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winter problem set 3 solutions

# winter problem set 3 solutions - Problem Set 3 Solutions...

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Quantity Price 100/3 P = 100/3 – Q/3 MPC = Q 30 25 A MSC = 3Q 10 25 B G K E F N H R M 75 Problem Set 3 Solutions 17.8 a) 100 – 3P = P P = 25 and Q = 25. c) MEC = 2Q, while MPC = Q. Thus, MSC = 3Q. To find the optimal quantity, we equate MSC to inverse demand, or 3Q = 100/3 – Q/3, or Q = 10. The socially optimal price would equal the marginal social cost at the optimal quantity, or P = 3(10) = \$30. e) The optimal emissions fee is equal to the difference between MSC and MPC at the socially optimal quantity. Since MSC = 3Q and MPC = Q, and Q = 10, the optimal emissions fee equals: 3(10) – 1(10) = \$20 per unit. Equilibrium price and quantity = P 2 and Q 2 Social optimum price and quantity = P 1 and Q 1 Difference between social optimum and equilibrium Consumer surplus A+B+G+K = \$104.167 A = \$1.67 -B-G-K = -\$102.50 Private producer E+F+R+H+N = B+E+F+R+H+G = B+G-N=

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surplus \$312.50 \$250 -62.5 - Cost of externality -R-H-N-G-K-M = - \$625 -R-H-G = - \$100 M+N+K= \$525 Net social benefits A+B+E+F-M = -\$208.33 A+B+E+F= \$151.67 M= \$360 Deadweight loss M = Zero M 17.12 The graph below demonstrates that the optimal toll would be less than \$5. Equilibrium with no toll occurs at Q 3 , although the socially optimal amount of driving would occur at Q 2 < Q 3 . Although the marginal external cost is \$5 at Q 3 , the socially optimal toll occurs at MEC ( Q 2 ), which is less than \$5 since MEC is upward sloping.
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winter problem set 3 solutions - Problem Set 3 Solutions...

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