Example 2

A company in Edmonton rents Segways to tour the river valley. The company charges $60 per day for each Segway rental. At this price, the company rents 48 Segways a week. An analyst predicts that for every $5 increase in rental price, 2 fewer Segways will be rented each week.

a. Determine the rental price that will optimize revenue and then determine the maximum revenue.

Step 1: Assess the information given.

  • Current: $60/Segway rental per day, 48 rentals/week
  • A rental increase of $5 each will decrease rental numbers by 2 per week.

Step 2: State what you are asked to find.

  • The rental price that will maximize revenue and that maximum revenue.

Step 3: Decide how to determine this value.

Revenue = rental price × number of rentals

Both the rental price and the number of rentals can be represented in terms of the number of $5 price increases. Let p represent the number of $5 price increases. This means

rental price = 60 + 5p
number of rentals = 48 − 2p

Using the new expressions for rental price and number of rentals, write the function that represents the revenue after the increase in price.

The number of price increases that will yield the maximum revenue will correspond to the coordinates of the vertex of the graph of this function (number of price increases, maximum revenue).

The function is in factored form, so you can determine the zeros of the function and then use them to determine the equation of the axis of symmetry, the new rental price, and then the maximum revenue.

The axis of symmetry is halfway between the zeros of the function (or x-intercepts of the graph of the function).

The equation of the axis of symmetry is p = 6 . This means the maximum revenue will occur after 6 price increases.

New Rental Price:

A rental fee of $90.00 per day will maximize revenue.

Substituting the p-value that yields the maximum revenue into the revenue function will give the maximum revenue.

The maximum revenue will be $3240 when the Segways are rented at a price of $90 per day.