Answer:
-100x^2 + 1,500x + 10,000 >= 12,000
The revenue will maximize at 12.50
The maximum profit the company can make is $15,625.00.
Explanation:
revenue =( ticket) * price
x is the price increase
for each 1 increase we lose 100 passengers, so for each x we increase, we lose 100x passengers
(2000-100x) (5+x)
-100x^2+1500x+10000
This must be greater than 12000
-100x^2 + 1,500x + 10,000 >= 12,000
Subtract 12000 from each side
-100x^2 + 1,500x + 10,000 -120000>= 0
-100x^2 + 1,500x -2,000 >= 0
Taking the derivative
-200x +1500 and setting it equal to zero
-200x +1500=0
-200x=-1500
x = 1500/200
x = 15/2
To find the max value of the ticket take x and add 5 to find the ticket price
The max is at 12.5
The revenue will maximize at 12.50
To find the maximum revenue, put 7.50 into the equation (x=7.5 which is the price increase)
(2000-100(7.5)) (5+7.5)
(2000-750) (12.5)
1250*12.5
15625