Let C(x) = 14 -x be the monthly charge for one single customer as the function of the projected decrease of "x" dollars.

Let N(x) = 6300 + 630x be the number of customers as the function of the same variable: the projected decrease of "x" dollars.

Then monthly revenue R is the product R = C*N, or

R = (14-x)*(6300 + 630x),    (1)

and we need to find the maximum of this function.

Write the function (1) in the general form for the quadratic function

R(x) = ,   or

R(x) = .


       Now, let me remind you that for the general quadratic function f(x) =  with the negative coefficient a < 0  
       the theory predicts the maximum  at x = .


In our case the maximum will be at  x =  = 2.

It means that the maximum is predicted at the $2 dollars decreased charge of $14 - $2 = $12.

The number of customers then will be 6300 + 2*630 = 7560,  and the total revenue will be $12*7560 = $90720.

       You can check that this revenue is the same as calculated in accordance with the function R(x) =   at   x = 2:
       R(2) =  = 90720.

Answer.  The value of monthly charge that results in maximum monthly revenue is $12 dollars.