This is a quadratics word problem and i need a little help solving it.?

Question

A movie theatre sells tickets for $8.50 each. The manager is considering raising the prices but knows for every 50 cents the price is raised, 20 fewer people go to the movies. The equation R=-40c^2+84c describes the relationship between the cost of tickets, c dollars and the amount of revenue, R dollars, that the theatre makes. What price should the theatre charge to maximize revenue?
This question has me really stumped and I don't know quite where to begin any help would be greatly appreciated.

Answer 1

To maximize the revenue, you need to differentiate the function once, and set R = 0 to obtain c.

Hence,
R = -40c^2 + 84c
R' = -80c + 84
To check for clarification of maximum value, differentiate for the second time, R'' to get:
R'' = -80 (<0) indicate a maximum value

Thus, maximising R: Let R' = 0
0 = -80c + 84
-84 = -80c
c = 1.05

Substitute c = 1.05 into original equation, R = -40c^2 + 84c
R = -40*(1.05^2) + (84*1.05)
R = $44.10
This is the maximum revenue that the theatre will get.

Answer 2

Well, all you have to do is find c so that R=-40c^2+84c is maximum. This is a polynomial of degree 2 and since it's leading term is negative, i has a maximum at c* = - (84)/(2*(-40)) = =84/80 = 1.05 US$.

It's not true that for every 50 cents the price is raised, 20 fewer people go to the movies. R is a quadratic function of c, and so variationa in R are not prportional to variations in c.

Answer 3

R=-40c^2+84c
R'=80c+84=0
80c=-84
c=-84/80=-1.05
which makes absolutely no sense. There has to be an error in the problem statement. Check & send me the correction.

Answer 4

try to graph the equation
it should face down/open downward
the max point is the max revenue
to get that point, set R=0, find C using quadratic formula
you will have 2 pts, say a and b.
(a+b)/2=d
put d in the equation and get max R

i think...