A township in our city is rich in citrus, village A has 200 tons of citrus, village B has 300 tons o

1) According to the meaning of the topic, use the known data to represent the space in the above ** with x, then the functional relationship between ya, yb and x is easy to find

2) The amount of freight is related to the weight of citrus transported from village A to warehouse C, i.e., x, which needs to be discussed in different situations, when ya=yb, yayb and yayb

3) From the meaning of the question yb 4830, the solution is x 50, the sum of the freight of the two villages ya+yb=-2x+9680, which can be solved by using the property of the primary function

Answer: Solution: (1).

ya=-5x+5000（0≤x≤200），yb=3x+4680（0≤x≤200）．

2) When ya=yb, -5x+5000=3x+4680, x=40;

When ya yb, -5x+5000 3x+4680, x 40;

When ya yb, -5x+5000 3x+4680, x 40;

When x=40, ya=yb, that is, the freight of the two villages is equal;

When 0 x 40, YA YB i.e. B village freight is less;

When 40 x 200, ya yb i.e. a village costs less

3) From YB 4830, get 3x+4680 4830

x 50 Let the sum of the freight costs of the two villages be y, y=ya+yb

That is: y=-2x+9680

When 0 x 50, y decreases with the increase of x, and when x=50, y has a minimum value, and the minimum value of y = 9580 (yuan).

Answer: When the weight of citrus transferred from village A to warehouse C is 50 tons, 150 tons to warehouse D, 190 tons to warehouse C from village B, and 110 tons to warehouse D, the sum of the freight costs of the two villages is the smallest, and the minimum cost is 9,580 yuan

YA = 15X + 25 (200-X) = -10X + 5000YB = 15 (240-X) + 18 (X + 60) = 3X + 4680 When the freight of AB and the two villages is the same, -10X + 5000 = 2X + 4680X = 80 3

When x 80 3, Otomura shipping is cheaper.

When x 80 3, Komura shipping is cheap.

It is not possible to be equal to, because x is an integer).

1）ya=-5x+5000（0≤x≤200），yb=3x+4680（0≤x≤200）

2) When ya=yb, -5x+5000=3x+4680, x=40;

When ya yb, -5x+5000 3x+4680, x 40;

When ya yb, -5x+5000 3x+4680, x 40 When x=40, ya=yb, that is, the freight of the two villages is equal;

When 0 x 40, YA YB i.e. B village freight is less;

When 40 x 200, ya yb i.e. a village costs less

The title of this question is not very clearly written, and there is no problem (1). Question (2) Ranlu: Please find out the transportation cost of the oranges transported to the two warehouses in Piduanqing A and B respectively, and the cost of transportation A to warehouse C is:

The cost of transporting the remaining 20x to library D is: (200-x) *15 = 3000 - 15x; Therefore, the shipping cost of a mandarin orange to the two warehouses is: 20x + 3000 - 15x = 3000 + 5x; The cost of transporting B to the remaining C warehouse is as follows:

240 - x) *15 = 3600 - 15x;The remaining cost of transportation to library D is: (300-(240-x)) 18 = 1080 + 18x; Therefore, the shipping cost of a mandarin orange to the two warehouses is: 3600 - 15x + 1080 + 18x = 4680 + 3x; Question (3):

When x = 100, try to find the sum of the transportation costs of the citrus transported to the two warehouses by 2, and the total cost = 3000 + 5x + 4680 + 3x = 7680 + 8x = 7680 + 800 = 8480 yuan;

1) Solution: fill in the table from top to bottom, from left to right: (200-x) tons, (240-x) tons, (60+x) tons;

Therefore, the answer is: (200-x) tons, (240-x) tons, (60+x) tons 2) Solution: according to the meaning of the question:

ya=20x+25(200-x)=5000-5x, yb=15(240-x)+18(60+x)=3x+4680, the value range of x is 0 x 200, answer: ya, yb and x are ya=20x+25(200-x)=5000-5x, yb=15(240-x)+18(60+x)=3x+4680, and the value range of the independent variable x is 0 x 200

3) Solution: from yb 4830, get 3x+4680 4830,x 50, let the sum of the freight of a and b villages be y, then y=ya+yb=-2x+9680, y decreases with the increase of x, and 0 x 50, when x=50, y has a minimum value The minimum value is y=9580 (yuan), 200-50=150, 240-50=190, 60+50=110, answer: if the citrus freight of village B shall not exceed 4830 yuan, in this case, The weight of citrus transported from village A to factory C is 50 tons, the weight of citrus transported to factory D is 150 tons, the weight of citrus transported from village B to factory C is 190 tons, and the weight of citrus transported to factory D is 110 tons, so that the freight cost of the two villages is the smallest, and this minimum value is 9,580 yuan

1) YA = 20X + (25-X) = 5000-5X (0YB, X<320 23, the freight of Village B is less.

When YA320 23, Village A has less freight.

When ya=yb, x=320 23, the freight of village A and B is the same.

2）yb<=4830

x=25/3

ya+yb=28715/3

What variety of citrus is it?

It is noted that the number of tons of citrus that can be stored in warehouses C and D is the same as that of citrus in villages A and B, and that the mass of citrus transported from village A to warehouse C is x tons, then the citrus transported from village A to warehouse D is 200-x, the citrus transported from village B to warehouse C is 240-x, and the citrus transported to warehouse D is 300-(240-x)=60+x. The requirement is that the freight of A and B is equal, then the freight of Village A: 20x+25 (200-x) The freight of Village B:

15 (240-x) + 18 (60 + x) and then find x.

Let the mass of citrus transported from village A to warehouse C be x tons, price = 20x, the quality of citrus transported from village A to warehouse D is (200-x) tons, price = 25 (200-x), the quality of citrus transported to warehouse C from village B is (240-x) tons, price = 15 (240-x), and the quality of citrus transported to warehouse D from village B is = (100 + x) tons, price = 18 (100 + x), 20x + 25 (200-x) = 15 (240-x) + 18 (100 + x).

1) 15 * (240-x) + 18 (260 - (200-x)) < 4830 so x < 50 tons (2) total transportation virtual brother and bird loss fee: dust 20x + 25 (200-x) + 15 * (240-x) + 18 (260-(200-x)) = 9680-22x When x = 50, the total cost is taken as the worst early small value = 8580

First question. The freight rate from village A to the two warehouses is more expensive than that of village B, and no matter how they distribute it, the freight rate of village A is greater than that of village B!

Therefore, the freight cost of Village B is less.

Second question. ya=20x+25(200-x)=5000-5xyb=15*(240-x)+18(260-240+x)=3600+360+3x=3960+3x

yb=3960+3x<=4830

x<=290, C warehouse can load up to 240 tons, which fully meets the requirements!

ya+yb=8960-2x>=8960

When Xiaoliang x=0, ya+yb=8960 is the least.

The first question is that the freight rate from village A to the two warehouses is more expensive than that of village B, no matter how they distribute it, the freight rate of village A is greater than that of village B!

Therefore, the freight cost of Village B is less.

The second question is ya=20x+25(200-x)=5000-5xyb=15*(240-x)+18(260-240+x)=3600+360+3x=3960+3x

yb=3960+3x<=4830

x<=290, C warehouse can load up to 240 tons, which fully meets the requirements!

ya+yb=8960-2x>=8960

When x=0, ya+yb=8960 is the least.

When village A does not ship to warehouse C, but all of them are shipped to warehouse D, the sum of the freight costs of the two villages is the least, which is 8,960 yuan.

(1) When ya=yb, -5x+5000=3x+4680,x=40;

When ya yb, -5x+5000 3x+4680, x 40;

When ya yb, -5x+5000 3x+4680, x 40 When x=40, ya=yb, that is, the freight of the two villages is equal;

When 0 x 40, YA YB i.e. B village freight is less;

When 40 x 200, ya yb i.e. village A has less cost (7 points) (2) from YB 4830 to get 3x+4680 4830 x 50

Let the sum of the freight costs of the two villages be y, y=ya+yb

That is: y=-2x+9680

When 0 x 50, Y decreases with the increase of X, when X=50, Y has a minimum value, and Y Minimum=9580 (yuan) Answer: When the weight of citrus transferred to warehouse C in village A is 50 tons, 150 tons in warehouse D, 190 tons in village B to warehouse C, and 110 tons in warehouse D, the sum of the freight of the two villages is the smallest, and the minimum cost is 9580 yuan (10 points).

Solution: (1).

2 points) ya=-5x+5000 (0 x 200), yb=3x+4680 (0 x 200).

2) When ya=yb, -5x+5000=3x+4680, x=40;

When ya yb, -5x+5000 3x+4680, x 40;

When ya yb, -5x+5000 3x+4680, x 40 When x=40, ya=yb, that is, the freight of the two villages is equal;

When 0 x 40, YA YB i.e. B village freight is less;

When 40 x 200, YB is less expensive for Village A (3) From YB 4830, we get 3x+4680 4830 x 50

Let the sum of the freight costs of the two villages be y, y=ya+yb

That is: y=-2x+9680

When 0 x 50, Y decreases with the increase of X, when X = 50, Y has a minimum value, and Y Min = 9580 (yuan) Answer: When the weight of citrus transferred to warehouse C in village A is 50 tons, 150 tons in warehouse D, 190 tons in village B to warehouse C, and 110 tons in warehouse D, the sum of the freight of the two villages is the smallest, and the minimum cost is 9580 yuan

1) YA=20X+(25-X)=-5X+5000 (0YB,X<40, the freight of Village B is less.

When YA40, Village A has less freight.

When ya=yb, x=40, the freight rate of villages A and B is the same.

2）yb<=4830

x<=50

yb+ya=-2x+9680

x is the maximum, i.e., 50

The minimum shipping fee is 9580 RMB.

1) YA = 20X + (25-X) = 5000-5X (0YB, X<320 23, the freight of Village B is less.

When YA320 23, Village A has less freight.

When ya=yb, x=320 23, the freight of village A and B is the same.

2）yb<=4830

x=25/3

ya+yb=28715/3

c d Total.

a x ton (200-x) ton 200 tons.

b (240-x) tons (60+x) tons 300 tons total 240 tons 260 tons 500 tons.

ya=-5x+5000（0≤x≤2000），yb=3x+4680（0≤x≤200）．

2) When ya=yb, -5x+5000=3x+4680, x=40;

When ya yb, -5x+5000 3x+4680, x 40;

When ya40, when x=40, ya=yb means that the freight of the two villages is equal;

When 0 x 40, YA YB i.e. B village freight is less;

When 40 When x = 50, y has a minimum value, y minimum value = 9580 (yuan) Answer: When the weight of citrus transferred to warehouse C in village A is 50 tons, 150 tons in warehouse D, 190 tons in village B to warehouse C, and 110 tons in warehouse D, the sum of the freight costs of the two villages is the smallest, and the minimum cost is 9580 yuan