Suppose that in Horsehead, Massachusetts, the cost of operating a lobster boat is $4,000 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(28x  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  24 and X₂  24.
B) X₁  28 and X₂  16.
C) X₁  24 and X₂  12.
D) X₁  12 and X₂  10.
E) None of the above.

To maximize profits, we need to find the number of lobster boats that will maximize revenue minus costs, i.e., maximize (1000(28x - x^2)) - 4000x. Taking the derivative of this expression with respect to x and setting it to zero, we get:
28000 - 2000x = 0
x = 14
This means that 14 lobster boats will maximize profits. Therefore, X₂ = 14.
To find X₁, we need to find the point at which new entrants will no longer make a profit. Setting the revenue equal to costs, we get:
1000(28x - x^2) = 4000x
28000 - 1000x = x^2
x^2 + 1000x - 28000 = 0
Using the quadratic formula, we get:
x = 24 or x = -116
Since we can't have negative lobster boats, we only take x = 24. Therefore, X₁ = 24.
The only choice that matches these values is C.