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The Application of Mathematical Inequality Group in Grade One of Junior Middle School
Solution 1: If the number of vehicles transporting A navel orange is x and the number of vehicles transporting B navel orange is y, then the number of vehicles transporting C navel orange is (20-x-y); According to the meaning of the question, this relationship can be listed:

6x+5y+4(20-x-y)= 100

6x+5y+80-4x-4y= 100

2x+y+80= 100

y=-2x+20

Solution 2: According to the meaning of the question, x≥6, and y=-2x+20≥6, 20-x-y=20-x+2x-20=x≥6, an inequality group can be listed:

x≥6

-2x+20≥6

The solution set of inequality group is 6≤x≤7.

The integer solution of the inequality group is x=6 and x=7.

When x=6, y=-2x+20=8, 20-x-y=6.

When x=7, y=-2x+20=6, 20-x-y=7.

There are two delivery schemes:

Scheme 1: The number of vehicles transporting A navel orange is 6, the number of vehicles transporting B navel orange is 8, and the number of vehicles transporting C navel orange is 6.

Scheme 2: The number of vehicles transporting A navel orange is 7, the number of vehicles transporting B navel orange is 6, and the number of vehicles transporting C navel orange is 7.

Scheme 3: The total profit (100 yuan) is:

12x+ 16y+ 10(20 x-y)

= 12x+ 16(-2x+20)+ 10x

= 12x-32x+320+ 10x

=- 10x+320

When x takes the minimum value, the total profit (-10x+320) has the maximum value, and the minimum value of x is x=6.

When x = 6,-10x+320 =-10x6+320 = 260.

In order to maximize the profit of this sale, the scheme 1 should be adopted, and the maximum profit is 26,000 yuan.