SOLUTION: A ticket to a film showing costs ₱20. At this price, the organizer found out that all 300 seats are filled. The organizers estimates that if the price is increased, the number of

To say that the number of viewers will fall by 50 for every ₱5 increase, is the same as to say that

the number of viewers will fall by 10 for every ₱1 increase.


So, we can write  

    n(p) = 300 - 10(p-20)      (1)

for the number of viewers n as a function of the price of each ticket p.


The revenue R(p) is the product of the ticket's price by the number of viewers

    R(p) = p*(300 - 10(p-20)) = p*(300-10p+200) = p*(500-10p) = -10p^2 + 500p.    (2)


Thus the revenue is a quadratic function of the ticket's price. 

Notice that the function (2) represents a parabola opened downward, which has the maximum.


There are different methods to find the optimal ticket's price.


In this case, I will factor function (2) in this way

    R(p) = -10p*(p-50).    (3)


Formula (3) tells us that the zeroes of the function R(p) are p= 0  and  p= 50.


It is well known fact that a downward parabola has the vertex at the point on x-axis, which is exactly
half way between the zeroes.


So, the parabola (3) has the maximum at  p = 25.


Thus the optimal ticket's price is ₱25.


At this price, the number of viewers is 

    n(25) = see formula (1) = 300 - 10*(25-20) = 300 - 10*5 = 300 - 50 = 250


and the revenue is  

    R(25) = see formula (3) = -10*25*(25-50) = -10*25*(-25) = 6250.


Compare it with the revenue  20*300 = 6000 at the ticket price p= 20.


ANSWER.  The optimal price is ₱25, and it provides the revenue value of ₱6250.