装-900-a, gives 9000-200Xg=0.0rX2=45. Substituting X into the profit equation for Zone B gives: πB=(100)(100)45)-45).(1,000)(45=$202,500 Total profit from both zones is $653,750.with 47.5 boats in Zone A and 45 boats in Zone B.Because each additional boat above 92.5 decreases total profit,the government should not grant any more licenses d dX X  B 2 = 9 000 − 200 2 , . Setting d dX  B 2 equal to zero to find the profit-maximizing level of output gives 9,000 - 200X2 = 0, or X2 = 45. Substituting X2 into the profit equation for Zone B gives: B = (100)((100)(45) - 452 ) - (1,000)(45) = $202,500. Total profit from both zones is $653,750, with 47.5 boats in Zone A and 45 boats in Zone B. Because each additional boat above 92.5 decreases total profit, the government should not grant any more licenses