Optimally packing $160 \times 80 \times 40$ boxes into a $1250 \times 230 \times 260$ space

What is the largest number of boxes of sizes $160 \times 80 \times 40$ that can be located in a space of $1250 \times 230 \times 260$ (long, wide, high)?

In a very crude way, I managed to locate $3\times7\times5+3\times3+1=115$ boxes, but I think more boxes can be located. Dividing the respective volumes, I get aproximately $146$ boxes and my answer is far from that (I know that it's not possible that amount) but I want to know how many more I can locate.

You've done very well. If we divide everything by $40$ your boxes are $1 \times 2 \times 4$ and your space is $31.5 \times 5.75 \times 6.5$. If you aren't willing to tilt the boxes the fractions are useless. Then the volume comparison becomes $$\frac {31 \cdot 5 \cdot 6}{1 \cdot 2 \cdot 4}=\frac {930}8=116.25$$ so you are only one box short of ideal.
• @JohnMa: I don't know for sure. I strongly suspect it is $115$. My point was that OP thought there was a lot of volume left, but a lot of that come from the fractional parts which don't matter unless you are willing to rotate the boxes a bit. He seemed unwilling to do that. Commented May 12, 2018 at 2:39