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.