# Managerial Economics Chapter 5 and 6 Homework

Topics: Marginal cost, Economics, Costs Pages: 6 (1305 words) Published: September 16, 2013
Chapter 5 Question 6 Page 218
Q = Dresses per week
L= Number of labor hours per week
Q = L –L2/800
MCL=\$20
P= \$40= therefore MR=\$40

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, α+β. Decreasing returns exist if α+β˂ 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=...