Quadratics - Revenue and Distance

Transcription

1 9.10 Quadratics - Reveue ad Distace Objective: Solve reveue ad distace applicatios of quadratic equatios. A commo applicatio of quadratics comes from reveue ad distace problems. Both are set up almost idetical to each other so they are both icluded together. Oce they are set up, we will solve them i exactly the same way we solved the simultaeous product equatios. Reveue problems are problems where a perso buys a certai umber of items for a certai price per item. If we multiply the umber of items by the price per item we will get the total paid. To help us orgaize our iformatio we will use the followig table for reveue problems First Secod The price colum will be used for the idividual prices, the total colum is used for the total paid, which is calculated by multiplyig the umber by the price. Oce we have the table filled out we will have our equatios which we ca solve. This is show i the followig examples. Example 1. A ma buys several fish for S56. After three fish die, he decides to sell the rest at a profit of S5 per fish. His total profit was S4. How may fish did he buy to begi with? Buy p 56 Sell Usig our table, we do t kow the umber he bought, or at what price, so we use varibles ad p. Total price was S56. Buy p 56 Sell 3 p+5 60 Whe he sold, he sold 3 less ( 3), for S5 more(p +5). Total profit was S4, combied with S56 spet is S60 p = 56 Fid equatios by multiplyig umber by price ( 3)(p +5) = 60 These are a simultaeous product p= 56 ad p +5= 60 3 Solvig for umber, divide by or( 3) 1

2 56( 3) +5( 3)= = 3 60( 3) 3 Substitute 56 for p i secod equatio Multiply each term by LCD: ( 3) 56( 3) +5( 3)=60 Reduce fractios = = = 0 = 19 ± ( 19)2 4(5)( 168) 2(5) = 19 ± = 19 ± Combie like terms Move all terms to oe side Solve with quadratic formula Simplify = 80 = 8 This is our 10 We do t wat egative solutios, oly do + 8 fish Our Solutio Example 2. A group of studets together bought a couch for their dorm that cost S96. However, 2 studets failed to pay their share, so each studet had to pay S4 more. How may studets were i the origial group? Deal p 96 Paid Deal p 96 Paid 2 p+4 96 S96 was paid, but we do t kow the umber or the price agreed upo by each studet. There were 2 less that actually paid ( 2) ad they had to pay S4 more (p +4). The total here is still S96. p = 96 Equatios are product of umber ad price ( 2)(p +4) = 96 This isasimultaeous product p= 96 ad p +4= = 96 2 Solvig for umber, divide by ad 2 Substitute 96 for p i the secod equatio 2

3 96( 2) 96( 2) +4( 2)= 2 Multiply each term by LCD: ( 2) 96( 2) +4( 2)=96 Reduce fractios = = 96 Distribute Combie like terms Set equatio equal to zero = 0 Solve by completig the square, Separate variables ad costat = 192 Divide each term by a or ( 2 2 = 48 Complete the square: b 1 ) 2 2 ( 2 1 ) 2 = 1 2 = 1 Add to both sides of equatio =49 Factor ( 1) 2 = 49 Square root of both sides 1=±7 = 1 ± 7 =1+7=8 Add 1 to both sides 8 studets Our Solutio We do t wataegative solutio The above examples were solved by the quadratic formula ad completig the square. For either of these we could have used either method or eve factorig. Remember we have several optios for solvig quadratics. Use the oe that seems easiest for the problem. Distace problems work with the same ideas that the reveue problems work. The oly differece is the variables are r ad t (for rate ad time), istead of ad p (for umber ad price). We already kow that distace is calculated by multiplyig rate by time. So for our distace problems our table becomes the followig: First Secod rate time distace Usig this table ad the exact same patters as the reveue problems is show i the followig example. Example 3. Greg wet to a coferece i a city 120 miles away. O the way back, due to road costructio he had to drive 10 mph slower which resulted i the retur trip 3

4 takig 2 hours loger. How fast did he drive o the way to the coferece? rate time distace There r t 120 Back rate time distace There r t 120 Back r 10 t We do ot kow rate, r, or time, t he traveled o the way to the coferece. But we do kow the distace was 120 miles. Comig back he drove 10 mph slower(r 10) ad took 2 hours loger (t+2). The distace was still 120 miles. rt = 120 Equatios are product of rate ad time (r 10)(t+2)=120 We have simultaeous product equatios t= 120 r ad t + 2= 120 r r + 2= 120 r 10 Solvig for rate, divide by r ad r 10 Substitute 120 r forti the secod equatio 120r(r 10) r +2r(r 10)= 120r(r 10) r 10 Multiply each term by LCD:r(r 10) 120(r 10)+2r 2 20r = 120r Reduce each fractio 120r r 2 20r = 120r Distribute 2r r 1200 = 120r Combie like terms 120r 120r Make equatio equal to zero 2r 2 20r 1200 = 0 Divide each term by 2 r 2 10r 600 = 0 Factor (r 30)(r + 20)=0 Set each factor equal to zero r 30 = 0 ad r + 20 = 0 Solve each equatio r = 30 ad r = 20 Ca t have a egative rate 30 mph Our Solutio World View Note: The world s fastest ma (at the time of pritig) is Jamaica Usai Bolt who set the record of ruig 100 m i 9.58 secods o August 16, 2009 i Berli. That is a speed of over 23 miles per hour! Aother type of simultaeous product distace problem is where a boat is travelig i a river with the curret or agaist the curret (or a airplae flyig with the wid or agaist the wid). If a boat is travelig dowstream, the curret will push it or icrease the rate by the speed of the curret. If a boat is travelig upstream, the curret will pull agaist it or decrease the rate by the speed of the curret. This is demostrated i the followig example. 4

5 Example 4. A ma rows dow stream for 30 miles the turs aroud ad returs to his origial locatio, the total trip took 8 hours. If the curret flows at 2 miles per hour, how fast would the ma row i still water? 8 Write total time above time colum rate time distace We kow the distace up ad dow is 30. dow t 30 Put t for time dowstream. Subtractig up 8 t 30 8 t becomes time upstream rate time distace dow r +2 t 30 up r 2 8 t 30 Dowstream the curret of 2 mph pushes the boat (r + 2) ad upstream the curret pulls the boat(r 2) (r +2)t = 30 Multiply rate by time to get equatios (r 2)(8 t) = 30 We have a simultaeous product t= 30 r + 2 ad 8 t= 30 r r +2 = 30 r 2 Solvig for rate, divide by r + 2 or r 2 Substitute 30 for t i secod equatio r +2 8(r + 2)(r 2) 30(r + 2)(r 2) 30(r + 2)(r 2) = r + 2 r 2 Multiply each term by LCD: (r +2)(r 2) 8(r +2)(r 2) 30(r 2)=30(r +2) Reduce fractios 8r r + 60 = 30r + 60 Multiply ad distribute 8r 2 30r + 28 = 30r + 60 Make equatio equal zero 30r 60 30r 60 8r 2 60r 32 =0 Divide each term by 4 2r 2 15r 8 =0 Factor (2r + 1)(r 8) =0 Set each factor equal to zero 2r +1=0 or r 8 =0 Solve each equatio r = 1 or r =8 2 2 r = 1 or r =8 2 Ca t have a egative rate 8 mph Our Solutio Begiig ad Itermediate Algebra by Tyler Wallace is licesed uder a Creative Commos Attributio 3.0 Uported Licese. ( 5

6 9.10 Practice - Reveue ad Distace 1) A merchat bought some pieces of silk for S900. Had he bought 3 pieces more for the same moey, he would have paid S15 less for each piece. Fid the umber of pieces purchased. 2) A umber of me subscribed a certai amout to make up a deficit of S100 but 5 me failed to pay ad thus icreased the share of the others by S1 each. Fid the amout that each ma paid. 3) A merchat bought a umber of barrels of apples for S120. He kept two barrels ad sold the remaider at a profit of S2 per barrel makig a total profit of S34. How may barrels did he origially buy? 4) A dealer bought a umber of sheep for S440. After 5 had died he sold the remaider at a profit of S2 each makig a profit of S60 for the sheep. How may sheep did he origially purchase? 5) A ma bought a umber of articles at equal cost for S500. He sold all but two for S540 at a profit of S5 for each item. How may articles did he buy? 6) A clothier bought a lot of suits for S750. He sold all but 3 of them for S864 makig a profit of S7 o each suit sold. How may suits did he buy? 7) A group of boys bought a boat for S450. Five boys failed to pay their share, hece each remaiig boys were compelled to pay S4.50 more. How may boys were i the origial group ad how much had each agreed to pay? 8) The total expeses of a campig party were S72. If there had bee 3 fewer persos i the party, it would have cost each perso S2 more tha it did. How may people were i the party ad how much did it cost each oe? 9) A factory tests the road performace of ew model cars by drivig them at two differet rates of speed for at least 100 kilometers at each rate. The speed rates rage from 50 to 70 km/hr i the lower rage ad from 70 to 90 km/hr i the higher rage. A driver plas to test a car o a available speedway by drivig it for 120 kilometers at a speed i the lower rage ad the drivig 120 kilometers at a rate that is 20 km/hr faster. At what rates should he drive if he plas to complete the test i hours? 10) A trai traveled 240 kilometers at a certai speed. Whe the egie was replaced by a improved model, the speed was icreased by 20 km/hr ad the travel time for the trip was decreased by 1 hour. What was the rate of each egie? 11) The rate of the curret i a stream is 3 km/hr. A ma rowed upstream for 3 kilometers ad the retured. The roud trip required 1 hour ad 20 miutes. How fast was he rowig? 6

7 12) A pilot flyig at a costat rate agaist a headwid of 50 km/hr flew for 750 kilometers, the reversed directio ad retured to his startig poit. He completed the roud trip i 8 hours. What was the speed of the plae? 13) Two drivers are testig the same model car at speeds that differ by 20 km/hr. The oe drivig at the slower rate drives 70 kilometers dow a speedway ad returs by the same route. The oe drivig at the faster rate drives 76 kilometers dow the speedway ad returs by the same route. Both drivers leave at the same time, ad the faster car returs 1 hour earlier tha the 2 slower car. At what rates were the cars drive? 14) A athlete plas to row upstream a distace of 2 kilometers ad the retur to his startig poit i a total time of 2 hours ad 20 miutes. If the rate of the curret is 2 km/hr, how fast should he row? 15) A automobile goes to a place 72 miles away ad the returs, the roud trip occupyig 9 hours. His speed i returig is 12 miles per hour faster tha his speed i goig. Fid the rate of speed i both goig ad returig. 16) A automobile made a trip of 120 miles ad the retured, the roud trip occupyig 7 hours. Returig, the rate was icreased 10 miles a hour. Fid the rate of each. 17) The rate of a stream is 3 miles a hour. If a crew rows dowstream for a distace of 8 miles ad the back agai, the roud trip occupyig 5 hours, what is the rate of the crew i still water? 18) The railroad distace betwee two tows is 240 miles. If the speed of a trai were icreased 4 miles a hour, the trip would take 40 miutes less. What is the usual rate of the trai? 19) By goig 15 miles per hour faster, a trai would have required 1 hour less to travel 180 miles. How fast did it travel? 20) Mr. Joes visits his gradmother who lives 100 miles away o a regular basis. Recetly a ew freeway has oped up ad, although the freeway route is 120 miles, he ca drive 20 mph faster o average ad takes 30 miutes less time to make the trip. What is Mr. Joes rate o both the old route ad o the freeway? 21) If a trai had traveled 5 miles a hour faster, it would have eeded hours less time to travel 150 miles. Fid the rate of the trai. 22) A traveler havig 18 miles to go, calculates that his usual rate would make him oe-half hour late for a appoitmet; he fids that i order to arrive o time he must travel at a rate oe-half mile a hour faster. What is his usual rate? Begiig ad Itermediate Algebra by Tyler Wallace is licesed uder a Creative Commos Attributio 3.0 Uported Licese. ( 7

8 9.10 1) 12 2) S4 3) 24 4) 55 5) 20 6) 30 7) S18 8) S6 Aswers - Reveue ad Distace 9) 60 mph, 80 mph 10) 60, 80 11) 6 km/hr 12) 200 km/hr 13) 56, 76 14) km/hr 15) 12 mph, 24 mph 16) 30 mph, 40 mph 17) r = 5 18) 36 mph 19) 45 mph 20) 40 mph, 60 mph 21) 20 mph 22) 4 mph Begiig ad Itermediate Algebra by Tyler Wallace is licesed uder a Creative Commos Attributio 3.0 Uported Licese. ( 8