Math Question need Help Thank You?

Question

i have this word problem that i dont understand its asking ( a manager at a movie theathre sells tickets for 8.50 $ and at this price 600 people come. He notices that a 0.25 $ decrease in price will result in 30 more people coming. What whould be the ticket price that will give the largest revenue, and what will the revenue be?) Help would be appreciated Thank You

Answer 1

simplest thing you can do is compare

600 * 8.50
630 * 8.25
660 * 8.00

etc., and see which is largest

but if calculus is to be involved, let's think of the revenue function:

p = number of people
c = cost of ticket

F = p * c

now we know that

p = 600 + 30 * x
c = 8.50 - 0.25 * x

so in all,

F = p * c = (600 + 30x) * (8.50 - 0.25x) =
5100 - 150x + 255x - 7.5x^2 =
-7.5x^2 + 105x + 5100

F'(x) =-15x + 105

for maximum of function, F'(x)=0 ==> x=7

since F''(x) = -15 < 0, it's a maximum indeed.


SO!!!!

x=7, so

price of ticket is 8.50 - 7*0.25 = 6.75
number of people = 600 + 30*7 = 810

revenue = 6.75 * 810 = 5467.5


cheers!

Answer 2

Presumably the intent is that every $0.25 decrease in the price will result in 30 more people coming.
Suppose we lower the price by $0.25 some number x of times. E.g. if x = 2 we lower the price by $0.50.
Then the ticket price will be $8.50 - x ($0.25).
For each decrease in price we get 30 more people attending, so we will have x times 30 extra people. So the number of patrons is 600 + 30x.

The revenue is the number of patrons multiplied by the ticket price, so it is
R = (600 + 30x) (8.5 - 0.25x)
= 5100 + 105x - 7.5x^2

This is a quadratic with negative x^2 term, so we know it will have a maximum value at x = -b/2a = -105 / -15 = 7.
So the ticket price is $8.50 - x ($0.25) = $8.50 - 7($0.25)
= $6.75
and the number of patrons is 600 + 30×7 = 810
for a total revenue of 810×$6.75 = $5467.50

Answer 3

Look at it this way.
We have total revenue now of 600*$8.50.
Let's suppose we lower the price in n lots of $0.25.
Then we have revenue of (600+30*n) * ($8.50-$0.25*n).
We need to maximise this.
You can do this by calculus (that's the best), or, if you aren't at that level, by numerical approximation, that is, trying different values for n.

Answer 4

Carefull they only asked which ticket price the $8.50 or $8.25 will give the most revenue, in this case it is $8.25.

Also if it is a Kevin Costner movie then your revenue will drop significantly.

Answer 5

basically it's saying that every time the price drops $.25, 30 more people will come. you just need to find the greatest product of (8.5 - .25x)*(600 + 30x), x being the number of price drops

the answer is when the price is $6.75 and 810 people will come causing the net profit to be $5467.50

Answer 6

just multiply 8.50 by 600 and multiply 8.25 by 630. the higher number is the right one. but saving .25 cents wouldn't get me into a movie. i have no idea what town this guy lives in.

Answer 7

revenue =price*number
r=p*n
n=-(30/0.25)p+A
600=-120*8.5+A
600=-1020+A
A=1620
n=-120p+1620
r=p*n
r=p*(-120p+1620)
r=-120p^2+1620p
r'=-240p+1620 for max, r'=0
0=-240p+1620
240p=1620
p=6.75
optimal price is $6.75
Max revenue is
r=6.75*(-120*6.75+1620)
r=6.75*(-810+1620)
r=6.75*810
r=$5,467.50

There will be 810 people at this price so it assumes that there are at least 810 seats available.

Answer 8

ITS SUMMER,ISNT IT?