a. Suppose you want to remove ten fish of an exotic species that have illegally

Question

a. Suppose you want to remove ten fish of an exotic species that have illegally been introduced to a lake. You have three possible removal methods. Assume that q1, q2 and q3 are, respectively, the amount of fish removed by each method that you choose to use so that the goal will be accomplished by any combination of methods such that q1 +q2 +q3=10. If the marginal costs of each removal method are, respectively, $10q1, $5q2, and $2.5q3, how much of each method should you use to achieve the removal cost-effectively?

b. Why isn't an exclusive use of method 3 cost-effective?

c. Suppose that the three marginal costs were constant (not increasing as in the previous case) such that MC1=$10, MC2=$5, and MC3=$2.5. What is the cost-effective outcome in that case?

Explanation / Answer

A.) You want to have the marginal costs be as equal as possible to keep total cost down, so once you need to remove 4 fish, you would want to use q2 as the removal method for the 4th fish because it would only cost $5, instead of $7.50, so to keep marginal costs equal, you would want to use 6 q1 ($15), 3q2($15), and 1q1 ($10) for a total cost of ($2.50+$5+$7.50+$10+$12.50+$15+$5+$10+$10)= $77.50

B.) Using method 3 exclusively would cost ($2.50+$5+$7.50...+$25) = $137.50, far more than the $77.50 from using a combination of the methods

C.) If the methods were constant in cost, then exclusively use Method 3, as it would cost ($2.50*10) which is only $25, compared to $50 for method 2 or $100 for method 3

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