try showing this on your own using the second derivative test. Then see if you can convince yourself that there is neither a global maximum nor a global minimum. We will do this in the next chapter. ◆ ◆ EXAMPLE 10.9 Find and classify the critical points of f(x)=e x (x 2 + x − 5). Find the x- and y- coordinates of all local extrema. Is there an absolute maximum value? An absolute minimum value? SOLUTION Begin by identifying critical points. f (x) = e x (x 2 + x − 5) + e x (2x + 1) By the Product Rule = e x (x 2 + 3x − 4) Factor = e x (x − 1)(x + 4) f is always defined, so the critical points are simply the stationary points. f (x) = 0atx=1and x =−4 364 CHAPTER 10 Optimization Both first and second derivative tests are simple to apply. We’ll use the second derivative test. f (x) = e x (x 2 + 3x − 4) f (x) = e x (x 2 + 3x − 4) + e x (2x + 3) = e x (x 2 + 5x − 1) f (1) = e(1 + 5 − 1) = 5e>0⇒f has a local minimum at x = 1. f (−4) = e −4 (16 − 20 − 1) = −5 e 4 < 0 ⇒ f has a local maximum at x =−4. When x = 1, y = f(1)=e(1 + 1 − 5) =−3e local minimum at (1, −3e) When x =−4, y = f(−4)=e −4 (16 − 4 − 5) = 7 e 4 local maximum at (−4, 7 e 4 ) Now we consider the global extrema. From the local information we have gathered it is unclear what happens globally. A bird’s-eye view will help. We need to look at lim x→∞ e x (x 2 + x − 5) and lim x→−∞ e x (x 2 + x − 5). As x grows without bound, both e x and (x 2 + x − 5) grow without bound. Therefore, lim x→∞ e x (x 2 + x − 5) =∞; f(x) has no absolute maximum. As x →−∞,e x →0and (x 2 + x − 5) →∞.Therefore it is not immediately clear what happens to the product. But e x is always positive and (x 2 + x − 5) is positive for |x| large. Therefore, the product is positive and f(x)takes on an absolute minimum value of −3e. In fact, it can be shown that lim x→−∞ e x (x 2 + x − 5) = 0, the behavior of the exponential dominating that of the cubic. f x –5 Figure 10.25 Look at the graph of f(x)on your calculator or computer. Notice which features of the function are easy to discern from the graphs and which are more difficult. ◆ Exploratory Problems for Chapter 10 365 Exploratory Problems for Chapter 10 Optimization 1. Let f(x)=xe x . (a) Find and classify all critical points of f . (b) Identify all points of inflection. (c) Does f have an absolute maximum value? If so, what is it and where is it attained? (d) Does f have an absolute minimum value? If so, what is it and where is it attained? (e) Find lim x→∞ xe x . (f) What is lim x→−∞ xe x ? This one is harder. Why? Argue convincingly that you have found this limit with error less than 10 −50 . (g) Graph f(x),labeling all critical points. Optional variation on the theme: Answer the same questions for the function f(x)=x 2 e x or h(x) = x 3 e −x . 2. Let g(x) = x + 4 x . (a) Find and classify all critical points of f .Whyisx=0not a critical point? (b) Identify all points of inflection. (c) Does f have an absolute maximum value? If so, what is it and where is it attained? (d) Does f have an absolute minimum value? If so, what is it and where is it attained? (e) Find lim x→∞ g(x). (f) What is lim x→−∞ g(x)? (g) Graph g(x), labeling all critical points. (h) Is g an even function? i.e., does g(x) = g(−x)? Is g an odd function? i.e., does g(x) =−g(−x)? Optional variation on the theme: Answer the same questions for the function g(x) = x + 4 x 2 . 366 CHAPTER 10 Optimization PROBLEMS FOR SECTION 10.3 1. Let f(x)= e x x 2 +1 . Find and classify the critical points. 2. A gardener has a fixed length of fence to fence off her rectangular chili pepper garden. Show that if she wants to maximize the area of her garden, then her garden should be square. 3. A gardener needs 90 square feet of land for her tomato plants. She will fence in a rectangular plot. The cost of the fencing increases with the length of the perimeter. Show that the cost of the fencing is minimum if she uses a square plot. 4. An open box with a rectangular base is to be constructed from a rectangular piece of cardboard 16 inches wide and 21 inches long by cutting a square from each corner and then bending up the resulting sides. Let x be the length of the sides of the corner squares. Our ultimate goal is to find the value of x that will maximize the volume of the box. (a) Express the volume V of the box as a function of x and determine the appropriate domain. (b) Use the sign V to make a very rough sketch of the graph of V on (−∞, ∞). Identify the portion of the graph that is appropriate for the context of the problem. (c) Find the value of x that will maximize the volume of the box. 5. You want to cut a rectangular wooden beam from a cylindrical log 14 inches in diameter. The strength of the beam is proportional to the quantity h 2 w, where h and w are the height and width of the cross section of the beam; the larger the quantity h 2 w, the stronger the beam. Find the height and width of the strongest beam that can be cut from the log. (Hint: You will need to find a way of relating h and w. Sketch the circular cross section and sketch in a line denoting the diameter of the log. By placing the diameter line appropriately, you should be able to produce a right triangle made of w, h, and the diameter. This will enable you to relate the width and height by using the Pythagorean Theorem.) 6. Let f be a function defined and continuous on the interval [−5, 5]. The graph of f (NOT f ) is given on the following page. Exploratory Problems for Chapter 10 367 f′ x –2 –1–5 –4 –3 2341 2 3 1 –1 –2 –3 5 (a) Identify all critical points. (b) For what value of x does f take on its maximum value? (c) For what value of x does f take on its minimum value? 7. Graphs of f(x)and its derivative f (x) are shown. Sketch graphs of the following and their derivatives. In each case (a) Describe in words how the new function relates to f(x). (b) Describe in words how the new derivative relates to f (x). (c) Identify the x-coordinates of all local minima. i. g(x ) = 2f(x) ii. j(x)=f(x)−3 iii. m(x) =|f(x)| iv. k(x ) = f(x −2) v. h(x) = f(2x) f′ x –2 –1 2 3 41 2 1 –1 f x –2 –1 2 3 41 1 –1 8. For each of the functions in parts (i)–(v) in the previous problem, consider the function restricted to the domain [−2, 3] and answer the following. (a) What are the critical points of the function on [−2, 3]? (b) At what value of x does the function attain its maximum value on [−2, 3]? (c) At what value of x does the function attain its minimum value on [−2, 3]? 368 CHAPTER 10 Optimization 9. (a) Use your knowledge of shifting, flipping, and stretching to graph the function f(x)=−2|x−2|+4. (b) At what value of x does f(x) attain its maximum value? At this point, what is f (x)? (c) Does f(x)have a minimum value? (d) Where on the interval 3 ≤ x ≤ 8 does f take on its maximum value? Its minimum value? 10. For each of the following functions, determine where the function is increasing and where it is decreasing. Find the x-coordinates of all local maxima and minima. (Give exact answers, not numerical approximations.) (a) f(x)=2x 3 −24x + 4 (b) f(x)=x 3 −3x 2 −9x +2 11. Below is a graph of f (x). The questions that follow are questions about f . f ′ x 246– 6 – 4 –2 (a) Identify all critical points of f . Are these critical points also stationary points? (Recall: A stationary point is a point at which f = 0.) (b) On a number line plot all the critical points of f . Use the number line to indicate the sign of f ; above this indicate where the graph of f is increasing and where it is decreasing. graph of f sign of f (c) Some of the x-intercepts of f correspond to the local extrema of f . How can you determine which ones do? (d) On a number line plot all the critical points of f . Use the number line to indicate the sign of f ; above this indicate where the graph of f is concave up and where it is concave down. graph of f sign of f (e) Where is the graph of f increasing and concave up? Where is the graph of f increasing and concave down? Where is the graph of f decreasing and concave up? Where is the graph of f decreasing and concave down? (f) Explain why the local extrema of f correspond to the points of inflection of f . (g) Suppose f(0)=0.Draw a rough sketch of the graph of f . 12. For each of the two graphs given on the following page, answer questions (a)–(d) from Problem 11 above. Exploratory Problems for Chapter 10 369 (a) f′ x 5 5 –5 (b) 2–2– 44 f′ x 13. A tin can for garbanza beans is designed to be a cylinder with volume of 300 cubic centimeters. Denote the radius by r and the height by h. The top and bottom are thicker than the sides; for the purposes of our model, we’ll assume that they are made with a double thickness of aluminum. (a) Give an expression for the volume of the can. (b) Give an expression for the amount of material used. (Remember that the top and bottom of the can are two layers thick.) (c) Make the expression from part (b) into a function of r alone. (d) What radius minimizes the material used? (e) What are the dimensions of the 300 cubic-centimeter can that require the least amount of material ? 14. A can for mandarin oranges is a cylinder with volume of 250 cubic centimeters. Denote the radius by r and the height by h. The material used for the top and bottom is stronger than that used for the sides. There is wasted material in constructing the top and bottom because they need to be cut from squares of metal and the scrap metal is not used. The manufacturers must pay for the material for the whole square from which the circle is cut. Suppose that the material for the top and bottom is three times as expensive as the material for the sides. What are the dimensions of the can that minimize the cost of the materials? 15. You are designing a wooden box for raspberries. It is to have the following specifica- tions: a square base, no lid, and a volume of V cubic centimeters. What is the ratio of the height to the length of the base that minimizes the amount of wood required? Your answer should not involve V . 16. A box of Maine blueberries is packaged in a cardboard box with a square base and a see-through plastic top. The plastic costs three times as much per square inch as does 370 CHAPTER 10 Optimization the cardboard. Assuming the volume of the box is fixed, what is the ratio of the height to the length of the base that minimizes the cost of the material for the box? 17. In its first printing, the printed material on a typical page of Frank McCourt’s Angela’s Ashes was 4 1 2 inches by 7 1 2 inches, with 1 2 -inch margins on the top and sides of the page and a 1-inch margin on the bottom. Assuming pages must hold 33.75 square inches of printed matter and have the margins specified, was this book laid out in such a way as to minimize the amount of paper per page? If not, what page dimensions would minimize the page area? 18. Q-Tips ® are a brand of cotton swabs each 3 inches long. You can purchase a pack of 300 of them in a plastic rectangular container backed in cardboard. In other words, the plastic forms an open box and the “lid” is cardboard. The width of the box is 3 inches. What should the length and depth be if the goal is to minimize the amount of plastic used? In order to hold 300 Q-Tips the box must have a volume of 33.75 square inches. In reality, such a box is 7.5 inches long and 1.5 inches deep. Has the amount of plastic been minimized? 19. f(x)= 1 3 x 3 − 2x − 1 x (a) Find f . (b) Find f . (c) Find all critical points. Which of these critical points are also stationary points? (d) Analyze the critical points. Are they local extrema? Global extrema? Points of inflection? (e) Is x = 0 the x-value of a critical point? Why or why not? (f) What is the absolute maximum value of the function? The absolute minimum value? 20. An artist wants to frame an 8-inch by 10-inch watercolor landscape with a mauve mat. She wants to use a total of 200 square inches of matting material, with x inches of matting above and below the painting and y inches of matting to the left and right of the painting. (Note: The matting is indicated by the shaded region in the accompanying figure; there is no matting underneath the painting.) The framing material is sold by the linear foot and is quite expensive. Therefore, she wants to minimize the perimeter of the frame. x y (a) Express the perimeter of the frame as a function of x and y. (b) Express the perimeter of the frame as a function of x alone. (c) Find the dimensions of the frame that minimize its cost. . watercolor landscape with a mauve mat. She wants to use a total of 200 square inches of matting material, with x inches of matting above and below the painting and y inches of matting to the left and. used for the top and bottom is stronger than that used for the sides. There is wasted material in constructing the top and bottom because they need to be cut from squares of metal and the scrap. 44 f′ x 13. A tin can for garbanza beans is designed to be a cylinder with volume of 300 cubic centimeters. Denote the radius by r and the height by h. The top and bottom are thicker than the sides;