Question

A ferry operator takes tourists to an island. The operator carries an average of 500 people per day for a round-trip fare of $20. The operator estimates that for each$1 increase in fare, 20 fewer people will take the trip. What fare will maximize the number of people taking the ferry?

Solution

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Let xx be the number of $1\$1 increase in fare. So, the fare is represented by 20+x20+x and the number of people taking the ferry is 50020x500-20x.

The revenue is the product of number of people taking the ferry and the fare:

R=(50020x)(20+x)R= (500-20x) (20+x)

Write in y=ax2+bx+cy=ax^2+bx+c form:

R=10000+100x20x2R= 10000+100x-20x^2

R=20x2+100x+10000R=-20x^2+100x+10000

The maximum revenue occurs at the vertex which has an xx-coordinate of b2a-\frac{b}{2a}, Here, a=20a=-20 and b=100b=100 so:

x=1002(20)=2.5x=-\dfrac{100}{2(-20)}=2.5

So, the fare that will maximize the number of people taking the ferry is:

20+2.5=22.5$22.5020+2.5=22.5\to \color{#c34632}\$22.50

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