Managerial Economics Chapter 5 and 6 Homework Essay

deviate AA pixilated maximizes turn a usefulness when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 L/ cdTherefore (40)*(1-L/ cd) = 20. The solution is L = two hundred.In turn, Q = 200 (2002/800). The solution is Q = 150.The firms advantage is= PQ (MC)L= ($40) (150) ($20) (200) = $2,000 incite B Price ontogenesis to $50Q = Dresses per weekL= Number of struggle hours per weekQ = L L2/800MCL=$20P= $50A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 L/400Therefore (50)*(1-L/400) = 20. The solution is L = 240.In turn, Q = 240 (2402/800). The solution is Q = 168.The firms profit is ($40) (168) ($20) (240) = $1,920Optimal output of the firm would amplify from 150 to 168, and labor would accession from 200 to 240, resulting in a decrease in profit to $1,920. Part B inflation in labor and output priceAssuming a 10% ontogenesis IN LABOR COST AND OUTPUT PRICEQ = Dresses per weekL= Number of labor hours per weekQ = L L2/800MCL=$20.20 (20*.10 )P= $40.40 ($40*.10)A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 L/400Therefore (40.40)*(1-L/400) = 20.20. The solution is L = 200. In turn, Q = 200 (2002/800). The solution is Q = 150.The firms profit is ($40.40) (150) ($20.20) (200) = $2,020 Optimal output of the firm would remain the same at 150, and labor would remain the same at 200, however, there would be an increase in profit to $2,020 to correspond to the percentage increase in output price and labor personify (in this example 10%). Part C 25% increase in MPLThe marginal cost of labor would increase by the same percentage amount as price (25%), therefore the Marginal address of labor would increase from 20 to 25. Therefore 50 L/8 =25 and L=200Output and hours of labor remain unchanged due to the fact that price and cost of labor increase by same percentage amounts ALSO SEE PART B ABOVE INFLATION exercising I MADE DENOTING 10 PERCENT INCREASE IN LABOR AND OUTPUT. Chapter 5 Question 12 pageboy 220Part AQ = 100(1.01).5(1).4 = 100.50. Compare this to the original of Q=100 and we can determine that Output increases by .5%. The power coefficient measures the elasticity of the output with respect to the input. A 1% increase in labor buzz offs a (.5)(1) = .5% increase in output.Part BDr. Ghosh- per my e-mail I was a bit confused with this distrust based on your lecture notes (as your notes state that some(prenominal) inputs must change for a returns to overcome to be determined) , so I have two different opinions. Opinion 1- The nature of returns to scale in return depends on the sum of the exponents, +. Decreasing returns exist if + 1. The sum of the power coefficients is .5 + .4 1, the production function exhibits decreasing 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- BOTHinputs 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 being changed. Does this concept not apply, and is my original answer in settle? I dont see any scale where only one of the inputs are changedAs such, if both(prenominal) 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)Demand is P = 48 Q/200 bes are C = 60,000 + .0025Q2.Therefore the TR= 48Q-Q2/200, and the derived function MR function would be MR = 48 Q/100. The firm maximizes profit by setting MR = MC. Therefore, MR = 48 Q/100 and MC = .005Q. Setting MR = MC (48 Q/100) = .005Q results in Q* = 3,200. In turn, P* = $32 (where 48-3200/200). Chapter 6 Question 8 Page 265CE= 250,000 +1,000Q + 5Q2$2,000= Cost of Frames and assemblyP= 10,000-30QPart AMarginal Cost of producing an additional locomotive engineCE = 250,000 +1,000Q +5Q2MCE = d/dQ (250,000 +1,000Q + 5Q2)=10Q + 1,000MCCycle=MCEngine +MCframes and assembly thereforeMCCylce = 1,000+ 2,000 +10QThe inverse demand function provided in the text was P= 10,000-30Q TR = (P)*(Q)= (10,000-30Q)*Q=10,000Q 30Q2Obtain the derivative of this function to find MRMR=d/dQ=(10,000Q 30Q2)MR=10,000 60QMR = MC10,000 60Q = 1,000 + 2,000 +10Q7,000 = 70QQ=100 (profit maximizing output)P= 10,000 30Q=10,000 -30(100) clear Maximizing Price=7,000Therefore the Marginal Cost of producing an engine=1,000 + 10Q (q=100 from solving above)=2,000 MCEngineMarginal Cost of Producing a CycleFrom correspond developed aboveMCCycle = 1,000 +2,000 +10Q=1,000 +2,000 + 10(100)=$4,000 MCCyclePart BSince 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=$1,400MR = MC10,000 60Q = 2,000 +1,40010,000- 60Q = 3400Q=110 (profit maximizing output)P = 10,000 30(110)=6,700 profit maximizing priceTherefore the firm should buy the engine since the engine produced by the firm is much than the engine provided by the other firm. Chapter 6 Question 10 Page 266Part A taxation is P*Q.Obtain Marginal Cost function throughclx + 16Q + 0.1Q2FOC (derivative of above equation)16 + 0.2Q= MCFrom the P= 96 .4Q we can determine that total revenue = 96Q .4Q2 and the derivative or FOC is thus 96 .8Q= MRSet MC = MR16 + 0.2Q = 96 0.8QQ=80We solve for P by plugging this into our original equationP= 96-.4(80)P=64Profit = 5,120 (80*64) 2,080 (160 + 16*80 + .1(80)2) = $3,040Part BC =160 + 16Q + .1Q2AC= (160+16Q+.1Q2)/QMC=d/dQ(160 + 16Q + .1Q2)MC=16 + .2QAC=MC160/Q + 16 + .1Q = 16 + .2Q160/Q = .1Q.1Q2 =160Q= 40Average cost of production is minimized at 40 units, she is correct as AC = MC (see below). AC = 960/40 =24MC = 16 + (.2) ($40) = $24However, optimal output is Q=80 where MR = MC, therefore her second claim of 40 units as the firms profit maximizing level of output is in correct. P =96 .4 (40)P=$80TR = 80*40 =3,200C = 160 + 16Q + .1Q2=960Profit = Revenue Cost = 3,200 960 = 2,240 therefore output at 80 is greater than the profit at 40. Part CWe learned from part a the single plant cost is $2,080 or (160 + 16*80 + .1(80)2). If two plants were open each producing the minimum level of output detailed 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.