Question 928652: The girls’ softball team is sponsoring a fund-raising trip to see a professional baseball game. They charter a 60-passenger bus for $525. In order to make a profit, they will charge $15 per person if all seats on the bus are sold, but for each empty seat, they will increase the price by $1.50 per person.
What is the minimum number of passengers needed in order for the softball team not to lose money?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! The girls’ softball team is sponsoring a fund-raising trip to see a professional baseball game.
They charter a 60-passenger bus for $525.
In order to make a profit, they will charge $15 per person if all seats on the bus are sold, but for each empty seat, they will increase the price by $1.50 per person.
What is the minimum number of passengers needed in order for the softball team not to lose money?
:
let x = no. empty seats on the bus
Also
let x = no. of $1.50 increases for each empty seat
:
f(x) = revenue - expense
f(x) = (60-x)(15+1.5x) - 525
FOIL
f(x) = 900 + 90x - 15x -1.5x^2 - 525
combine like terms to form a quadratic equation
f(x) = -1.5x^2 + 75x + 375
Breakeven points: f(x) = -0
-1.5x^2 + 75x + 375 = 0
Simplify, divide by -1.5
x^2 - 50x - 250 = 0
Using the quadratic formula, got a positive solution:
x = 54.58 we have to round it down to not lose money, therefore
x = 54 empty seats, only 6 paying passengers the minimum to not lose money
:
:
We can confirm this, 54 empty seats means the cost for the remaining 6 will be
15 + (54*1.5) = $96 per person
6 * 96 = $576, they would make $51
However 55 empty seats, only 5 passengers
15 + (55*1.5) = $97.50 per person
5 * 97.5 = $487, they would lose about $38
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Did all this make sense to you? Any questions? ankor@att.net
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