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A simple mathematical model
Solution: How can we save materials to the greatest extent? Don't leave any scraps on every piece of wood.

You can start:

1 root (0.8→8, 1→6, 1.2→4, 1.5→3).

6m: 0.8m * (quantity 1)+ 1m* (quantity 1)+ 1.2m* (quantity 1)+ 1.5m* (quantity 2).

: 0.8+ 1+ 1.2+3=6m (just used up)

The second root (0.8→7, 1→5, 1.2→3, 1.5→ 1).

6m: 0.8m * (quantity 5)+ 1m* (quantity 2)+ 1.2m* (quantity 0)+ 1.5m* (quantity 0)

: 4+2=6m (just used up)

The third root (0.8→2, 1→3, 1.2→3, 1.5→ 1).

6m: 0.8m * (quantity 1)+ 1m* (quantity 0)+ 1.2m* (quantity 3)+ 1.5m* (quantity 1).

: 0.8+3.6+ 1.5=5.9m (even if it can't be completely cut, choose the minimum residual material of 0. 1m).

4th root (0.8→ 1 root, 1→3 root, 1.2→0 root, 1.5→0 root)

6m: 0.8m * (quantity 1)+ 1m* (quantity 3)+ 1.2m* (quantity 0)+ 1.5m* (quantity 0).

: 0.8+3=3.8m (all necessary materials have been cut)

Finally, 3 pieces of 3.8m timber were used, and 6 pieces of 2.2m timber were left, and 0. 1m timber was discarded.