maximize p=x(30-x)-100-6x subject to
x >= 10
x <= 20
the profit is revenue-cost, so
p(x) = x(30-(x-10)) - 100 - 6x = -x^2+34x-100
dp/dx = 34-2x
dp/dx=0 at x=17
You own a small airplane that holds a maximum of 20 passengers. It costs you $100 per flight from
St. Thomas to St. Croix for gas and wages plus an additional $6
per passenger for the extra gas
required by the extra weight. The charge per passenger is $30 each if 10 people charter your plane
(10 is the minimum number you will fly), and this charge is reduced by $1 per passenger for
each passenger over 10 who travels (that is, if 11 fly they each pay $29, if 12 fly they each pay $28,
etc.). What number of passengers on a flight will maximize your profit?
1 answer
1 answer