# Open box Problem.

Extracts from this document...

Introduction

Open box Problem

Introduction

Now I am going to investigate the size of the cut out square which makes an open box of the largest volume. Let the length of the square be 24cm, the card is shown as below:

Method

Now, I am going to set the length of the square to be 24cm and the other one will be 12cm. Then I will find out the relation between the two experiment and find out a formula to calculate the max. volume for any size of square. Then I will explain why the formula works for all squares.

I have to find out the value of X to give the maximum of volume.

First, I have to find out the formula to calculate the volume of the open box. Base on the formula, the volume= height x length x width.

Let the height be X.

Then the length is 24cm – X- X .

Therefore we get X(24cm-2X)

With a cut out of 1cm,X=1cm, 11cm is the maximum for there to be a box left.

Here are the results:

X(cm) | X(24cm-2X)(cm) |

1 | 484 |

2 | 800 |

3 | 972 |

4 | 1024 |

5 | 980 |

6 | 864 |

7 | 700 |

8 | 512 |

9 | 324 |

10 | 160 |

11 | 44 |

We found out that when X=4, we can get the maximum volume in integer.

Now I am going to find out the volume when X is between 3.5 and 4.5.

Here are the results:

X(cm) | X(24cm-2X)(cm) |

3.5 | 1011.5 |

3.6 | 1016.064 |

3.7 | 1019.572 |

3.8 | 1022.048 |

3.9 | 1023.516 |

4.0 | 1024 |

4.1 | 1023.524 |

4.2 | 1022.112 |

4.3 | 1019.788.4.4 |

4.4 | 1016.516 |

4.5 | 1012.5 |

We found out that 4cm is still the biggest value whereas we can get the biggest volume.

Middle

X(cm)

(2W – X –X )*(W – X - X)*X(cm)

2.1

324.324

2.2

327.712

2.3

330.188

2.4

331.776

2.5

332.5

2.6

332.384

2.7

331.452

2.8

329.728

2.9

327.236

3

324

We can see that 2.5cm is the maximum value of X. To get further confirmation of the result, I will try X between 2.41cm and 2.6cm.

X(cm) | (2W – X –X )*(W – X - X)*X(cm) |

2.41 | 331.8869 |

2.42 | 331.9892 |

2.43 | 332.0828 |

2.44 | 332.1679 |

2.45 | 332.2445 |

2.46 | 332.3125 |

2.47 | 332.3721 |

2.48 | 332.4232 |

2.49 | 332.4658 |

2.5 | 332.5 |

2.51 | 332.5258 |

2.52 | 332.5432 |

2.53 | 332.5523 |

2.54 | 332.5531 |

2.55 | 332.5455 |

2.56 | 332.5297 |

2.57 | 332.5056 |

2.58 | 332.4732 |

2.56 | 332.4327 |

2.57 | 332.384 |

## We can see that the maximum value of X is 2.54cm

Now I am going to try between 2.535 – 2.545

X | (2W – X –X )*(W – X - X)*X |

2.535 | 332.5537 |

2.536 | 332.5538 |

2.537 | 332.5537 |

2.538 | 332.5536 |

2.539 | 332.5534 |

2.54 | 332.5531 |

2.541 | 332.5527 |

2.542 | 332.5522 |

2.543 | 332.5517 |

2.544 | 332.551 |

2.545 | 332.5503 |

We can see that the max. value of X is 2.536cm, that means 2.54cm correct to 2 decimal places.

Then we divide the width by X, 12/2.54, we get 4.724

I expect this will be the result for all rectangles in the same ratio.

Now I am going to try W=6cm and 2W=12cm, with also the same ratio 1:2.

## With a cut out of 1cm,X=1cm, 2cm is the maximum for there to be a box left.

X(cm) | (2W – X –X )*(W – X - X)*X(cm) |

1 | 40 |

2 | 32 |

## We can see that the maximum value of X is 1cm

I will try the value X between 1 and 2.

X(cm) | (2W – X –X )*(W – X - X)*X(cm) |

1 | 40 |

1.1 | 40.964 |

1.2 | 41.472 |

1.3 | 41.548 |

1.4 | 41.216 |

1.5 | 40.5 |

1.6 | 39.424 |

1.7 | 38.012 |

1.8 | 36.288 |

1.9 | 34.276 |

2 | 32 |

## 1.3cm is the maximum value of X. To further confirm the result, I will try X between 1.21cm and 1.3cm.

X(cm) | (2W – X –X )*(W – X - X)*X(cm) |

1.21 | 41.49864 |

1.22 | 41.52099 |

1.23 | 41.53907 |

1.24 | 41.5529 |

1.25 | 41.5625 |

1.26 | 41.5679 |

1.27 | 41.56913 |

1.28 | 41.56621 |

1.29 | 41.55916 |

1.3 | 41.548 |

From the results we can see, 1.27cm is the maximum value of X.

Now I am going to try between 1.265 and 1.275.

Conclusion

2.915

2053.41

2.916

2053.413

2.917

2053.415

2.918

2053.417

2.919

2053.4180

2.92

2053.4193

2.921

2053.4192

2.922

2053.4191

2.923

2053.418

2.924

2053.417

2.925

2053.416

We can see 2.92cmis the biggest value for the maximum volume of the box.

We then divide the width by X, 12cm/2.92cm= 4.10958

As I say above, for this ratio 1:10 of a rectangle, I will expect the formula will be

X=W/4.10958

Ratios | Formula |

1:1 | X=W/6 |

1:2 | X=W/4.72 |

1:3 | X=W/4.43 |

1:4 | X=W/4.11 |

As the ratio of sides increase, the value of the width/max. value will decrease, as the ratio is bigger, the value will get closer to 4, as I predicted on the graph, so we assume that the value is 4, then we get this formula:

X=W/4

## This is the formula to calculate the approximate maximum value of X for a rectangle. So if the ratio get bigger, you can get a approximate value of X

The formula for calculating the box is:

V=X(W-X-X)(L-X-X)

We can simplify this formula like that:

V=X(W-2X)(L-2X)

V=(XW-2X)(L-2X)

V=XWL- 2XL- 2XW +4X

We put X= into this formula, we will get:

V=()WL- 2()L- 2()W +4()

After simplifying it, and then we will get the final formula:

Here is the formula to calculate the maximum volume of a rectangle box. But this formula can only find out the approximate volume of the box. As the ratio get bigger, you can get a more accurate result. Because the value of X is not the same on different ratio of rectangle and the value is getting nearer to 4 when at a bigger ratio. By this formula, you can very easily find out the approximate maximum volume of a rectangle box.

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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