Suppose that in Horsehead, Massachusetts, the cost of operating a lobster boat is $6,000 per month.Suppose that if x lobster boats operate in the bay, the total monthly revenue from lobster boats in the bay is $1,000(18x  x²) .If there are no restrictions on entry and new boats come into the bay until there is no profit to be made by a new entrant, then the number of boats who enter will be X₁.If the number of boats that operate in the bay is regulated to maximize total profits, the number of boats in the bay will be X₂.

A) X₁  6 and X₂  4.
B) X₁  12 and X₂  6.
C) X₁  16 and X₂  10.
D) X₁  12 and X₂  12.
E) None of the above.

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Luisa Bernia · Accepted Answer

Final Answer : B, Explanation :  To find the profit-maximizing number of boats, we need to find the number that maximizes total revenue minus total cost.Total revenue is given by:$1,000(18x - 10x^2)$Total cost is given by:$6,000x$So, total profit is:$1,000(18x - 10x^2) - 6,000x$Simplifying, we get:$12,000x - 10,000x^2$To find the profit-maximizing number of boats, we need to take the derivative with respect to x and set it equal to 0:$\frac{d}{dx} (12,000x - 10,000x^2) = 12,000 - 20,000x = 0$Solving for x, we get:$x = 0.6$So, the optimal number of boats is 0.6 boats, which we can round up to 1 boat. This means that X 2  = 1.To find X 1 , we need to find the number of boats that would enter the bay if there were no restrictions on entry. This would occur when the revenue earned by the last boat entering the bay is equal to the cost of operating the boat. So, we set $1,000(18x - 10x^2) = 6,000$ and solve for x:$10x^2 - 18x + 6 = 0 (5x - 3)^2 = 9 x = \frac{3}{5}$ or $x = 1.2$Since we only want the positive value for x, X 1  = 2 (rounded up from 1.2).Therefore, the best choice is B as X 1  = 2 and X 2  = 1.

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