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

VerifiedLet $x$ be the number of $\$1$ increase in fare. So, the fare is represented by $20+x$ and the number of people taking the ferry is $500-20x$.

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

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

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

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

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

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

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

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

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

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