Cubes in Cuboid

(Rant to explain why my solution is so long)

A common assumption is that "We can do no better than maximizing the optimization along each side", hence the maximum number of cubes would be

$\lfloor \frac{400}{3} \rfloor \times \lfloor \frac{300}{3} \rfloor \times \lfloor \frac{90}{3} \rfloor = 399000$

As it turns out, for this assumption to be valid, we actually require that the faces of the unit cubes are parallel to the faces of the larger rectangular box, which I've since added to the problem. For example, try this problem of packing 5 unit squares , which has an answer that is smaller than 3. For more information (on the 2-D case), see Erdos's paper On packing squares with equal squares .

---

(The rest of the solution establishes why with the assumption of parallel faces, then the naive optimization is the best possible. Note that this step is skipped in most solutions.)

Notice that since we are dividing the figure up into cubes, it doesn't matter what the side lengths are as we can simply scale the figure. Henceforth, we will assume that we're using unit cubes.

Before we get started, let's analyze the 1-dimensional and 2-dimensional situation first. Understanding how to approach this problem would greatly help with generalizing the argument further to 3 dimensions.

In a line segment of side length $l$ , what is the maximum number of unit line segments that we can place within the original line segment? The unit line segments are allowed to touch, but not overlap.

In this case, since each unit line segment has length 1, thus we are clearly allowed to have a maximum of $\lfloor l \rfloor$ of them.

Let's move on to the 2-dimensional case:

In a rectangle of side length $l \times m$ , what is the maximum number of unit squares that we can place within the original rectangle? The unit squares are allowed to touch, but not overlap.

If we repeated the previous argument, which was based on the measure of "volume", we would say that the rectangle has area $l \times m$ , and so we can fit at most $\lfloor l \times m \rfloor$ squares. Note that while this is a valid upper bound, it is not the best upper bound in all cases. For example, with $l = m = 1.99$ , we cannot fit 3 unit squares without overlap. (Can you prove this?)

Notice that not all area of the square is useful to us. In particular, for the counter example of $l = m = 1.99$ , we see that after fitting in 1 square, the rest of the area does not allow us to fit in another square. At this point, the mathematician in you should realize that considering "area/length" as we did in the 1-dimensional case, isn't the right way to generalize this problem. Instead, we might have to find a better characterization.

An observation in the $l = m = 1.99$ case is that the center of the large square must lie in any unit square that is contained within, hence we can have at most 1 square. More generally, if we consider the various lattice points in the rectangle, our hope is to associate each point to a square, which then (might) gives us a bound on the number of squares.

Let's see how to rigorize this. For any $l \times m$ rectangle, place it on a grid with the lower left corner at the origin. Consider any placement of unit squares.
Define the interesting lattice points $(i, j)$ with $1 \leq i \leq \lfloor l \rfloor$ and $1 \leq j \leq \lfloor m \rfloor$ . For each lattice point, we map it to the unit square which touches or contains it, and has the smallest "upper right" corner by lexicographic ordering. For example, in a $2 \times 2$ squares, though all 4 grid squares touch the lattice point $(1, 1)$ , the "upper right corners" are $(1, 1), (2, 1), (1, 2), (2, 2)$ , and so we map the lattice point to the unit square whose upper right corner is $(1,1)$ .
Conversely, let's show that each unit square has to touch at least one of these lattice points. If the upper right corner of the unit square is $(x, y)$ , then the lattice point $(\lfloor x \rfloor , \lfloor y \rfloor )$ will be mapped to it. Note that this mapping need not be unique, so we have a surjection from the interesting lattice points to the unit squares. This establishes that the number of unit squares is at most equal to the number of interesting lattice points, of which there are $\lfloor l \rfloor \times \lfloor m \rfloor$ .

Now, going back to the 1-D case, it is clear that the map of lattice point to unit line segments also provides us with a simple way to perform the length counting argument. Furthermore, it is clear how we can extend this argument to 3-D, by considering the interesting lattice points $(i, j , k)$ with $1 \leq i \leq \lfloor l \rfloor , 1 \leq j \leq \lfloor m \rfloor , 1 \leq k \leq \lfloor n \rfloor$ , hence we are done.

---

Do you see where in the argument we used the fact that the unit squares are parallel to the sides of the rectangle? This also helps to explain why tilting squares might eventually allow for a larger number of unit squares to be fitted in.

The maximum number of cubes along the length= $\frac {400}{3}$ but since we cannot put fractions of cubes inside it, so

instead of $133.33$ we have to take it as $133$ . For width and height, $\frac {300}{3}$ and $\frac {90}{3}$ are already integers, so we can

now multiply all to get our result, which is $133×100×30=\boxed {399000}$

Answer = floor(400/3) x floor(300/3) x floor(90/3) = floor(133.333) x floor(100) x floor(30) = 133 x 100 x 30 = 399000