Revenue maximization (optimization)

Question

A ferry service can carry up to 450 cars/day but presently carries 400 cars daily at $12/cars. Marketing thinks that each decrease in price of $1 will increase ridership by 20 cars. If this is true what price will maximize ferry revenue? How will this revenue compare to their present revenue?

what do you suggest that they do with excess capacity?

Answer 1

Both the price and the ridership will be determined by how many $1 decreases in price we have, so let's let x = the number of $1 decreases in price. The revenue is determined by price · number of cars, and where each quantity will change depending on the value of x ( i.e. price will decrease by 1x and ridership will increase by 20x), we can write our general revenue function as follows: R(x) = ( 12 - 1x )( 400 + 20x ). If we eliminate the (), we have R(x) = 4800 - 160x - 20x² or, if we write it in standard form, R(x) = -20x² - 160x + 4800.

This is a downward-opening parabola, so our maximum should be at the vertex. Let's take the derivative of R to find where the curve "levels off". R '(x) = -40x - 160. Setting this equal to 0, we have -40x - 160 = 0, and solving this for x, we have x = -4. To confirm this as a maximum, let's check R '(x) at convenient values on either side of x = -4. At x = -5, the derivative is 40, so a positive slope there, and checking at x = -3, the derivative is -40, so a negative slope there, so that confirms the maximum at x = -4. Checking this value in our R(x) function produces a revenue of $5120, where as the current revenue is $4800.

So with x = -4 and our price represented as ( 12 - 1x ), this gives us a price of $16 and a ridership of 320 cars, producing a maximum revenue of $5120.

With the extra capacity, perhaps they could allow bikers to ride the ferry for a reduced rate, or use the space for cargo transport for private business. Just a couple of thoughts on that last part!