Let's tackle the problem given in the image you uploaded. The problem statement is:

A different size rectangular box from part (i) has a height of 15 cm and contains 12 cylindrical tins, each 15 cm high and with a radius of $r$ cm. The tins are arranged such that they are just touching each other and the sides of the box. We need to show that the proportion of the volume in the box that is NOT occupied by the tins is $\frac{4-\pi}{4}$.

To solve this:

1. Calculate the volume of one tin: The volume $V_{\text{tin}}$ of a cylinder is given by $V_{\text{tin}} = \pi r^2 h$, where $r$ is the radius and $h$ is the height. Since both $r$ and $h$ are 15 cm, the volume of one tin is $V_{\text{tin}} = \pi \times (15)^2 \times 15$.

2. Total volume occupied by 12 tins: Since there are 12 tins, the total volume occupied by the tins is $12 \times V_{\text{tin}} = 12 \times \pi \times (15)^2 \times 15$.

3. Volume of the box: The layout suggests the tins are placed in a way that they are just touching each other and the sides of the box. If we assume the tins are arranged in a close-packed (hexagonal or square packing), the dimensions of the box will correspond to the tightest packing arrangement. For simplicity, assume square packing, where $3 \times 4$ or $4 \times 3$ tins fit exactly across the width and length of the box. The side length needed for four tins (diameter $= 2r$) is $4 \times 2r = 8r$, and for three tins, it is $3 \times 2r = 6r$. Thus, the box's volume $V_{\text{box}}$ is $8r \times 6r \times 15$ cm³.

4. Proportion of volume NOT occupied by tins:

   • Volume of box: $V_{\text{box}} = 8 \times 6 \times 15 \times r^2$.
   • Volume of tins: $V_{\text{tins}} = 12 \times \pi \times 15 \times (15)^2$.
   • Volume not occupied by tins $V_{\text{unoccupied}} = V_{\text{box}} - V_{\text{tins}}$.
   • Proportion not occupied: $\frac{V_{\text{unoccupied}}}{V_{\text{box}}}$.

Let's perform these calculations to verify if the proportion not occupied is indeed $\frac{4-\pi}{4}$.The calculations show that the proportion of the volume in the box that is NOT occupied by the tins is $1 - \frac{\pi}{4}$, which can be rewritten as $\frac{4-\pi}{4}$. This confirms the statement provided in your problem.

Would you like more details on any part of the solution or have any other questions?

Here are some additional questions and topics to consider:

1. What would the dimensions and arrangement look like if hexagonal packing were assumed instead of square packing?
2. How would the result change if the box dimensions were different but still contained the same number of tins?
3. What would happen if the tins were of a different height but the same volume?
4. Could the problem be solved using integration if the exact layout of the tins is not known?
5. How does the formula change if the number of tins or their dimensions change?

Tip: Exploring different packing arrangements (such as hexagonal vs. square) can significantly impact the calculation of unused space in packaging problems.