Question

A paper company dumps nondegradable waste into a river. The firms production function is q=6KL, where...

A paper company dumps nondegradable waste into a river. The firms production function is q=6KL, where q= annual paper production measured in pounds; K= machine hours; L=gallons of polluted water dumped. The firm currently faces no environmental regulations. The rental rate of capital is $30 per hour and it costs the firm $7.5 per gallon to dump the water.

a) Write down the cost function that the firm faces

b) to minimize cost, the ratio of wastewater to capital should be kept at

c)Suppose the company's goal is to produce 60,000 pounds of paper per year, in order to minimize cost how many machine hours and gallons of wastewater to minimize its cost.

d)Suppose the state environmental protection agency plans to impose $22.50 pollution fee for each gallon dumped. To produce same output of 60,000 pounds the should use how many machine hours and dump how many gallons of wastewater

e)The pollution fee increases the cost of dumping water. Graphically illustrate how this pollution fee would affect the firms decision on machine hours and wastewater. Your graph should contain isoquants and isocosts and firms choices before and after pollution fee

Homework Answers

Answer #1

a) the cost function the form faces is C = 30K + 7.5L

b) The objective function the consumer faces is:

Maximise q = 6KL

Subject to C=30K +7.5L

Now differentiate there w.r.t K, L and Lambda once each while equating the partial differentials to 0. This will give you K in terms of lambda from the first equation and L in terms of lambda from the second equation. Now divide the two i.e K/L to get the ratio which would be 1/4.

c)Given K = L/4. Substitute this in the equation q=6KL. This becomes 60000=(L^2)/4. Solve for L will get you L= 200 and subsequently K = 50.

d)Here now the production function changes to q=6KL - 22.5L

Now our new objective function will be.

Follow the same procedure as in 2. This will give you the new ratio of K=(22.5+3L)/6

Now follow the procedure as in 3. This will give you L as 141 and subsequently K can be calculated.

e) This is a normal isoquant, isocosts graph.

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