| |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

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-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