Question 611245
x = number of $10 tickets sold
y = number of $14 tickets sold

x = 3y

10x = revenue from cheaper tickets
14y = revenue from expensive tickets

Substituting:

10(3y) = 30y

30y > 14y for all real non-zero positive y, cheaper tickets will always generate more revenue, since you sell 3 times as many, the problem is unsolvable