3 questions included in description

Question

Optimal Chapter-Flight Fare

 If exactly 208 people sign up for a charter flight, Leisure World Travel Agency charges $310/person. However, if more than 208 people sign up for the flight (assume this is the case), then each fare is reduced by $1 for each additional person. Determine how many passengers will result in a maximum revenue for the travel agency. Hint: Let x denote the number of passengers above 208. Show that the revenue function R is given by R(x) = (208 + x)(310 − x).

1.    passengers
2. What is the maximum revenue?

$  

3.What would be the fare per passenger in this case?

  dollars per passenger

Answer 1

Given x passengers over 208, there will be 208+x passengers.

Each passenger gets a discount of $x, so each pays 310-x.

The total revenue is then the product of passengers and price:

R(x) = (208 + x)(310 - x) = 208×310 + 102x - x²

R(x) is a quadratic function with negative leading coefficient, therefore

R(x) is convex up, therefore its maximum occurs where its derivative is 0.

R'(x) = 102 - 2x = 0

x = 51

The maximum revenue is reached when there are 51 extra passengers, i.e, 259 passengers in total.

Each passenger will pay $310 - $51 = $259,

for total revenue of $259×259 = $67081

By the way, the fact that the number of passengers ended up same as the price is not a coincidence.

In general, quadratic forms like R(x) = a(x)×b(x) are maximized when a(x)=b(x).

So we could have seen the result immediately without any differentiation.