Bricks in a Box

Solution

The brick is \(15\times5\times8=600 cm^3\)

The box is \(130\times80\times100=1 040 000 cm^3\)

A common misconception is then \(1 040 000\div 600=1733.33...\) so the box can hold \(1733\) because we have to round down. The problem there is that some bricks might need cutting up to fit in the remaining space.

Here's another question that serves as a prompt to reconsider the first part.

 

One brick doesn't even fit in the box, despite the volumes being \(40 cm^3\) and \(2000 cm^3\) suggesting that \(50\) boxes should fit.

So how do we tackle the first part?

This is an example of a packing problem and they get exceptionally difficult. Even with 11 squares in 2D, this is incredibly complicated, as Walter Trump showed in 1979 with his diagram below showing an optimal packing.

  

Our problem involves 3 dimensions and I'm almost certain I don't have the optimal packing.

We could arrange the bricks in different orientations, keeping faces of the bricks parallel to those of the box. For each orientation we can include a maximum number of rows, columns and layers. Here are those results.

LWH - 8, 16, 12 - 1536
LHW - 8, 10, 20 - 1600
WLH - 26, 5, 12 - 1560
WHL - 26, 10, 6 - 1560
HWL - 16, 16, 6 - 1536
HLW - 16, 5, 20 - 1600

The LHW orientation allows us to fit 8 rows of 15cm into the 130cm length, 10 columns of 8cm into the 80cm width and 20 layers of 5cm into the 100cm height, resulting in 1600 boxes. This is a new lower bound, but I suspect there are ways of stacking bricks diagonally once the box is nearing capacity to optimise the number.

 

 I'm going no further with this. I'll leave that as an exercise for you.

Ultimately, the point of this was to get students thinking about whether their calculations and the mathematical findings apply to the real-life situation. Sometimes we need to stop and look at the bigger picture.