Respuesta :
Answer:
$320000
Step-by-step explanation:
Given that on a certain route, an airline carries 5000 passengers per month, each paying $60.
A market survey indicates that for each $1 increase in the ticket price, the airline will lose 50 passengers
Let 1 dollar be increased x times (say)
Original revenue = 60(5000) =300000
After increase of x dollars, revenue would be a function of x
R(x) = [tex]R(x) =(60+x) (5000-50x)\\\\R(x) = 300000+2000x-50x^2[/tex]
Use derivative test to find maximum
[tex]R(x) = 300000+2000x-50x^2\\R'(x) = 2000-100x\\R"(x) = -100<0[/tex]
So maximum when I derivative is 0
i.e. when x = 20
Maximum revenue
[tex]R(20) = (60+20)(5000-1000)\\= 320000[/tex]
The ticket price that will maximize the airline's monthly revenue for the route is $20 and the maximum monthly revenue is $320000
The given parameters are:
- Passengers = 5000
- Unit rate = $60
From the question, a $1 increase will cost a loss of 50 passengers.
So, the revenue function is:
[tex]R(x) = (60 + x) \times (5000 - 50x)[/tex]
Expand the function
[tex]R(x) = 300000 - 3000x + 5000x - 50x^2[/tex]
[tex]R(x) = 300000 +2000x - 50x^2[/tex]
Rewrite the equation as:
[tex]R(x) = - 50x^2+2000x +300000[/tex]
Differentiate
[tex]R'(x) = - 100x+2000[/tex]
Set to 0
[tex] - 100x+2000 = 0[/tex]
Collect like terms
[tex]100x=2000 [/tex]
Solve for x
[tex]x=20[/tex]
Substitute 20 for x in R(x)
[tex]R(20) = - 50 \times 20^2+2000 \times 20 +300000[/tex]
[tex]R(20) = 320000[/tex]
Hence, the ticket price that will maximize the airline's monthly revenue for the route is $20 and the maximum monthly revenue is $320000
Read more about revenue functions at:
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