An airline finds that if it prices a cross-country ticket at $300, it will sell 200 tickets per day. it estimates that each $10 price reduction will result in 50 more tickets sold per day. find the ticket price (and the number of tickets sold) that will maximize the airline's revenue

Ticket price = $170

Number of tickets sold = 850

Step-by-step explanation:

Let X= the number of $10 price reduction and the number of 50 tickets increase

Let p= price per ticket

Let q= number of tickets sold

Let R= revenue

Let p(x) = price of ticket now sold

Let q(x) = number of tickets now sold

R(x) = p(x) * q(x)

P(x) = 300 - 10x

q(x)= 200 + 50x

R(x) = (300 -10x) (200 - 50x)

= 60000 + 15000x - 2000x - 500x^2

= 60000 + 13000x - 500x^2

Differentiate with respect to x

dR(x)/dx = 13000-1000x

Put dR(x)/dx = 0

0= 13000 - 1000x

1000x = 13000

Divide through by 1000

1000x/1000 = 13000/1000

x = 13

Therefore x = 13 for $10 price reduction and 50 tickets increase

p(x) = 300 - 10x

= 300 - 10*13

= 300 - 130

= $170

q(x) = 200 + 50x

= 200 + 50*13

= 200 + 650

= 650

Maximum Revenue = p(x) * q(x)

= 170 * 850

= $144,500

For 13 reductions, we have $170 price reduction and 850 ticket sales

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.