If 3x is the increase in quantity corresponding to a decrease in price of 250x, then we can see that the profit, P=(30+3x)(4500-250x)=(3(10+x))(250(18-x))=
750(10+x)(18-x)=750(180+8x-x²).
We need the maximum value of 180+8x-x². We can use calculus to find the maximum value, but we can also work it out by writing this as -(-180-8x+x²) = -(x²-8x+16-16-180) = 196-(x-4)². This has a maximum value of 196 when x=4. (The maximum profit corresponding to x=4 is 750×196=147000.)
Therefore the selling price is 4500-1000=3500.00 (and the number of tickets sold is 30+12=42).