1. Stacking Footballs How many footballs can you fit in a classroom?

2. Initial information Room 26 is 8m long by 8m wide by 3m tall A football has a diameter of 0.2m so a radius of 0.1m 3m 8m Room 26 r = 0.1m

3. Method 1 Volume of classroom Volume of classroom What is wrong? Space is always wasted so you cannot ever fit that many in! Volume of football Volume of football

4. Method 2 – box stacking > box size Each football fits in a box. 0.2m

5. Method 2- box stacking > putting it together Number of balls in room Number of balls in room How many balls (in boxes) fit along each side? 3m 8m What is wrong? What is wrong? Wastes half the space! Wastes half the space!

6. Method 3 – orange stacked block > height Using Pythagoras’ Theorem: 0.2m 0.1m x

7. Method 3 – orange stacked block > height We now have the basis for a vertical line: We now have the basis for a vertical line: Using this line we work out how many footballs high the orange stacked block is  0.1m 0.173m

8. Method 3 – orange stacked block > height Rule for height: Rule for height: On the diagram:

9. Method 3 – orange stacked block > width We already know that 40 balls fit in the width of the back wall. We already know that 40 balls fit in the width of the back wall. However, alternate rows will only be 39 balls wide so that the rows fit together. However, alternate rows will only be 39 balls wide so that the rows fit together. Therefore in 17 rows there will be: Therefore in 17 rows there will be: = 672 balls per orange stacked block One block is as wide as one ball so 40 blocks fit along the length of the room

10. Method 3 – orange stacked block > evaluation 672 balls per block x 40 block = 672 balls per block x 40 block = 26880 balls fit in the room using method 3 26880 balls fit in the room using method 3 What is wrong? Space wasted between blocks

11. Method 4 – complete orange stacked > width The room sideways on The room sideways on As before, we can calculate x to be 0.173m 0.2m 0.1m x

12. Method 4 – complete orange stacked > width Rule for number of blocks (width) = Rule for number of blocks (width) = So the room will contain 46 rows of orange stacked blocks However, row 2 must fit into row1 so cannot have the same number of balls in it

13. Method 4 – complete orange stacked > block 2 Block 2 Block 2 In block 2, one row must be missed out to allow block 2 to fit into block 1 Block 1 contains 17 rows so block 2 contains 16 rows

14. Method 4 – complete orange stacked > putting it together The room sideways on The room sideways on Room is 46 blocks wide = Total balls = 14536 + 15456 = 299992 balls Total balls = 14536 + 15456 = 299992 balls +

15. Method 4 – complete orange stacked > evaluation This is the best answer as complete orange packing is proved to be the most efficient way (Kepler) This is the best answer as complete orange packing is proved to be the most efficient way (Kepler) What is wrong? The value of 0.173 was rounded to 3DP so we could round the answer to 30000 balls However, you cannot have 16.5 rows so rounding is accurate. 29992 should be an accurate answer