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# Mathematical Methods for Economists I AMS 11

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This 31 page Class Notes was uploaded by Milton Sawayn DVM on Monday September 7, 2015. The Class Notes belongs to AMS 11 at University of California - Santa Cruz taught by Staff in Fall. Since its upload, it has received 39 views. For similar materials see /class/182151/ams-11-university-of-california-santa-cruz in Applied Math And Statistics at University of California - Santa Cruz.

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chc ECONAMS 11B Review Questions 9 Solutions 1 a Q20155 z 575866 b First compute the marginal products of labor and real estate at the given point QL 15K25L 12R14 QL20 15 5 x 19195 QR 75K25L12Pr34 QR20155 x 28793 Now7 use linear approximation AQ z QL20155ALQR20155AR z 19195075 28793O5 x 28793 2 a We want to maximize the output7 Q 30K25L12Rl47 subject to the budget constraint K L R 697 since the inputs are all being measured in millions The Lagrangian for this problem is FK L R 30K25L12R14 7 K L R 7 69 and the rst order conditions are 12K 35L12R14 15K25L 12R14 FR 0 75K25L12R 34 17 gt K L R 69 The rst two equations imply that 12L12 15K25 73512147 2571214 12K LR715KL R gt Km 7 LUZ after canceling the common factor of 13514 Clearing denominators gives 12L 15K Likewise7 comparing the rst and third equations implies that 12R14 75125 7351214 2512734 12K LR775KLR gt K357 347 and clearing denominators gives 12R75K gt R0625K Substituting for R and L in the fourth equation the constraint gives K125K 0625K 69 gt 2875K 69 3 5 Thus the critical values for the inputs are K 24 U 30 and F 15 and the hotel chain s maximum output is Q Q2430 15 z 1152894 b When output is maximized the critical value of is x 12K 35L12R14 x 19215 c By the envelope theorem dc A dB 7 7 where B is the budget It follows that A6 z AB x 9607 since we measure the budget in the same units as the inputs so an increase of 500000 means that AB 05 The equation 7 F 7013 gives a differential equation for the demand function namely dq p 701191 dp First separate the variables dq 701p712dp Then integrate both sides lnq 702191 C 701p712dp Next exponentiate to solve for q q fowl20 Aei S where A 50 Finally use the data q16 200 to solve for A 200 14545 gt A 200545 q 200545a 5 2005lt4 gt5 so q25 200515 x 16375 he rst step is to nd the point of market equilibrium First we nd the equilibrium demand q 706 i 036 22 001q25p60706q 001q206q7550 q 002 so q 50 since the other solution q 7110 is negative The equilibrium price is then 13 607 06g 30 Next consumers7 surplus CS f0 demand 7 13 dq 50 50 CS 60 7 06g 7 30 dq 30g 7 03q2 750 0 0 and producers7 surplus PS fO jQS 7 supplydq 50 50 PS 30 7 001q2 7 5 dq 25g 7 q3 g 0 a We need to optimize the utility function U 2 23y 322 5 subject to the constraint 102 20y 252 1500 The Lagrangian for this problem is Fa w A m 7 23y 7 322 7 5 7 A10z 7 20y 7 252 71500 and the rst order conditions for constrained optimization are F 0 35 7 22y 7 322 7 5 10A Fy 0 25 7 23y 7 32 7 5 20A F 0 gt m 7 23y 7 32 25A FA 9 gt 102 20y 252 1500 Solving the rst three equations for A gives A 035 7 22y 7 32 2 7 5 A 012 23y 32 5 A 00495 7 23y 7 3 Setting the rst and second expressions for A equal to each other and canceling the common factor 012 22y 32 5 gives 7 Setting the rst and third expressions for A equal to each other canceling the common factor 2 22y 22 from both gives 2 71 5 2 Substituting these expressions for y and 2 into the constraint gives y 2 7 2 71 1 1 102 20lt 7 g 25 E 7 E 1500 gt 202 1665 Thus the critical values for m y and 2 are 305 191 7 7 7 z 7 z x 78325 y 7 12 25 42 and 2 30 637 and the maximum utility is U Um y2 z 5689731532 m b By the envelope theorem dU dB x 0306 22y 3 5 z 20016646 where B is the budget By linear approximation AU z AB so if AU 1 then AB z 1 z 000000005 le to increase her utility by one util the consumer will have to increase her budget by next to nothing Comment I corrected the solution to match the given problem but the given problem is not very realistic The present value is given by the integral folo 2002525 039045 dt which we compute using formulas 39 and 38 from appendix C in the book 10 20052 70045 10 2 10 2002sze 039045 dt 7 m 200te 039045tdt 0 i o i 0 400 570045 10 728339029 7 700452 71 0045 00452 0 400 7045 728339029 e 71451 728339029 33114592 4775563 The rst integral was computed using formula 39 a 70045 and n 2 and the second integral was computed using formula 38 a 70045 again The rm s total daily output is QA 623 so the rm s daily cost is C 40QA 23 2500 The rm s revenue from market A is RA PAQA and the revenue from market B is RB PBQB The rm s pro t function is H PAQA PBQB C i PA100 7 04PA PB120 7 05PB 7 40100 7 04PA 120 7 05PB 2500 704133 7 05P 116PA 140133 711300 The rst order conditions for an optimum value give nPA 0 708PA1160 P 145 nPB 0gt7PB1400gt Pg 140 The second order conditions for a maximum are HPAPAPX7 PE HPBPB P127 PE 7 HPAPB PEPE gt 0 and HPAPAPZ7PE lt 039 F7 In this case we have HpApA 708 lt 0 and HPAPAPZ7PE HPBPBP17PE 7 HPAPBPZ7PE2 0398 gt 0 for all PAPB so that the second order conditions are satis ed and because the conditions hold everywhere the critical value of pro t l HPZP 6910 is the absolute maximum daily pro t To summarize the rm s pro t is maximized when P 145 and PE 140 at which point the maximum pro t is l 6910 The sales tax does not affect the rm s cost but it does change the rm s revenue in each market since it raises the price of the rm s product and so affects the demand Speci cally the price with tax to consumers is 11PA in market A and 11PB in market B so the new demands in each of these markets will be 5A 100 7 0411PA 100 7 044PA and 53 120 7 0511PB 120 7 055PB respectively This means that the rm s new pro t function is PAQA 7 PBQB 7 CQA 7 B 70441337 0551257117731211 142133 711300 The rst order conditions are now HPA 0 gt 7088PA71176 0 gt 314 13364 HPB 0 gt 711PA1420 gt 13 12909 and since the second order conditions for a maximum are still satis ed checkl 11 7 11033133 x 572327 is the rm s new maximum pro t The government s daily revenue from the tax is 10 of the rm s mvenue not pro t 7 that s taxed elsewhere ie GR 0113100 7 044133 7 g120 7 055133 z 118313 The cost to the rm is the difference between the old maximum pro t and the new l 7 11 6910 7 572327 118687 The effect on consumers is that the total price per unit has increased The costs per unit to consumers in each market are now 1113147 and 1113142 respectively ie the cost is now 2 more per unit to consumers in both markets This leads to a slight decrease in overall daily consumption of the rm s product of about 374 UCSC ECONAMS 11A WINTER 2008 Review Questions 2 Limits Continuity and Differentiation 1 Compute the following limits 5quot F 9quot Sa What can you say about liH i ma 1 953 i 8 d 1 5y4 300y2 i 60y 1000 a 1m 7 1m was z i 2 yam 20000 7 3y 4y3 i 001y4 3 7 8 t7 2 7 b hint factor 378 839 1322 m i x h2 7 2 f lim i c lim maO haO h express your answer in terms of gquot 133 m Explain brie y why the limit in 1e must be one sided z 7 W Without consulting your notes or book7 give the mathematical de nitions for the two bold faced concepts below You should use limits in both de nitions 0 The function fz is continuous at the point z a o The function fz is differentiable at the point z a Consider the function 3 1 0 4 Wt 472 L gt0 ls this function continuous at z 0 Justify your answer Find the slope ofthe graph ofthe function y z23z71 at the point where z 27 using the definition of the derivative ie7 using limits7 and nd the equation of the tangent line to this graph at the point 29 UCSC ECONAMS 11B Review Questions 8 Solutions Note The formula numbers below refer to the table of integral formulas in Appendix C 1 a C7 0 4dz 4 1 xz2 73 4 V22 73 7 in C 7ln C 5229 5 3 z 15 z Formula 28 with a 3 4 2zdz 24z71894z 4 472 425 4718 7 2 7 0 M94495 48 0 48 48 11 37 Formula 15 with a 9 and b 4 First using formula 39 with n 2 followed by formula 38 and sim plifying we have 670061 dt 57006 7 2 1270061 dt 7006 7006 t267006t 2 67006t 7 7 7 7006t71 O 7006 006 0062 H 670061 0 06 0062t2 012t 2 0 So 10 7006t 10 200 200125005 i dt 76076 0062t2 012t 2 0 39 0 i 20067020 36 1 2 2 200 2 42806 09 7 0063 39 39 0063 N 39 39 3d1 i 3 du i 412 4 25 F 2 u2 4 25 F 3 Eln2141225C substitute u 21 so 412 u2 and d1 du then use formula 27 with a 5 3 Eln uu225 0 5x4 lnx 5x4 4 16 0 use formula 42 with n 3 You can compute this integral by splitting the numerator and computing two integrals using formulas 2 and 3 or you can make a substitution as follows Substitute u 27s then x U727 so 35z 35u727 11 5u7 and dx du Remember that the limits of integration also change ie z 0 i u 2 and z 2 i u 16 Putting all of this together e 5zglndx h gives 2 16 3 5 1 ll 5 z Ch i u du 0 2 7x 49 2 u 1 16 1llnu5u 2 11 l 8 70 n 189538483589 49 Note When I leave out steps as above you should ll them in on your own to make sure you understand whats going on 2 First nd the equilibrium price and quantity by setting demand equal to supply 2 p01q54071 07 001q202q7350 150 orq770 Since quantity must be positive the equilibrium quantity is j 50 and the equilibrium price is 15 01j 5 10 Now compute the Consumers7 Surplus and Producers7 Surplus if Consumers7 Surplus demand 715 dq 0 7 5N95833 i 3 N if Producers7 Surplus f 7 supply dq 0 50 10 7 01q 5dq 0 3 F 9quot Gini coef cient7 y 1 5 0 6039 02z3 052 3 2 1 4 03 7172 03x302x205xdx172 495 0 See section 153 in the text T fte dt 0 20 25015410475t dt 0 2506700473 Woommi 1 Present Value 20 m 2724155 0 The average value of the cost function7 c 005q2 35g 120007 on the interval 0100 is 35q2 100 T 12000q w13916666 0 1 Avgc m 100 3 1 005 005q235q12000 dq lt q 0 m 3 The value of the average cost function E cq7 when q 100 is 12000 5100 005 100 35 W 160 These two quantities are different because they are measuring completely different things7 and take different things into account The value of B when q 100 only takes into account the cost of producing 100 units7 and then divides this by 100 The average of the cost function on the interval 01007 on the other handl7 takes all of the values of the cost function on 07 100 into account7 and this give much more weight to the cced cost Geornetrically7 the average value of the cost function can be thought of as the average height of the graph of the cost function7 and since this height is always bigger than 12000 why7 the average height will also be bigger than 12000 6 Consider the sum 500 Z 0 4k 0001 k1 UCSC ECONAMS 11B Review Questions 1 Solutions 1 Compute the indicated partial derivatives of the functions below a 23z24my75y274m7y72 2m 6m 4y 7 4 2y 4m 7 10y 7 Fu7 U w 60u23U16w12 2 40u713U16w12 82F 810814 w zzzlny2 l 23 wm szlny2 l 23 7 Qnyz y2 23 wm Qzlny2 23 7 2x2yy2 23 7 6x2y23 7 2m2y3 7 23423 E7 20u713U16w712 O wy wyz yz 232 yz 1 232 4my3 7 23423 i wmyz W Since wmyz my 2 dz 782 gtltdz gtlt82 gtltdy gt7 dt t0 89 t0 dt t0 99 t0 dt t0 CO 6zy3 5y 7 3 1H 3 My 10 1 1H 2 71257333672012730344 Note z0 72 and y0 1 The monthly cost function for ACME Widgets is C 002023 001QAQB 00303 35QA 28623 5000 where QA and Q3 are the monthly outputs of type A widgets and type B widgets7 respectively7 measured in 100 s7 so7 for example7 if 3000 type A widgets are produced in a month7 then QA 3000100 30 The cost is measured on dollars a CQA 004QA 001623 35 and CQB 001QA 006623 287 so CQA2507 360 486 and CQB2507 360 521 8C b Approximation formula AC z QA250 8623 Q 7360 B this case that AQA 0 Now AQB 50100 05 so AC z 52105 2605 QA250 QB360 4 SO AQB since we are assuming in 80 80 C GeneralApproximation formula AC z 862714 gfm AQA 8Q B 360 A623 In this case we have AQA 60100 06 and A623 40100 0 AC 3 870 AQA AQB 4860652104 50 8 333 3 80 6 QB QA250 QB360 4 The annual production function for SlugTools Inc is Q 2394K13L35 3K12L14 where K and L are the capital and labor inputs respectively and where Q K and L are all measured in 1000 s a QK 08K 23L35 15K 12L14 and QL 144K13L 25 075K12L 34 so QK200320 x 11936 and QL200320 x 09783 b Recall that K L nQKQK Q and nQLQL Now we have K 200 and L 320 so Q Q 626446 Plugging these numbers and te partial derivatives that we computed in part a into the formulas above we obtain 200 320 x 11936 7 x 0381 d x 09783 7 z 0 5 77 gt626446 an 77 gt626446 c General Approximation Formula for percentage change AQ nQK AK 77 AL In this case AK 0 so AK 0 and AL w 100 156257 AQ z 0 0515625 0723125 5 The demand function for a monopolist rm s product is given by Yd132P9 5W2 P1 where UCSC ECONAMS 1113 Review Questions 5 Solutions 1 Compute the following integrals Substitute u x2 1 du 2x dx then 3x dx 32du and 3zdz 3 13 3 1123 9 2 23 7 d777 07 1 O 3z21 2u u 4ltgt Substitute u 33x2 71 du 3x2 6x dz then 2 2x dx 13 du and 22x33z2713dz ug du T12u40 z33z27140 9 C7 Substitute 1 lnm d1 idx then d 11dzlnlvl0lnllnxl0 1 O zlnz lnx z CL Substitution does not work here why but the integrand is a polynomial x2zx32712d2zx62z5472x372x21dz 9 8 37 6 45 z82x73z6iz574x472z32zdz777 x4 3 2 7 7 7 O 232 Substitute u x x2 1 giving du 3x e V 2 d 2 1 Sometimes the ugliest7 integrals are very easy Substitute u 700512 du 705 dt then dt 720du and dz Checkll so D x xxz 1 x3e du36 C36Vm210 h 10005 03905tdt 7200005 du 7200005 O 7200005 005t 0 Substitute u 3t 1 du 3dt so dt du and t 71 Then 2 l 7 2 dtlMdullui ldu 0 3t1 3 u 3 9 9 9U 1 u 3t12 23t1 46 7 772 46l 0777 fl 3t 1 C 27lt2 u nu 54 27 27nl 1 t2 t 46 7 Elnl3t1l0 Constants have been absorbed by C 2 03 h Substitute u 2x 1 du 2dz This implies that d du and z L1 2 2 1 2 7 1 1 3 1 3 d7 2 du7 u du7 17du 21 2 u 4 u 4 u 21 u 3 3 3 1111H UHO Zlnl21lC Zlnl21lo 2 1 1 How did become 3 In other words what happened to the Z First integrate 4 c 25q 10012dq E5q 100 K using the substitution u 5g 100 du 5dq so dq Next use the initial data namely 00 xed cost 10000 to solve for K 4 4000 146000 10000c0 E010032KYK K 15 45q 100 146000 and the cost function is c 15 The rms pro t is the difference between their revenue and their cost 7139 r 7 c so the Change in the rms pro t is the difference between the Change in their revenue and the Change in their cost le 020077T100 r200702007r1007c100 7 2007T100l7C2007C100l To nd the Changes in revenue and cost we use the idea discussed in Class Namely if r fQ 01 and c 9a 027 then r200 7 r100 f200 Cl 7 f100 Cl f200 7 f100 Cl 7 01 f200 7 f1007 and C200 7 C100 9200 02 7 9100 02 9200 i 9100 02 i 02 9200 4 9100 In other words7 to nd the total change in the value of r for exarnple7 we can use my antiderivative of drdq 7 the constant of integration does not play a role in this problem To nd the change in the value of the revenue function7 we rst nd the antideriva tives of drdq7 using the substitution u 2g 87 dq du2 200 7 2g 523 dq 200g 7 2 8 53 01 200g 7 032q 853 01 1 2 So 7 200g 7 032q 853 01 for some constant 01 and therefore r200 i r100 Now we repeat this 7 200 200 i 03 40853 01 i 200 100 i 03 20853 01 3326680 7 1780941 1545739 22 procedure for the cost function 02q 65 dz 01q2 65g 02 so 0 01q2 65g Cg and the change in the value of the cost function is C200C100 012002765200701701100265100Oll 1700077500 9500 Finally7 the change in pro t is 1545739 7 9500 595739 4 First7 we integrate the marginal revenue function dr 5qu so7 r Now7 use the 1 1 15 5Oinq dq q1 50dq7 50q7u5du 6 u 1nq 1 16 6 1 5 nq11 dq 11 1 substituteuln 1 17 so du 7d lt q 11 q 50q7 0 50g 7 1nq6116 0 initial value r0 0 to solve for the constant of integration 0 111116 7 1 7 7 0r05007 0 80 UCSC 1a ECONAMS 1113 Review Questions 3 Solutions Note In these problems7 you may generally assume that the critical points you nd produce the required optimal values At the same time7 you should see if you can nd an argument to justify this assumption in each example Find the minimum value of ay 2 y2 subject to 3x By 68 Lagmngian Fy7 A x2 y2 7 A3x By 7 68 Stmctuml equations Fm 2a 7 3A 0 Fy 2y 7 5A 0 Solving these equations for A gives 2x 2y By A i 3 i E gt z i Substituting this into the constraint gives 3 5y68 34y340 y010and06 The minimum value of x2 y2 subject to 3x By 68 is therefore obtained at the point 6107 giving 62 102 136 Find the maximum value of gxyz 20z l2y l32 l67 subject to the constraint 5x 4y 72 1680 Lagrangian Fy7 z A 20z12y13216 7 A5x 4g 72 7 1680 Stmctuml equations Fm 10x 12y1321675A 0 Fy 203z12y 2321674A 7 0 F1 103z12y132 5677A 0 Solving these equations for A yields the triple equation 2y13216 A gt W 512216 i 10x12y13 3y23 21256 39 Comparing the rst and second expressions canceling the factor of 216 from both and clearing denominators gives 5 6y5x gt y Comparing the rst and third expressions canceling the factor of 9 from both and clearing denominators gives 52 42 10 gt 7 z z z 21 Substituting the expressions for y and 2 into the constraint gives 5 5 5z4 395 1680 420x 70560 950 168 yo 140 20 40 H 0 Thus the maximum value of f2yz 20212y13216 subject to the con straint 5x 4y 72 1680 is given by f16814040 2489262 Find the maximum and minimum values of the function h2y 3x 5y subject to the constraint 22 y2 136 Lagrangian F2y A 3x 5y 7 M22 y2 7 136 S tructural equations 3722A 0 572y Fm Fy Solving these equations for A gives 3 5 5 gt 6y102 gt y 2x 7 E Substituting this expression for y into the constraint gives 2 5 2 2 2 2 3 136gt34 1224gt36 There are two critical x values 21 6 and 22 76 so there are two critical points 2191 610 and zgy2 76710 The two critical values are h6 10 68 which is the constrained maximum value and h76 710 768 which is the constrained minimum value The objective function is the utility Uzy z 5 lnz 7lny 18lnz and the constraint is the budget or income constraint we obtain from the prices and the budget 2pm ypy zpz B gt 42 By 302 1200 a Lagrangian F2y2 A 5 lnz 7 lny 18ln2 7 M42 By 302 71200 S tructural equations 5 FE 7 7 4A 0 2 7 Fy 7 7 8A 0 y 18 F1 7 7 30A 0 2 Solving these equations for A gives the triple equation 5 i 7 i 3 4x 7 8y i 5239 Comparing the z term and the y term and clearing denominators gives 72 40y282 gt y Comparing the z term and the 2 term and clearing denominators gives 25 12 gt 1295 2 x 2 7 25 Substituting the expressions for y and 2 that we found into the budget constraint gives 7 12 4z8 17990 1200 s 120095 60000 zo 50 yo 35 20 24 Thus Jack maximizes his utility by consuming 50 fast food meals 35 diner meals and 24 fancy7 restaurant meals in a month resulting in a utility of U503524 101652 0quot Since the utility function and the prices of meals are not changing the maximum possible utility Um is a function of the budget 6 le increasing the budget increases Umax and decreasing the budget decreases Um The envelope theorem tells us that dUmax 0 d5 where A0 is the critical value of the multiplier A In this case A0 5 i 0025 47 200 03 99 Hence by the approximation formula7 AUmax A0 AB 0025 50 125 In other words7 if Jack7s food budget increases by 50007 then his maximum possible utility will increase by approximately 125 The objective function in this case is the cost function CK L 1280K 14580L which is the cost of using K units of capital input and L units of labor input The constraint in this case is given by the equation 10K25L35 20480 since the task here is to minimize the cost of producing 20480 drills Lagmng ian FK LA 1280K 14580L 4 A10K25L35 4 20480 S tructuml equations 128074AK 35L35 0 1458076AK25L 25 0 243 gt 320K2430L gt K EL Plugging this expression for K into the constraint gives the critical L value7 90 2 8192 10 LZ5 L35 20480 IL 20480 L E 20480 T and the critical value capital input is obtained using the relationship boxed above7 FK FL Solving these equations for A gives 320K35 2430L25 Lax5 25 This is easier than it may appear at rst glance7 since we have done most of the work already in part a All we need to do is replace 20480 by q in our formula for Li ie7 in the second boxed equation in part a If we denote by L q and Kq the cost minimizing levels of labor and capital input necessary to produce q drills7 then UCSC ECONAMS 11B Review Questions 6 Solutions 1 Compute the following integrals a 5z772xdx5xv772zdz 1076 714 772 32 3 7 772 32 96 63 96 0 963 96 0 23bu 7 2a a bu32 Use the formula uxa budu C with a 7 and b 72 15b2 712 3t71 tzdt tdt dt b dt7 3 77 7 25t 25t 25t 25t 7 t2 2t 41 l25tl 3 t 21l25tl 11l25tl 0 777 7n 777m 7 7n 10 25 125 5 25 5 7 1 27 7 7771 2 5t 0 10 25 125 ml H d l d Use the formulas aubu glnlawLbuHC7 ii auu 7blnlabulC and uzdu u au 12 111abu77b72 lnlaanulnLC7 all witha2andb5 t2 7004t 2 c 500t25 03904 dt 500 5 7 te70v04tdt 7004 004 t2670V04t 2 670V04t 500 7 7 70041171 0 lt 7004 004 00016 7125005700 t2 50t 1250 O uneau Use the formulas u e m du ii ue m du 6 a2 a 7 g 1L quotle mdu7 with n 2 and a 70047 and an au 71 C with a 7004 362m 61 d 7d3 7emdx 4ew V4em u 3 du7 substitutm u em and du em dx 7 m g 372u 8v4u0 2em7847O d 2 b 2 V bu Use the formula u u M C with a 4 and b 1 1 bu 352 300 7001t71 1 025 00 dt300 nl e lo 1 0255701 7001 30000 001t ln 1 02550010 o d k il b k Use the formula u w C with a 17 b 025 and k 701 1 belm ak 2 2 f 4lnm Ch lnx Ch 3zx27lnx 3 xV27lnx 4 2d g substituting u ln x du idm 4 2147u2756u32x27u 0 7 3 5145 8147ln z2 7 56 lnx 322 7 lnx 15435 O 39 2 2 2 7 2 Use the formula u du 7 23b 4abu 8a xa bu if Cwitha2andb7 W 1553 2 Let y fx satisfy 37y Bxyz and ii y1 2 Find the function d Separate 7y 3x dz 112 d l 3 2 Integrate 3 dx gt 717 C This is the implicit solution Solve for 39 i y y i C 7 32239 2 SolveforC y12 gt 20 12 gt 07121gt 013 S l t39 2 o u ion 7 y 13 7 3x2 03 The income elasticity of demand for a rm7s product is proportionalto the square root of income Find the demand as a function of income7 given that q100 50 and q400 90 We are told that 774 ix7 where k is the unknown constant of proportionality Expressing the elasticity in terms of dqdY7 q and Y gives the differential equation dq Y nqYW39Ek7 which is separable F dq dY Separate i k Integrate ky12dy lnq2kY12OkY12O q Notes I dropped the absolute value from q in ln q because q gt 0 since q is output ii The factor of 2 was absorbed by the unknown constant k Solve for q Exponentiation gives q ek c Aek where A 60 gt 0 Solve for the parameters k and A This is where we use the data q100 50 and q400 907 leading to the pair of equations AelOk AEZOk 50 90 Dividing 90 by 50 on the left and AEZOk by Aelok on the right7 give the equation ln1 8 18510k 10kln18 kW005877m and pugging k ln1810 not the approximate value back into the rst equation gives 250 50 Aeh l39s 18A A Thus the demand function is given by 250 q 6h118371039 The population of a tropical island grows at a rate that is proportional to the third mot of its size In 19507 the islands population was 1728 and in 19807 the islands population was 2744 What will the Islands population be in 2020 First7 translate the description of the growth rate into a differential equation lf Pt the size of the population at time t in years7 then the description above leads to the differential equation dP 3 7 k P dt f7 where k is the unknown constant of proportionality dP Separate the variables PUB k dt 3 Integrate both sides P 13dP kdt gt EPzS kt C Solve for P Multiplying by 23 gives 1323 kt 0 because the factor of 23 is absorbed by both k and C Next7 raise both sides to the power 32 to see that P kt 0W2 9quot b Solve for the parameters k and C First set It 0 for the year 1950 so 1728 P0 0 O 032 s C 172823 144 Next the year 1980 corresponds to t 30 so 32 23 26 2744 P3 30k 144 gt 30k 144 2744 196 gt k T Thus Pt gt 144 2 and in 2020 the population will be 26 3 2 P70 E 70 144 m 4322 The population of bass in a large lake grows according to the logistic model dY 7 005Y 10 7 Y d lt gt where Yt is the size of the bass population measured in thousands of sh 1f the bass population in 1990 was 1500 sh what will the population be in 2010 kYM 7 Y in general and obtained the solution dY We solved the logistic equation E M y 1brkMt7 where b comes from the constant of integration We can use that knowledge and conclude that the solution here is given by 10 1 b570517 since M 10 and k 005 in this example Next use the initial data Y0 15 remember Y is measured in 1000s to solve for b Y 10 7 bi 1b 15 gt gt 7 10 1 b 7 15 so the precise solution here is 10 Y 17 7 1 6 05t Now we can nd the population in 2010 by computing Y20 710 7 w m 999742 Y20 le in 2010 the bass population will be about 9974 sh This means that by 2010 the bass population will be very close to the carrying capacity of the lake When willdid the bass population reach 5000 Using the solution we found in a7 we solve the equation Yt 5 for the variable t 10 17 705 10 17 705 705 3 5w gt1 e 3 gt 36 1 gt e 17 Taking natural logs of both sides of the equation on the right gives 05til 3 l17 gtt2l1734692 7 n 17 7 n 3 7 n 3 N In other words7 the bass population reached 5000 a little less than three and half years after the initial population count was done If the initial data was from January 17 19907 then the bass population reached the 5000 sh mark on about June 20th7 1993 O Once the population reaches 30007 bass are harvested7 from the lake at the constant rate of 1000 sh per year Describe what will happen to the sh population over time If the lake is opened to bass shing harvesting7 then the rate at which the population grows or shrinks changes If YH denotes the size of the bass population subject to harvesting7 then dY dH 71 005YH10 7 YH 7 E 005YH10 7 YH 71 7005Y2 051 717 because the harvesting rate dHdt is constant and equal to 1000 sh per year7 and the popu lation is measured in 1000s7 so dHdt 1 D7005YZ 05Y71 Figure 1 Graph of DY 7005Y2 051 7 1 To describe what will happen to the bass population over tirne7 we need to know when the population is increasing or decreasing under the new conditions7 and that means we need to gure out the values of Y for which the quadratic function DY 700552 051 71 is positive and for which values of Y it is negative The graph of this function is depicted in Figure 1 To nd where D gt 0 and where D lt 07 we need to nd the solutions YH1 and YHZ of the quadratic equation 7005Y2 051 7 1 07 which we do using the quadratic formula 705 V025 7 02 705 7 v025 7 02 x 2764 and YHZ y H1 701 701 m 7236 lThis degree of accuracy is unrealistic because the population of sh in a lake does not grow in a nice continuous manner7 but rather in discrete jumpsi UCSC ECONAMS 1113 Review Questions 7 Solutions 1 Compute the following de nite integrals Use the Fundamental Theorem 0f Calculus7 ie dont use limits of right hand sums 4 Bxdm 9ltz2123 a 7 0 V321 4 3 b x22zz33xz713dx 1 2 5 d c z lnllnzl 5 zlnz Note Problems a7 b and c7 above7 involve the same integrals that appeared in 1abc of RQ 5 See the solutions of RQ 5 to see how the antiderivatives were found 4 9 11723 7123 m 1262585 0 1 3 1 7 72w 3x2 714 E 534 i 34 657533333 1 H lnln e2 7 lnln e ln2 7 ln1 ln 2 82 e d Substitute u 70051t7 du 7005 dt dt 720 du7 and also7 change the limits of integration according to the substitution t 0 i u 7005 0 0 and t 10 i u 7005 10 705 Then 10 705 705 1000e 03905 dt 7200005 du 7 200005 0 720000 5 05 7 50 m 7869387 0 0 e Substitute u x2 17 du 2x dx so mdx 05du Also7 z 0 i u 1 and z1 u2so 1 z 1 2du lnu d7 77 0 x21 2 1 u 2 f Substitute 1 em 717 e m7 then d1 em 7 e w dz and mi 7w d dzlnvClneme mC 27111271111713 2 2 39 1 ex e m This gives 2 z 7 7m 2 2 72 ldz lnew e m lne2e 27lne0e0 ln 6 6 x 1325 0 ex e m 0 2 Note In problems d and e l substituted directly in the de nite integral7 and the limits of integration changed accordingly In f I rst computed the inde nite integral of the integrand7 and then used this to compute the de nite integral in terms of the original variable of integration7 so there was no need to change the limits of integration This was also the case in ab and c7 above 1 Conclusion If you change the variable of integration in a de nite integral Via substitution then the limits of integration must change accordingly 2 Compute 20 20 9 a First note that Z 2k122k1722k1 k10 k1 k1 Next notethat 22k1 2 nn1nn22n as we k1 k1 C7 0 k1 see by using the formulas j k 1 V L and 2171 k4 H 20 This means that 2 2k 1 202 40 i 92 18 341 k10 100 100 100 100 100 100 100 Z 23 74Z 2Z3 iz42 232 7421 i0 i1 i1 i0 i1 i1 i0 1001212013lt100 101 7404 35309639 I split the sum again pulled out constant factors again and in addition to the summation formulae that I used in a I also used if i nn 12n 1 i1 6 The only other point of interest here is that whenz 0 the terms 2392 and 3239 are also 0 so I omitted these terms from the corresponding sums on the right but I didn7t do this for the constant term 4 ln problem 7 from the exercises in SN 1 you were asked to show that n1 n1 q q 171 m1 In this problem all you have to do is use the formula To do this set q 6 03905 then 0 100 1006751 67505 7 67005 70 05m 7 7005 m 7 N Z We 7 2 m 5 gt 7 67005 71 7 67005 712 N 3847212 m1 3 The change in consumption is given by 15 15 d0 BY 13 O idy 7 dY 10 dY 10 9Y21 To compute this integral we make the substitution u 9Y21 This entails the following changes 1 o du 9dY gt dY du YL2139 9 0 And the limits of integration change Y 10 gt u 111 and Y 15 gt u 156 Thus the change in consumption is 15 O 8Y13dY du 1 156 8u i 219 13 0 91 21 1 11 u 156 8 7 g 9 3 du 111 u 11568 17 1 7777du 9 m 9 3 u 1 9 1 9 7Kln156 7 8887 ln111 423 81 27 81 27 When income increases from 10 billion to 15 billion consumption increases by about 423 billion and consequently savings increase by 077 billion S Y 7 C d A rm7s marginal revenue function is d7r Q11000 7 01q2 Suppose that the rms output increases from q 40 to q 50 Then the total change in revenue is 50 dr 50 Ar 7dq qx1000701q2dq 40 d 40 To compute the integral substitute u 1000 7 01q2 du 702qu i qdq 75 du chc ECONAMS 11B Review Questions 1 Partial Derivatives and Partial Elasticities 1 Compute the indicated partial derivatives of the functions below a 23z24my75y274m7y72 c wmzzlny223 21 wm 2y w 7 b Fu 1 w 607L231116u12 y 7 8F 7 wmm 66 my 810814 wmyz 7 2 Suppose that z 3z2y3 5zy2 7 3x y 71 where z 3t 7 2 and y 2t 1 1 Use the chain rule to compute 72 dt t0 9 The monthly cost function for ACME Widgets is C 00262124 001QAQB 003ng 35QA 28623 5000 where QA and Q3 are the monthly outputs of type A widgets and type B widgets respectively measured in 100 s so for example if 3000 type A widgets are produced in a month then QA 3000100 30 The cost is measured on dollars a Compute the marginal cost of type A widgets and the marginal cost of type B widgets ifthe monthly outputs are 25000 type A widgets and 36000 type B widgets Suppose that production of type A widgets is held xed at 25000 and production of type B widgets is increased from 36000 to 36050 Use your answer to part a to estimate the change in cost to the rm E7 0 Suppose that production of type A widgets is increased from 25000 to 25060 and production of type B widgets is increased from 36000 to 36040 Use your answer to part a to estimate the change in cost to the rm 4 The annual production function for SlugTools Inc is Q 24K13L35 3K12L14 where K and L are the capital and labor inputs respectively and where Q K and L are all measured in 1000 s a Compute the marginal products of capital and labor when capital input is 200000 and labor input is 320000 UCSC ECONAMS 1113 FALL 2008 Review Questions 6 Solutions 1 Use implicit differentiation to nd the indicated derivative at the given point dy i i a Flnd d7 at the p01nt 27y 17 2 on the graph of the equatlon 2 23y Qxys 7 422y2 2 Differentiate both sides of the equation with respect to 2 and solve for dyd2 d d 3 3 22 Ch 2 y 2xy 42 y dz2 d d d i 322g 237y 2y3 6zy27y 7 Sxyz 7 Szzyi 0 d2 d2 d2 d i d7y 23 6xy2 7 822g 7 32 2y3 7 Szyz x i dy i Z wzy 2y3 7 8y2 d2 7 3 62y 7 53ny i dy 7 616732 710 dz 21 124716 939 y d b Find d7u at the point um 27 e on the graph of the equation 1 u2 lnv 1267 5 Differentiate both sides of the equation with respect to 2 and solve for dudv d 2 27M 7 d u lnvve idv5 du u2 7 2 7 du 2ulnv772veu7ve 70 d1 1 d1 du U2 72172 7727 dUunv 116 waW 2 du u72ve g i 1 d1 2u1n1 7 1126 4 71 d 726 2 g l 67 7 70735759 e 2 Since the expression on the left hand side of the equation in this problem is the same as the one in 1a and the right hand side here is still a constant albeit a different constant the formula that we found in 121 for dydx is correct for this equation too le dy 7 32y 2y3 7 Smyz T z5 31L6xy278x2y7 dx 7 122716 7 1 27 812732 7 1239 m 211 dy dx Thus the equation of the tangent line to the graph of the given equation at the point 21 is 17 1 2 7 7 y 7 12 95 y 7 12 6 39 3 The demand equation for a rm7s product is given by pzq 2pq32 3g2 42500 where p is the price per unit for the rms product and q is the quantity demanded of the rms product d a Recall price elasticity of demand is given by nqp dig B We use implicit 1390 q d differentiation to nd dig at the given point 1390 d 2 3 d 7 2 2 3 2 7 42500 dppq pq q dplt d d d 7 2pqp2 2Q 310123 6F 0 dp dp dp d 7 6721 3quZ 6g 7 2m 29132 n E 1 dp p2 3pq1 2 6Q 7 771000200077m dp F5 25 150 600 775 q100 3000 5 30 75 7 7 77 And we see that 117300 775 100 155 demand is inelastic at this point on the demand curve This means that nqp b Use the approximation formula for percentage change Aq nqp Ap The percentage change in p is 525 7 5 Ap T 100 5 30 Aq if 5 m 70968 5 3 d9 77410 155 125 5 1 7 7 77 lt 0 175 30 6 q100 dr When p increases q decreases and since d7 lt 0 at the point in question this 1 Recall p lt1 1 This means that g dq implies that 7 increases P d The approximation formula tells us that Ar CTT Ap We know that p 5 Ap 025 so it remains to compute 6 10 175 To do this we note that Cl d dq dq 7 7 1 7 7 dp dp10 I qp dp qp dp7 by the product rule and implicit differentiation This means that dr 71005 3000 7 2500 dp PS 775 31 7 since 15 100 and therefore 2500 Ar 7 025 m 2016 31 4 The degree 10 Taylor polynomial for fx em centered at 0 0 is given by 0 10 0 f f 7000 2 10 2 3 4 5 6 7 8 9 10 1 7 7 7 7 7 7 x 2 6 24 120 720 5040 40320 362880 36288007 T1006 f0f0z70 702 since f 95 W 5 5 for all k and 60 1 lSee SN 7 9quot Sa Comment If we use T101 to approximate the value of e 61 then we obtain the estimate 1 1 1 1 1 1 1 1 1 1 7 7 7 7 7 7 7 7 z 2 6 24 120 720 5040 40320 362880 3628800 6 m 27182818011463845 The rst 7 digits after the decimal point are correct If g 5 xlS then g z x 233 and g z 72x 539 Evaluating g and its rst two derivatives at z 10007 we nd that 2 47 1 900000 739 45000039 1 jummigi 91000 10 7 300 and g 1000 The second degree Taylor polynomial for g centered at z 1000 is therefore equal to x 7 1000 ziimmz 900000 Using this polynomial we obtain the approximation 3 1001 m 10 10003332222 71 777J77 300 900000 the decimal expansion has nothing but 27s from the seventh point on The rst 10 digits after the decimal point are correct here First7 we compute the derivatives up to and including order 4 of the natural log function fm 71 f z 72 fHx 2x73 and f4x 76x74 i Evaluating lnx and its derivatives at z 17 we nd that no1mrkain 7m2477m44y zihi we1Vi1fiiiD5 The graphs of y lnz and y T4z in red appear below taken from SN 7 Next7 we evaluate i 1673 750039 since 125 108 nmsymsini msiif msiifiimsin4 1673 This means that ln 08 x 71673 75007 m and ln 125 7 ln08

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