# Open Box Problem.

Extracts from this document...

Introduction

Robiul Matin 11S MA5

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.

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 10cm. Using this side length, the maximum whole number I can cut off each corner is 4cm, 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 * Width * Height

I will use a formula to test for all box sizes.

Formula

I will do a test to see if my formula works. I will use a 10*10 square and the length cut is 2cm. this should equal to 72.

xl2 – 4x2l + 4x3

= 2 * 102 – 4 * 22 * 10 + 4 * 23

= 2 * 100 - 40 * 4 + 4 * 8

= 200 – 160 + 32

= 72

I can see that my formula does work. Therefore I can use this formula when I do 3 boxes of different sizes.

Middle

392.04

1999.404

5.2

19.6

384.16

1997.632

The table shows us that as the cut is increased so did the volume, but when it reached 5cm, the volume decreased. Therefore the maximum volume was at 5cm where it had reached 2000cm. I should put the table in 2 decimal place to find an accurate result.

Length cut from each side ( x ) | Length of Base ( l ) | Area of Base ( a ) | Volume ( v ) |

4.91 | 20.18 | 407.2324 | 1999.5111 |

4.92 | 20.16 | 406.4256 | 1999.614 |

4.93 | 20.14 | 405.6196 | 1999.7046 |

4.94 | 20.12 | 404.8144 | 1999.7831 |

4.95 | 20.20 | 404.01 | 1999.8495 |

4.96 | 20.08 | 403.2064 | 1999.9037 |

4.97 | 20.06 | 402.4036 | 1999.9459 |

4.98 | 20.04 | 401.6016 | 1999.976 |

4.99 | 20.02 | 400.8004 | 1999.994 |

5 | 19.98 | 400 | 2000 |

5.02 | 19.96 | 399.2004 | 1999.994 |

4.03 | 19.94 | 398.4016 | 1999.976 |

The maximum volume I found in this table was at 5cm with its volume at 2000cm

Size of square cut = 40 by 40 (cm)

Length cut from each side ( x ) | Length of Base ( l ) | Area of Base ( a ) | Volume ( v ) |

1cm | 38 | 1444 | 1444 |

2cm | 36 | 1296 | 2592 |

3cm | 34 | 1156 | 3468 |

4cm | 32 | 1024 | 4096 |

5cm | 30 | 900 | 4500 |

6cm | 28 | 784 | 4704 |

7cm | 26 | 676 | 4732 |

8cm | 24 | 576 | 4608 |

9cm | 22 | 484 | 4356 |

10cm | 20 | 400 | 4000 |

11cm | 18 | 324 | 3564 |

12cm | 16 | 256 | 3072 |

13cm | 14 | 196 | 2548 |

14cm | 12 | 144 | 2016 |

15cm | 10 | 100 | 1500 |

16cm | 8 | 64 | 1024 |

17cm | 6 | 36 | 612 |

18cm | 4 | 16 | 288 |

19cm | 2 | 4 | 76 |

From this result I can tell the highest volume is the length cut is from 6cm – 7cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.

Length cut from each side ( x ) | Length of Base ( l ) | Area of Base ( a ) | Volume ( v ) |

6.5 | 27 | 729 | 4738.500 |

6.6 | 26.8 | 718.24 | 4740.384 |

6.7 | 26.6 | 707.56 | 4740.652 |

6.8 | 26.4 | 696.96 | 4739.328 |

6.9 | 26.2 | 686.44 | 4736.436 |

7 | 25.8 | 676 | 4732 |

The table shows us that as the cut is increased so did the volume, but when it reached 6.7cm, the volume decreased. Therefore the maximum volume was at 6.7cm where it had reached 4740.652cm. I should put the table in 2 decimal place to find an accurate result.

Length cut from each side ( x ) | Length of Base ( l ) | Area of Base ( a ) | Volume ( v ) |

6.65 | 26.7 | 717.1684 | 4740.7185 |

6.66 | 26.68 | 716.0976 | 4740.7372 |

6.67 | 26.66 | 715.0276 | 4740.7399 |

6.68 | 26.64 | 713.9584 | 4740.7265 |

6.69 | 26.62 | 712.89 | 4740.6972 |

7 | 26.6 | 711.8224 | 4740.652 |

The maximum volume I found in this table was at 6.67cm with its volume at 4740.7399cm

I made a table below to show the maximum volume of each square more clearly.

Size of square cut | Cut for volume maximum | Max volume |

10 by 10 | 1.7 | 74.07385 |

20 by 20 | 3.3 | 592.59215 |

30 by 30 | 5 | 2000 |

40 by 40 | 6.7 | 4740.7399 |

I created another table to find the relationship between the 4 squares.

Size of square cut | Cut for volume maximum | Volume/Square |

10 by 10 | 1.7 | 0.17 |

20 by 20 | 3.3 | 0.16 |

30 by 30 | 5 | 0.16 |

40 by 40 | 6.7 | 0.16 |

From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/6.

Differentiation

I am checking my results using differentiation. Differentiation is a useful way of finding the gradient at any point. If y = xn then dy/dx = nxn-1.

If the relationship is 1/6 then I choose a square divisible by 6. In this case I choose a 6 by 6 square.

Volume = y

Volume = (6 - 2x)2x

= (36 + 4x2 - 24x) x

= 36 + 4x3 - 24x2

Then dy/dx = 36 + 12x2 - 48x

Factorising = 12 (x2- 4x +3) -3 * -1 = 3

-3 + -1 = -4

= 12((x - 1) (x - 3)) = 0

x = 1 or x = 3

x = 3 isn’t possible because if u cut 3cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2

If dy/dx = 12x2 – 48x + 36 d2y/dx2 = 24x – 48

So d2y/dx2 < 0 so it is at its maximum when x = 1

(24 * 1) – 48

= 24 – 48 = -24

I choose another square divisible by 6. I choose a 12 by 12 square.

Volume = y

Volume = (12 - 2x)2x

= (144 + 4x2 - 48x) x

= 144 + 4x3 - 48x2

Then dy/dx = 144 + 12x2 - 96x

Factorising = 12 (x2- 8x +12) -6 * -2 = 12

-6 + -2 = -8

= 12((x - 2) (x - 6)) = 0

x = 2 or x = 6

x = 6 isn’t possible because if u cut 6cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2

If dy/dx = 12x2 – 96x + 144 d2y/dx2 = 24x – 96

So d2y/dx2 < 0 so it is at its maximum when x = 2

(24 * 2) – 96

= 48 – 96 = -48

I choose another square divisible by 6. I choose an 18 by 18 square.

Volume = y

Volume = (18 - 2x)2x

= (324 + 4x2 - 72x) x

= 144 + 4x3 - 72x2

Then dy/dx = 324 + 12x2 - 144x

Factorising = 12 (x2- 12x +27) -9 * -3 = 27

-9 + -3 = -12

= 12((x - 3) (x - 9)) = 0

x = 3 or x = 9

x = 9 isn’t possible because if u cut 9cm from the side it would cut the entire square leaving nothing.

If I differentiate again I can tell if the gradient at the turning point is a maximum or minimum.

dy/dx if I differentiate again it will be D dy/dx / dx = d2y/dx2

If dy/dx = 12x2 – 144x + 324 d2y/dx2 = 24x – 144

So d2y/dx2 < 0 so it is at its maximum when x = 3

(24 * 3) – 144

= 72 – 144 = -72

Further Investigation

To find the length of the cut-out corner squares that gives the maximum open box volume for rectangular pieces of card of different sizes. The length of the cut-out corner squares will be to 3 significant figures of accuracy.

Method- I will investigate the size of the corner cut-out squares that give the largest open box volume for rectangular shaped pieces of card that have the width to length ratio of 1:2, those being 12 x 24, 24 x 48 and 48 x 96. I will then produce a formula for the maximum open box volume, for all rectangles that have width to length ratios of 1:2.

Size of square cut = 10 by 20 (cm)

Length cut from each side ( x ) | Area of Base ( a ) | Volume ( v ) |

1 | 144 | 144 |

2 | 96 | 192 |

3 | 56 | 168 |

4 | 24 | 96 |

Conclusion

Area of Base

( a )

Volume

( v )

5

2100

10500

6

1904

11424

7

1716

12012

8

1536

12288

9

1364

12276

10

1200

12000

From this result I can tell the highest volume is the length cut is from 8cm – 9cm. I am going to find out where the highest volume by using trail and improvement and breaking the two numbers down into 1 decimal place.

Length cut from each side ( x ) | Area of Base ( a ) | Volume ( v ) |

8.3 | 1483.56 | 12313.55 |

8.4 | 1466.24 | 12316.42 |

8.5 | 1449 | 12316.5 |

8.6 | 1431.84 | 12313.82 |

8.7 | 1414.76 | 12308.41 |

The table shows us that as the cut is increased so did the volume, but when it reached 8.5cm, the volume decreased. Therefore the maximum volume was at 8.5cm where it had reached 12316.5cm. I should put the table in 2 decimal place to find an accurate result.

Length cut from each side ( x ) | Area of Base ( a ) | Volume ( v ) |

8.43 | 1461.0596 | 12316.73 |

8.44 | 1459.3344 | 12316.78 |

8.45 | 1457.61 | 12316.8 |

8.46 | 1455.8864 | 12316.8 |

8.47 | 1454.1636 | 12316.77 |

8.48 | 1452.4416 | 12316.7 |

8.49 | 1450.7204 | 12316.62 |

The maximum volume I found in this table was at 8.46cm with its volume at 12316.8cm

I made a table below to show the maximum volume of each square more clearly.

Size of square cut | Cut for volume maximum | Max volume |

10 by 20 | 2.1 | 192.444 |

20 by 40 | 4.2 | 1539.552 |

30 by 60 | 6.3 | 5195.988 |

40 by 80 | 8.5 | 12376.5 |

I created another table to find the relationship between the 4 squares.

Size of square cut | Cut for volume maximum | Volume/Square |

10 by 20 | 2.1 | 0.2 |

20 by 40 | 4.2 | 0.2 |

30 by 60 | 6.3 | 0.2 |

40 by 80 | 8.5 | 0.2 |

From each of the four squares I found out the relationship from each of them with the highest volume divided by the same size. I found out at the relationship is 1/5. I found the relationship by finding the difference of the Size of square cut divided by the Cut for volume maximum.

This student written piece of work is one of many that can be found in our GCSE Comparing length of words in newspapers section.

## Found what you're looking for?

- Start learning 29% faster today
- 150,000+ documents available
- Just £6.99 a month