300hw4answer - 1 / 3 . d. With a fee of 60, farm 1 sets 60...

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Eco 300 HW 4 Answer Key 16.6 a. Setting MB = MC yields 100 – R = 20 + R or R = 40. b. The fee should be set so that farmers choose R = 40. So, Fee = MC = 20 + R = 20 + 40 = 60. A fee of 60 for each percent not reduced would prompt farmers to choose R = 40. It is cheaper to pay the fee than reduce methane by 41 percent, however, since that would cost 61. c. With a mandate of a 40 percent reduction, the average reduction will be 40 percent. MC 1 = 20 + 2 / 3 (40) = 46 2 / 3 MC 2 = 20 + 2(40) = 100. Total cost is given area under MC up to the required level. For farm 1 this amounts to 20 × 40 + (46 2 / 3 – 20)(40) = 800 + 533 1 / 3 = 1,333 1 / 3 . For farm 2, this is 20 × 40 + 1 / 2 (100 – 20)(40) = 800 + 1,600 = 2,400. So total costs of achieving the 40 percent reduction are 3,733
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Unformatted text preview: 1 / 3 . d. With a fee of 60, farm 1 sets 60 = 20 + 2 / 3 R , and calculates R 1 = 60. Farm 2 calculates 60 = 20 + 2 R 2 or R 2 = 20. Again, the average reduction is 40. Total costs now for farm 1 are 20 60 + 1 / 2 (60 20)(60) = 1,200 + 1,200 = 2,400 and for farm 2 20 20 + 1 / 2 (60 20)(20) = 400 + 400 = 800 so total costs are 3,200. The fee achieves the same average reduction at a much lower total cost. e. The fee procedure recognizes the differential costs of methane reduction. By having the low cost farm (farm 1) make relatively greater reductions, overall costs are reduced....
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