| |Iowa |Nebraska |
|Wheat |20 million bushels |120 million bushels |
|Corn |120 million bushels |20 million bushels |
a. Explain how, with trade, Nebraska can wind up …show more content…
-0.012Q^2 +0.000002Q^3 +100000
MC=d(TC)\dQ
MC=2000 -0.024Q +0.000006Q^2 ---(3)
In profit maximizing situation, MC=MR,,from eq 2 and 3,
4240 –Q\250 = 2000 -0.024Q +0.000006Q^2
Taking 250 l.c.m,
1060000 – Q =500000 -6Q +0.0015Q^2
1060000 -500000 –Q +6Q=0
560000 +5Q -0.0015Q^2=0
Solving we get Q=21060
PART 2)
Profit maximizing price
P=4240-Q\250
=4240-21060\250
=$ 4155.8
PART 3)
Max profit
TR=P*Q
=4155.8*21060 =$ 9,37,54,908
PART 4) If M=30000
TR=500P*Q=112000Q+5QM-Q^2
500PQ=112000Q+150000Q-Q^2
500PQ=262000Q-Q^2
MR=500P=262000-2Q
P=5240-Q/250
Equating MR=MC
SMC
e AR = MR
Revenue/Cost
Output
e: denotes equilibrium in above graph
5240-Q/250=2000-.024Q+.00006Q^2
Q=24964
• Maximum amount of Profit if M=30000
TR=P*Q
TR=4155.8*24964=$103745391
Answer 3:
Given:
MP for 4 month is $115
FC is $3500
Part a:
SMC=125 - 0.42Q+0.0021Q^2
Integrating SMC to get TC
TC=125Q-0.21Q^2+0.0007Q^3+3500
TVC=125Q-0.21Q^2+0.0007Q^3
AVC=TVC/Q
Hence AVC=125-0.21Q+0.0007Q^2
Part b:
AVC reaches its minimum when d(AVC)/d(Q)=0
d(125-0.21Q+0.0007Q^2)/d(Q)=0
0.21+0.0014Q=0
Hence Q=150
Substituting Q in AVC equation:
Min