This is an optimization problem, linear programming. There are three schemes: A, B, C and A, C. Only the models of Party A and Party B are listed below.

Set an x, b, y,

Constraint: 3x+2y

x+Y & lt； =4000,

x+3Y & lt； =5000,

x，Y & gt=0。

Objective function: max z = (300x-50y-2000)+(320y-80y-2500).

=250X+240Y-4500 .

X=0 and Y= 1720 are obtained by graphic method. At this time Z=408300.

In the same way, listing the E-C and E-C models separately, we can get

Ethylene and propylene: X= 1000, Y= 1500. At this time Z=729500.

A and C: X=600, Y= 1700. At this time Z=706000.

To sum up, scheme B and scheme C should be selected, and max Z=729500.