a) Let $x be the price of a ticket.
Price increase = $(x - 20).
For $1 increase in price, 15 fewer tickets are sold.
For $1(x - 20) increase in price, 15(x - 20) fewer tickets are sold.
Therefore if the price is $x, then the number of tickets sold would be 750 - 15(x - 20), which is 1050 - 15x upon simplification.
The revenue is equal to (number of tickets sold) * (price of each ticket), which is $x(1050 - 15x) = $15x(70 - x).
You can find the maximum revenue by completing the square.
b) is just similar.