(1) According to the equivalence relation: required funds = number of Class A equipment × unit price+number of Class B equipment × unit price, the functional relation between W and X can be obtained;

Total sewage treatment = number of Class A equipment × sewage treatment capacity of each set+number of Class B equipment × sewage treatment capacity of each set, and the functional relationship between Y and X can be obtained;

(2) Using w≤ 106 and y≥2040, find out the value range of X, and then judge which scheme is the most economical and how much it costs.

Solution: Solution: (1) To purchase X sets of Type A equipment, the required capital * * * is W million yuan, and the total amount of sewage treatment per month is Y tons.

Then the functional relationship between W and X is: W =12x+10 (10-x) = 2x+100;

The functional relationship between y and x: y = 240x+200 (10-x) = 40x+2000.

(2) According to (1), {2x+100 ≤10640x+2000 ≥ 2040,

If {x≤3x≥ 1, then x= 1 or 2 or 3.

So all the purchase schemes are:

When x= 1, w= 102 (ten thousand yuan);

When x=2, w= 104 (ten thousand yuan);

When x=3, w= 106 (ten thousand yuan).

Therefore, it is the most economical to buy 1 set of A-type equipment and 9 sets of B-type equipment, and it needs 1.02 million yuan.