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 =

*…show more content…*

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