① formula: x+y+z = 20 ② formula: x+2y+3z = 50.
②-① Formula ③:y+2z = 30 can be obtained.
3× ①-② can get ④ formula: 2x+y = 10.
④-③ The formula ⑤:2z-2x = 20→z = x+ 10 can be obtained.
The following formula can be obtained: ③ Formula: y+2z = 30 (available: Z≤ 15).
④ Formula: 2x+y = 10
⑤ formula: x+10 = z.
So there is the following answer:
1. Suppose X = 1, then Y = 8, Z = 1 1, and the cost =1×120+8×160+65438.
2. If X = 2, then Y = 6, Z = 12, and cost = 2×120+6×160+12×180 = 3360;
3. Suppose X = 3, then Y = 4, Z = 13, and the cost = 3x120+4x160+13x180 = 3340;
4. Suppose X = 4, then Y = 2, Z = 14, and the cost = 4×120+2×160+14×180 = 3320;
5. Suppose X = 5, then Y = 0, Z = 15, and the cost = 5×120+0×160+15×180 = 3300;
Therefore, the freight car scheduling that can hold 1, 2, and 3 pieces at a time: 5, 0, and 15 respectively.
* * * There are five arrangements.
The fifth kind of freight is the least, and the lowest freight is 3300.
(In fact, according to formulas ③, ④ and ⑤, it can be concluded that every time X increases 1 vehicle, Y decreases by 2 vehicles and Z increases 1 vehicle).
The unit marginal cost MC =/kloc-0 /×120-2×160+/kloc-0 /×180 =-20.
Therefore, when y is the smallest, the cost is the lowest.