This algebra problem has me stumped:
A plane with 125 seats may be chartered for $250 per person plus a fee of $6 per person for each empty seat. What number of passengers will provide the maximum income.
I have 250x + 6x as my initial equation, but that is not leading anywhere. What am I not thinking of?
Try 250x + 6*(125-x).
Trick question…? Wouldn’t it be 0 passengers (or 1 to be practical), since each empty seat earns $256?
Or does each empty seat earn only $6?
Obviously the latter. The problem wasn’t stated very well, but that was the intended meaning.
Maybe this is what the problem is looking for? It leads to a more interesting result, but is a very weird way to charge for the empty seats.
250x + 6x*(125-x)
(in which case, the maximum income would come from simply filling up all the seats with people at $250 each, no?)
Is there some other condition/limit?
(Sorry, my second reply was meant to reply to my first reply.)
Anyway, thanks nivlac, but in that case… why would you ever want less than a full load of passengers (125 people)?
I was interpreting the problem to mean that if 123 seats fill then each seat costs $250 + 12 = 262. If 120 seats sell then each passenger pays 250 + 30 = 280.
hmmm.
= 250x + 750x - 6x[sup]2[/sup]
= 1000x - 6x[sup]2[/sup]
= - 6x[sup]2[/sup] + 1000x
Factor and set equal to zero:
0 = x(1000 - 6x)
x = 0 and x = 166(.666)
(Here comes the half-assed part.)
These are the zeroes of an inverted parabola (where it crosses the y-axis). Thus the max point is halfway between, or at 83.333.
83 passengers: 1000(83) - 6(83)(83) = $41,666
Let’s check either side:
82 passengers: 1000(82) - 6(82)(82) = $41,656
84 passengers: 1000(84) - 6(84)(84) = $41,664
So 83 is the max (since you can’t have a fractional passenger).
This solution will not earn you any points with an algebra teacher who wants you to show your work (she said from bitter experience).
I’ll welcome any holes that anyone would care to shoot in it. I forget how you’re SUPPOSED to solve these.
The answer is 83 if you want to check your answer. Pochacco had the right expression for the price per ticket, but you need to multiply it with the number of passengers to get a revenue function. Now find its maximum. Hint: It’s a parabola.
Beat ya.
I got a different answer.
P(n) = (250 + 6n)(125 - n)
= 31,250 + 725n -6n^2
Maximize with -b/2a = -725/-6 = 60.4 or 60
P(60) = -6(60)^2 +725(60) +31,250
= 53,150
I think that is right. Am I correct?
Yeah, but my solution is more elegant. 
(Once you know it’s an upside down parabola, the optimal solution occurs at the vertex. In this case, it’s x = 83.3. The optimal integral solution has to be 83 since it’s the closest integer to 83.3, and you’re dealing with a parabola.)
Nope, your formula’s wrong.
125 - n is the number of empty seats.
You’re charging $250 + $6(number of paying passengers) for each empty seat.
That’s just wrong, son.
(Should be 250n + 6n(125 - n). You’ve got one extra set of parens.)
Touché.
Yah, that’s what I said! 
No. If n = 60, you get 65 paying seats, which leads to 606 + 250 for each of the 65 passengers, which is 61065 = $39,650.
Scarlett67, the formula’s right if ‘n’ is empty seats, which I think it is.
No, that’s still not right.
You get a max cost of $39,650 for 60 passengers.
I get a max cost of $41,666 for 83 passengers.
$41,666 > $39,650.
Look again at the correct formula and what it represents.
$250 per passenger = 250n
PLUS
$6 per passenger per empty seat = 6(n)(125 - n)
EQUALS
250n + 6n(125 - n)
Note that the 250n is not grouped with the 6n, as dauerbach had it. The number of empty seats (125 - n) is multiplied by the $6/passenger (6n) only. That’s completely separate from the base price of $250 per passenger, which is why they’re added, not multiplied.
Boy, I REALLY want to procrastinate tonight. 
The formula’s right. dauerbach’s maximization isn’t.
Cost per seat = $250 base + $6 per empty seat.
Let number of empty seats be ‘n’.
Therefore, number of occupied seats = ‘125 - n’
Total price = Price for each seat x no. of occupied seats
= ($250 + $6 x n) x (125 - n)
Ok, I see where my error was. And Scarlett67, as far as my teacher is concerned showing the work is maximizing with -b/2a which comes to 83.3 and then substituting in p(83) = -6(83)^2 + 1000(83) which just happens to give the exact same answer as you got. Thank you very much for the help. Best of all, I actually understand it now.
Aargh, OK, I see.
Part of the problem is dauerbach’s choice of what n represents. Normally you would choose n to be the value asked for in the problem (passengers) but he’s chosen it to represent the number of empty seats. Teacher might ding for that.
<scribble scribble>
OK, dauerbach, check your expansion of that polynomial again:
Your outside and inside terms are (-250n) + (125)(6n) = -250n + 750n = 500n.
So -b/2a = -500/-12 = 41.666 = 42 empty seats.
Or 83 occupied ones.
Thanks for the mental gymnastics session!