Minnesota has two oil refineries that emit methane gas. The Bend refinery emits 300 units of methane, and the Park refinery emits 900 units. Government regulators would like to reduce total methane emissions from 1,200 to 900 units. Bend's total abatement and marginal abatement costs are TC_B = 400Q_B + 4Q_B² and MAC_B = 400 + 8Q_B, where Q is units of methane. Park's total abatement cost and marginal abatement costs are TC_P = 40Q_P + 2Q_P² and MAC_P = 40 + 4Q_P.
Suppose that the government regulators evenly split 900 methane pollution permits between the two refineries. The total cost of removing 300 units of methane, using a tradable permit system is $____.

Question

Quizplus · Accepted Answer

The marginal abatement cost for Bend is 400 + 8Q_B, and the marginal abatement cost for Park is 40 + 4Q_P. To minimize the total cost of abatement, the marginal abatement costs should be equal. So, we have:400 + 8Q_B = 40 + 4Q_PSolving for Q_B, we get:Q_B = 45So, Bend should abate 45 units of methane. The total cost of abatement for Bend is:TC_B = 400Q_B + 4Q_B²= 400(45) + 4(45)²= 101,250The marginal abatement cost for Park is 40 + 4Q_P, and the total abatement for Park is 300 - 45 = 255 units. So, the total cost of abatement for Park is:TC_P = 40Q_P + 2Q_P²= 40(255) + 2(255)²= 150,750The total cost of abatement for both refineries is:TC_B + TC_P = 101,250 + 150,750= 252,000Since the government regulators evenly split 900 methane pollution permits between the two refineries, each refinery will receive 450 permits. The cost of each permit is:TC_P / 450 = 252,000 / 450= 560So, the total cost of removing 300 units of methane, using a tradable permit system is:560 * 300= 172,200

Minnesota Has Two Oil Refineries That Emit Methane Gas