Managerial Economics Chapter 5 and 6 Homework Essay

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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 = …show more content…
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

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