In 07, a county in our city was preparing for the 20th anniversary, and the county celebration garde

1) Solution: If there are x models of A, then the B shapes are (50-x), which is obtained according to the topic.

80x+50(50-x)≤3490

40x+90(50-x)≤2950

Solving this inequality group yields: 31 x 33, x is an integer, x can take 31, 32, 33 can be designed three combinations:

There are 31 horticultural modeling A and 19 horticultural modeling B, 32 horticultural modeling A and 18 horticultural modeling B, 33 horticultural modeling A and 17 horticultural modeling B, 2) because the cost of B modeling is higher than the cost of A modeling, so the less B modeling, the lower the cost, so the scheme should be selected, the lowest cost and the lowest cost is.

33 800 + 17 960 = 26400 + 16320 = 42720 (yuan).

Therefore, the answer is: 42,720 yuan.

Hope! Thank you!

The topic is that in 2009, a county in our city was preparing for the 20th anniversary of the county celebration, and the Ministry of Gardens decided to use the existing 3490 pots of A kind of flowers and 2950 pots of flowers that have been planted into two kinds of horticultural shapes, a total of 50, placed on both sides of Yingbin Avenue. It is known that 80 pots of A flowers and 40 pots of flowers are required to match a type A shape: 50 pots of A flowers and 90 pots of flowers are required to match a B shape.

1) The extracurricular activity group of the ninth grade of a school undertook the design of this gardening modeling scheme, and asked how many collocation schemes are there that match the topic? Please help design it:

2) If you want to match a type A shape the cost is 800 yuan, with a type B style cost is 960 yuan, try to explain (1) which scheme has the lowest cost, the lowest cost is how much yuan?

It's just a middle school exam question that hasn't been completed.

Solution: (1) If there are x models of A type, then (50 x) of B shapes. According to the title:

80x+50(50-x)≤3490

40x+90(50-x)≤2950

Solving this inequality yields:

31≤x≤33

Because x is an integer, x can take 31, 32, 33

Therefore, three matching schemes can be designed:

1. There are 31 A horticultural shapes and 19 B horticultural collocations.

2. There are 32 gardening styles of A species and 18 gardening styles of B species.

3. There are 33 A horticultural shapes and 17 B horticultural collocations.

2) Method 1:

Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, and the lowest cost is.

33 800 + 17 960 = 42720 (yuan) Method 2: Scheme cost 31 800 + 19 960 = 43040 (yuan) scheme Cost 32 800 + 18 960 = 42880 (yuan) scheme Cost 33 800 + 17 960 = 42720 (yuan) should choose the scheme, the lowest cost, the lowest cost is 42720 yuan Agree with 8|Comment (1).

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Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, and the lowest cost is.

33 800 + 17 960 = 42720 (yuan) Method 2:

Scheme cost 31 800 + 19 960 = 43040 (yuan) scheme cost 32 800 + 18 960 = 42880 (yuan) scheme cost 33 800 + 17 960 = 42720 (yuan) should choose the scheme, the lowest cost, the lowest cost is 42720 yuan

Solution: (1) If there are x models of A type, then (50 x) of B shapes. According to the title:

80x+50(50-x)≤3490

40x+90(50-x)≤2950

Solving this inequality yields:

31≤x≤33

Because x is an integer, x can take 31, 32, 33

Therefore, three matching schemes can be designed:

1. 31 A horticultural shapes, 19 B horticultural shapes, 32 A horticultural shapes, 18 B horticultural shapes, 33 A horticultural shapes, and 17 B horticultural shapes (2) Method 1:

Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, and the lowest cost is.

33 800 + 17 960 = 42720 (yuan) Method 2:

Scheme cost 31 800 + 19 960 = 43040 (yuan) scheme cost 32 800 + 18 960 = 42880 (yuan) scheme cost 33 800 + 17 960 = 42720 (yuan) should choose the scheme, the lowest cost, the lowest cost is 42720 yuan

Solution: (1) If there are x models of A type, then (50 x) of B shapes. According to the title:

Solving this inequality yields:

Because x is an integer, x can take 31, 32, 33

Therefore, three matching schemes can be designed:

1. 31 A horticultural shapes, 19 B horticultural shapes, 32 A horticultural shapes, 18 B horticultural shapes, 33 A horticultural shapes, and 17 B horticultural shapes (2) Method 1:

Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, and the lowest cost is.

33 800 + 17 960 = 42720 (yuan) Method 2:

Scheme cost 31 800 + 19 960 = 43040 (yuan) scheme cost 32 800 + 18 960 = 42880 (yuan) scheme cost 33 800 + 17 960 = 42720 (yuan) should choose the scheme, the lowest cost, the lowest cost is 42720 yuan

If there are x models in type A, then (50-x) models in type B.

According to the title, yes.

Solve this group of inequalities and get 31 x 33

x is an integer, x can be taken as 31, 32, 33

Three matching schemes can be designed:

There were 31 horticultural shapes of A type and 19 horticultural shapes of type B;

There are 32 horticultural shapes of A type and 18 horticultural shapes of B type;

33 A horticultural shapes 17 B horticultural shapes (2) Method 1: Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, the lowest cost is 33 800 + 17 960 = 42720 (yuan).

Method 2: Scheme cost: 31 800 + 19 960 = 43040 (yuan);

Scheme cost: 32 800 + 18 960 = 42880 (yuan);

Scheme cost: 33 800 + 17 960 = 42720 (yuan);

The lowest cost should be selected, and the lowest cost is 42,720 yuan.

Solution: (1) If you match X models of A, then the shape of B is (50-x) from the title, and you get 80x+50 (50-x) 3490 40x+90 (50-x) 2950

Solve 31 x 33

x is an integer, x=31,32,33

Three matching schemes can be designed:

Scheme 1: 31 horticultural shapes of A species and 19 horticultural shapes of type B;

Scheme 2: 32 horticultural shapes of A type and 18 horticultural shapes of type B;

Scheme 3: 33 gardening styles of A and 17 gardening shapes of B (2) The cost of B modeling is higher than the cost of modeling A, and the less modeling B, the lower the cost, so plan 3 should be selected, and the lowest cost The lowest cost is: 33 800 + 17 960 = 42720 (yuan) Answer:

Option 3 should be selected with the lowest cost, and the lowest cost is 42,720 yuan

Solution: (1) If there are x shapes in A, then (50-x) in B shapes, and the inequality group is solved according to the meaning of the problem, 31 x 33

x is an integer, x can be taken as 31, 32, 33

Three matching schemes can be designed:

There were 31 horticultural shapes of A type and 19 horticultural shapes of type B;

There are 32 horticultural shapes of A type and 18 horticultural shapes of B type;

There are 33 horticultural models of A type and 17 horticultural models of B species

2) Method 1:

Because the cost of B modeling is higher than the cost of A modeling, the less B modeling, the lower the cost, so the scheme should be selected, the cost is the lowest, and the lowest cost is.

33 800 + 17 960 = 42720 (yuan) Method 2:

Scheme cost 31 800 + 19 960 = 43040 (yuan) scheme cost 32 800 + 18 960 = 42880 (yuan) scheme cost 33 800 + 17 960 = 42720 (yuan) should choose the scheme, the lowest cost, the lowest cost is 42720 yuan

Solution: (1) If you match X models of A, then the shape of B is (50-x) from the title, and you get 80x+50 (50-x) 3490 40x+90 (50-x) 2950

Solve 31 x 33

x is an integer, x=31,32,33

Three matching schemes can be designed:

Scheme 1: 31 horticultural shapes of A species and 19 horticultural shapes of type B;

Scheme 2: 32 horticultural shapes of A type and 18 horticultural shapes of type B;

Scheme 3: 33 gardening styles of A and 17 gardening shapes of B (2) The cost of B modeling is higher than the cost of modeling A, and the less modeling B, the lower the cost, so plan 3 should be selected, and the lowest cost The lowest cost is: 33 800 + 17 960 = 42720 (yuan) Answer:

Option 3 should be selected with the lowest cost, and the lowest cost is 42,720 yuan

Solution: If there are x models of A, then the B shapes are (50-x)80x+50*(50-x)<=3490....140x+90*（50-x）<=2950...

2 solution 1 yields 30x<=990(x<=33).

Solution 2 yields 50x>=1550(x>=31).

So x=31,32,33 these 3 collocations.

Solution: A type X shapes, then B type (50-x) pieces.

Solution: If there are x models of A, then there are (50-x) of B shapes.

According to the title: 80x+50(50-x) 349040x+90(50-x) 2950

Solving this equation gives 31 x 33, so the value of x is 31, 32, 33.

Answer: There are three ways: 31 A style pendulums, 19 B style pendulums;

32 A shapes and 18 B shapes;

There are 33 A style possessions and 17 B style possessions.

2) Solution: 31x800 + 19x960 = 43040 (yuan) 32x800 + 18x960 = 42880 (yuan) 33x800 + 17x960 = 42720 (yuan) Answer: The cost of the program is the lowest, which is 42720 yuan.

1) Set X horticultural shapes of A and Y of B species. Then x+y=,y satisfies :

80*x+50*y<=3490,40*x+90y<=2950.There are 3 types of x,y that meet the conditions:

31,19),(32,20),(33,17）（80*x+50*(50-x)<=3490,40*x+90(50-x)<=2950.Launched 31<=x<=33)

2) Obviously, the more A-shapes, the lower the cost. Therefore, the scheme should be selected: 33 A shapes, 17 B shapes. The lowest cost is y0 = 33 * 800 + 960 * 17 = 42,720 yuan.

Solution: (1) There are three ways: 31 A shapes, 29 B shapes;

32 A shapes and 28 B shapes;

There are 33 pendulums in type A and 27 pendulums in type B.

2) Solution: 31x800 + 29x960 = 56240 (yuan) 32x800 + 28x960 = 55480 (yuan) 33x800 + 27x960 = 52320 (yuan) Answer: The cost of the program is the lowest, which is 52320 yuan.

1.Three kinds, the first one, A31 disc, B29 disc, the second A32, 28B, the third A33, B27

2 Minimum cost 52320

Solution: If there are x models of A, then the B shapes are (50-x) from the title, and 80x+40(50-x) 349050x+90(50-x) 2950

Solve 31 x 33

x is an integer, x=31,32,33

Three matching schemes can be designed:

Scheme 1: 31 horticultural shapes of A species and 19 horticultural shapes of type B;

Scheme 2: 32 horticultural shapes of A type and 18 horticultural shapes of type B;

Scheme 3: 33 horticultural shapes of A species and 17 horticultural shapes of B species