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The Open Box Investigation

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Introduction

Mitul Patel 5P

The Open Box Investigation Part 1

The aim of this investigation is to find the largest volume within for an open box with any size square cut out

I will be increasing the square cut out by 1cm until I reach a point where the volume decreases. At this point I will decrease the square cut out by 0.1cm until I reach the maximum volume. This will be done on several different grids until I see a pattern which I will then use to create a formula.

I will record my results in a table for the different grids and record the peaks to try and establish a pattern.

My initial grid size will be 12cm x 12cm and I will increase this as I continue my investigation. The volume will be calculated by multiplying the length by the width by the height. When I appear to reach a maximum volume I will try cut sizes 0.1cm smaller and larger than the cut size that appears to give the maximum volume.

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Middle

484cm3

2cm

20 x 20 x 2

800cm3

3cm

18 x 18 x 3

972cm3

4cm

16 x 16 x 4

1024cm3

5cm

14 x 14 x 5

980cm3

3.9cm

16.2 x 16.2 x 3.9

1023.5cm3

4.1cm

15.8 x 15.8 x 4.1

1023.5cm3

Shaded numbers are ones that give the maximum volume.

With the square that has a grid size of 12cm x 12cm the cut size that gives the maximum volume is 2cm and I have tested this by trying cut sizes slightly smaller and slightly larger than the cut size that appears to give the maximum volume.

The grid size of 18cm x 18cm has a maximum volume which comes from the cut size of 3cm and again this has been tested by trying cut sizes slightly smaller and slightly larger than the cut size that appears to give the maximum volume.

With a grid size of 24cm x 24cm the cut size of 4cm gives the maximum volume and this has been tested again by trying cut sizes slightly smaller and slightly larger than the cut size that appears to give the maximum volume.

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Conclusion

For three rectangles I have done where the length is twice the width I have the following results:

For the 20:10 the cut size of 2.1cm gave the maximum volume, for the 40:20 it was a cut size of 4.2cm and for the 80:40 a cut size of 8.5cm gave the maximum volume. Each of these cut sizes are approximately 1/5 of their original box width which is quite different to the square where you could find the exact maximum volume by calculating 1/6 of the grid size.

x = length

y = width

z = cut size

z = height

Length of open box = x-2z

Width of open box = y-2z

image06.png

image07.png

image13.pngimage10.pngimage03.pngimage11.pngimage08.pngimage09.pngimage02.png

The cut size that gives the maximum volume is about 1/5 of the width so this can be written as y/5. For the exact maximum volume I have made a formula which is z((y-2z)(x-2z)).

I have also investigated other ratios of, 1:3, 1:4 and 1:5 and have discovered that that the ratio of length to width doesn’t make a difference to how the maximum volume is found and the same rules as before apply to these rectangles as well.


Ratios of 2:1

image14.png

image15.png

image16.png

Ratios of 3:1, 4:1 and 5:1

image17.png

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