Math 111 Midterm I, Lecture A, version 1 -- Solutions January 30 th, 2007

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1 NAME: Suden ID #: QUIZ SECTION: Mah 111 Miderm I, Lecure A, version 1 -- Soluions January 30 h, 2007 Problem 1 4 Problem 2 6 Problem 3 20 Problem 4 20 Toal: 50 You are allowed o use a calculaor, a ruler, and one shee of noes. Your exam should conain 5 pages in oal and 4 problems. Check ha your es is complee! You mus explain how you ge your answers. Correc (or incorrec) answers wih no supporing work may resul in lile or no credi. On problems in which you use a graph, draw any lines you use, label hem, and mark poins clearly. Wrie your final answer in he indicaed spaces. If you need more room, use he backs of pages and indicae o he reader ha you have done so. Raise your hand if you have a quesion. GOOD LUCK! Do you wan me o pos your grade so far on he class websie under he las 4 digis of your suden number? Yes, please pos my grade. Sign o give permission: No, please don pos my grade so far.

2 1 (4 poins) Draw he graph of he following funcion: f ( x) = 0.5x 2. Label is y-inercep and he coordinaes of one oher poin on he graph. No need o show any oher work. (4,0) -2 2 (6 poins) Le D() represen he disance raveled by a car (in miles), up o ime (in hours), saring from he car s iniial posiion (i.e. D(0)=0). a) Translae he following saemen ino English (including he appropriae unis): D ( 5) D(3) > 150. Answer: The car raveled more han 150 miles from 3 hours o 5 hours. b) Translae he following saemen ino funcional noaion: The average rip speed of he car over he firs half an hour was 60 mph Answer: D(0.5) 0.5 = 60

3 3 (20 poins) Mah 111 Miderm Winer 2007, Nichifor, Lecure A, version 1 The graph below is of he oal amoun of he waer O() which was drawn ou of a reservoir by various imes, over a 12 hour inerval, saring a noon. gallons O() hours a) How much waer was drawn ou of he reservoir from = 3 pm o = 6 pm? O(6)-O(3)= =1500 Answer: 1500 gallons. b) Find a 3-hour ime inerval over which he (incremenal) average rae of waer being drawn ou of he reservoir was 400 gallons per hour. Draw a reference line of slope 400 (hru poin (10, 4000)). Move ruler parallel o i unil i inersecs he O() graph in wo poins which are 3 hours apar. There are muliple correc answers: from abou 0.3 o 3.3 from abou 2.6 o 5.6 from abou 5.2 o 8.2 Answer: From = 0.3 unil = 3.3 hours pas noon.

4 The following quesions coninue he problem from he previous page. For your convenience, here is he same graph again. Recall ha O() is he amoun of waer ou of he reservoir up o ime. gallons I() c) Wha is he lowes value of? A wha ime is i achieved? (d) O() hours (c) corresponds o slopes of diagonal lines hru he graph of O(). Draw he lowes diagonal line ha ouches O(). Is slope =3000/8=375 gallons per hour, And he ime his rae occurs is abou =10.4 hours. Answer: The lowes value of is 375 gallons per hour, a =_10.4 hours. d) A pipe brings waer ino he reservoir a he consan rae of 800 gallons per hour. How much waer should here be in he reservoir a noon in order no o run ou a any ime before midnigh? Recall ha his amoun equals he larges shorage. Draw I(), he oal amoun of waer in amoun: diagonal line of slope 800 (poin (4000, 5)). Find he larges verical gap beween I() and O(), wih O above I. I s abou 600 gallons. Answer: We need a leas 600 gallons of waer in he reservoir a noon.

5 4 (20 poins) The graphs below represen he marginal cos (MC), he marginal revenue (MR), and he average cos (AC) for he Seale Rain Company, which is producing and selling Umbrellas. Dollars 20 per Umbrella MR MC AC Umbrellas a) Find he change in he oal revenue if you sell 101 Umbrellas insead of 100 Umbrellas. Recognize his as MR(100) & read i from graph b) Wha quaniy of Umbrellas produced and sold maximizes he profi? Answer: TR(101)-TR(100)= 16 dollars. The q for where MR=MC, ransiioning from MR>MC o MR<MC (in our case,ha s he second crossing poin of he graphs MR and MC.) c) Find he breakeven price (BEP). The y-coordinae of crossing poin of AC and MC Answer: Maximum profi is achieved a q = 260 Umbrellas. Answer: BEP = 7 Unis: $ per Umbrella. d) The fixed coss are FC=$150. Wha is he average variable cos (AVC) for producing 100 Umbrellas? AVC(100)=VC(100)/100. VC(100)=TC(100)-FC=TC(100)-150. TC(100)=AC(100) x 100=10 x 100=1000 So AVC(100)=( )/100=8.5 Answer: AVC(100)= 8.5 dollars per Umbrella.