Distribute (40-x) air conditioners and 60-(70-x)=(x- 10) refrigerators to chain store b,

Then y = 200 x+170 (70-x)+160 (40-x)+150 (x-10),

That is y = 20x+ 16800.

∫x≥070-x≥040-x≥0x- 10≥0

∴ 10≤x≤40.

∴y=20x+ 16800( 10≤x≤40)；

(2) From the meaning of the question: y = (200-a) x+170 (70-x)+160 (40-x)+150 (x-10),

That is y = (20-a) x+ 16800.

∫200-a > 170，

∴a 0, the function y increases with the increase of x,

Therefore, when x=40, the total profit is the largest, that is, 40 air conditioners, 30 refrigerators, 0 air conditioners and 30 refrigerators are allocated to chain A;

When a=20, the profits of all schemes with x value within 10≤x≤40 are the same;

When 20 < a < 30 and 20-a < 0, the function y decreases with the increase of x,

Therefore, when x= 10, the total profit is the largest, that is, 10 air conditioners, 60 refrigerators, 30 air conditioners and 0 refrigerators are allocated to a chain store.