Question

1) When a theater owner charges $5 for admission, 100 people attend. For every $.20 increase in admission, there is a loss of 1 customer. What admission should be charged in order to maximize revenue? (Only a calculus solution with appropriate derivative testing is acceptable) How much is the maximum revenue and how many people attend?

Answer 1

let x = number of increases in admission

Answer 2

so if there are no increases in admission, then the cost is y = 5

if there is one increase in admission (of $0.20), then the cost is now y = 5+0.2 = 5.2

if there are two increases in admission(of $0.20), then the cost is now y = 5+2*0.2 = 5.4

etc etc

Answer 3

in general, the cost y is

y = 5 + 0.2x

or

y = 0.2x + 5

where x is the number of increases in admission

Answer 4

now if there are 100 people when the price is $5

and 1 person leaves everytime you bump the price up $0.20, then

after x price increases you will have 100-x people left

Answer 5

So the revenue is

R = (number of people)*(price)

R = (100-x)*(0.2x + 5)

R = 100*(0.2x + 5) - x*(0.2x + 5)

R = 20x + 500 - 0.2x^2 - 5x

R = -0.2x^2 + 15x + 500

Answer 6

Let me know if this is making sense so far or not

Answer 7

The revenue function is

R(x) = -0.2x^2 + 15x + 500

the derivative of the revenue function is

R(x) = -0.2x^2 + 15x + 500

R ' (x) = -0.4x + 15

Answer 8

Set the revenue function equal to zero and solve for x

R ' (x) = -0.4x + 15

0 = -0.4x + 15

-15 = -0.4x

-15/(-0.4) = x

37.5 = x

x = 37.5

So the max revenue occurs when you increase the price 37.5 times




Note: In the real world, you cannot increase the price a fraction of the time, so you would round to the nearest integer to get 37 or 38.


So increasing it 37 or 38 times will lead you to the max revenue

Answer 9

okay so far so good!

Answer 10

But let's say we go with x = 37.5

When x = 37.5, R(x) is...

R(x) = -0.2x^2 + 15x + 500

R(37.5) = -0.2(37.5)^2 + 15(37.5) + 500

R(37.5) = 781.25

This means that theoretically your max revenue is $781.25

However, you can only increase it a whole number of times, so you would either increase it x = 37 or x = 38 times

If x = 37, then

R(x) = -0.2x^2 + 15x + 500

R(37) = -0.2(37)^2 + 15(37) + 500

R(37) = 781.2

This means that the more realistic max revenue is $781.20

Answer 11

So if you were making a presentation on this to your boss, then you would recommend that s/he raise the price by $0.20 a total of 37 times and that will lead to a max revenue of $781.20

If your boss raises it 38 times, then the max revenue will still be $781.20...but the revenue will go down after this point. So 38 is the highest you should go

Answer 12

everything making sense?

Answer 13

yes it is making sense so far I am just trying to see what the admission price that should be charged is? I see the max revenue just not the price we should charge per ticket

Answer 14

37*.20+5 for price ticket?

Answer 15

well if you increase the price 37 times, then you start out at 5 dollars and you add on 37(0.2) dollars

ie

y = 0.2x + 5

y = 0.2(37) + 5

Answer 16

exactly you got it

Answer 17

evaluate that out to find the price per ticket

Answer 18

okay!

Answer 19

12.60

Answer 20

close

Answer 21

12.40

Answer 22

off by 20 cents

Answer 23

much better

Answer 24

I did 38 the first time!!

Answer 25

well it turns out that x = 37 and x = 38 give the same R(x) value

Answer 26

so that explains why you're able to use x = 37 or x = 38

I guess to be fair to the customer (and make the company's image look better), you should go with x = 37

Answer 27

it's the smarter of the two decisions

Answer 28

although technically you can still use x = 38 and still get the max revenue just fine

Answer 29

okay perfect

Answer 30

So the first two questions are answered in order to get how many people attend all I do is divide the max revenue by the price per ticket?

Answer 31

or you can plug x = 37 (or x = 38, depending on which x you use) into 100 - x

Answer 32

remember you start with 100 people when the price is $5

Answer 33

well that is easier, haha

Answer 34

then for every increase, 1 person leaves

so x increases leaves you with 100 - x people

Answer 35

either way works

Answer 36

wonderful, okay ill post the next

Answer 37

alright