A mobile phone dealer plans to purchase a total of 60 mobile phones of a certain brand: Type A, Type

60-x-y

2.From the title: 900x+1200y+1100(60-x-y)=61000

y=2x-50

3.①p=1200x+1600y+1300(60-x-y)-61000-1500

100x+300y+15500

100x+300(2x-50)+1550500x+500

The next little question is not very good.

Because every phone is bought at least 8

Ministry, so 8 2x-50 60-8-8

Solve 29 x 47

So when x=47, p has a maximum.

Give me a bounty.

60-x-y

y=(61000-900x-1100(60-x-y) 1200, i.e.: y=2x-50

x≤82x-50≤8

110-3x≤8

Solution. 29≤x≤34

Because k=500 is greater than 0

So when x=34, p is maximum.

p max = 34 times 500 + 500 = 17,500 yuan.

When x=34, y=18, 60-x-y=8

Therefore, at this time, 34 Type A mobile phones, 18 Type B mobile phones, and 8 Type C mobile phones were purchased.

1）60-x-y.

2) From the title, 900x+1200y+1100(60-x-y)= 61000, and y=2x-50

3) From the title, p= 1200x+1600y+1300(60-x-y)- 61000-1500, finishing, get p=500x+500

The number of mobile phones purchased in Type C is 60-x-y = 110-3x according to the inequality group of the title.

x≤82x-50≤8

110-3x≤8

Solve 29 x 34

x ranges from 29 x 34 and x is an integer

p is a primary function of x, k = 500 0, p increases with the increase of x When x takes the maximum value of 34, p has a maximum value, and the maximum value is 17500 yuan At this time, 34 type A mobile phones, 18 type B mobile phones, and 8 type C mobile phones can be purchased I hope to solve your problem.

Summary. Dear, the store bought 90 A brand mobile phones, according to the title, it can be calculated that A mobile phone originally had 630 * (1 5) = 126, so it can be listed as (126 + x) (630 + x) =, you can calculate x

Hello, I'm Furutani Wind and Rain, this question will be answered by me, it takes a little time to query and type, please be patient.

The store originally had a total of 630 mobile phones of two brands of AB, of which one-fifth of the mobile phones of brand A did not come and bought one, sorry, please describe in detail your problem.

The store originally had a total of 635 mobile phones of two brands of A and B, of which A brand mobile phones accounted for one-fifth, and some of the brand mobile phones that were purchased later tested the two kinds of hungry cavity of the A brand mobile phones and the total number of mobile phones of the brand.

Pro, the store bought 90 Qi Li of a brand mobile phone, according to the title spine touch can be calculated A mobile phone originally had 630 * (1 5) = 126 units, so it can be listed as Gao Ye Search (126 + X) (630 + X) =, you can calculate X

126+x is OK.

126+x) what is a slash. Except.

60-x-y.

2) From the meaning of the title, the imaginary width is 900x+1200y+1100(60-x-y)= 61000, and the arrangement is y=2x-50

3) From the title, p= 1200x+1600y+1300 (trillion60-x-y)- 61000-1500, finishing, p=500x+500

The number of mobile phones purchased in Type C is 60-x-y = 110-3x according to the title of the inequality group.

x≤82x-50≤8

110-3x≤8

Solve 29 x 34

x ranges from 29 x 34 and x is an integer

p is a primary function of x, k = 500 0, p increases with the increase of x When x takes the maximum value of 34, p has a maximum value, and the maximum value is 17500 yuan At this time, 34 type A mobile phones, 18 type B mobile phones, and 8 C type mobile phones were purchased

110 3x is less than or equal to 8, and the solution is x greater than or equal to 35.

(1) Type B: 2X-50 Type C: 110-3X

2) There are a total of 6 purchase schemes which are:

3) Type A into 34, Type B into 18, Type C into 8 most profitable, can make a profit of 19,000 yuan.

1. b+c=60-x

2.6 types: 29A+8B+23C, 30A+10B+20C, 31A+12B+17C, 32A+14B+14C, 33A+16B+11C, 34A+18B+8C

3.34a+18b+8c is the most profitable, with a profit of 19,000 yuan.

Solution: (1) Set up a shopping mall to plan to purchase X part of a mobile phone, and a kind of mobile phone Y, which can be solved by the title: x = 20

y=302) If the A mobile phone is reduced by A, then the B mobile phone is increased by 2A to solve a 5

Let the profit obtained after all sales be w yuan, and w== from the title

w increases with the increase of a.

When a=15, w max=

The mall plans to purchase X units of mobile phone A and mobile phone Y unit of B mobile phone.

From the meaning of the question, the solution: x 20

Y 30 A: The mall plans to purchase 20 mobile phones A and 30 mobile phones B

2) If the A mobile phone is reduced by A, then the B mobile phone will be increased by 3A, and the title will be 4000 (20-A) + 2500 (30 + 3A) 172500

Solution A 5 set the gross profit after all sales to be w yuan.

w=300（20-a）+500（30+3a）=1200a+21000．

1200 0, w increases with the increase of a, when a=5, w has a maximum, w max = 1200 5 + 21000 = 27000

Answer: When the mall buys 15 mobile phones of type A and 45 mobile phones of type B, the gross profit is the largest after all sales, and the maximum gross profit is 10,000 yuan

Are you an elementary school student? If so, use your own brains to think about it, and you can't be opportunistic.

Solution: (1) Set up a shopping mall to plan to purchase X part of a mobile phone, and a kind of mobile phone Y, which can be solved by the title: x = 20

y=302) If the A mobile phone is reduced by A, then the B mobile phone is increased by 2A to solve a 5

Let the profit obtained after all sales be w yuan, and w== from the title

w increases with the increase of a.

When a=15, w max=

(1) The mall plans to purchase X units of mobile phone A and mobile phone Y unit of B to be purchased.

From the meaning of the question, the solution: x 20

Y 30 A: The mall plans to purchase 20 mobile phones A and 30 mobile phones B

2) If the A mobile phone is reduced by A, then the B mobile phone will be increased by 3A, and the title will be 4000 (20-A) + 2500 (30 + 3A) 172500

Solution A 5 set the gross profit after all sales to be w yuan.

w=300（20-a）+500（30+3a）=1200a+21000．

1200 0, w increases with the increase of a, when a=5, w has a maximum, w max = 1200 5 + 21000 = 27000

Answer: When the mall buys 15 mobile phones of type A and 45 mobile phones of type B, the gross profit is the largest after all sales, and the maximum gross profit is 10,000 yuan

Solution: (1) Set up a shopping mall to plan to purchase a kind of mobile phone x, B kind of mobile phone Y, by the title, get, solution:

Answer: The mall plans to purchase 20 mobile phones of type A and 30 mobile phones of type B;

2) If the A mobile phone is reduced by A, then the B mobile phone will be increased by 2A, and the solution is obtained by the title: A 5

Let the gross profit obtained after all sales be w yuan, from the title, w= = k=, w increases with the increase of a, when a=5, w is the maximum =

Answer: When the mall purchased 15 A mobile phones and 40 B mobile phones, the maximum profit after all sales was 10,000 yuan.

This is not easy to play with the computer, you set x and y, very simple.

That should be the maximum when a=5.

The amount of the amount reaches the Oriental International Financial Law.

(1) Set up a shopping mall to plan to purchase X part of a mobile phone A and a mobile phone Y part of B type, and establish an equation system to find the solution according to the purchase amount of two kinds of mobile phones of 10,000 yuan and the sales profit of two kinds of mobile phones of 10,000 yuan.

2) Set a type of mobile phone to reduce part A, then type B mobile phone increases 2a, indicating the total amount of money purchased, by the total capital department of more than 160,000 yuan to establish an inequality can find the value range of a, and then set the total profit after sales to w yuan, indicating the relationship between the total profit and a, and the maximum profit can be found by the nature of a function.

(1) The number of type C is 60-x-y

The total amount of pre-sale is 1200x+1600y+1300(60-x-y)=1300*60-100x+300y=78000-100x+300y

2)p=78000-100x+300y-61000-15000=2000-100x+300y