On a certain route, an airline carries 8000 passengers per month, each paying $70.A market survey indicates that for each $1 increase in the ticket price, the airline will lose 50 passengers. Find the ticket price that will maximize the airline's monthly revenue for the route. What is the maximum monthly revenue?

The maximum revenue is $661,250.

Step-by-step explanation:

The total revenue (R) is the product between the number of passengers (N) and the price of each ticket (P), so the inicial revenue is:

R = N * P = 8000 * 70 = $560,000

For each 1$ increase in the ticket, the airline loses 50 passengers, so if we call "x" the increase in the ticket price, we have that the revenue equation will be:

R = (8000-50*x) * (70+x) = 560000 + 8000*x - 50*70*x -50x2

R = -50*x2 + 4500*x + 560000

The value of x that gives the maximum value of a quadratic function is found using the formula:

x = -b/2a

where a and b are coefficients of the quadratic equation (in our case, a = -50 and b = 4500)

So the value of x that gives the maximum revenue is:

x = -4500 / (-100) = 45

Using this value in the revenue equation, we have that the maximum R is:

R = -50*45^2 + 4500*45 + 560000 = $661,250

(To get this revenue, the ticket will cost 70+45 = $115 and there will be 8000-50*45 = 5750 passengers)

439400

step-by-step explanation:

130^2 = 16900

16900*26 = 439400

1/4 because there is 4 grades.