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The Open Box Problem

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Introduction

GCSE Maths Investigation

The Open Box Problem

An open box is made form a sheet of card. Identical squares are then cut from each corner, making a cross shape. The card is then folded to make an open-lid box.

The Yellow squares are the shapes, which are removed. The box is made by folding along the dotted lines.

AIM: The main aim of this investigation is to find the relationship between the size of the rectangle cut and the volume of the box. The size of the rectangle cut which makes the volume of the box as large as possible must be determined. Remembering that a square is also a special form of a rectangle.

As well as the general aim there are two other aims:

  1. For any sized square sheet of card, investigate the size of the cut out square, which makes an open box of the largest volume.
  2. For any sized rectangular sheet of card, investigate the size of the cut out square, which makes an open box of the largest volume.

AIM 1

First I will be looking at aim 1 which uses a square sheet of card.

A square is being cut from each corner.

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Middle

A = 10

10cm

                                               8cm

                                                       10cm

                                              2cm

                                                            2cm          8cm

Y = A – 2X

The volume of the box = X x Y x Y (length x width x height)

Substitute Y with A – 2X in the equation V = X x Y x Y.

X (A – 2X) ² = V

Check:  

2 (10 – 2x2) ²

2(10 – 4) ²

2 (6) ²

2 x 36

72          

I have produced a graph to show how the volume changes according to the size of the squares cut from each corner of the original. A 10cm x 10cm piece of card was used. 1.1>X>2.2 because I know from previous calculations that the maximum point lies between X = 1cm and X = 2cm

X =  Length of one side of the squares cut out

A = length of one side of the original square before the corners are cut out.

image01.png

This graph shows that the volume reaches its maximum when X = 1.6

I noted down the maximum volume for each size of original card:

...read more.

Conclusion

Before I differentiated I put L in terms of W so the formula is simpler and doesn’t contain too many different letters, which could be confusing.

The following example demonstrates the route to answering the aim. This process can be carried out using any size rectangle but I will be using a rectangle in which L = 2W.

Therefore X (W – 2X) (2W – 2X) will be differentiated with respect to X.

Before differentiation can occur the formula must be expanded then simplified - getting rid of the brackets.

EXPAND:

V = (WX – 2X²) (2W – 2X)

V = 2W²X – 2WX² - 4X²W + 4X³

SIMPLIFY:

2W²X – 6WX² + 4X³

DIFFRENTIATE:

dV/dX must equal 0 for the maximum volume.

dV/dX = 2W² - 12WX + 12X²

This can be simplified by dividing by 2

dV/dX = W² - 6WX + 6X²

I recognized this formula as quadratic so I used the general formula:

To find out what X equaled.

A =  6

B = - 6

C = W

In a rectangle 10cm x 20cm the size of X needed to make the maximum volume is 2.1.

The area of this rectangle is 200cm². 2.1² is cut out from each corner. 2.1² = 4.41

2.205% is cut out from each corner to make the largest possible volume of box.

...read more.

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