Question

Suppose it costs $36 for each lobster trap set. Lobsters sell for $12. If X traps are set, the harvest rate of lobsters, L, as a function of the number of traps, is given by:

L=42X - X^2

a. With no restrictions (competitive markets exist) on the number of traps, and open access to the waters, how many traps will be set? How many lobsters will be harvested? How much profit will be realized from lobstering?

b. Suppose lobstermen could limit the number of traps permitted. How many traps should be permitted, if Australia wanted to maximize overall profits from lobstering? How many lobsters would be harvested? What would total profits be from lobstering? Note– prices are fixed at $12

Answer 1

So, here the price of lobster is “$12” and the price of lobster trap is “$36”, => the profit function is give by.

=> A = 12*L - 36*x, => A = 12*(42*x-x^2) - 36*x, => A = 504*x - 12*x^2 - 36*x = 468*x - 12*x^2.

=> A = 468*x - 12*x^2, => the FOC require “dA/dx=0”, => 468 - 24*x = 0, => x = 468/24 = 19.5. So, the optimum number of lobster trap is “x=19.5”, => L = 42*x-x^2 = 42*19.5 – 19.5^2 = 438.75 = 439 = L.

So, the optimum numbers of lobsters harvested is “L*= 439”.

=> A = 468*x - 12*x^2 = 468*19.5 - 12*19.5x^2 = 4,563, => “A* = 4,563”.

b).

Now, let’s assume that there are “N” lobsterman, => the new profit function is given by.

=> A = 12*N*L - 36*x = 12*N*L - 36*x, => A = 12*N*(42*x - x^2) - 36*x.

A = 504*N*x – 12*N*x^2 - 36*x, => FOC require “dA/dx=0”.

=> 504*N - 24*N*x – 36 = 0, => x = (504*N – 36)/24N = (126*N – 9)/6N = (42*N – 3)/2N.

=> x = 21 – 3/2N, be the optimum level of trap. So, the total lobster harvested is “N*L”.

=> NL = N*(42*x - x^2) = 42N*x – N*x^2, => NL = 42N*(21- 3/2N) – N*(21- 3/2N)^2.

=> NL = (882*N- 63) – N*(21- 3/2N)^2 = 838N-9/4N, be the optimum level of harvesting.

Now, the total profit is given by.

=> A = 12*NL – 36*x = 12*(383*N-9/4N) – 36*(21-3/2N) = (4,596*N-36/N) – (756-54/N).

=> A= 4,596*N - 36/N - 756 + 54/N = 4,596*N - 756 + 18/N = A, be the optimum profit.