Packing Cylindrical Pencils in a Square Stand

The problem asks for the maximum number of cylindrical pencils, each with a diameter of $0.5\text{ cm}$, that can fit inside a square stand with an inner cross-section of $5\text{ cm} \times 5\text{ cm}$. This is a 2D circle packing problem where we need to fit circles of diameter $d = 0.5\text{ cm}$ into a square of side length $S = 5\text{ cm}$.

Grid Packing Arrangement

First, consider arranging the pencils in a simple square grid. The number of pencils that can fit along one side of the square stand is the floor of the stand's side length divided by the pencil's diameter:

Number per side = $\lfloor \frac{S}{d} \rfloor = \lfloor \frac{5\text{ cm}}{0.5\text{ cm}} \rfloor = \lfloor 10 \rfloor = 10$.

In a grid arrangement, the total number of pencils would be the square of the number per side:

Total Pencils (Grid) = $10 \times 10 = 100$.

Hexagonal Packing Arrangement

Circles can often be packed more densely using a hexagonal arrangement. In this method, rows are offset, allowing circles in one row to nestle into the gaps of the row below.

1. Pencils per Row: The maximum number of pencils that can fit side-by-side along the $5\text{ cm}$ width remains $10$. The total width occupied by $10$ pencils is $10 \times 0.5\text{ cm} = 5\text{ cm}$.
2. Vertical Spacing: In hexagonal packing, the vertical distance between the centers of circles in adjacent rows is $h = d \frac{\sqrt{3}}{2}$.

   $h = 0.5\text{ cm} \times \frac{\sqrt{3}}{2} \approx 0.5 \times 0.866 = 0.433\text{ cm}$.

3. Number of Rows: The total height required for $k$ rows is the diameter of the first row plus the cumulative spacing for the subsequent rows: $d + (k-1)h$. This must be less than or equal to the stand's height ($5\text{ cm}$).

   $0.5\text{ cm} + (k-1) \times 0.433\text{ cm} \le 5\text{ cm}$

   $(k-1) \times 0.433 \le 4.5$

   $k-1 \le \frac{4.5}{0.433} \approx 10.39$

   $k \le 11.39$. Therefore, the maximum number of rows ($k$) is $11$.

4. Pencils per Staggered Row: When rows are staggered, alternate rows might accommodate fewer pencils if perfectly centered in the gaps. If the first row fits $10$ pencils perfectly ($5\text{ cm}$ width), the second (staggered) row fits $9$ pencils centered in the gaps. The total width occupied by $9$ pencils is $9 \times 0.5\text{ cm} = 4.5\text{ cm}$, which fits within the $5\text{ cm}$ stand width.
5. Total Calculation: With $11$ rows, assuming rows alternate between fitting $10$ and $9$ pencils:
   • Number of rows with 10 pencils = $\lceil \frac{11}{2} \rceil = 6$ (Rows 1, 3, 5, 7, 9, 11)
   • Number of rows with 9 pencils = $\lfloor \frac{11}{2} \rfloor = 5$ (Rows 2, 4, 6, 8, 10)
   Total Pencils (Hexagonal) = $(6 \times 10) + (5 \times 9) = 60 + 45 = 105$.

Comparing the grid packing ($100$ pencils) and the calculated hexagonal packing ($105$ pencils), the hexagonal arrangement allows for more pencils.