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Unformatted text preview: tth. three of these :sed thus far in , :s? at are available :tion. r this activity ZD-ROM and < nal incarcera. .' nces of more Jder state and from 1977 I )rtheastincar- i, x, the number 5' equation you N exhibit the he incarcera- ‘apidly was it est incarcera- :iVity (located 1nd web site) ther benefici- nd 2001. 'mine during l the number aries showed verage rate of er of mother action of the :he inflection : in context. Instructor Note This section is optional. Instructor Note Students will find setting up the equation the most difficult part of this section. 5.4 Derivatives in Action 341 Derivatives in Action By now you should have a solid understanding of the basic method of finding both relative and absolute maximum and minimum values of a function. We now extend the techniques of optimization to situations in which you must first set up an equa— tion to optimize and then apply optimization techniques to that equation. Many of the problems presented in this section are different from most of those in this book because no data or equation is given. You must find an equation from a word description. We begin by outlining some steps that are helpful in setting up equations and optimizing their outputs. Refer to this list often until these steps become second nature to you. Step 1: Read the question carefully, and identity the quantity to be maximized or min— imized (the output variable) and the quantity or quantities on which the output quantity depends (the input variable or variables). This information is often found in the final sentence of the problem statement. Because this step is the most crucial one for correctly directing your thinking as you approach the problem, we present four questions in Example 1 to give you some practice in identifying input and output quantities. EXAMPLE 1 Determining Output and Input When Optimizing For each of the following statements, determine the quantity to be maximized or minimized and the corresponding input quantity or quantities. a. Find the order size that will minimize cost. b. Determine the dimensions that will result in a box of greatest volume. c. For what size dose is the sensitivity to the drug greatest? d. What is the most economical way to lay the pipe? Solution a. Output quantity to be minimized: cost Input quantity: order size b. Output quantity to be maximized: volume Input quantity: dimensions of box (height, length, and width) c. Output quantity to be maximized: sensitivity Input quantity: dosage d. Output quantity to be minimized: cost Input quantity: Not enough information is given in this sentence, but we can conclude that the input is probably some measure of distance related to how the pipe is laid. O 342 CHAPTER 5 WEB CD Strengthening the Concepts: Formulas for Geometry- Related Problems Instructor Note Students should not assume when there is one relative extremum that this value is automatically the answer to the question, because the answer could occur at one of the endpoints. Encouraging your students to graph the function as a means of visually checking their answers will help them avoid this mistake. Instead of graphing the function, a student could check the sign of the first derivative on either side of the relative extreme point. Instructor Note A good short quiz in the class following your discussion of this section is to have your students list these seven steps in their own words. Analyzing Change: Applications of Derivatives Step 2: Step 3: Step 4: Step 5: Step 6: Step 7: Sketch a picture if the problem situation is geometric in nature—that is) if the problem involves constructing containers, laying cable or Wire, build'mg enclosures, or the like. The picture should be labeled With appropriate vari. ables representing distances or sizes. You should find that at least one of the variables represents the input quantity identified in Step 1. Begin building the equation. From the results of Step 1, you will know the quantity for which you need an equation, such as cost, volume, or drug sensitivity. If the quantity is a geometric one (such as volume of a right circular cylinder, area of a trapezoid, hypotenuse of a right triangle) you may need to look up the formula in some reference source. Rewrite the equation as a function of only one variable if it is not already in this form. This step may require the use of other secondary equations relat— ing the input variables in the equation in Step 3. For example, if the equation in Step 3 is for volume of a box as a function of length and width, you need to find an equation relating length and width. You will then use this equation to solve for one variable in terms of the other and substitute for that variable in the equation in Step 3. Determine any limitations on the input variable. State the limitation as an interval in which the input must lie. Apply optimization techniques to the final equation. This step includes finding all relative extrema and determining all absolute extrema over the interval identified in Step 5. This step should also include graphing the equation or some other method of verifying that the answer you found is indeed a maximum or minimum on the interval. Read the problem again to be sure you have answered the question asked. Some- times the question asks you to find the output quantity, sometimes the input quantity, sometimes both, and occasionally some other quantity that can be cal- culated using the input and output values that correspond to the extreme point. Before demonstrating this multistep problem—solving method with three exam- ples, we summarize the seven steps. Problem-Solving Strategy for Optimization Problems Step 1: Identify the quantity to be maximized or minimized (the output) and the quantity or quantities on which the output quantity depends (the input). Step 2: Sketch and label a picture if the problem is geometric in nature. Step 3: Build a model for the quantity that is to be maximized or minimized. Step 4: Rewrite the equation in Step 3 in terms of only one input variable. Step 5: Identify the input interval. Step 6: Apply optimization techniques to the final equation. Step 7: Answer the question or questions posed in the problem. —that is,“ ‘ EXAMPLE 2 re, building " priate vari~ One of the : lknowthe 3 1e, or drug Of a right €)Y011 may i t already in tions relat_ 1e equation 1) You need is equation tat variable 1ti0n as an p includes 1a over the 1phing the u found is ked. Some- 5 the input can be cal- eme point. [1'66 exam- 5.4 Derivatives in Action 343 Applying the Problem-Solving Strategy Ranching A rancher removed 200 feet of wire fencing from a field on his ranch. He wishes to reuse the fencing to create a rectangular corral into which he will build a 6—foot-wide wooden gate. What dimensions will result in a corral with the greatest possible area? What is the greatest area? Solution Step 1: Output variable to maximize: area of rectangular corral Input variables: length and width of the corral Step 2: A sketch of the corral is shown in Figure 5.23. (Note that you can put the gate Step 3: Step 4: on any side of the corral.) }——— lfeet ——| FIGURE 5.23 Recall that the area of a rectangle is A = lw square feet, where l is the length of the rectangle in feet and w is its width in feet. This equation is the one we seek to maximize. Because the equation in Step 3 has two input variables, we need to find a sec- ond equation relating the length and the width. A clue to this equation is found in the statement “A rancher removed 200 feet of wire fencing.” If the area of the corral is to be a maximum, it makes sense that the rancher will use all of the fencing. Referring to Figure 5.23, we see that if we calculate the dis— tance around the corral and subtract the 6—foot—wide gate, we should obtain 200 feet. This leads to the equation 2w+2l~6=200 Solving this equation for one of the variables (it doesn’t matter which one) results in the equation _ 206 — 2w 1 2 =103—w Substituting this expression for 1 into the area equation gives A = lw = (103 — w)w = 103w — w2 square feet where w is the width of the corral in feet. 344 CHAPTER 5 Analyzing Change: Applications of Derivatives Step 5: Look again at Figure 5.23. If the length of the corral were 0, then the width would be 100 feet. We also know that width cannot be zero or negative. Thlls we have the input interval 0 < w < 100. Step 6: Note that the graph of the area A is a concave—down parabola. We find the maximum point on the parabola by determining where the derivative of A is zero (that is, where its graph crosses the w—axis). dA l ll; E=103—2w=0 W: 51.5 M we have no endpoints to check. Thus the corral with maximum area should be 51.5 feet wide. l 1 , Step 7: The statement of the problem asked for the dimensions of the corral of i greatest area as well as for the area. We know the width and need to deter- l ‘ mine the length of the corral. To do so, we use the equation from Step 4 that l l l l l l l j l j This value lies in the interval in Step 5. Because the interval is an open interval, I l l l l l expresses length in terms of width. Z: 103 — w I: 103 — 51.5 = 51.5 feet This calculation tells us that the corral with maximum area is a square with sides that are 51.5 feet long. To determine the corresponding area, simply square 51.5 feet. A = (51.5 feet)(51.5 feet) = 2652.25 square feet 0 The next example shows the solution to a more involved geometric problem. -' 1, ‘ EXAMPLE 3 Optimizing in a Geometric Setting I . ——————————_“_ Popcorn Tins In an effort to be environmentally responsible, a confectionery com- “ : pany is rethinking the dimensions of the tins in which it packages popcorn. Each ‘ ‘ cylindrical tin is to hold 3.5 gallons. The bottom and lid are both circular, but the lid must have an addltional 1g inch around it 1n order to form a lip. (Consrder the amount of metal needed to create a seam on the side and to join the side to the bottom to be negligible.) What are the dimensions of a tin that meets these specifications but uses the least amount of metal possible? Solution Step 1: Output quantity to be minimized: amount of metal Input quantities: dimensions of the tin Step2: Figure 5.24 shows the tin and its components. We have chosen to label the height of the tin h and the radius of the bottom of the tin r. (It is also possi- ble to use the diameter, instead of the radius, of the top or bottom as the other variable.) then the width negative. Thlls ’ la. We find the I *rivative of A is i l open interval, In area should ' the corral of need to deter— 3m Step 4 that a square with g area, simply )blem. :tionery com- »opcorn. Each ar, but the lid :Consider the to the bottom :ifications but [1 to label the is also possi— Iottom as the Step 4: S = 27rr< 5.4 Derivatives in Action 345 r inches r + ginches P0pcorn Side 2 T: r inches h inches FIG U R E 5 . 24 Step 3: The amount of metal used to construct the tin is measured by finding the combined surface area of the side, top, and bottom. Thus the total surface area is determined by the equation 8 = area of side + area of bottom + area of top The area of the side is anh. The area of the bottom is 1T72. The radius of the . . 9 . 9 2 Circle used to make the top 18 r + g; thus the area of the top is 17(r + g) . Using these formulas, we have the model for the surface area of the container: 9 2 . S = 27rrh + 'rrrz + 17(r + square 1nches where r is the radius of the bottom of the container in inches and h is the height of the container in inches. We need to rewrite the formula for surface area in terms of either radius or height. To do so, we need an equation relating these two quantities. Recall from the statement of the problem that the volume of the container needs to be 3.5 gallons. This information allows us to set up a second equation. The volume of a cylinder is given by the equation V = 1TT2h. Because we will be using dimensions in inches, we must express the volume in cubic inches. One gallon is 231 cubic inches (the authors found this value in a dictionary), so 3.5 gallons is 3.5(231) = 808.5 cubic inches. Thus we have the equation V = "lTI‘ZI’l = 808.5 We must solve for one variable in terms of the other. It makes sense to choose the variable for which this can be done most simply. In this case, we solve for h in terms of r. = 808.5 1T7‘2 I1 forreéO Next, we substitute this expression for 11 into the surface area equation in Step 3. 808.5 "lTTz 9 2 1617 2 > + "lTTz + 11(r + = T + 1172 + 17(r + square inches 346 CHAPTER 5 Analyzing Change: Applications of Derivatives Step 5: Because we are given no indication that the dimensions are restricted (exCe by the volume of the container, which was taken into account in Step 4) only restriction is that the radius must be positive: r > 0. Pt ) 0111‘ Step 6: Next we find the minimum of S by determining where a graph of g crosses the horizontal axis. To do so, we find the derivative of S with respect to r, . d8 -1617 9) —1617 9w 1 E: r2 +21'rr+2'rr(r+§(1)= r2 +4TFT+T 50) Setting this expression equal to zero and solving for r gives r k 4.87 inches, Surface area which lies in the interval in Step 5. To confirm that this value of r does indeed (Square inches) give a minimum surface area, we examine a graph of the surface area func; 2000 tion. See Figure 5.25. Step 7: Our final step is to answer the question posed. In this case, we must provide the dimensions (height and radius) of the tin with minimum surface area. r We know the radius. In order to find the height, we substitute the unrounded 2000 4 87 17 Radius value of the radius into the equation that we found in Step 4 for height in ' (mChCS) terms of radius. FIGURE 5.25 8085 h z *3 10.86 inches In order to use the least amount of metal possible, the popcorn tins should be constructed with height approximately 10.86 inches and radius approxi- mately 4.87 inches.* 0 Our final example illustrates optimization in a nongeometric setting. EXAMPLE 4 Optimizing Revenue Cruise A travel agency offers spring—break cruise packages. The agency advertises a cruise to Cancun, Mexico, for $1200 per person. In order to promote the cruise among student organizations on campus, the agency offers a discount for student groups selling the cruise to over 50 of their members. The price per student will be discounted by $10 for each student in excess of 50. For example, if an organiza- tion had 55 members go on the cruise, each of those 55 students would pay $1200 — 5($10) = $1150. a. What size group will produce the largest revenue for the travel agency, and what is the largest possible revenue? b. If the travel agent limits each organization to 75 tickets, what is the agent’s maxi- mum revenue for each organization? c. If you were the travel agent, would you set a limit on the number of students per organization? *Point of Interest: The authors found that the popcorn tin they purchased as a reference tool for this example did have dimensions that matched those of a tin with minimum surface area. stricted (except _ :in Step 4), our :1 d5 ? )h of a crosses I respect to r. E 4 z 4.87 inches, )f r does indeed '7. face area fund- a" e must provide ' in surface area. the unrounded 4 for height in )rn tins should 'adius approxi— ency advertises iote the cruise int for student er student will if an organiza— 1ts would pay ency, and what e agent’s maxi- of students per rence tool for this Solution a. Step 1: Step 2: Step 3: 5.4 Derivatives in Action 347 Output quantity to be maximized: travel agency’s revenue from student group Input quantity: size of the group Because this scenario is not geometric, there is no need for a picture. We need an equation for the travel agency’s revenue from the students. In this example, revenue is the number of students traveling on the cruise multiplied by the price each student pays: Revenue = (number of students)(price per student) In Step 1, we determined that the input quantity is the size of the group. Because the factor affecting price is the number of students in excess of 50, we choose to use a variable to represent that quantity. (It is also pos- sible to use a variable representing the total number of students.) If s is the number of students in excess of 50, then the total number of students is 50 + s. The price is $1200 minus $10 for each student in excess of 50. Converting this statement into math symbols, we have the price per stu— dent as 1200 — 105 dollars. Thus Revenue = R(s) = (50 + s)(1200 — 105) = 60,000 + 7005 — 1052 dollars Step 4: Step 5: Step 6: Step 7: where there are 50 + 5 students on the cruise. Because the revenue function has only one input variable, this step is not necessary. We are not given any restriction concerning the number of students. Our interval is s > 0. The revenue equation is a concave—down parabola, so we can be confi- dent that it has a maximum. We simply set the derivative of the revenue function equal to zero and solve for s. R’(s) = 700 — 205 = 0 s = 35 Revenue is maximized when 5 = 35. The question asks for the number of students and the corresponding revenue. Recall that the variable 5 represents the number of students in excess of 50. Thus the total number of students is 50 + 35 = 85. The price per student is $1200 — 35(10) = $850. The revenue is (85 stu— dents)($850 per student) = $72,250. Revenue is maximized at $72,250 for a campus group of 85 students. If the number of students is limited to 75, then the interval in Step 5 is 0 < s S 25. The solution to part a is no longer valid because it does not lie in this interval. In this case, the maximum revenue occurs at the endpoint of the interval (when there are 75 students). Examine a graph of R, and observe that this is true. The price each student pays is 1200 — 25(10) = $950. The travel agent’s revenue in this case is (75 students)($950 per student) = $71,250. 348 CHAPTER 5 Analyzing Change: Applications of Derivatives c. In part a, we found that when 85 students bought cruise tickets, the travel agent’s revenue was a maximum. If more than 85 students bought tickets, the agent’s rev. enue would actually decline because of the low price. (Recall that the revenue function graph was a concave—down parabola.) Therefore, it would make sense for the agent to limit the number of students per organization to 85. 0 Optimization problems such as those discussed in this section are often challeng_ ing, but as long as you approach them using the seven steps outlined here, you should be able to solve them—or at least identify the point at which you need help. As we have said before, the more problems you work, the more skilled and confident you will become. 5.4 Concept Inventory 0 Seven-step problem-solving strategy for optimization Activities Key @ 1—16 if 17—18 See page xvi of the preface for icon explanations. 5.4 Activities 1. Popcorn Tins Refer to Example 3, and repeat the solution using the diameter of the bottom of the tin, instead of the radius, as an input variable. Show that the dimensions you obtain for minimum surface area are equivalent to those found in Example 3. 2. Cruise Refer to Example 4, and repeat the solu— tion using the total number of students as the input variable. Show that the maximum revenue and the corresponding number of students and price are equivalent to those found in Example 4. 3. Garden Area A rectangular—shaped garden has one side along the side of a house. The other three sides are to be enclosed with 60 feet of fencing. What is the largest possible area of such a garden? 4. Floral Frame A florist uses wire frames to sup- port flower arrangements displayed at weddings. Each frame is constructed from a wire of length 9 feet that is cut into six pieces. The vertical edges of the frame consist of four of the pieces of wire that are each 12 inches long. One of the remaining pieces is bent into a square to form the base of the frame; the final piece is bent into a circle to form the top of the frame. See the figure. a. How should the florist cut the wire of length 9 feet in order to minimize the combined area of the circular top and the square base of the frame? b. Verify that the answer to part a minimizes the combined area. c. What is the answer to part a if the frame must be constructed so that the area enclosed by the square is twice the area enclosed by the circle? . Game Show You have been selected to appear on the Race for the Money television program. Before the program is aired on television, you and three other contestants are sent to Myrtle Beach, South Carolina. Each of you is given a piece of cardboard that measures 8 inches by 10 inches, 21 pair of scis— sors, a ruler, and a roll of tape. The contestant Wh0 carries away the most sand from the beach is eligi- ble to win the grand prize of $250,000. Edges of I vire that maining se of the to form .ength 9 . area of of the izes the must be by the ircle? pear on . Before d three , South 'dboard of scis— nt who is eligi- i There are two restrictions in the race. The sand must be carried away in an open box that each per- son makes from the piece of cardboard by cutting equal squares from each corner and turning up the sides. A Second restriction is that the sand cannot be piled higher than the sides of the box. a, What length should the corner cuts be for a con- testant to carry away the most sand? What is the largest volume of sand that can be carried away in such a box? b. What length should the cuts be if there is a third restriction that the box can hold no more than 50 cubic inches of sand? c. Do you feel you would have a better chance of winning the sand race with the two restrictions given in part a or with all three restrictions on the amount of sand that can be carried away? 6. Ferrying Supplies A portion of the shoreline of a Caribbean island is in the shape of the curve y = 2\/J_c. A hut is located at point C, as shown in the figure. You wish to deliver a load of supplies to the shoreline and then hire someone to help you carry the supplies to your hut. The helper will charge you $10 per mile. At what point (x, y) on the shoreline should you land in order to minimize the amount you must pay to have the supplies carried to the hut? How much currency should you have on hand to pay the helper? 6——-— 100 miles ——| 7. Costs During one calendar year, a year-round elementary school cafeteria uses 42,000 Styrofoam plates and packets, each containing a fork, spoon, and napkin. The smallest amount the cafeteria can order from the supplier is one case containing 1000 5.4 Derivatives in Action 349 plates and packets. Each order costs $12, and the cost to store a case for the Whole year is approxi- mately $4. In order to find the optimal order size that will minimize costs, the cafeteria manager must balance the ordering costs incurred when many small orders are placed with the storage costs incurred when many cases are ordered at once. a. How many cases does the cafeteria need over the course of 12 months? b. If x is the number of cases in each order, how many times will the manager need to order dur- ing one calendar year? c. What is the total cost of ordering for 1 calendar year? (1. Assume that the average number of cases stored throughout the year is half the number of cases in each order. What is the total storage cost for 1 year? e. Write a model for the combined ordering and storage costs for 1 year. f. What order size minimizes the total yearly cost? (Note that only full cases may be ordered.) How many times a year should the manager order? What will the minimum total ordering and stor— age costs be for the year? . Costs A situation similar to the one in Activity 7 occurs when a company uses a machine to produce different types of items. For example, the same machine may be used to produce popcorn tins and 30-gallon storage drums. The machine is set up to produce a quantity of one item and then be recon- figured to produce a quantity of another item. The plant produces a run and then ships the tins out at a constant rate so that the warehouse is empty for storing the next run. Assume that the number of tins stored on average during 1 year is half of the number of tins produced in each run. A plant manager must take into account the cost to reset the machine (similar to the ordering cost in Activity 7) and the cost to store inventory. Although it might otherwise make sense to produce an entire year’s inventory of popcorn tins at once, the cost of storing all the tins over a year’s time would be prohibitive. Suppose a company needs to produce 1.7 mil— .lion popcorn tins over the course of a year. The cost 350 10. 11. CHAPTER 5 to set up the machine for production is $1300, and the cost to store one tin for a year is approximately $1. What size production run will minimize set-up and storage costs? How many runs are needed dur— ing 1 year, and how often will the plant manager need to schedule a run of popcorn tins? Iuice Cans Frozen juice cans are constructed with cardboard sides and metal top and bottom. A typi— cal can holds 12 fluid ounces. If the metal costs twice as much per square inch as the cardboard, what dimensions will minimize the cost of a 12—fluid-ounce can? Booth Design As a sales representative, you are required to travel to trade shows and display your company’s products. You need to design a display booth for such shows. Because your company gen— erally must pay for the amount of square footage your booth requires, you want to limit the floor size to 300 square feet. The booth is to be 6 feet tall and three-sided, with the back of the booth a display board and the two sides of the booth made of gath- ered fabric. The display board for the back of the booth costs $30 per square foot. The fabric costs $2 per square foot and needs to be twice the length of the side to allow for gathering. 21. Find the minimum cost of constructing a booth according to these specifications. What should be the dimensions of the booth? b. In order to accommodate your . company’s display, the back of the booth must be at least 15 feet wide. Does this restriction change the dimensions necessary to minimize cost? If so, what are the optimal dimensions now? Dog Run A dog kennel owner needs to build a dog run adjacent to one of the kennel cages. See the figure. Kennel A4 0 o __. .0 L4 0. u I: .... U Cinder block Chain link side 12. Duplex Rent 13. Analyzing Change: Applications of Derivatives The ends of the run are to be cinder block, which costs 50 cents per square foot. The side is to be chain link, which costs $2.75 per square foot. Local regulations require that the walls and fence be at least 7 feet high and that the area measures no less than 120 square feet. a. Determine the dimensions of the dog run that will minimize cost but still meet the regulatory standards. b. Repeat part a for conditions where two identical dog runs are to be built side by side, sharing one cinder block wall. c. Do the answers to parts a and 17 change if there is a mandatory minimum width or length of 6 feet? A set of 24 duplexes (48 units) was sold to a new owner. The previous owner Charged $700 per month rent and had a 100% occupancy rate. The new owner, aware that this rental amount was well below other rental prices in the area, im- mediately raised the rent to $750. Over the next 6 months, an average of 47 units were occupied. The owner raised the rent to $800 and found that 2 of the units were unoccupied during the next 6 months. a. What was the monthly rental income made by the previous owner? b. After the first rent increase, what was the monthly income? c. After the second increase, how much did the new owner collect in rent each month? d. Find an equation for the monthly rental income as a function of the number of $50 increases in the rent. e. If the occupancy rate continues to decrease as the rent increases in the same manner as it did for the first two increases, what rent amount will maximize the owner’s rental income from these duplexes? How much will the owner collect in rent each month at this rent amount? f. What other considerations besides rental income should the owner take into account when deter— mining an optimal rent amount? Bus Trip A sorority plans a bus trip to the Great Mall of America during Thanksgiving break. The bus they charter seats 44 and charges a flat rate of $350 plus $35 per person. However, for every empty seat, the charge per person is increased by i units) was 1er charged occupancy .tal amount .e area, im- 3r the next :upied. The :hat 2 of the months. 1e made by t was the ch did the 1? ital income ncreases in lecrease as er as it did mount will from these 7 collect in ital income 'hen deter— D the Great Dreak. The flat rate of for every creased by 14. $2. There is a minimum of 10 passengers. The sorority leadership decides that each person going on the trip will pay $35. The sorority itself will pay the flat rate and the additional amount above $35 per person. For example, if there are only 40 pas- sengers, the fare per passenger is $43. Each of the 40 passengers will pay $35, and the sorority will pay $350 + 40($8) = $670. a. Find a model for the revenue made by the bus company as a function of the number of passen- gers. b. Find a model for the amount the sorority pays as a function of the number of passengers. c. For what number of passengers will the bus company’s revenue be greatest? For what num— ber of passengers will the bus company’s rev- enue be least? ‘ d. For what number of passengers will the amount the sorority pays be greatest? For what number of passengers will the amount the sorority pays be least? Cable Line A cable television company needs to run a cable line from its main line ending at point P in the figure to point H at the corner of the house. The county owns the roads shown in the figure, and it costs the cable company $25 per foot to run the line on poles along the county roads. The area bounded by the house and roads is a privately owned field, and the cable company must pay for an ease- ment to run lines underground in the field. It is also more costly for the company to run lines under— ground than to run them on poles. The total cost to run the lines underground across the field is $52 per foot. The cable company has the choice of running the line along the roads or cutting across the field. 351 5.4 Derivatives in Action a. Calculate the cost for the company to run the line along the roads to point H. b. Calculate the cost to run the line directly across the field from point P to point H. c. Calculate the cost to run the line along the road for 50 feet from point P and then cut across the field. (1. Set up an equation for the cost to run the line along the road a distance of x feet from point P and then cut across the field. If x = O, the line will cut directly across the field, and the value of the cost function should match your answer to part a. e. Using calculus and your equation from part d, determine whether it is less costly for the com— pany to cut across the field. If so, at what dis— tance from point P should the company begin laying the line through the field? 15.* Costs A trucking company wishes to determine the recommended highway speed for its truckers to drive in order to minimize the combined cost of driver’s wages and fuel required for a trip. The aver— age wage for the truckers is $15.50 per hour, and average fuel efficiencies for their trucks as a func— tion of the speed at which the truck is driven are shown in the table. Fuel consumption (mp9) 21. Find a model for fuel consumption as a function of the speed driven. b. For a 400—mile trip, find formulas for the follow- ing quantities in terms of speed driven: i. Driving time required ii. Wages paid to the drive Gallons of fuel used iv. Total cost of fuel (use a reasonable price per gallon based on current fuel prices) v. Combined cost of wages and fuel c. Using equation 1/ in part 17, find the speed that should be driven in order to minimize cost. *Activities 15 and 16 are time consuming. You may wish to assign one or both of them as projects. 352 CHAPTER 5 d. Repeat parts 17 and c for 700—mile and 2100-mile trips. What happens to the optimal speed as the trip mileage increases? e. Repeat parts b and c for a 400-mile trip, increas- ing the cost per gallon of fuel by 20, 40, and 60 cents. What happens to the optimal speed as the cost of fuel increases? f. Repeat parts 17 and c for a 400-mile trip, increas— ing the driver’s wages by $2, $5, and $10 per hour. What happens to the optimal speed as the wages increase? 163‘ Diskettes A standard 3.5-inch computer diskette stores data on a circular sheet of mylar that is 3.36 inches in diameter. The disk is formatted by magnetic markers being placed on the disk to divide it into tracks (thin, circular rings). For design reasons, each track must contain the same number of bytes as the innermost track. If the innermost track is close to the center, more tracks will fit on the disk, but each track will contain a relatively small number of bytes. If the innermost track is near the edge of the disk, fewer tracks will fit on the disk, but each track will contain more bytes. For a standard double-sided, high-density 3.5-inch disk, there are 135 tracks per inch between the innermost track and the edge of the disk, and about 1400 bytes per inch of track. This type of disk holds 1,440,000 bytes of information when formatted. a. For the upper disk in the accompanying figure, calculate the following quantities. The answers to parts ii and iii must be positive integers, and the rounded values should be used in the remaining calculations. i. Distance in inches around the innermost track Number of bytes in the innermost track 1:: Total number of tracks iv. Total number of bytes on one side of the disk v. Total number of bytes on both sides of the disk b. Calculate the quantities in part a for the lower disk shown in the figure. *Activities 15 and 16 are time consuming. You may wish to assign one or both of them as projects. 17. 18. Analyzing Change: Applications of Derivatives c. Repeat part a for a disk whose innermost track begins r inches from the center of the disk. Use the optimization techniques presented in this chapter to find the optimal distance the inner- most track should be from the center of the disk in order to maximize the number of bytes stored on the disk. What is the corresponding number of bytes stored? How does this number compare to the total number of bytes stored by a standard double—sided, high-density 3.5-inch diskette? d. If you are willing to destroy a used disk for the sake of curiosity, open a 3.5-inch disk and meas- ure from the center to where the tracks begin. How does this measurement compare with the optimal distance you found in part c? Explain how the problem-solving strategy pre- sented in this section can be applied to situations in which data are given. How can Step 2 of the problem—solving stratng help in building a model? How could it be mislead— ing? Explain how you can keep from being misled while implementing Step 2. Inst] If ya secti tWO' the l Rela intrt equa equa of d< Inst] In S< terrr depe mea outp this betw qua] tern we a digit did i Inst You Cha befo secti A-38 CHAPTER 5 23. a. The first differences are greatest between 6 and 10 minutes, indicating the most rapid increase in activity. 1.930 1 + 31.7208-0AS9118M m minutes after the mixture reaches 95°C; The inflec— tion point is (7.872, 0.965). After approximately 7.9 minutes, the activity was approximately 0.97 U/100uL and was increasing most rapidly at a rate of approxi— mately 0.212 U/ 100 uL/min. Because the graph of g” crosses the t—axis at 13.68, we know that input corresponds to an inflection point of the graph of g. Because g” (t) is a minimum at that same value, we know that it corresponds to a point of slowest increase on the graph of g. M (1 + 0.1376e0-2854x)2 b. A(m) = U/lOOuL . L’(x) = million tons per year x years after 1970 I . 1970: decreasing by 7.63 million tons per year 1995: decreasing by 0.41 million tons per year . Emissions were declining most rapidly in 1977 at a rate of 17.9 million tons per year. At that time, yearly emis- sions were 124.7 million tons. 27. The graph of f is always concave up. A concave—up parabola fits this description. 29. a. The graph is concave up to the left of x = 2 and con— cave down to the right of x = 2. b- f,(x) 130:) 31. Cubic and logistic models have inflection points. In some cases product, quotient, and composite functions may have inflection points. 33. Answer located on the Calculus Concepts CD-ROM and web site. Section 5.4 1. The equation for surface area in terms of diameter d is d2 d 92 S=Trdh+n§ +1T§+§ ,andthevolume equa— 2 tion is V='rr(g) h = 808.5. Solving the volume Analyzing Change: Applications of Derivatives . . . _ 808.5 _ 3234 equat1on for h in terms of d glves h — ;(d—)2 — “d2. 2 Substituting this expression into the surface area equation results in the equation we seek to optimize: 3234 d 2 d 9 2 = —— + — + — + - S d “(2) “(2 8) Setting the derivative equal to zero and solving for d gives d z 9.73568, which is twice the optimal radius found in Example 3. . 450 square feet, occurring when one side is 30 feet and each of the other two sides is 15 feet. . a. A corner cut of approximately 1.47 inches will result in the maximum volume of 52.5 cubic inches of sand. b. Corner cuts of approximately 1.12 inches and 1.86 inches will result in boxes with volume 50 cubic inches. . Answers will vary. . 42 cases 42 . . —x- orders rounded up to the nearest integer 42 504 . ;($12 per order) = 7 dollars to order . cases to store)($4 per case to store) = 2x dollars to store 504 . C(x) = —x— + 2x dollars when x cases are ordered . C(x) is a minimum when x g 15.9 cases. The man- ager must order 3 times a year. Because the order sizes are not all the same (16 cases ordered 3 times a year are more cases than the cafeteria needs), we cannot just substitute the optimal value of x into the cost equation to obtain the total cost. There are two most cost-effective ways to order: 15, 15, and 12 cases, or 16, 16, and 10 cases (assuming that the storage fees can be calculated as $1.33 per case for 4 months). The total cost associated with each option is $63.93. . A radius of approximately 1.2 inches and a height of approximately 4.8 inches will minimize the cost. 11. a. The chain link side should be approximately 6.6 feet, and the cinder block sides should be approximately 18.2 feet. . b. The chain link side should be approximately 5.7 feet, and the cinder block sides should be approximately 21 feet. . The answer to part a does not change. The answer to part 17 becomes: The chain link side should be 6 feet and the cinder block sides should be 20 feet. romv-lr-rr-rr-rr—x if 13. 15. a. a. P f. R(n) = 350 + 2n(44 — n) + 3511 dollars when n stu- dents go on the trip S(n) = 350 + 211(44 — 11) dollars when 11 students go on the trip The bus company’s revenue is maximized when 31 stu- dents go on the trip and is minimized when 10 stu— dents go on the trip. The sorority pays the most when 22 students go and pays the least when 44 students go. m(s) = —0.001552 + 0.10435 + 3.5997 mpg at a speed of 5 mph . 400 1. Thours .. 6200 u. Tdollars ;% gallons, where m(s) is the output of the model in part a iv. $603 dollars (assuming a price of $1.15 per gallon) 6200 460 v. C(s) = T + m dollars 60.3 miles per hour 700—mile trip: 1200-mile trip: i. 1:29 hours i. @ hours 10’350 dollars 18’5600 dollars 700 ... 1200 m gallons m. m gallons 805 iv. — dollars 1380 m(s) iv. m dollars (assuming a price of $1.15 per gallon) v. 700—mile trip: C(s) = 10’350 2L?) dollars 1200—mile trip: C(s) = 18’5600 + 5%; dollars Optimal speed: Optimal speed: 60.3 miles per hour 60.3 miles per hour The optimal speed remains constant. It does not depend on trip length. Price of gas: Optimal speed for 400-mile trip: $1.35/gallon 58.8 mph $1.55/gallon 57.5 mph $1.75/gallon 56.4 mph As the gas price increases, the optimal speed decreases. Wages of driver: Optimal speed for 400—mile trip: $17.50/hour 61.4 mph $20.50/hour 62.9 mph $25.50/hour 64.9 mph As the wages increase, the optimal speed increases. 17. Answers will vary. Chapter 5 Answers to Odd Activities A—39 Section 5.5 7 9. —=6—-—+6lna—=6(1 +lna) 11 l3 l7. 19. 21. 23. 25. 27. 29. 31. df ' E dh dy é ' dt .0:— Vr2+h2 15. a. b. b. C. _a a- a a a _ dt 3' dy _ 12x dy ' dt _ dt x dx — 62(ln 1.02)(1.02 ) dt da da fl dy dy dy _ 1Trh dh _Vfl+hZE W (r£+h%)+7r\/r2+h2fl Approximately 52.4 gallons per day dw Ez0.4323; The amount of water transpired is increasing by approximately 0.4323 gallon per day per year. In other words, in 1 year, the tree will be transpir- ing about 0.4 gallon more each day than it currently is transpiring. 45 . 0.00064516112 pom“ dB _ -90 dh E 7 000064516713 E dB B: . — z -0.2789 point per year dt Approximately 0.0014 cubic foot per year Approximately 0.0347 cubic foot per year L = 5,3 = _IVI_ 5/3 K-2/3 48.10352 K‘"4 48.10352 d_L = (__.M—)5/3 K'5/3) fi dt 48.10352 3 dt The number of worker hours should be decreasing by approximately 57 worker hours per year. The balloon is approximately 1529.7 feet from the observer, and that distance is increasing by approximately 1.96 feet per second. The runner is approximately 67.08 feet from home plate, and that distance is decreasing by about 9.84 feet per second. a. b. Approximately 4188.79 cubic centimeters By about -167.6 cubic centimeters per minute Approximately 5.305 centimeters per second V V Begin by solving for h: h = — = — r"2 2 TI'T' 1T Differentiate with respect to t (V is constant): %_X_.a dt _ ‘1T ( 2r ) dt dh nrzh dr ‘ 2 . _ ___ __v _ ~3 _ Substltute 1T1’ h for V. dt 1T ( 2r ) dt ...
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This note was uploaded on 11/29/2010 for the course MATH 1743 taught by Professor Davidson-rossier during the Spring '08 term at The University of Oklahoma.

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optimize (5) - tth. three of these :sed thus far in , :s?...

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