The Open Box Problem

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.(fig 1)

FIG 1

The card is then folded along the dotted lines to make the box.

The main aim is to 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.

I am going to begin by investigating a square with a side length of 20 cm. 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: ( * = multiply)

Volume = Length * Width * Height

If I am to use a square of side length 20cm, then I can calculate the side lengths minus the cut out squares using the following equation.

Volume = Length -- (2 * Cut Out) * Width -- (2 * Cut Out) * Height

Using a square, both the length & the width are equal. I am using a length/width of 20cm. I am going to call the cut out ""x."" Therefore the equation can be changed to:

Volume = 20 -- (2x) * 20 -- (2x) * Height

If I were using a cut out of length 1cm, the equation for this would be as follows:

Volume = 20 -- (2 * 1) * 20 -- (2 * 1) * 1

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

(20 -- 2) * (20 -- 2) * 1

8 * 18 * 1

= 324cm³

I used these formulae to construct a table, which would allow me to quickly and accurately calculate the volume of the box.

Cut out (cm²)

formula

volume (cm³)

Volume = 20 -- (2*1) * 20 -- (2*1) * 1

324

2

Volume = 20 -- (2*2) * 20 -- (2*2) * 2

512

3

Volume = 20 -- (2*3) * 20 -- (2*3) * 3

588

4

Volume = 20 -- (2*4) * 20 -- (2*4) * 4

576

5

Volume = 20 -- (2*5) * 20 -- (2*5) * 5

500

6

Volume = 20 -- (2*6) * 20 -- (2*6) * 6

384

7

Volume = 20 -- (2*7) * 20 -- (2*7) * 7

252

8

Volume = 20 -- (2*8) * 20 -- (2*8) * 8

28

9

Volume = 20 -- (2*9) * 20 -- (2*9) * 9

36

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. (See below fig 2)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 cm². This is because the volume the second highest volume is 4cm² cutout, so the highest has got to be between 3 and 4 cm²

FIG 2

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.

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

Cut out (cm²)

formula

Volume (cm³)

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.(fig 1)

FIG 1

The card is then folded along the dotted lines to make the box.

The main aim is to 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.

I am going to begin by investigating a square with a side length of 20 cm. 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: ( * = multiply)

Volume = Length * Width * Height

If I am to use a square of side length 20cm, then I can calculate the side lengths minus the cut out squares using the following equation.

Volume = Length -- (2 * Cut Out) * Width -- (2 * Cut Out) * Height

Using a square, both the length & the width are equal. I am using a length/width of 20cm. I am going to call the cut out ""x."" Therefore the equation can be changed to:

Volume = 20 -- (2x) * 20 -- (2x) * Height

If I were using a cut out of length 1cm, the equation for this would be as follows:

Volume = 20 -- (2 * 1) * 20 -- (2 * 1) * 1

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

(20 -- 2) * (20 -- 2) * 1

8 * 18 * 1

= 324cm³

I used these formulae to construct a table, which would allow me to quickly and accurately calculate the volume of the box.

Cut out (cm²)

formula

volume (cm³)

Volume = 20 -- (2*1) * 20 -- (2*1) * 1

324

2

Volume = 20 -- (2*2) * 20 -- (2*2) * 2

512

3

Volume = 20 -- (2*3) * 20 -- (2*3) * 3

588

4

Volume = 20 -- (2*4) * 20 -- (2*4) * 4

576

5

Volume = 20 -- (2*5) * 20 -- (2*5) * 5

500

6

Volume = 20 -- (2*6) * 20 -- (2*6) * 6

384

7

Volume = 20 -- (2*7) * 20 -- (2*7) * 7

252

8

Volume = 20 -- (2*8) * 20 -- (2*8) * 8

28

9

Volume = 20 -- (2*9) * 20 -- (2*9) * 9

36

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. (See below fig 2)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 cm². This is because the volume the second highest volume is 4cm² cutout, so the highest has got to be between 3 and 4 cm²

FIG 2

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.

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

Cut out (cm²)

formula

Volume (cm³)