<PRE><B>Question 74196</B>
Can you please help me solve: A ferry service transporting passengers to an island charges a fare of $10 and carries 300 persons per day. The manager estimates that the company will lose 15 passengers for each increase of $1 in the fare. Find the fare that yields the greatest income.
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Let &quot;x&quot; be the number of dollars the fare increases.
Then Income to the company =(10+x)(300-15x)
I(x) = -15x^2 +285x+3000
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This is a quadratic with a=-15 ; b=285 ; c=3000
Maximum income occurs when x=-b/2a = -285/-30=9.5
The fare when x=$9.50 is $19.50.
The Income will be maximum when fare is $19.50
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Cheers,
Stan H.</PRE>