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.