How Do You Calculate Marginal Revenue for Maximum Bus Fare Efficiency?

[salemchic05]

We're in the derivatives chapter, so we assume finding the answer to this requires the use of a derivative, but we are completely lost.

A bus will hold 60 people. The fare charged (p dollars) is related to the number x of people who use the bus by the formula p=[3-(x/40)]^2.

(a) Write a formula for the total revenue per trip received by the bus company. ***Here we got the answer-R(x)=[3-(x/40)]^2(x)

(b) What number of people per trip will make the marginal revenue equal to zero? What is the corresponding fare? ***this is where we thought we should use the derivative and set it equal to zero, but our answer is amillion miles off from the answer given in the book. The answer we are given is 40 and a $4 fare.

 

[Sirus]

Your method is correct and gives the right answer. You must have made an arithmetic error. Check your work.

 

[Andrew Mason]

(a) Write a formula for the total revenue per trip received by the bus company. ***Here we got the answer-R(x)=[3-(x/40)]^2(x)

Would it not be:

Revenue = no. passengers x fare per passenger

$$R = x(3-x/40)^2 = 9x -6x^2/40 + x^3/1600$$

(b) What number of people per trip will make the marginal revenue equal to zero? What is the corresponding fare? ***this is where we thought we should use the derivative and set it equal to zero, but our answer is amillion miles off from the answer given in the book. The answer we are given is 40 and a $4 fare.

You are trying to find the point where the change in revenue per change in unit passenger = 0. How does this relate to the derivative of Revenue with respect to x? For what values of x does the derivative = 0?

AM