Respuesta :
Answer:
The ticket price that maximizes revenue is $50.
The maximum monthly revenue is $250,000.
Step-by-step explanation:
We have to write a function that describes the revenue of the airline.
We know one point of this function: when the price is $30, the amount of passengers is 7000.
We also know that for an increase of $1 in the ticket price, the amount of passengers will decrease by 100.
Then, we can write the revenue as the multiplication of price and passengers:
[tex]R=p\cdot N=(30+x)(7000-x)[/tex]
where x is the variation in the price of the ticket.
Then, if we derive R in function of x, and equal to 0, we will have the value of x that maximizes the revenue.
[tex]R(x)=(30+x)(7000-100x)=30\cdot7000-30\cdot100x+7000x-100x^2\\\\R(x)=-100x^2+(7000-3000)x+210000\\\\R(x)=-100x^2+4000x+210000\\\\\\\dfrac{dR}{dx}=100(-2x)+4000=0\\\\\\200x=4000\\\\x=4000/200=20[/tex]
We know that the increment in price (from the $30 level) that maximizes the revenue is $20, so the price should be:
[tex]p=30+x=30+20=50[/tex]
The maximum monthly revenue is:
[tex]R(x)=(30+x)(7000-100x)\\\\R(20)=(30+20)(7000-100\cdot20)\\\\R(20)=50\cdot5000\\\\R(20)=250000[/tex]
