= 26000+ 100 x(0 & lt； = x & lt= 10)

(2) According to the meaning of the question: 2600+100 x > =26800, and because 0

∴8<； = x<= 10, because x is an integer.

∴ x = 8,9, 10, there are three schemes.

Scheme 1: A dispatches type a vehicles 12 vehicles and type b vehicles 8 vehicles; B sent 8 A's and 2 B's;

Option 2: Dispatch a type car 1 1 car and 9 type cars in Area A; Nine A's and 65,438+0 B's were sent to B's;

Option 3: Dispatch 10 A car and 10 B car in Area A; B car delivery 10.

(3)∵y=26000+ 100x is a linear function and k = 100 > 0,

Y increases with the increase of x,

When x= 10, these 30 cars get the most rent every day.

∴ The reasonable allocation scheme is to send car A 10 and car B10 in Area A; B car delivery 10