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= 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.