<PRE><B>Question 201313</B>
an airport shuttle has an average of 1200 passengers per day and charges 15 per ride. Each 1 dollar increase in fare results in a loss of 50 riders. What should the fare be to maximize the revenue?
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Revenue = (# of passengers)(charge per passenger)
Let the dollar increase be x dollars.
Equation:
R(x) = (1200-50x)(15+x)
R(x) = 50(24-x)(15+x)
R(x) = 50(360 + 9x - x^2)
Maximum R occurs when x = -b/2a = -9/-2 = 4.5
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Maximum Revenue occurs when the price is 15 + 4.5 = $19.50
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Cheers,
Stan H. </PRE>