Dawn Shop distributes two kinds of products, A and B, and the following information is available

Solution: 1) Purchase price: A + B = 5

Price: A +1, 2 B -1

Spend 19 yuan to buy the selling price, 3 (A + 1) + 2 (2 B - 1) = 19 3 A + 4 B = 18 to solve the purchase price: A = 2 yuan, B = 3 yuan, selling price: A = 3 yuan, B = 5 yuan.

2) Reduce the relationship between the sales price and the sales amount: 100, the sales volume of commodity A = (500 + 1000m) (3-m).

Sales of goods B = (300 + 1000m) (5-m) profit r = (500 + 1000m) (3-m-2) + (300 + 1000m) (5-m-3) = 1100 + 400m-2000m

To maximize the daily profit, find the derivative of r.

r'=400-4000m

Ream'=0, then m=

Substitute m=,r=1120 yuan.

Your 1-m is the multiplier for calculating the profit of commodity A.

Solution: (1) Assuming that the purchase unit price of A and B commodities is x,y yuan respectively, according to the question: x+y=53(x+1)+2(2y-1)=19, solution: x=2y=3;

Answer: The purchase unit price of A and B is 2 yuan and 3 yuan respectively;

2) The store sells an average of 500 pieces of product A and 300 pieces of product B per day After investigation, it was found that for every time the retail unit price of goods A and B decreased, each of these two goods could sell 100 more pieces per day

When the retail unit price of both commodities A and B decreases by m yuan, A and B sell each day: (500+ pieces, (300+ pieces, the profits obtained from the sale of A and B commodities are: The profits of A and B are respectively

3-2 = 1 yuan, 5-3 = 2 yuan, the profit of each piece after the price reduction is: (1-m) yuan, (2-m) yuan;

W=(1-m) (500+,-2000m2+2200m+1100,when m=-b2a=-22002 (-2000)=yuan,w maximum,maximum value:4ac-b24a=1705 yuan,when m is set as yuan,in order to make the store sell a and b two kinds of goods every day to obtain the largest profit,the maximum profit per day is 1,705 yuan

Analysis: (1) According to the information on the figure, the equivalent relationship between goods A and B can be obtained, and the system of equations can be obtained;

2) According to the price reduction, A and B sell each day: (500+ pieces, (300+ pieces, the profit of each piece after the price reduction is: (1-m) yuan, (2-m) yuan; The total profit can be obtained, and the maximum value of the quadratic function of the wax mill can be found Answer:

Solution: (1) Assuming that the purchase unit price of A and B commodities is x,y yuan respectively, according to the question: x+y=53(x+1)+2(2y-1)=19, the solution:

x=2y=3；

Answer: The purchase unit price of A and B is 2 yuan and 3 yuan respectively;

2) The store sells an average of 500 pieces of product A and 300 pieces of product B per day After investigation, it was found that for every time the retail unit price of goods A and B decreased, each of these two goods could sell 100 more pieces per day

When the retail unit price of both commodities A and B drops m yuan, A and B sell each day: (500+ pieces, (300+ pieces in the air, the profit obtained from selling A and B commodities is: The profit of A and B per piece is:

3-2 = 1 yuan, 5-3 = 2 yuan, the profit of each piece after the price reduction is: (1-m) yuan, (2-m) yuan;

w=(1-m) (500+, 2000m2+2200m+1100, when m=- b2a=- 22002 (-2000)=yuan, w is maximum, the maximum value is: 4ac-b24a=1705 yuan, when m is set as yuan, douju mountain can make the store sell a and b two kinds of goods every day to obtain the largest profit, the maximum profit per day is 1705 yuan Comments: This question mainly examines the application of binary linear equations and the application of the maximum-value method of quadratic functions. This question is more typical and is also a hot question type in the high school entrance examination in recent years, pay attention to the total profit when expressing the single profit of the product and the number of goods sold is the key to solving the problem

Solution: 1) Purchase price: A + B = 5

The price is quiet: A +1, 2 B - 1

Spend 19 yuan to buy the price, 3 (A + 1) collapse + 2 (Qixiang 2B - 1) = 19 3 A + 4 B = 18

Solution purchase price: A = 2 yuan, B = 3 yuan, selling price: A = 3 yuan, B = 5 yuan.

2) Reduce the relationship between the sales price and the sales amount: 100, the sales volume of commodity A = (500 + 1000m) (3-m).

Sales of goods B = (300 + 1000m) (5-m) profit r = (500 + 1000m) (3-m-2) + (300 + 1000m) (5-m-3) = 1100 + 400m-2000m 0 5

To maximize the daily profit, find the derivative of r.

r'=400-4000m

Ream'=0, then m=

Substitute m=,r=1120 yuan.

Your 1-m is the multiplier for calculating the profit of commodity A.

1) Let the purchase unit price of the goods of the first merchant be x yuan, and the purchase unit price of the B commodity is y yuan According to the title, x+y=53(x+1)+2(2y-1)=19 The solution is x=2y=3

Answer: The purchase unit price of commodity A is 2 yuan, and the unit price of commodity B is 3 yuan 2) The profit obtained by setting up a store to sell two kinds of goods A and B every day is S yuan, then.

s=(1-m)(500+100×

i.e. s = -2000m2 + 2200m + 1100 = -2000 (when m = , s has a maximum value and the maximum value is 1705

1. Set the purchase unit price of A to x, and B's to y, then there is x+y=5,3(x+1)+2(2y-1)=19, and the solution is rolling, x=2, y=32, and there is a problem with the question, so there is no need to reduce the price, and it seems that the price has to be increased.

(1) Solution: (1) Assuming that the purchase unit price of A and B commodities is X and Y yuan respectively, according to the title:

x+y＝33(x+1)+2(2y?1) 12, solution:

x 1y 2, the retail unit price of A and B is 2 yuan and 3 yuan respectively;

So the answer is: 2, 3;

2) According to the meaning of the title:

1?m)(500+100×m

That is, 2m2-m=0, the solution is m= or m=0 (rounded), answer: when m is set as yuan, the store can sell a total of 1700 yuan of profits from the sale of goods A and B every day

1. Solution: If the purchase price of A is X yuan, and the purchase price of B is Y, the retail unit price of commodity A is X+1The retail unit price of commodity B is 2y-1

Now the column equation x+y=5

3(x+1)+2(2y-1)=19

The solution is x=2y=3A: The purchase price of A and B is 2 yuan and 3 yuan.

2.(1-m)×(500+100×10m)+(2×3-1-3-m)×(300+100×10m)

1-m)(500+1000m)+(2-m)(300+1000m)

500+1000m-500m-1000m²+600+2000m-300m-1000m²

2000m²+2200m+1100a

2000,b=2200,c=1100

m=-b/(2a)=2200/4000=

The maximum profit per day = (4ac-b) (4a) = 1705 yuan.

When m is set as yuan, the store can make the maximum profit obtained by selling two kinds of goods A and B every day, and the maximum profit per day is 1705 yuan.

Analysis: (1) According to the information on the figure, the equivalent relationship between goods A and B can be obtained, and the system of equations can be obtained;

2) According to the price reduction, A and B sell each day: (500+ pieces, (300+ pieces, the profit of each piece after the price reduction is: (1-m) yuan, (2-m) yuan; The total profit can be obtained, and the maximum value of the quadratic function can be found by using the solution:

Solution: (1) Assuming that the purchase unit price of A and B commodities is x,y yuan respectively, according to the question: x+y=53(x+1)+2(2y-1)=19, solution: x=2y=3;

Answer: The purchase unit price of A and B is 2 yuan and 3 yuan respectively;

2) The store sells an average of 500 pieces of product A and 300 pieces of product B per day After investigation, it was found that for every time the retail unit price of goods A and B decreased, each of these two goods could sell 100 more pieces per day

When the retail unit price of both commodities A and B drops m yuan, A and B sell each day: (500+ pieces, (300+ pieces, the profits obtained from the sale of A and B commodities are: The profits of A and B per piece are: 3-2 = 1 yuan, 5-3 = 2 yuan, and the profits of each piece after the price reduction are: (1-m) yuan, (2-m) yuan;

W=(1-m) (500+, 2000m2+2200m+1100, when m=- b2a=- 22002 (-2000)=yuan, w is maximum, the maximum value is: 4ac-b24a=1705 yuan, when m is set as yuan, in order to make the store sell a and b two kinds of goods every day to obtain the largest profit, the maximum profit per day is 1705 yuan Comments: This question mainly examines the application of binary linear equations and the application of the maximum-value method of quadratic functions. This question is more typical and is also a hot question type in the high school entrance examination in recent years, pay attention to the total profit when expressing the single profit of the product and the number of goods sold is the key to solving the problem

1. Set: the purchase unit price of commodity B is x yuan, and the retail price of B is: 2x-1 yuan; The purchase unit price of commodity A is 3-x yuan, the retail price is: 4-x, and the column equation: 3 * (4-x) + 2 * (2 x -1) = 12

x+10=12

x=2 yuan. The purchase unit price of commodity A: 3-2 = 1 yuan.

2. The retail price of A: 2 yuan; B's retail price: 3 yuan, when A sells 500 pieces and B sells 1200 pieces, it makes a profit; 1700 yuan, when the yuan is reduced, profit:

1700+200)*RMB; When the yuan is reduced: profit (1700 + 400) * yuan. m should be set as the yuan, and the maximum profit is:

1710 yuan.

(1) Assuming that the purchase unit price of commodity A and type B is x,y yuan, and the purchase unit price of commodity B is y yuan, according to the title, it can be obtained:

x+y＝33(x+1)+2(2y?1)＝12

Solution: x 1y 2

Therefore, the retail unit prices of A and B are 2 yuan and 3 yuan respectively;

2) According to the meaning of the title:

1-m）（500+100×m

That is, 2m2-m=0, the solution is m= or m=0 (rounded).

Answer: When M is set as yuan, the store can make a total profit of 1,700 yuan from selling goods A and B every day

I want to ask you how you solved the equations, they were all wrong.

According to information 1, let's set: the purchase price of commodity A is X yuan. B is $5-x.

According to information 2, the retail price of A is x+1 yuan, and B is (5-x) 2-1 According to information 3, x+1) 3+[(5-x) 2-1] 2=19 yuan, and x, that is, the purchase price of commodity A is 2 yuan, and the purchase price of commodity B is 3 yuan.

Solution: If the purchase price of A is x yuan and the purchase price of B is y, then the retail unit price of commodity A is x+1The retail unit price of commodity B is 2y-1

Now the column equation x+y=5

3(x+1)+2(2y-1)=19

x=2 y=3

Alas, it's not easy to get a fortune, Khan.

Solution: Let B be the purchase price of x yuan, then the purchase price of A is (5-x)(5-x+1) 3+(2x-1) 2=1918-3x+4x-2=19

16+x=19

x=35-3=2 yuan.

A: A 2 yuan, B 3 yuan.

If the purchase price of B is x, then the purchase price of A is 5-x, the retail price of A is 5-x+1=6-x, and the retail price of B is 2x-1

From 3 to 3(6-x)+2(2x-1)=19, x=3 is solved

Therefore, the purchase price of B is 3 yuan, and the selling price is 5 yuan, and the purchase price of A is 2 yuan, and the selling price is 3 yuan.

Let the purchase unit price of commodity A be x yuan, the selling price of commodity B is 5-x yuan, the selling price of commodity A is x+1 yuan, the selling price of commodity B is 2 (5-x)-1=9-2x yuan, and the selling price of 3 pieces of commodity A and 2 pieces of commodity B is 19 yuan, so that 3 (x+1) + 2 (9-2x) = 19 yuan is obtained, then x = 2 yuan, so the purchase unit price of commodity A is 2 yuan, and the purchase unit price of commodity B is 5-2=3 yuan.

Let the purchase price of A be X yuan, then B is (5-x) yuan.

The retail unit price of A is x+1 yuan, and B is 2*(5-x)-13*(x+1)+2*[2*(5-x)-1]=19 to obtain x=2

The purchase unit price of A is 2 yuan, and B is 3 yuan.