Question

A movie theater estimates that for each $0.50 increase in ticket price, the number of tickets sold decreases by 40. the current ticket price of $8.50 yields 100 tickets sold. Set up an equation to represent the daily revenue as a function of the ticket price. What price should the theater charge for tickets in order to maximize daily revenue?

Answer 1

where are you stuck at, do you have an equation or.. ?

Answer 2

I completed it just not sure if my answer is correct. @DemolisionWolf

Answer 3

awh, ok,
what did you get for " What price should the theater charge for tickets in order to maximize daily revenue?" and i'll work it out real quick to see if its what I get ^_^

Answer 4

I got $5.50 just not sure if it's correct :o and thanks so much!

Answer 5

back! sorry, sooo I get.....

Answer 6

I got $5.0..
what equation did you use?

Answer 7

I'm not too good a business calculus so I might be wrong ^_^
! maybe i'll just do it in a spreadsheet

Answer 8

I used (8.50+x)(100-40x) distributed that and got -40x^2-340x+850

Answer 9

And then I found the vertex using -b/2a and got -3 so I plugged -3 back into (8.50+ -3) and got 5.50

Answer 10

ur probably right! I just worked it out by hand.. boy, sorry for not being much help!

Answer 11

by hand, at 5.50 I get $1870
and at 5.0 i get $1900...

Answer 12

Hehe it's all good! Thanks though! ^_^

Answer 13

wait, should it be:
(8.50+x)(100-40(2x))?

Answer 14

Where do you get the 2x?! :o

Answer 15

see,
if x = 0 then,
(8.50+x)(100-40(2x)) = $850
if x = .5 then = $540
and if we did it by hand, 8.50 + .50 = 9.00 and 100-40 = 60, then 9.00*60=540

Answer 16

yaya, it works it works.
if x = -$1.0 then
(8.50+x)(100-40(2x)) = $1350
which is the same as
8.50-1 = $7.50 and
100+40+40 = 180
$7.5*180 = $1350

Answer 17

well, you're probably over this question, buuut thanks for not being upset with how little i know about business calculus ^_^

Answer 18

Of course! Thanks for trying! :)