SOLUTION: Alex is selling tickets. If she charges $100 she can sell 2000 tickets. For every $3 increase in price she sells 20 fewer tickets. What should she charge to maximize her revenue?

After increasing the price n times, the price per ticket is 100+3n dollars,
but the number of sold tickets is 2000-20n.


Thus, after changing the price n times, the revenue is

      R(n) = (2000-20n)*(100+3n) = 200000 - 2000n + 6000n - 60n^2 = -60n^2 + 4000n + 200000.    (1)


We want to find "n" in a way to maximize quadratic function (1).


   +-------------------------------------------------------------------+
   |       The maximum is at  n = "  ",  where "a"               |
   |    is the coefficient at n^2 and "b" is the coefficient at n.     |
   +-------------------------------------------------------------------+


In our problem,  a= -60,  b= 4000,  so the desired n is 

    n =  =  =  =  = 33.



Next, the number of n in this problem/solution not necessary should be integer,

but the numbers 100+3n and 2000-3n must be integer numbers, due to their meaning.

         (100+3n must be integer number of cents, at least.)



We have 100+3n = 200  is still integer,  but 200-20n  is not integer at n= 33.


THEREFORE, we should test the closest integers n= 33 and n= 34 as possible candidates.


We have  the revenue R(33) = -60*33^2 + 4000*33 + 200000 = 266660 dollars,

         the revenue R(34) = -60*34^2 + 4000*34 + 200000 = 266640 dollars.


The revenue R(33) is higher, so we choose n= 33 as the number of increases.


It gives the optimum price  100 + 33*3 = 199 dollars.


ANSWER.  Optimum price is 199 dollars per ticket.