A tour bus carries tourists through the historic district of Savannah Georgia, serving 300 customers a day. The charge is $8.00 per person. The company owner estimates that the company would lose 20 passengers a day for each $1 increase in the fare. What charge would be most profitable for the tour bus company?
This is a problem of finding the optimal price that maximizes the revenue for the tour bus company. The revenue is the product of the price and the number of passengers. Let x be the number of dollars that the price is increased from $8.00. Then the price is 8 + (1)x and the number of passengers is 300 - 20x. The revenue function is R(x) = (8 + x)(300 - 20x). To find the maximum value of R(x), we can use calculus or complete the square. I will use the latter method.
R(x) = (8 + x)(300 - 20x) = -20x2 + 140x + 2400
R(x) = -20(x2 - 7x - 120)
R(x) = -20(x2 - 7x + ) + 2400 +
Remember, to complete the square, you take half of b, which is -7, and square it and add accordingly further so the completion and the addition are zero:
R(x) = -20(x2 - 7x + 49/4) + 2400 + 20(49/4))
R(x) = -20(x - 7/2)2 + 2400 + 245
R(x) = -20(x - 3.5)2 + 2645
The revenue function is a downward parabola with a vertex at (3.5, 2645). This means that the maximum revenue is $2645 when x = 3.5. Now remember that x is the price increase. Therefore, the most profitable charge for the tour bus company is calculated by adding the inital price with the increase: $8.00 + $3.50 = $11.50. So $11.50 per person is the most profitable charge.