Solution: Suppose that each group originally produced X products every day.
According to the meaning of the question, we can get 3 * 10x < 500.
3* 10(X+ 1)>500
The solutions are 15 and 2/3 < x <; 16 and two thirds
The value of x should be an integer,
X= 16
A: Each group initially produced 65,438+06 products per day.
2.
Analysis: If you buy an X-type sewage treatment equipment, then buy a B-type (10-X) sewage treatment equipment.
(1) From price condition to inequality, find non-negative solution;
(2) From the perspective of sewage treatment, several purchase schemes in (1) are optimized;
(3) Compare the two sewage treatment methods, and calculate the annual savings of 10.
Solution: (1) If you buy an X-type sewage treatment equipment, then buy a B-type (10-x) sewage treatment equipment. According to the meaning,12x+10 (10-x) ≤105, and x ≥ 2.5.
∵x is a non-negative integer, ∴x can be 0, 1, 2.
∴ There are three purchase schemes: 0 sets of type A and 0 sets of type B 10; Buy type a 1 set; Nine for type B, two for type A and eight for type B..
(2) 240x+200 ( 10-x) ≥ 2040，
X≥ 1,∴x is 1 or 2.
When x= 1, the purchase capital is12×1+10× 9 =102 (ten thousand yuan).
When x=2, the purchased capital is12× 2+10× 8 =104 (ten thousand yuan).
In order to save money, you need to buy 1 model A and 9 model B.
(3) In 10, the total amount of funds for enterprises to treat sewage by themselves is:
102+10×10 = 202 (ten thousand yuan)
If the sewage is discharged to the sewage plant for treatment, the cost of 10 is:
2040×12×10×10 = 2448000 (yuan) =244.8 (ten thousand yuan)
244.8-202=42.8
It can save 428,000 yuan.
3.
Suppose that the price for customers to buy electrical appliances is X yuan;
The same is true when X≤500.
When 500
When X≥ 1000, the charge of mall A is:1000+0.9 * (x-1000) = 0.9x+100, and the charge of mall B is: 500+0.95 * (x
(0.9X+ 100)-(0.95 x+25)= 75-0.05 x
Order 75-0.05X >0, x < 2500.
Namely 1000 ≤ x
4.
Solution: (1) If you rent X A-class cars, you rent 8-x B-class cars.
40x+30(8-x)≥290。
10x+20(8-x)≥ 100
Solution: 5≤x≤6
That is * * * has two car rental schemes:
The first one is to rent 5 Class A cars and 3 Class B cars;
The second is to rent six A-class cars and two B-class cars.
(2) The cost of the first car rental scheme is
5 * 2000+3 *1800 =15400 yuan;
The cost of the second car rental scheme is 6 * 2000+2 *1800 =15600 yuan.
The first rental scheme is more cost-effective.
5.
Set x vehicles for vehicle a and (8-x) vehicles for vehicle B.
40x+20(8-x)≥200
10x+20(8-x)≥ 120
40x+ 160-20x≥200
20x≥40
x≥2
10x+ 160-20x≥ 120
- 10x≥-40
x≤4
The design scheme is as follows:
(1) A car 2, B car 6;
(2) There are 3 vehicles in Car A and 5 vehicles in Car B;
There are 4 vehicles in A and 4 vehicles in B. 。
(3) The freight rates of the three schemes are as follows:
①2×4000+6×3600=29600;
②3×4000+5×3600=30000;
③4×4000+4×3600=30400.
Scheme ① has the least freight, with the lowest freight of 29,600 yuan.