
Bradford T. answered 10/17/21
Retired Engineer / Upper level math instructor
Revenue = quantity×price = QP
Let x = the extra passengers over 194
Quantity, Q = 194+x
The price, P, is reduced by $1 for extra passenger or 306-x
R(x) = QP = (194+x)(306-x)
To find the maximum, take the derivative of the revenue, set that to zero and solve for x.
R'(x)=112-2x
112-2x=0
x=56
1) 56 passengers
2) R(56) = (250)(250)= $62500
3) R(56)/(194+56) = 62500/(194+56) = $250 per passenger