On a certain route, an airline carries 5000 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?

$85

Step-by-step explanation:

5000 x $70 = $350,000

4950 x $71 = $351,450

4900 x $72 = $352,800

etc.

Create an equation for the monthly revenue:

f(n) = (5000 - 50n) x (70 + n)

     = 350000 + 5000n - 3500n - 50n²

     = 350000 + 1500n - 50n²

where n is the amount the ticket price is increased from $70.

Differentiate, equal to zero and solve for n to find the value of n when the monthly revenue is at its maximum:

f'(n) = 1500 - 100n

1500 - 100n = 0

1500 = 100n

15 = n

Therefore, the maximum monthly revenue is when n = 15.

So the cost of the ticket will be 70 + n = 70 + 15 = 85

  y= -3x is the required function.

step-by-step explanation:

we have been given a table:

x             y

0             0

1             -3

3             -9

-2             6

we can see that the values of y is -3 times the value of x

hence, the function would be: y= -3x.

put the values of x will give the same y as in the table that means function is satisfying the given table.