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MATH312-Extralistofexercises 1. As responsible for urban planning at the town council, you have to completely reinstall all the electricity,gasandwaterlinesinthecity. Reinstallingonemileofelectricitylinecosts$12(million), takes10hoursandrequires13workers. Reinstallingonemileofgaslinecosts$7(million),takes4 hoursandrequires7workers. Reinstallingonemileofwaterlinecosts$4(million),takes6hours andrequires6workers. Ifyouhaveamonthlybudgetof$195(million),162hourspermonthand ateamof216workers,howmanymilesofeachlinecanyoureinstallinamonth? Solution. Thevariables x, y and z will represent, respectively, x= milesofelectricitylineinstalledinamonth. y = milesofgaslineinstalledinamonth. z = milesofwaterlineinstalledinamonth. Thenumberofmilesofelectricity line times the cost of one mile, plus the number of miles of gas line times the cost of one mile, plus the number of miles of water line times the cost of one mile has to be equaltothe$195(million)thatIhaveasbudget. Thisgivestheequation 12x+7y+4z=195. If we do the same for hours of work, we get the equation 10x+4y +6z = 162; similarly, for number of workersneededgivestheequation13x+7y+6z=216.Thuswehavetosolvethesystem 12x+7y+4z=195 12 7 4 195 10x+4y+6z=162 whosematrixis 10 4 6 162 13x+7y+6z=216 13 7 6 216 Let’s solve the systemusingGauss-Jordanelimination. Ateachstepthepivotwillbethecirclednumber. Forinstance,forthefirststepthenumber 12 ,inthetopleftcornerofthematrix,willbethepivot. 12 7 4 195 R3−R1 12 7 4 195 R1↔R3 1 0 2 21 R2−10R1 −−−−−→ 10 4 6 162 −−−−→ 10 4 6 162 −−−−→ 10 4 6 162 1 0 2 21 12 7 4 195 R3−12R1 13 7 6 216 12 7 4 195 2R2−R3 12 7 4 195 R3−7R2 12 7 4 195 0 4 -14 -48 −−−−→ 0 1 -8 -39 −−−−→ 0 1 -8 -39 0 7 -20 -57 0 7 -20 -57 0 0 36 216 Thustheinitialsystemofequationsisequivalent(inotherwords,ithasthesamesolutions)tothesystem 12x+7y+4z=195 y−8z=−39 36z=216 Fromthebottomequationweget z = 216 =6milesofwaterline. 36 1 Replacingthisinthemiddleequationweget y =−39+8z=−39+48=9milesofgasline. Finally, replacing the values of y and z in the top equation, we get x=195−7y−4z =195−63−24=108=9milesofelectricityline. 12 12 12 2. Yourunadeliverycompany,deliveringinthreedifferentareasofManhattan,A,BandC.Inaver- age, a trip to the area A takes 4 hours, 5 gallons of fuel and you deliver 3 tons of goods. A trip to areaBtakes6hours,4gallonsoffuelandyoudeliver1tonofgoods. Finally,atriptoareaCtakes3 hours,2gallonsoffuelandyoudeliver3tonsofgoods. everydayyourcompanydeliversthrough all 24 hours, and your budget allows to spend 16 gallons in total, to deliver 24 tons of goods. How manytripstoeachareacanyoudoeveryday? Solution. Thevariables x, y and z will represent, respectively, x= tripstoareaA. y = trips to area B. z = trips to area C. Theequationsaresetupastheywereinthepreviousexercise. Thatis, withrespect to time, we have 4 hoursforeachtriptoA,6hourspertriptoBand3hourspertriptoC,andthetotalmustbeequalto24, sowegettheequation 4x+6y+3z=24. Doingthesameforgallonsoffuel,wegettheequation5x+4y+2z=16;anddoingthesamefortonsof goodsdeliveredyieldstheequation3x+y+3z=24. Thesystemtosolveis 4x+6y+3z=24 4 6 3 24 5x+4y+2z=16 whosematrixis 5 4 2 16 3x+y+3z=24 3 1 3 24 Againlet’ssolveitbyGauss-Jordanelimination. 4 6 3 24 1 5 0 0 1 5 0 0 1 5 0 0 R1−R3 R2−5R1 3R3−2R2 5 4 2 16 −−−−→ 5 4 2 16 −−−−→ 0 -21 2 16 −−−−→ 0 -21 2 16 3 1 3 24 3 1 3 24 R3−3R1 0 -14 3 24 0 0 5 40 Sotheoriginalsystemisequivalenttothesystem x+5y =0 −21y+2z=16 5z =40 Wegetthenz=8tripstoareaC.Replacinginthemiddleequationweget−21y =16−16=0,so y =0 trips to area B, and then we see from the top equation, x+5y =0,that x =0tripstoareaA. 2 3. Youwanttorunfourdifferentexperimentsinalab. Thefirstexperimentwillrequire,persample,2 hoursoflabwork,2hoursofprocessingtime,3labassistantsand$5,000. Asampleofthesecond experimentrequires2hoursoflabwork,noprocessingtime,3labassistantsand$1,000. Thethird experiment requires, per sample, 2 hours of lab work, 3 hours of processing time, 3 assistants, and $7,000. Finally, each sample of the fourth experiment takes 1 hour of lab work, 2 hours of processingtime,1labassistantand$5,000. Perday,yourscheduleandbudgetallowyoutospend3hoursonlabwork,4hoursonprocessing time, the use of 4 lab assistants and $11,000. How many samples of each experiment can you do perday? Solution. Let x, y, z and t denote the following x= numberofsamplesofexperiment1. y = numberofsamplesofexperiment2. z = numberofsamplesofexperiment3. t = numberofsamplesofexperiment4. Thesystemtosolveisthenthefollowing 2x+2y+2z+t=3 2 2 2 1 3 2x +3z+2t=4 2 0 3 2 4 inmatrixform 3x+3y+3z+t=4 3 3 3 1 4 5000x+1000y+7000z+5000t=11000 5000 1000 7000 5000 11000 Onceagain,let’suseGauss-Jordanelimination. 2 2 2 1 3 2 2 2 1 3 1 3 0 -1 0 2 0 3 2 4 R3−R2 2 0 3 2 4 R1↔R3 2 2 2 1 3 −−−−→ −−−−→ 1 3 0 -1 0 3 3 3 1 4 R4/1000 5 1 7 5 11 2 0 3 2 4 5000 1000 7000 5000 11000 5 1 7 5 11 1 3 0 -1 0 1 3 0 -1 0 1 3 0 -1 0 R2−2R1,R3−2R1 0 -4 2 3 3 R4−2R3 0 -4 2 3 3 R2↔R4 0 -2 1 2 3 −−−−−−−−→ −−−−→ −−−−→ 0 -6 3 4 4 0 -6 3 4 4 R4−5R1 0 -6 3 4 4 0 -2 1 2 3 0 -4 2 3 3 0 -14 7 10 11 1 3 0 -1 0 1 3 0 -1 0 R3−3R2 0 -2 1 2 3 R3−2R4 0 -2 1 2 3 −−−−→ −−−−→ 0 0 0 -2 -5 0 0 0 0 1 R4−2R2 0 0 0 -1 -3 0 0 0 -1 -3 The third row of the last matrix corresponds 0 = 1, which means that the system does not have any solutions. 4. You are designing a new CPU combining four different microchips. Each of the first type of mi- crochip performs 2,000 operations per second, costs $3,000, takes 4 hours to be installed and requires 3 engineers to be set up. A microchip of the second type performs 2,000 operations per 3 second, costs $1,000, takes 4 hours to be installed, and requires 3 engineers to be set up. Each of the third type performs 2,000 operations per second, costs $4,000, takes 4 hours to be installed, andrequires 3 engineers to be set up. Finally, a microchip of the fourth type performs 1,000 op- erations per second, costs $3,000, takes 2 hours to be installed, and requires 2 operators to be set up. If youwantyourCPUtoperform3,000operationspersecond,youhave$10,000tospend,6hours and14engineersworkinginthelab,howmanychipsofeachtypedoyouneed? Solution. Thevariables x, y, z and t denote x= numberofchipsoftype1.y= numberofchipsoftype2.z= numberofchipsoftype3.t = numberofchipsoftype4. Thesystemthatwehavetosolveis 2000x+2000y+2000z+1000t=3000 3000x+1000y+4000z+3000t=10000 4x+4y+4z+2t=6 3x+3y+3z+2t=14 AndagainwesolveitusingGauss-Jordanelimination. 2000 2000 2000 1000 3000 2 2 2 1 3 2 2 2 1 3 3000 1000 4000 3000 10000 R1/1000 3 1 4 3 10 R4−R1 3 1 4 3 10 −−−−→ −−−−→ 4 4 4 2 6 4 4 4 2 6 3 3 3 2 14 R2/1000 4 4 4 2 6 1 1 1 1 11 3 3 3 2 14 1 1 1 1 11 1 1 1 1 11 1 1 1 1 11 R1↔R4 3 1 4 3 10 R2−3R1,R3−4R1 0 -2 1 0 -23 R3−2R4 0 -2 1 0 -23 −−−−→ −−−−−−−→ −−−−→ 0 0 0 -2 -38 0 0 0 0 0 4 4 4 2 6 R4−2R1 2 2 2 1 3 0 0 0 -1 -19 0 0 0 -1 -19 Sincealltheentriesarezerointhethirdrow,weknowthatthesystemhasinfinitelymanysolutions. The original systemisequivalentto x+y+z+t=11 −2y+z =−23 −t =19 Fromthebottomequationweseethat t =19microchipsoftype4. Nextwehavetochooseoneofthe variables to become a parameter to express the solutions of the system. We have to be careful at this point because we cannot choose just any variable: the variables x and t cannot become parameters, sincethesecondequationdoesnotcontainanyofthesetwovariables! Thuswehavetochoosebetween y andz. Let’schoose y. Thisway, z =−23+2y microchipsoftype3. Thetopequationthengives x=11−y−z−t=11−y−(−23+2y)−19=15−3ymicrochipsoftype1. 4
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