Step 2: The profit from selling each commodity is (13.5-2.5-x) yuan, that is, (1 1-x) yuan.
Step 3: The sales volume of each commodity minus X yuan is (500+ 100x).
Step 4: When the daily profit of the store selling this small commodity is Y yuan, the relationship between Y and X is
Y = (11-x) * (500+100x) =-100x squared +600x+5500.
Step 5: Because the unit selling price of the commodity is reduced on the basis of 13.5 yuan, x≥0. And because the reduction range is between 2.5 ~ 13.5 yuan, X≤ 1 1. In this way, the value range of x is 0≤x≤ 1 1.
(2) Using the method of complete leveling, we can get:
Y=- 100(x-3) squared +6400.
So when x=3, y gets the maximum value of 6400 yuan. Because x=3 is within the range of 0≤x≤ 1 1, it is possible to maximize the sales profit, that is, when the sales unit price is = 13.5-3= 10.5 yuan, the daily profit of shops selling this small commodity is the largest, which is 6400 yuan.