Managerial Economics Chapter 5 and 6 Homework Essay

Part A: A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 – L/400 Therefore (40)*(1-L/400) = 20. The solution is L = 200. In turn, Q = 200 – (2002/800). The solution is Q = 150. The firms profit is= PQ – (MC)L= ($40) (150) – ($20) (200) = $2,000 Part B Price increase to $50: Q = Dresses per week L= Number of labor hours per week Q = L –L2/800 MCL=$20 P= $50 A firm maximizes profit when it equates MRPL = (MR) *(MPL) = MCL MPL= dQ/dL =1 – L/400 Therefore (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) = $1,920 Optimal output of the firm would increase from 150 to 168, and labor would increase from 200 to 240, resulting in a decrease in profit to $1,920. Part B inflation in labor and output price: Assuming a 10% increase IN LABOR COST AND OUTPUT PRICE†¦ Q = Dresses per week L= Number of labor hours per week Q = L –L2/800 MCL=$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/400 Therefore (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 cost (in this example 10%). Part C 25% increase in MPL: The marginal cost of labor would increase by the same percentage amount as price (25%), therefore the Marginal Cost of labor would increase from 20 to 25. Therefore 50 – L/8 =25 and L=200 Output 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 EXAMPLE I MADE DENOTING 10 PERCENT INCREASE IN LABOR AND OUTPUT. Chapter 5 Question 12 Page 220 Part A: Q = 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 produces a (.5)(1) = .5% increase in output. Part B: Dr. Ghosh- per my e-mail I was a bit confused with this question based on your lecture notes (as your notes state that BOTH inputs must change for a returns to scale to be determined) , so I have two different opinions. Opinion 1- The nature of returns to scale in production depends on the sum of the exponents, ÃŽ ±+ÃŽ ². Decreasing returns exist if ÃŽ ±+ÃŽ ²Ã‹â€š 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- BOTH inputs 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 incorrect? I don’t see any 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) Demand is P = 48 – Q/200 Costs 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 265 CE= 250,000 +1,000Q + 5Q2 $2,000= Cost of Frames and assembly P= 10,000-30Q Part A: Marginal Cost of producing an additional engine†¦ CE = 250,000 +1,000Q +5Q2 MCE = d/dQ (250,000 +1,000Q + 5Q2) =10Q + 1,000 MCCycle=MCEngine +MCframes and assembly; therefore MCCylce = 1,000+ 2,000 +10Q The inverse demand function provided in the text was P= 10,000-30Q TR = (P)*(Q) = (10,000-30Q)*Q =10,000Q – 30Q2 Obtain the derivative of this function to find MR: MR=d/dQ =(10,000Q – 30Q2) MR=10,000 – 60Q MR = MC 10,000 – 60Q = 1,000 + 2,000 +10Q 7,000 = 70Q Q=100 (profit maximizing output) P= 10,000 – 30Q =10,000 -30(100) Profit Maximizing Price=7,000 Therefore the Marginal Cost of producing an engine =1,000 + 10Q (q=100 from solving above) =2,000 MCEngine Marginal Cost of Producing a Cycle From equation developed above†¦ MCCycle = 1,000 +2,000 +10Q =1,000 +2,000 + 10(100) =$4,000 MCCycle Part B: Since 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 MR = MC 10,000 – 60Q = 2,000 +1,400 10,000- 60Q = 3400 Q=110 (profit maximizing output) P = 10,000 – 30(110) =6,700 profit maximizing price Therefore 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 Part A: Revenue is P*Q. Obtain Marginal Cost function through 160 + 16Q + 0.1Q2 FOC (derivative of above equation) 16 + 0.2Q= MC From the P= 96 – .4Q we can determine that total revenue = 96Q – .4Q2 and the derivative or FOC is thus 96 – .8Q= MR Set MC = MR 16 + 0.2Q = 96 – 0.8Q Q=80 We solve for P by plugging this into our original equation P= 96-.4(80) P=64 Profit = 5,120 (80*64) – 2,080 (160 + 16*80 + .1(80)2) = $3,040 Part B: C =160 + 16Q + .1Q2 AC= (160+16Q+.1Q^2)/Q MC=d/dQ(160 + 16Q + .1Q2) MC=16 + .2Q AC=MC 160/Q + 16 + .1Q = 16 + .2Q 160/Q = .1Q .1Q2 =160 Q= 40 Average cost of production is minimized at 40 units, she is correct as AC = MC (see below). AC = 960/40 =24 MC = 16 + (.2) ($40) = $24 However, optimal output is Q=80 where MR = MC, therefore her second claim of 40 units as the firm’s profit maximizing level of output is incorrect. P = 96 – .4 (40) P=$80 TR = 80*40 =3,200 C = 160 + 16Q + .1Q2 =960 Profit = Revenue – Cost = 3,200 – 960 = 2,240 therefore output at 80 is greater than the profit at 40. Part C: We 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.