Then the daily traffic volume from May 1 to May k constitutes a arithmetic progression {an} with 1000 as the first term and 100 as the tolerance.

Let the daily traffic volume from May (k+ 1) to May 3 1 day form another arithmetic progression {bn} in turn, and its first term is b1= AK-100 = [1000+(k-/kloc)

∴s3 1-k=(3 1-k)( 100k+800)+(3 1-k)(30-k)(- 100)=- 150 k2+5350k-2 1700。

From sk+s3 1-k = 59300, k2-63k+8 10 = 0.

The solution is k = 18 or k = 45 (truncation).

When k = 18, AK =1000+(18-1) ×100 = 2700 (ton).

∴ May18th, the maximum transportation capacity is 2700 tons.