Managerial Economics Chapter 5 and 6 Homework Essay

lineament AA quick maximizes do good when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 L/400Therefore (40)*(1-L/400) = 20. The antecedent is L = 200.In turn, Q = 200 (2002/800). The solution is Q = 150.The unbendables value is= PQ (MC)L= ($40) (150) ($20) (200) = $2,000 take leave B outlay append to $50Q = Dresses per weekL= tour of fag out 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 product of the firm would growth from 150 to 168, and press would increase from 200 to 240, resulting in a diminution in profit to $1,920. Part B inflation in labor and turnout priceAs summarizeing a 10% increase IN LABOR COST AND OUTPUT PRICEQ = Dresses per weekL= Number of labor hours per weekQ = L L2/800MCL=$20.20 (20*.10)P= $40.4 0 ($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 akin at 150, and labor would remain the homogeneous at 200, however, there would be an increase in profit to $2,020 to gibe to the theatrical role increase in output price and labor address (in this example 10%). Part C 25% increase in MPLThe marginal personify of labor would increase by the same percentage amount as price (25%), therefore the fringy Cost of labor would increase from 20 to 25. Therefore 50 L/8 =25 and L=200Output and hours of labor remain unchanged overdue to the fact that price and cost of labor increase by same percentage amounts ALSO SEE PART B ABOVE lump EXAMPLE I MADE DENOTING 10 PERCENT INCREASE IN LABOR AND OUTPUT. Chapter 5 Question 12 p aginate 220Part AQ = 100(1.01).5(1).4 = 100.50. Compargon this to the original of Q=100 and we green goddess 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 produces a (.5)(1) = .5% increase in output.Part BDr. Ghosh- per my e-mail I was a bit confused with this question ground on your lecture notes (as your notes state that both inputs must change for a returns to scale to be determined) , so I have devil disparate opinions. Opinion 1- The nature of returns to scale in production depends on the sum of the exponents, +. Decreasing returns exist if + 1. The sum of the power coefficients is .5 + .4 1, the production wait on exhibits decrease 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 sa me proportion (according to your lecture notes). Therefore, in this question I am confused. Only one of the inputs are universe changed. Does this concept not apply, and is my original answer incorrect? I wear thint see any scale where only one of the inputs are changedAs 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)Demand is P = 48 Q/200 be are C = 60,000 + .0025Q2.Therefore 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. 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 A borderline Cost of producing an additional railway locomotiveCE = 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)Profit Maximizing toll=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 equation 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=one hundred ten (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 more than the engine provided by the other firm. Chapter 6 Question 10 Page 266Part ARevenue 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 entire revenue = 96Q .4Q2 and the derivative or FOC is gum olibanum 96 .8Q= MRSet MC = MR16 + 0.2Q = 96 0.8QQ=80We put to work 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= 40 intermediate 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 outcry of 40 units as the firms profit maximizing direct of output is incorrect. 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.