# HW 6a - AEM 250 Spring 2007 HW #6 Solutions 1. 2. The...

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AEM 250 Spring 2007 HW #6 Solutions 1.

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2. The marginal control costs for each power plant are plotted in the graph below as a function of the number of units of pollution reduction. The marginal control costs for Plant 2 are higher than those for Plant 1 as given by the MC c functions. The total control costs for Plant 1 to reduce pollution by 5,000 tons are given by the red-hatched triangle in the graph above. The area of the triangle is: (0.5)*(5,000)*(100) = \$250,000 The total control cost for Plant 2 to reduce their emissions by 5,000 tons is given by the sum of the red and blue hatched triangles as:
(0.5)*(5,000)*(150) = \$375,000 The total control costs for the plants are simply the sum of these two values, or \$625,000. The total pollution reduction is 10,000 tons. b. With a pollution tax, each plant will reduce emissions as long as the tax amount exceeds the marginal control costs. We again plot each plant’s marginal control costs in the graph below. Plant 1 reduces its pollution by 6,000 tons until the marginal control costs equal the tax level. Note that we can solve the problem algebraically by finding when the marginal control costs for Plant 1 equal the tax of \$120 per unit: 120 = 0.02Q Q = 6,000 The total control costs for Plant 1 is the sum of the purple and red areas in the graph above, or: (0.5)*(6,000)*(120) = \$360,000 As Plant 1 is still producing 2,000 tons of pollution, it pays a total tax of \$240,000. To solve for the pollution reduction by Plant 2, solve for a MC c of 120 as:

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120 = 0.03Q Q = 4,000 The total control costs are the purple and blue areas in the graph, or: (0.5)*(4,000)*(120) = \$240,000 As Plant 2 is still emitting 4,000 tons of pollution, it pays a tax of \$480,000. The
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## HW 6a - AEM 250 Spring 2007 HW #6 Solutions 1. 2. The...

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