Write inequalities to represent the following problem, and then solve to answer the question. 1. The Rent-A-Lemon Car Rental Company charges $60 a day to rent a car and an additional $0.40 per mile. Alex can spend at most $800 on his car. He needs to rent the car for at least 2 days and needs to drive at least 1000 miles to visit his customers. Give a recommendation of how many days Alex should rent the car and how many miles he can drive. Make sure that your recommendation allows Alex to stay within his budget and that it makes practical sense. (Note: As you consider what makes practical sense, think about things such as the amount of time spent driving each day, the time Alex needs to meet with customers, and the possibility of Alex needing to run personal errands or get meals while traveling.) Let x be the number of days the car is rented and let y be the number of miles driven. Then we have the following equations: x 2 Alex needs the car for 2 days or more y 1000 He needs to drive at least 1000 miles 60x + 0.40y 800 Total cost cannot exceed $800 If we put the last equation in slope-intercept form, we get: 60x + 0.40y 800 0.40y 60x + 800 60x + 800 y 0.40 y 150x + 2000 The three equations are graphed on the next page. The first equation, x 2, is shaded red. The second equation, y 1000, is shaded blue. The third equation, y 150x + 2000,is shaded yellow. The triangle in the middle is the intersection of the three graphs. Any of the points on or in the triangle would satisfy each equation, however, there are other parameters to consider. For Page 1 of 6
instance, Alex could not rent a car for a fraction of a day, so the x-values can only be integers. In order to stay within budget, Alex could rent a car for 2, 3, 4, 5, or 6 days. The corresponding data points from the graph (indicating the maximum mileage Alex could drive) are: ( 2,1700 ), ( 3,1550 ), ( 4,1400 ), ( 5,1250 ), ( 6,1100 ) ( 2,1700 ) This would have Alex renting a car for 2 days. He has to drive a minimum of 1000 miles. This would average 500 miles a day. At 60 mph, it would take him 8 hours and 20 minutes each day in driving time. ( 3,1550 ) This would have Alex renting a car for 3 days. He would average driving 1 3 333 miles a day. At 60 mph, it would take him approximately 5½ hours each day in driving time. Page 2 of 6
( 4,1400 ) Alex could rent a car for 4 days. He would average driving 250 miles a day. At 60 mph, it would take him approximately 4¼ hours each day in driving time. ( 5,1250 ) This option would have Alex renting a car for 5 days. He would average driving 200 miles a day. At 60 mph, it 1 would take him approximately 3 hours each day in driving 3 time. However, Alex only has an extra 250 miles. If he plans on doing lots of additional driving, this might not be the best option for him. ( 6,1100 ) Alex could rent the car for 6 days, but he would only have an extra 100 miles to drive. He could drive the minimum of 1000 miles, but would have to be careful that he didn t drive too much and go over his budget. Renting the car for 2, 3 or 6 days would not be a good option because of driving time (2 and 3 days) and limited mileage (6 days). The best options are renting the car for 4 or 5 days. The 4-day option might prove the best one since it gives more flexibility in miles traveled. 2. A total of $21,000 is invested in two funds. One fund pays 4.25% annual interest, and the other pays 3% annual interest. If the combined annual interest is $811.25, how much is invested in each account Let x represent the amount invested at 4.25%, and let y represent the amount invested at 3%. Then we get the following equations: x + y 21000 0.0425x + 0.03y 811.25 We can solve this system of equations by elimination. We multiply the first equation by -3 and the second equation by 100. ( x y) ( ) ( ) ( x y) ( ) 3 + 21000 3 100 0.0425 + 0.03 811.25 100 Page 3 of 6
3x 3y 63000 4.25x + 3y 81125 1.25x 18125 1.25x 18125 1.25 1.25 x 14,500 x + y 21000 14500+ y 21000 y 6500 Check: $14500 invested at 4.25%; 6500 invested at 3%. I x 0.0425(14500) $616.25 I y 0.03(6500) $195.00 Total Interest $616.25 + $195.00 $811.25 3. The area of a rectangle is 2 less than twice the perimeter. If the width of the rectangle is 5, find the length of the rectangle. Assume that the sides of the rectangle are measured in centimeters. Area l w Perimeter 2l + 2w w 5 A 5l P 2l + 10 A 2P 2 Using substitution, we get: A 2P 2 ( l ) 5l 2 2 + 10 2 Solving for l gives: Page 4 of 6
( l ) 5l 2 2 + 10 2 5l 4l + 20 2 l 18 cm Check: If the length is 18 and the width is 5, then the area is 90 and the perimeter is 46. A 2P 2 ( ) 90 2 46 2 90 92 2 90 90 4. Flying into a head wind, it takes a plane 2½ hours to fly miles. If the plane were flying with a tailwind, it could make the trip in 2 hours and 9 minutes. Find the speed of the plane and the speed of the wind. Let p be the speed of the plane, and let w be the speed of the wind. The speed the plane traveled into the head wind is given by r p w. The speed the plane travels with the tailwind is given by 1 r p+ w. The time it takes to fly against the wind is t 1 2.5 hours, 2 and the time it takes to fly with the wind is approximately t 2 2.15 hours. For each leg of the trip, the distance is miles. Since D r t, we get the following equations: r t D 1 1 1 r t D 2 2 2 ( p w) ( p w) 2.5 + 2.15 2.5p 2.5w 2.15p+ 2.15w Page 5 of 6
We can solve this system of equations by elimination. We multiply the first equation by 2.15, and we get: 2.5 2.15p 2.15w 795.5 2.15p+ 2.15w 2.15p 2.15w 795.5 2.15p+ 2.15w 4.3p 1720.5 p 400 mph Since ( p w) 2.5, we substitute to get: ( w) 2.5 400 1000 2.5w 2.5w 75 w 30 Check: If the speed of the plane is 400 mph and the speed of the wind is 30 mph, the time needed to travel miles against the wind is: D t r 400 30 ( ) 370 2.5 hours with the wind is: D t r 400+ 30 ( ) 430 2.15 hours 2 hrs 9 min Answer: Plane speed is 400 mph; wind speed is 30 mph. Page 6 of 6