SOLUTION: Hello,the problem I'm trying to solve is, "A plane with a capacity of 120 passengers is to be chartered for an excursion. The price of one ticket is to be $60 if 100 people or f

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hello,the problem I'm trying to solve is, "A plane with a capacity of 120 passengers is to be chartered for an excursion. The price of one ticket is to be $60 if 100 people or f      Log On


   

Question 683794: Hello,the problem I'm trying to solve is,
"A plane with a capacity of 120 passengers is to be chartered for an excursion. The price of one ticket is to be $60 if 100 people or fewer buy tickets. The airline will reduce the price of every ticket by 50 cents for each ticket sold over 100. How many additional passengers over 100 will bring about maximum income?"
I know the first part of the equation will probably be (120 + 100x), but I'm not to sure about the rest of the equation which I need in order to graph the parabola. Help would be greatly appriciated. Thanks.



Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the income will go up linearly for each passenger until the 100th one, where the income would be $60+%2A+100+=+6000

after that, evidently, the next passenger would pay only $59.50, and the 100 other passengers would also pay $0.50 less
let's put that in equation:
Income+=+%2860+-%28n-100%29%2F2%29%2An
where "n" is the number of passenger
note that this equation applies only for "n" greater or equal to 100
the %28n-100%29%2F2 is the bit that takes away 50 cents for each passenger above the 100th; and the "n" is the number of passenger
A bit of algebra turns the income equation to
income=110n+-n%5E2%2F2+
Now, we want to maximize the income. Calculus to the rescue, as the local maximum implies that the derivative of the equation equals zero.
The derivative of the equation is
110+-+n
If we put this equal to zero, then n has to be 110.
That is thus the number of passengers that maximizes income.
Plugging n=110 into the income equation means that said income would be
$6050.
And we can easily verify that this is indeed the maximum, by seeing what the income would be for 109 passengers (it would be $6049.50) and 111 passengers ($6049.50 also).