'Managerial Economics Chapter 5 and 6 Homework Essay\r'

' check A:\r\nA steady maximizes benefit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 †L/400\r\n so (40)*(1-L/400) = 20. The resolving power is L = cc.\r\nIn turn, Q = 200 †(2002/800). The solution is Q = 150.\r\nThe fuddleds realise is= PQ †(MC)L= ($40) (150) †($20) (200) = $2,000 Part B Price increase to $50:\r\nQ = Dresses per week\r\nL= Number of roil hours per week\r\nQ = L â€L2/800\r\nMCL=$20\r\nP= $50\r\nA soused maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 †L/400\r\n wherefore (50)*(1-L/400) = 20. The solution is L = 240.\r\nIn turn, Q = 240 †(2402/800). The solution is Q = 168.\r\nThe upstandings profit is ($40) (168) †($20) (240) = $1,920\r\nOptimal widening of the firm would increase from 150 to 168, and crunch would increase from 200 to 240, resulting in a decrease in profit to $1,920. Part B inflation in fight and output price:\r\nAssuming a 10% increase IN LABOR COST AND issue PRICEà ¢â‚¬Â¦\r\nQ = Dresses per week\r\nL= Number of stab hours per week\r\nQ = L â€L2/800\r\nMCL=$20.20 (20*.10)\r\nP= $40.40 ($40*.10)\r\nA firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 †L/400\r\nTherefore (40.40)*(1-L/400) = 20.20. The solution is L = 200. In turn, Q = 200 †(2002/800). The solution is Q = 150.\r\nThe firms profit is ($40.40) (150) †($20.20) (200) = $2,020 Optimal output of the firm would remain the analogous at 150, and cranch would remain the same at 200, however, there would be an increase in profit to $2,020 to correspond to the region increase in output price and beat back comprise (in this example 10%). Part C 25% increase in MPL:\r\nThe marginal embody of labor movement would increase by the same piece amount as price (25%), consequently the marginal apostrophize of labor would increase from 20 to 25. Therefore 50 †L/8 =25 and L=200\r\n make and hours of labor remain unvaried due to the fact that p rice and cost of labor increase by same percentage amounts likewise SEE PART B ABOVE swelling EXAMPLE I MADE DENOTING 10 part INCREASE IN LABOR AND OUTPUT. Chapter 5 doubt 12 summon 220\r\nPart A:\r\nQ = cytosine(1.01).5(1).4 = 100.50. equivalence this to the reliable of Q=100 and we hind end buoy typeset that Output increases by .5%. The power coefficient measures the elasticity of the output with compliance to the input. A 1% increase in labor produces a (.5)(1) = .5% increase in output.\r\nPart B:\r\nDr. Ghosh- per my e-mail I was a bit entangled with this dubiousness based on your lecture notes (as your notes order that BOTH inputs must change for a returns to casing to be determined) , so I have cardinal different opinions. stamp 1- The nature of returns to racing shell in production depends on the sum of the exponents, α+β. Decreasing returns follow if α+β˂ 1. The sum of the power coefficients is .5 + .4 < 1, the production function exhib its diminish returns to scale where output increases in a smaller proportion than input. This is reflected in Part A of this problem where a 1% increase in labor (input) results in a .5% increase in output. Opinion 2- BOTH\r\ninputs must be changed in the same proportion (according to your lecture notes). Therefore, in this question I am confused. Only one of the inputs are be changed. Does this concept not apply, and is my original answer wrong(p)? I don’t see whatever scale where only one of the inputs are changed…As such, if both inputs MUST be changed then returns to scale can not be determined for this question as only L was originally changed. Chapter 6 Question 6 Part B Page 265 (part A not required)\r\nDemand is P = 48 †Q/200\r\n embodys are C = 60,000 + .0025Q2.\r\nTherefore the TR= 48Q-Q2/200, and the derivative MR function would be MR = 48 †Q/100. The firm maximizes profit by setting MR = MC. Therefore, MR = 48 †Q/100 and MC = .005Q. Sett ing MR = MC (48 †Q/100) = .005Q results in: Q* = 3,200. In turn, P* = $32 (where 48-3200/200). Chapter 6 Question 8 Page 265\r\nCE= 250,000 +1,000Q + 5Q2\r\n$2,000= follow of Frames and assembly\r\nP= 10,000-30Q\r\nPart A:\r\nmarginal appeal of producing an supernumerary engine…\r\nCE = 250,000 +1,000Q +5Q2\r\nMCE = d/dQ (250,000 +1,000Q + 5Q2)\r\n=10Q + 1,000\r\nMCCycle=MCEngine +MCframes and assembly; therefore\r\nMCCylce = 1,000+ 2,000 +10Q\r\nThe inverse demand function provided in the textbook was P= 10,000-30Q TR = (P)*(Q)\r\n= (10,000-30Q)*Q\r\n=10,000Q †30Q2\r\nObtain the derivative of this function to hold MR:\r\nMR=d/dQ\r\n=(10,000Q †30Q2)\r\nMR=10,000 †60Q\r\nMR = MC\r\n10,000 †60Q = 1,000 + 2,000 +10Q\r\n7,000 = 70Q\r\nQ=100 (profit maximizing output)\r\nP= 10,000 †30Q\r\n=10,000 -30(100)\r\n profit Maximizing Price=7,000\r\nTherefore the marginal Cost of producing an engine\r\n=1,000 + 10Q (q=100 from solving above)\r\n=2,000 MCE ngine\r\nMarginal Cost of Producing a Cycle\r\nFrom equation positive above…\r\nMCCycle = 1,000 +2,000 +10Q\r\n=1,000 +2,000 + 10(100)\r\n=$4,000 MCCycle\r\nPart B:\r\nSince the firm can produce engines at a Marginal Cost of $2,000, the opportunity to buy from another firm at a greatly reduced Marginal Cost of $1,400 would be sensible. MCEngine=$1,400\r\nMR = MC\r\n10,000 †60Q = 2,000 +1,400\r\n10,000- 60Q = 3400\r\nQ=110 (profit maximizing output)\r\nP = 10,000 †30(110)\r\n=6,700 profit maximizing price\r\nTherefore the firm should buy the engine since the engine produced by the firm is more than the engine provided by the other firm. Chapter 6 Question 10 Page 266\r\nPart A:\r\nRevenue is P*Q.\r\nObtain Marginal Cost function through\r\n160 + 16Q + 0.1Q2\r\nFOC (derivative of above equation)\r\n16 + 0.2Q= MC\r\nFrom the P= 96 †.4Q we can determine that full revenue = 96Q †.4Q2 and the derivative or FOC is therefore 96 †.8Q= MR\r\nSet MC = MR\r\n16 + 0.2Q = 96 †0.8Q\r\nQ=80\r\nWe solve for P by plugging this into our original equation\r\nP= 96-.4(80)\r\nP=64\r\nProfit = 5,120 (80*64) †2,080 (160 + 16*80 + .1(80)2) = $3,040\r\nPart B:\r\nC =160 + 16Q + .1Q2\r\nAC= (160+16Q+.1Q^2)/Q\r\nMC=d/dQ(160 + 16Q + .1Q2)\r\nMC=16 + .2Q\r\nAC=MC\r\n160/Q + 16 + .1Q = 16 + .2Q\r\n160/Q = .1Q\r\n.1Q2 =160\r\nQ= 40\r\n fairish cost of production is minimized at 40 units, she is correct as AC = MC (see below). AC = 960/40 =24\r\nMC = 16 + (.2) ($40) = $24\r\nHowever, optimal output is Q=80 where MR = MC, therefore her countenance claim of 40 units as the firm’s profit maximizing level of output is incorrect. P =\r\n96 †.4 (40)\r\nP=$80\r\nTR = 80*40 =3,200\r\nC = 160 + 16Q + .1Q2\r\n=960\r\nProfit = Revenue †Cost = 3,200 †960 = 2,240 therefore output at 80 is great than the profit at 40. Part C:\r\nWe erudite from part a the single plant cost is $2,080 or (160 + 16*80 + .1(80)2). If two plants were open(a) each producing the minimum level of output precise in part B (Q=40) then total cost would be (Q)*(AC) = 24*80 = $1,920. You can compare this to the cost in part A of $2,080 and determine it is cheaper to produce using the two plants.\r\n'