Economy

Chapter 11 fuss 2. a. MC = d(TC)/dQ = ?5000 + 200 Q b. P = MR = $20,000 c. MC = MR is acquire maximizing output, where ?5000 + 200 Q = 20,000. Therefore, 200 Q = 25,000, so Q* = 125. enigma 4. a. MC = dTC/dQ = 20 c.MR = 60(1 + 1/?1.5) = $20 conundrum 6. a.ED = %?QD / %?P, 2.0 = +15% / %?P. To get 15% more sales, it can %?P = 7.5%. victimisation the arc damage formula, we can get the new damage and new measuring rod: .075 = (P2 15.00)/ [(P2 + 15)/2] P2 = $13.92 ?P = $15 $13.92 = $1.08 in like manner: .15 = (Q2 30,000)/ [(Q2 + 30,000)/2] Q2 = 34,865 gallons b. i). On TR: forward: TR, = 15(30,000) = $450,000 later on: TR2 = 13.92(34,865) = $485,321, so ?TR = +$35,321 ii). On TC: Before: FC1 =$90,000 afterwards: FC2 = $90,000 VC/ social unit = $6.00 .60 = $5.40 VC2 = $5.40 × 34,865 = $188,271 TC2 = 9 0,000 + 188,271 = $278,271, so ?TC = +$8,271 iii). On Total Profits: Before: ?, = $450,000 $270,000 = $180,000 After: ?2 = $485,321 $278,271 = $207,050, or ?? = + 27,050 Chapter 12 Problem 1. a.

?C = PQC ? TCC = (600 ? QC ? QD)QC ? (25,000 + 100QC) = ?25,000 + 500QC ? QC2 ? QCQD ?D = (600 ? QC ? QD) QD ? (20,000 + 125QD) = ?20,000 + 475QD ? QD2 ? QCQD ??C/?QC = 500 ? 2QC ? QD and ??D/?QD = 475 ? 2QD ? QC Conditions for an optimum require that twain partials be set equal to zero and the resulting equations be single-minded sim ultaneously for optimal values of QC and QD:! QD* = 150 units and QC* = 175 units. P = 600 150 175 = $275, so P* = $275. b. ?C* = ?25,000 + 500(175) ? (175)2 ? 175(150) = $5,625. ?D* = ?20,000 + 475(150) ? (150)2 ? 175(150) = $2,500. Problem 2. b. ?A* = ?1500 + 145(30) ? 2(30)2 ? 30(25) = $300. ?B* = ?1200 +180(25) ? 3(25)2 ?30(25)= $675. Also ?* = 300 + 675 = $975. Problem 5. b....If you want to get a full essay, order it on our website: BestEssayCheap.com

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