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# Calculus for Business and Economics MATH 1071Q

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This 23 page Class Notes was uploaded by Mary Veum on Thursday September 17, 2015. The Class Notes belongs to MATH 1071Q at University of Connecticut taught by Ushani Gamage in Fall. Since its upload, it has received 24 views. For similar materials see /class/205815/math-1071q-university-of-connecticut in Mathematics (M) at University of Connecticut.

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Date Created: 09/17/15

Math 1071 Fall 2010 Review for Final Exam SOLUTIONS 1 Lot fr I2 7 3x 1 and 91 I4 7 17 Find each of the following a f91 f9I f191 I273z1 14717 z4z273z716 b 97 f1 y 7 f1 91 7 at 7 14717712 73171 1471771273171 I47z23z718 C MW MW 1 171 a g z 12 7 3x 1 I4 7 17 15 71712 7315 511714 717 15 7 315 14 7 1712 511 717 f 7 fr lt3 I gltzgt 12 7 31 1 4 W Where I 7 17 y 0 0 f 0 101 f 0 HM f 9 1 7 f I4 717 7 x4717273z4717 1 f y o fr y o fr 91f 11 912 7 31 1 2 See attached graphs 7 12731704717 3 Let 2quot zlt1 121 L4 fltzgt1 Calculate the following limits 737 I73 1 I fr 3 I73 73 b 1337 1133 f1 3 3 l39 l39 l39 3 051310 f 96 1313 fr Egg f I z 7 3 1317 W 7 1317 7121 7 00 l39 l39 2 1 2 1 1 3 311 W 311 I lt gt 011311 fI Does not eXist Let f 512 1 Use the limit de nition of the derivative to nd the equation of the tangent line at z 1 Things we need to liDOWI I y and m given 51 1 7 6 m f 1 Use the limit de nition of the derivative to nd f then evaluate f at z 1 hm 1 h 7 at h7gt0 h 5 z h 1 7 512 1 no h Hz 7 512 101 5h 1 7 512 1 hm h7gt0 h 512101h5h217512 71 hm h7gt0 h 101h 5h2 m ill gt0 h hm h 101 5h h7gt0 h lim 101 5h h gt0 7 101 50 101 Thus f 101 You can checli that you answer is correct by calculating the derivative the easy wayquot ie the way we ve been doing this past month or so We still need to nd the slope of the tangent line at z 1 m f 1 101 10 Thus I 1 y 6 and m 10 Plug these values into your favorite general equation for a line and solve for y NB If using y mm 12 don t forget to solve for b cn Thus the equation of the tangent line at z 1 is 9 91 7711 11 y76 10171 y 1017106 y 10174 Find the derivative of each of the following functions 14 a Mr F 71217 MI gt hz b z 7 321n 13 21 71 Let u L Then 1 Product Rule uv gt NI 14 F 712z7 141 10 71217 1 1 14710z1 1 712mm 1171 1 1 217 7140141 7 8415 132 1n 13 21 71 2W7HD 7 13472171 uvuv 312472 3 2 21761nz 2171z73 gt c 91 3614771 12 7 1 3614771 12 71 9 314 7 7I614 7T I 71 I 71 3 413 7 7 e144 12 7 0 21 1213 7 21 614771 I ltz271 2 63 5 k a z I 5 Let u e3125 v z 5 Then 1 312 5 e37 25 61637525 1 1 u 11 7 uv Quotient Rule 7 T lt6I 325gt I 5 7 53125gt 1 gt k z 2 I 5 e3 25 612 301 7 1 z 5 6 Find the absolute extrema of hr I3 7 612 on the following intervals Note For each of these problems you ll need to liDOW the critical values of hz 3x2 7 121 0 312 7 121 12 7 4x 7 11 7 4 gt z 07 z 4 a 7172 Mil 13 7 6712 717 6 77 Mo 7 03 7 60 0 M2 7 23 7 62 8 7 24 716 b 273 h2 716 M3 7 33 7 63 277 54 727 Absolute minimum at 37 727 Absolute maximum at 27 716 Mil h0 M4 M5 77 0 40376002 64796 732 537662 52576 254 725 Absolute minimum at 47 732 Absolute maximum at 07 0 7 Find the absolute extrema of hr I3 7 I on the following intervals Note For each of these problems you ll need to know the critical values of WI 0 1 l 3 1 if 3 I 31271 31271 a 7171 Absolute minimum at lt b 7170 Absolute minimum at 710 and 07 0 Absolute 8 Evaluate the following limits 112 7 101 a 132100 3 7 417 77 5 LL JV I 00 12 lim 7 xgt7oltgt 7417 Mil EV EV h1 h7 h0 15 1110 1 1 Absolute maximum at lt7 437 H 711 l0 737 137 1 0 170 72 3x 0 maximum at lt7 700 00 c 79y8 2y7 3y 9 7900 700 d 21 1 11 13207356723573 715130716 7131307 7 7 862 3 862 75E lt0 3111 726 9 7 3210 Ten 7 046 7 0 h4 7 h2 h 7 1 h4 f hilxIPoo h3 1 7 hgrlloo 7 hgrllooh 7 700 79 79 n 139 139 139 779 7 h 132 12 7 81 16 0232 z 7 4 I 7 4 023 I 7 42 00 h5 7 2h 1 h5 h 133 3h 7 711113 13110 711113 1313100 711107 0 9 Let fr 213 312 7121 6 Note 6 means negativequot and EB means positivequot a Determine Where f is increasing and decreasing f z 7 612 6x 712 0 7 612 6x 7 12 0 12 z 7 2 0 I 2 I 7 1 gt z 72 z 1 Thus the critical values are I 72 and z 1 I ll choose test values I 73 z 0 and z 2 H73 7 6732 673712 54 7 18 7 12 24 EB f 0 7 60260712 7 712 e f 2 7 62262712 7 2412712 7 24 EB Thus f is increasing on 7007 72 and 17 00 and f is decreasing on 721 b Find all critical points of f Identify all relative maxima and minima From a we liDOW z 72 and z 1 are critical values f2 f 1 2723 3722 71272 6 716 12 24 6 26 213 312 7121 6 71 Thus the critical values are 727 26 and 17 71 In particular 17 71 is the relative minimum and 727 26 is the relative maximum c Determine Where f is concave up or concave down f z 12z6 0 121 6 76 12x 1 7 ii 7 I I ll choose test values I 71 and z 0 f 1 12716 7 712 6 7 76 G f 0 7 120 6 6 EB interval 700 71gt lt7 00gt 7 2 27 test value I 71 z f quot95 G 69 f concavity concave down U concave up Thus f is concave down on 1 1 700 7E and concave up on lt7 700gt 1 1 Additionally lt75 f is the inflection point Note that flt 2 13 1 2 1 27 7 12lt gt6 2 3 77 7 6 6 84 1 g 4 4 4 0 4 25 2 1 25 That is lt7 3 is the inflection point d Slietch a graph of f Label any critical points and inflection points See attached graph 73 13712 10 Let x a Find all asymptotes A h Find all critical points of f Identify all relative maxima and minima c Determine Where f is increasing or decreasing d Slietch a graph of f Lahel any critical points 11 Find the following antiderivatives 5 a 2z3 7 7 z 710ed1 5 21377z710 dz 7 f I 2z3 7 517 z 7 loedz I4 1 I2 2 4 75 g 710 zc 4 2 7101710 zc h3157i32z77dz I 4 3157732z77dz 31574z 32z77dz z 15 1 2 x2 376gt74lt772gt2lt72gt77IC 5 z 2 2 E 41 c 213 7 I I4 7 x2 35 d1 Let u 14 712476 Then du 413 7 2x dx 2z3 7 IgtI47126gt5d1 2 2x3 71 I471265dz 413 7 2x I4 7 12 65 dz u lt 7 Wu 2 6 I471266 12 6 1 E 1 E 1 d61272d1 1371 Letux37z Then du 312 71dz 613L2d1 23271d1 z 71 1 ix du 7 Zln u Jrc 21n 1371 c 2 o ltx tgt dt lt tgtdt ttdt t2tg i i c 2 3 f e 72d1 ex 7x 2 7x dx e idx e 6 6 7 12 Edz z2 xc g M2731 dz Let u 2 7 3x Then du 73dz 827 3z5dz 73 273z5dz 72073305 73d1 h 3136 148173 dz Lotux48173 Then du 413 8 3zs6 148173 dz i 41 12 1 Let u x2 1 Then du Zrdx m Educ I Let u lnx Then du ldz I 12 See attached graph 13 Evaluat ethe following de nite integrals lnz dz I 2 2 3132 148173 dz 369 4 132 z48173dz 24134r8 148173 dz gugdu 3ltug 7 i c 4 5 gu 4 5 6 5 9z48173 20 6 5 3 a t 23t 4dt 1 1 b zexg1dz 71 Let u I2 1 Then du Zrdx c A4 Lot 1 7d WI 1 u211 3 t 23t 4dt 1 1 2 reg 1dz 71 71 1 7 73 l 1 1 71 1 1 11 lt1 79 1 W a 2 710 54 7 E 44 a 1 1 21 72161d1 271 l 2 7 eudu 1 1 61244 7 e7121 0 Then du 2dr 1 o 2x1dx 0 41 AW 1 I2 211dz 27z 0 dz 4 211 dz 0 1 4 1 7 2211 dz 2 0 iu du 7 1ltugt 7 2 Z 211 0 lt2lt4gt1 7 lt2lt0gt1gt lt9gt7lt1gt 371 2 4 1 74u 72 71 1 4u4 u 72 1 71 Few 4 4 71 1 T 1a2 716641 64 49 a 12170270 4 f1127E1dz 4 4 z27i1dz 127I1dz 1 1 4 zj z 3 g 1 7 3 21 4 7 3731 1 7 43 242 7 3 6416 1 7 7 4quot 771 3 3 3 49 7 3 3 53 7 3 1217112715dz Lltgttux27z Then du 21 7 1 dx 12 21 71 I2 7 z5dz t w 4 Then du 2zdx 41 1272 A 2f 2dr 3 2 A12 273 u 2111M 2ln z272 2 211132 72 7 211122 72 2ln772ln2 21115 14 Find the area of the enclosed region Within each of thhe following lists of curves Note Graphs of each of the following are included in the attached graphs a yx2 y1274 i Points of Intersection 1274 127176 173z2 1 3 172 Graph to determine which function is the topbottom Top y z 2 Bottom y x2 7 4 iii lntegrate to nd the area z27 1274gtd1 b y 712 y 731 3 217126dz 9 918727 12 2 27 7 16 304 6 E 6 105 2 Points of Intersection w 712 731 0 12 7 3x 0 11 7 3 gt 107 13 F Graph to determine which function is the topbottom Top y 71 Bottom y 731 N F Integrate to iind the area 3 3 712 7 73zdz 712 3zdz 0 0 13 3x2 3 7 T 0 33 332 03 302 77 77 7 727 27 T E 7 7227 327 7 f 7 27 7 F 7 9 7 E c y x3 7 31 y 212 i Points of Intersection 13 7 3x 212 0 IS 7 2x2 7 3x 0 112 7 2x 7 3 0 11 7 3 1 1 gt z 71 z i 0 z 3 7 7 Graph to determine which function is the topbottom Let A1 he the area of the region on 717 0 and let A2 he the area of the region on 07 3 A1 Top y x3 7 3x A1 Bottom y 212 A2 Top y 212 A2 Bottom y x3 7 3x iii Integrate to iind the area The total area A is A A1 A2 For A1 0 13 7 3x 7 212dz 71 For A2 3 212 7 13 7 31 dz 0 Thus the total area is dy142 y72 x0 z1 i Points of Intersection I4 3x2 213 4 2 3 71 9 9 9 714 34 2713 4 2 3 4 2 3 131 4 2 3 173 2 4 2 3 731373 3 z 3 3 2127133zdz 0 23 z E 7 3 4 2 0 233 34 33 0 0 0 T 7TT 37 31 27 187 i 4 7 7273154 7 4 7 45 7 4 A A1A2 7 7745 7 3 4 7 23135 7 12 7 163 7 12 Not needed since these two graphs never intersect and were given I 0 z 1 Graph to determine which function is the topbottom Top y x4 2 Bottom y 72 iii Integrate to iind the area 1 1 z427 72dz z44dz 0 0 5 1 I i 4 5 I 0 1 7 4 5 7 21 5 e y12 y712 z72 z2 i Points of Intersection 12 712 212 0 gt z 0 Graph to determine which function is the topbottom Top y 1 Bottom y 712 iii Integrate to iind the area Again the total area is going to he the area of the region on 72 0 plus the area of the region 0 0 12 7 712 dz 212dz 2 72 0 on 7 E 7 3 72 9 2W 7 3 3 E 7 3 2 2 12 7 712 dz 212dz 0 0 2 7 213 7 T 0 E 7 3 Thus the total area of the region is A A1 A2 L6 1 E 7 3 3 2 7 3 f ye y0 z71 z3 i Points of Inter ction Not needed since these two graphs never intersect and were given I 71 z 3 18 Graph to determine which function is the topbottom as Top y e Bottom y 0 iii Integrate t0 iind the area 15 Solve for I 1 7 1 a E b 7 2r 7 3 0 3 3 ex 7 0dr exdz 71 71 e 71 637671 63 e 6471 e O H m w w A M7 V i H H H Mwwrx cow 7 2x 7 3 0 2z7 7 37 0 2z7 7 377 21 3 3 I E The other way to look at this is that 7 gt 0 always and thus it must he that 2x 7 3 0 Then I c 53 12540 4 53x 1254174 531 53 4 53x 7 5121712 31 12x 7 12 12 9x 4 I g d e 4 e log81 4 logz 32 log81 4 101038144 81 4 7x I f ln4z 7 3 7 lnz 71 ln3 ln4 ln4 i ln4 logz 32 1010gm32 z 32 28 4 ln4z 7 3 7 lnz 71 ln3 ln lt41 7 3 ln3 z 7 1 1n43 elnS 4x 7 3 7 3 z 71 4x 7 3 3z 71 4x 7 3 3x 7 3 4x 31 z 0 x 63 A fence is to be built around a 200 squarefoot rectangular iield Two opposite sides are to be made of wood costing 10 per foot While ther remaining two opposite sides are to be made of stone costing 30 per foot Find the dimensions of the enclosure that minimize the total cost Let C total cost A area ft2 z length it y Width ft A my C 101 101 30y 30y gt C 20160y We re given that the area is 200 sqft thus A 7 200 gt my 200 200 gt y 7 Cl Plugging this value in for y in the equation C gives 0 find critical values 0 0 12000 12 12000 600 i600 I 201 60y 200 201 60 I 12000 201 7 z 2011200011 20 7 12000172 20 7 120200 I 20 7 120200 I 20 2012 12 I 110 We can t have a negative length so I 7 100 Check that this is a minimum That is z 10x0 is the relative minimum Recall y 20 when I 10x0 feet and y feet x l a 200 a z WW Thus the cost is minimized A farmer has 800 feet of fencing and will use it to construct a rectangular pasture using a river as one side What dimensions maximize the area of the pasture Let A area sqft P perimeter ft 1 P 21 y y A find critical values A Check that z 200 is a minimum length ft and y Width ft 800 800 800 7 21 4 my 1800 7 21 800177212 800 7 4x 800 7 4x 800 200 Thus I 200 is the relative maximum Since y 800 7 2x 800 7 2200 400 the area of the pasture is maximized when I 200 ft and y 400 ft 18 Suppose a logistic equation is given by P t a Find P0 130 b Find P10 P10 56707 1 c Find the point Where the concavity changes P c 34 34 m 34 F E 3 1 5670710 34 l 56 7 34 65 ei 3467 e7 5 wmwlmws Thus the point at which concavity changes ie the inflection point is 1n5717 d Find the limiting value of P L34 e Slietch a graph of P t maliing sure to label the limiting value and the inflection point See attached graphs 19 Suppose a logistic equation is given by P t 17 1 86 a Find P0 P0 b Find P5 P5 c Find the point Where the concavity changes 135 100 1 86745 100 1 86 20 100 m em 100620 620 8 Thus the point at which concavity changes ie the in ection point is i 1n87 50 d Find the limiting value of P L100 e Sketch a graph of P t making sure to label the limiting value and the in ection point See attached graphs

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