Tickets to a school dance cost $4 and the projected attendance is 300 people. for every $0.10 increase in ticket price, the dance committee projects that attendance will decrease by 5.
using formula r(4+.10t) (300-5t) where t is ticket price.
determine the dance committee's greatest possible revenue. ( show work as in #'s)
what ticket price will produce the greatest revenue? ( show work as in #'s)

Money made (R) = tickets sold * price of ticket
the number of $0.10 increases we make will be called t. so:
price of ticket =  $4 + $0.10t
tickets sold = 300 - 5t
R = (4 + .10t)(300 - 5t)
R = 1200 + 10t - .5t^2
R = -1/2t^2 + 10t + 1200
to find the maximum, we have to find the roots of this function.
0 = -1/2t^2 + 10t + 1200
0 = -t^2 + 20t + 2400
0 = t^2 - 20t - 2400
quadratic formula
t = 20 +- sqrt((-20)^2 - 4(1)(-2400)) all over 2(1)
t = 20+-sqrt(400+9600) all over 2
t = 20+-sqrt(10000) all over 2
t = 20+-100 all over 2
t = -80/2 or t=120/2
t = -40 or 60
now find the average of these roots. this will find the "middle" of the parabola, which is where the vertex (maximum) will be.
-40 + 60 all over 2
20/2
10
the vertex is at t=10. plug this into the formula:
R = -1/2t^2 + 10t + 1200
R = -1/2(10)^2 + 10(10) + 1200
R = 50 + 100 + 1200
R = 1350
therefore, the number of 10 cent price increases should be 10, because that will lead to total profits of $1350, the maximum possible. the ticket price will be $5.

Im sorry but what on