0.3x+0.6(2000？ x)≤9000.5x+0.2(2000？ X)≤850, and the solution is 1000 ≤ X ≤ 1500. (3 points)
∫x is a positive integer, ∴x is 1000 to 1500, and a * * has 50 1 production schemes; (4 points)
(2)y = 10x+20(2000-x)= 40000- 10x； (7 points)
(3) According to the conclusion in (2), y=40000- 10x,
When x= 1000, y is the largest,
So choose the most profitable scheme (x = 1000). (9 points)