Determine how many passengers will result in a maximum revenue for the travel agency.

Question

If exactly 206 people sign up for a charter flight, Leisure World Travel Agency charges $290/person. However, if more than 206 people sign up for the flight (assume this is the case), then every fare is reduced by $1 times the number of passengers above 206. Determine how many passengers will result in a maximum revenue for the travel agency. Hint: Let x denote the number of passengers above 206. Show that the revenue function R is given by
R(x) = (206 + x)(290 − x).

 

Answer 1

Assume that 206+x people sign up for the flight, where x is the number of people over 206.  For example, if 220 people signed up, x would be 14 (206+14 = 220).  The cost per person is $290 but goes down $1 for everyone for each person over 206; that is for x people.  Hence the cost per person is ($290 - $1*x) = (290-x).  Revenue is the number of people (206+x) times the cost per person (290-x):

 

R(x) = (206+x)($290-x)

R(x) = -x² + 84x + 59,740

 

You don't need calculus to solve this.  The revenue equation is a quadratic equation, so its graph is a parabola.  Since the coefficient of the x² term is negative, the parabola is inverted with the vertex at the top.  The vertex is thus the maximum value of the revenue.  The vertex is located at the point x = -b/2a where b is the coefficient of the x term (84) and a is the coefficient of the x² term (-1).  Hence x = -b/2a = -84/(-2) = 42.  Hence the optimal number of passengers is 206+42=248.

 

If you want to use calculus, take the derivative of R(x) wrt x, set it to zero, and solve for x:

 

R(x) = -x² + 84x + 59,740

dR(x)/dx = -2x + 84

0 = -2x + 84

2x = 84

x = 42

 

So the number of passengers with the max revenue is 206+42 = 248.