An Olympic franchise store plans to purchase 100 pieces of newly developed Olympic franchise products A, B, C, B and C. According to the regulations, each Olympic franchise product must purchase at least 18 pieces, and it just ran out of 9600 yuan. The purchase price and selling price of three Olympic franchise products are as follows.

Olympic licensed commodity model A B C

Purchase price (unit: RMB/unit) 80 100 1 10.

Sales price (unit: RMB/unit) 100 130 160

Suppose you buy X pieces of Class A Olympic licensed goods and Y pieces of Class B Olympic licensed goods.

(1) Use the formula containing x and y to express the quantity of licensed goods purchased for the C Olympic Games.

(2) y is represented by an algebraic expression containing x.

(3) Assuming that all purchased Olympic licensed goods can be sold, considering various factors, franchisees need to spend an extra 1 000 yuan in the process of purchasing and selling these Olympic licensed goods.

(1) Let the sales profit be p, express p with an algebraic expression containing x, and write the value range of x;

(2) Seek the maximum sales profit P, and write down how many pieces of each of the three Olympic licensed commodities are purchased at this time.

Solution:

1, let z be the number of c products, then z =100-x-y.

2.80x+100y+110 (100-x-y) = 9600, and simplified y= 140-3x.

3. 1, P=20*x+30*y+50*z- 1000 is substituted and simplified to get p = 30x+ 1200 (18 =

3.2. As can be seen from the above formula, P increases with the increase of X, which is the increasing function, that is, when X reaches the maximum, P is the maximum.

P(max)=3 120 when x=64.

At this time, y= 18 and z= 18.