Question

A football team can sell 10,000 tickets at $10 each. For each increase in price of $.25, the demand for tickets decreases by 200 tickets. How much should the team charge per ticket to maximize the revenue? (Only a calculus solution with appropriate derivative testing is acceptable) How much is the maximum revenue and how many tickets were sold to reach that maximum revenue?

Answer 1

same idea as last time, just different numbers

starting at $10 a ticket, we have 10,000 people that want them and are willing to pay for them

Answer 2

as we increase the price, the amount of people that want them drops by 200 each time

Answer 3

if you increase the price x times, then your final price will be

y = 0.25x + 10

at the same time, 200x people will leave if you increase the price x times

so you go from 10,000 people to 10000 - 200x people

Answer 4

Revenue = (# of tickets sold)*(price per ticket)

R(x) = (10000 - 200x)(0.25x + 10)

R(x) = 10000(0.25x + 10) - 200x(0.25x + 10)

R(x) = 2500x + 100000 - 50x^2 - 2000x

R(x) = -50*x^2+500x+100000

Answer 5

let me know when you want me to continue

Answer 6

Since it is the same general idea I can sort of follow a little faster, it seems fairly similar my problem just seems to be creating the equations to begin with

Answer 7

yeah it's all the same idea, just different numbers...that's all that's new

Answer 8

I think they just wanted you to practice with this sort of thing...or just give you busy work lol

Answer 9

so when you derive R(x) with respect to x, what do you get?

R(x) = -50x^2+500x+100000

R ' (x) = ??

Answer 10

-100x+500

Answer 11

good, now set that equal to zero and solve for x

Answer 12

to find your critical value

Answer 13

okay, I got 5!

Answer 14

same here

Answer 15

so if they increase the price exactly 5 times, they will max out the revenue

Answer 16

the tickets will be

y = 0.25x + 10

y = 0.25(5) + 10

y = 11.25

dollars each

Answer 17

and there will be

y = 10000 - 200x

y = 10000 - 200(5)

y = 9000

people that will buy the tickets

Answer 18

this will make the max revenue to be

R(x) = -50x^2+500x+100000

R(5) = -50(5)^2+500(5)+100000

R(5) = 101,250 dollars

Answer 19

okay and I already was able to answer the last question! I'll post the last one and then we will be on our way!!