# Open Box Problem.

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Introduction

Maths Coursework: Problem Solving Investigation

Tahamtan Pishgharavol

Open Box Problem

Aim

During this project I will be determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card.

What is an Open Box

An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below. The card is then folded along the dotted lines to make the box.

Names of things needed for investigation

I will write-up my investigation in Microsoft Word and all formulae shall be calculated on Microsoft Excel and all table and graph will be produce in spreadsheets again in Microsoft Excel.

Structure of investigation

- Evidence: ∙ Table

- Graphs
- Formulas

- Evaluation

1. Evidence

To obtain evidence I will be used a series of methods:

- Table
- Graphs
- Formulae

Part 1, Square

I am going investigate 3 different sizes for the square open box. Once I have obtained all information on the 3 sizes I will look for patterns and try to formulate a rule to work out the largest volume for an open box square. The sizes that I will be using are:

- 20 x 20
- 40 x 40
- 25 x 25

Because it is a square the length = width so, we can write this as L=1W therefore there is a ratio 1:1.

I am going to begin by investigating a square with a side length of 20cm. Using this side length, the maximum whole number I can cut off each corner is 9cm, as otherwise I would not have any box left.

I am going to begin by looking into whole numbers being cut out of the box corners.

The formula that needs to be used to get the volume of a box is:

Volume = Length x Width x Height

Middle

=C12*(32-2*C12)^2

11

=A13*(32-2*A13)^2

6

=C13*(32-2*C13)^2

12

=A14*(32-2*A14)^2

6.1

=C14*(32-2*C14)^2

13

=A15*(32-2*A15)^2

6.2

=C15*(32-2*C15)^2

14

=A16*(32-2*A16)^2

6.3

=C16*(32-2*C16)^2

15

=A17*(32-2*A17)^2

6.4

=C17*(32-2*C17)^2

As you can see from table, the nth is correct. Using the rule we found out that X = 5.3 and V = 2427.188.

From the table we see that X does equal 5.3 and V does equal 2427.188, therefore the nth rule is correct and can be used to find out the largest volume for a square open box, for any give number.

1. Evidence

To obtain evidence I will be used a series of methods:

- Table
- Graphs
- Formulae

Part 2, rectangle

I am now going to continue my investigation by looking at the shape of rectangles.

I am going investigate 3 different ratios for the rectangular open box. With each ratio I’m going consider 3 sizes.

Once I have obtained all information on the 3 ratios, I will look for patterns and try to formulate a rule to work out the largest volume for an open box square. The ratios that I will be using are:

- L=2W Ratio: 1:2

- 10 x 20
- 15 x 30
- 20 x 40

- L=3W Ratio: 1:3

- 10 x 30
- 15 x 45
- 20 x 60

- L=4W Ratio: 1:4

- 10 x 40
- 15 x 60
- 20 x 80

The formula that needs to be used to get the volume of a box is:

Volume = Length x Width x Height

The width of the rectangle is double the length (ratio 1:2) so width = 2W

As square cut out isX, the height will become X. Therefore the formula for volume will be:

V = X(l-2 X)(2w-2 X)

V = Volume

l = Length

X = Height and cut off

W = Width

I shall begin with a width of 10cm, and a length of 20cm, this is a ratio of 1:2, the width being twice as long as the length.

The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 4cm being the maximum for there to be a box left.

I’m to going cut out of length 1cm, the equation for this would be as follows:

Volume = 1(10-2x1)(2x20-2x1)

So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:

1(10-2x1)(2x20-2x1)

1 x 8 x 38

= 304³

I used the formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 4cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 10x20 volume | |

x | Volume |

1 | 304 |

2 | 432 |

3 | 408 |

4 | 256 |

As you can see by the table above, the largest volume is achieved when an area of 2cm² is cut off each corner of the box. I have also drawn a graph to show my results. By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 2 and 3cm².

In the table below I have shown the formulae I have inputted into the spreadsheet to attain the above results.

Results for rectangle size 10x20 volume | |

x | Volume |

1 | =A3*(10-2*A3)*(2*20-2*A3) |

2 | =A4*(10-2*A4)*(2*20-2*A4) |

3 | =A5*(10-2*A5)*(2*20-2*A5) |

4 | =A6*(10-2*A6)*(2*20-2*A6) |

To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place. We can see that the maximum box area is made from the cut size of 2cms. Now I will try between 2-3cms

Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.

Results for rectangle size 10x20 volume | |

x | Volume |

2 | 432 |

2.1 | 436.044 |

2.2 | 438.592 |

2.3 | 439.668 |

2.4 | 439.296 |

2.5 | 437.5 |

2.6 | 434.304 |

2.7 | 429.732 |

2.8 | 423.808 |

2.9 | 416.556 |

3 | 408 |

As you can see by the table above, the largest volume is achieved when an area of 2.3cm² is cut off each corner of the box. I have also drawn a graph to show my results.

Now I’m going to investigate a rectangle with a side length of 15x30cm. The maximum whole number I can cut off each corner is 6cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.

Using the formula:

V = X(l-2 X)(2w-2 X)

I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 6cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 15x30 volume | |

x | Volume |

1 | 754 |

2 | 1232 |

3 | 1458 |

4 | 1456 |

5 | 1250 |

6 | 864 |

As you can see by the table above, the largest volume is achieved when an area of 3cm² is cut off each corner of the box. I have also drawn a graph to show my results. By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 3 and 4cm².

To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place. We can see that the maximum box area is made from the cut size of 3cms. Now I will try between 3-4cms

Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.

Results for rectangle size 15x30 volume | |

x | Volume |

3 | 1458 |

3.1 | 1467.664 |

3.2 | 1475.072 |

3.3 | 1480.248 |

3.4 | 1483.216 |

3.5 | 1484 |

3.6 | 1482.624 |

3.7 | 1479.112 |

3.8 | 1473.488 |

3.9 | 1465.776 |

4 | 1456 |

As you can see by the table above, the largest volume is achieved when an area of 3.5cm² is cut off each corner of the box. I have also drawn a graph to show my results.

Now I’m going to investigate a rectangle with a side length of 20x40cm. The maximum whole number I can cut off each corner is 9cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.

Using the formula:

V = X(l-2 X)(2w-2 X)

I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 9cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 20x40 volume | |

x | Volume |

1 | 1404 |

2 | 2432 |

3 | 3108 |

4 | 3456 |

5 | 3500 |

6 | 3264 |

7 | 2772 |

8 | 2048 |

9 | 1116 |

As you can see by the table above, the largest volume is achieved when an area of 5cm² is cut off each corner of the box. I have also drawn a graph to show my results. By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 4 and 5.5cm².

To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place. We can see that the maximum box area is made from the cut size of 5cms. Now I will try between 4-5.5cms

Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.

Results for rectangle size 20x40 volume | |

x | Volume |

4 | 3456 |

4.1 | 3473.684 |

4.2 | 3488.352 |

4.3 | 3500.028 |

4.4 | 3508.736 |

4.5 | 3514.5 |

4.6 | 3517.344 |

4.7 | 3517.292 |

4.8 | 3514.368 |

4.9 | 3508.596 |

5 | 3500 |

5.1 | 3488.604 |

5.2 | 3474.432 |

5.3 | 3457.508 |

5.4 | 3437.856 |

5.5 | 3415.5 |

As you can see by the table above, the largest volume is achieved when an area of 4.6cm² is cut off each corner of the box. I have also drawn a graph to show my results.

In the table below is the records from results for l=2w ratio 1:2.

L=2W | Largest Volume - Integer | Largest Volume – 1 d.p | ||

10x20 | X = 2 | Volume = 432 | X = 2.3 | Volume = 439.668 |

15x30 | X = 3 | Volume = 1458 | X = 3.5 | Volume = 1484 |

20x40 | X = 5 | Volume = 3500 | X = 4.6 | Volume = 3517.344 |

To work out the proportion between the size of the rectangle and cut-off we divide:

10/2.4 = 4.17

15/3.5 = 4.29

20/4.6 =4.35

And then we find the average,

4.17 + 4.29 + 4.35

3

=4.27

Therefore if you multiply any of the 1 d.p results by 4.27 you will get a result which is close to the size.

2.4 x 4.27 = 10.248 to 2 significant figures this is 10

3.5 x 4.27 = 14.945 to 2 significant figures this is 15

4.6 x 4.27 = 19.642 to 2 significant figures this is 20

Rearrange this rule you can get the formula to work out the highest volume for a rectangular open box at the ratio 1:2:

X = L

4.27

So the rule to work out X for the ratio 1:2 is X = L

4.27

Now that we know the formula to get X we can substitute X in the formula V = X(l-2 X)(2w-2 X) to get the volume.

Now will start to investigate rectangles where L=3W.

I will begin with a width of 10cm, and a length of 30cm, this is a ratio of 1:3, the width being 3 times as long as the length.

The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 4cm being the maximum for there to be a box left.

I’m to going cut out of length 1cm, the equation for this would be as follows:

Volume = 1(10-2x1)(3x30-2x1)

So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:

1(10-2x1)(3x30-2x1)

1 x 8 x 88

= 704cm³

I used the formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 4cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 10x30 volume | |

x | Volume |

1 | 704 |

2 | 1032 |

3 | 1008 |

4 | 656 |

As you can see by the table above, the largest volume is achieved when an area of 2cm² is cut off each corner of the box. I have also drawn a graph to show my results. By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 2 and 3cm².

In the table below I have shown the formulae I have inputted into the spreadsheet to attain the above results.

Results for rectangle size 10x30 volume | |

x | Volume |

1 | =A3*(10-2*A3)*(3*30-2*A3) |

2 | =A4*(10-2*A4)*(3*30-2*A4) |

3 | =A5*(10-2*A5)*(3*30-2*A5) |

4 | =A6*(10-2*A6)*(3*30-2*A6) |

To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place. We can see that the maximum box area is made from the cut size of 2cms. Now I will try between 2-3cms

Results for rectangle size 10x30 volume | |

x | Volume |

2 | 1032 |

2.1 | 1045.044 |

2.2 | 1054.592 |

2.3 | 1060.668 |

2.4 | 1063.296 |

2.5 | 1062.5 |

2.6 | 1058.304 |

2.7 | 1050.732 |

2.8 | 1039.808 |

2.9 | 1025.556 |

3 | 1008 |

As you can see by the table above, the largest volume is achieved when an area of 2.4cm² is cut off each corner of the box. I have also drawn a graph to show my results.

Now I’m going to investigate a rectangle with a side length of 15x45cm. The maximum whole number I can cut off each corner is 6cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.

Using the formula:

V = X(l-2 X)(3w-2 X)

I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 6cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 15x45 volume | |

x | Volume |

1 | 1729 |

2 | 2882 |

3 | 3483 |

4 | 3556 |

5 | 3125 |

6 | 2214 |

As you can see by the table above, the largest volume is achieved when an area of 4cm² is cut off each corner of the box. I have also drawn a graph to show my results. By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 3 and 4.5cm².

To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place. We can see that the maximum box area is made from the cut size of 4cms. Now I will try between 3-4.5cms

Results for rectangle size 10x30 volume | |

x | Volume |

3 | 3483 |

3.1 | 3513.664 |

3.2 | 3539.072 |

3.3 | 3559.248 |

3.4 | 3574.216 |

3.5 | 3584 |

3.6 | 3588.624 |

3.7 | 3588.112 |

3.8 | 3582.488 |

3.9 | 3571.776 |

4 | 3556 |

4.1 | 3535.184 |

4.2 | 3509.352 |

4.3 | 3478.528 |

4.4 | 3442.736 |

4.5 | 3402 |

As you can see by the table above, the largest volume is achieved when an area of 3.6cm² is cut off each corner of the box. I have also drawn a graph to show my results.

Now I’m going to investigate a rectangle with a side length of 20x60cm. The maximum whole number I can cut off each corner is 9cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.

Using the formula:

V = X(l-2 X)(3w-2 X)

I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 9cm2. Below are the results I got through this spreadsheet.

By using a table all results are shown in a clear way and is easier to look for a pattern.

Results for rectangle size 20x60 volume | |

x | Volume |

1 | 3204 |

2 | 5632 |

3 | 7308 |

4 | 8256 |

5 | 8500 |

6 | 8064 |

7 | 6972 |

8 | 5248 |

9 | 2916 |

Conclusion

2.5 x 4.04 = 10.1 to 2 significant figures this is 10

3.7 x 4.04 = 14.948 to 2 significant figures this is 15

4.9 x 4.04 = 19.796 to 2 significant figures this is 20

Rearrange this rule you can get the formula to work out the highest volume for a rectangular open box at the ratio 1:4:

X = L

4.04

So the rule to work out X for the ratio 1:4 is X = L

4.04

Now that we know the formula to get X we can substitute X in the formula V = X(l-2 X)(2w-2 X) to get the volume.

- Evaluation for open box – Rectangles

Rectangle Size | X = L Y |

L=2W | Y = 4.27 |

L=3W | Y = 4.14 |

L=4W | Y = 4.04 |

X=L

4

Is the limit. There is no smaller denominator

We can use this to calculate the cut-off size then substitute it in the volume equation.

Nth rule

L:nW

X = L

4

V= X(l-2X)(nW-2X)

Putting the nth rule into action

Find the biggest volume for size square 32

Rule

n = L

6

V = n(L-2n)2

X = 32

6

X = 5.3

V = 5.3(32-2 x 5.3)2

V = 2427.188

Proving that this is correct

Results for square size 32 volume | |||

x | Volume Integer | x | Volume 1 d.p |

1 | 900 | 5 | 2420 |

2 | 1568 | 5.1 | 2423.724 |

3 | 2028 | 5.2 | 2426.112 |

4 | 2304 | 5.3 | 2427.188 |

5 | 2420 | 5.4 | 2426.976 |

6 | 2400 | 5.5 | 2425.5 |

7 | 2268 | 5.6 | 2422.784 |

8 | 2048 | 5.7 | 2418.852 |

9 | 1764 | 5.8 | 2413.728 |

10 | 1440 | 5.9 | 2407.436 |

11 | 1100 | 6 | 2400 |

12 | 768 | 6.1 | 2391.444 |

13 | 468 | 6.2 | 2381.792 |

14 | 224 | 6.3 | 2371.068 |

15 | 60 | 6.4 | 2359.296 |

Results for square size 32 volume | |||

x | Volume Integer | x | Volume 1 d.p |

1 | =A3*(32-2*A3)^2 | 5 | =C3*(32-2*C3)^2 |

2 | =A4*(32-2*A4)^2 | 5.1 | =C4*(32-2*C4)^2 |

3 | =A5*(32-2*A5)^2 | 5.2 | =C5*(32-2*C5)^2 |

4 | =A6*(32-2*A6)^2 | 5.3 | =C6*(32-2*C6)^2 |

5 | =A7*(32-2*A7)^2 | 5.4 | =C7*(32-2*C7)^2 |

6 | =A8*(32-2*A8)^2 | 5.5 | =C8*(32-2*C8)^2 |

7 | =A9*(32-2*A9)^2 | 5.6 | =C9*(32-2*C9)^2 |

8 | =A10*(32-2*A10)^2 | 5.7 | =C10*(32-2*C10)^2 |

9 | =A11*(32-2*A11)^2 | 5.8 | =C11*(32-2*C11)^2 |

10 | =A12*(32-2*A12)^2 | 5.9 | =C12*(32-2*C12)^2 |

11 | =A13*(32-2*A13)^2 | 6 | =C13*(32-2*C13)^2 |

12 | =A14*(32-2*A14)^2 | 6.1 | =C14*(32-2*C14)^2 |

13 | =A15*(32-2*A15)^2 | 6.2 | =C15*(32-2*C15)^2 |

14 | =A16*(32-2*A16)^2 | 6.3 | =C16*(32-2*C16)^2 |

15 | =A17*(32-2*A17)^2 | 6.4 | =C17*(32-2*C17)^2 |

As you can see from table, the nth is correct. Using the rule we found out that X = 5.3 and V = 2427.188.

From the table we see that X does equal 5.3 and V does equal 2427.188, therefore the nth rule is correct and can be used to find out the largest volume for a square open box, for any give number.

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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