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Problems in mathematics
(1) Let's assume that the store needs A yuan to buy a souvenir and B yuan to buy a souvenir.

10a+5b= 1000

5a+3b=550

∴ A = 50 b = 100 for solving equations

∴ It takes 50 yuan to buy a souvenir A and 100 yuan to buy a souvenir B.

(2) Suppose the store buys X pieces of A-type souvenirs and Y pieces of B-type souvenirs.

50x+ 100y= 10000

6y≤x≤8y

The solution is 20≤y≤25.

* * There are six purchase schemes.

(3) Let the total profit be W yuan.

w=20x+30y=20(200-2 y)+30y

=- 10 y+4000

W decreases with the increase of y

When y = 20, w has a maximum.

W Max =- 10× 20+4000 = 3800 (yuan)

∴? Buy 0/60 pieces of Class A souvenirs/KLOC-and 20 pieces of Class B souvenirs, and you can get the maximum profit, with a maximum profit of 3,800 yuan.