# Tbe Open Box Problem

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Introduction

The open box problemBy Luke Johnston

An open box is to be made from a sheet of card with identical squares cut of the corners to make a box:

y

x = corner length

y = box length

The card is then folded along the dashed lines to make the box. The aim of this activity is to determine the size of the cut out square which makes the largest volume for a square.

Firstly I will find this in a square of y = 10cm

To find the volume I will use this equation-Volume= Width x length x height

Cut off (cm) | width (cm) | length (cm) | height (cm) | volume (cm²) |

1 | 8 | 8 | 1 | 64 |

2 | 6 | 6 | 2 | 72 |

3 | 4 | 4 | 3 | 48 |

4 | 2 | 2 | 4 | 16 |

The maximum box volume is made from the cut out square of 2cms. I will now try between 1-2 cm because the highest volume is somewhere between them.

Cut off (cm) | width (cm) | length (cm) | height (cm) | volume (cm²) |

1.1 | 7.8 | 7.8 | 1.1 | 66.924 |

1.2 | 7.6 | 7.6 | 1.2 | 69.312 |

1.3 | 7.4 | 7.4 | 1.3 | 71.188 |

1.4 | 7.2 | 7.2 | 1.4 | 72.576 |

1.5 | 7 | 7 | 1.5 | 73.5 |

1.6 | 6.8 | 6.8 | 1.6 | 73.984 |

1.7 | 6.6 | 6.6 | 1.7 | 74.052 |

1.8 | 6.4 | 6.4 | 1.8 | 73.728 |

1.9 | 6.2 | 6.2 | 1.9 | 73.036 |

2 | 6 | 6 | 2 | 72 |

The highest value of x is 1.7, but once again I will find a volume even higher in the 10 by 10 box by going into two decimal places.

Cut off (cm) | width (cm) | length (cm) | height (cm) | volume (cm²) |

1.61 | 6.78 | 6.78 | 1.61 | 74.0091 |

1.62 | 6.76 | 6.76 | 1.62 | 74.0301 |

1.63 | 6.74 | 6.74 | 1.63 | 74.047 |

1.64 | 6.72 | 6.72 | 1.64 | 74.0598 |

1.65 | 6.7 | 6.7 | 1.65 | 74.0685 |

1.66 | 6.68 | 6.68 | 1.66 | 74.0732 |

1.67 | 6.66 | 6.66 | 1.67 | 74.0739 |

1.68 | 6.64 | 6.64 | 1.68 | 74.0705 |

1.69 | 6.62 | 6.62 | 1.69 | 74.0632 |

The highest value of x is therefore 1.67. I will now go into even smaller numbers between 1.66 and 1.67.

Cut off (cm) | width (cm) | length (cm) | height (cm) | volume (cm²) |

1.661 | 6.678 | 6.678 | 1.661 | 74.0734 |

1.662 | 6.676 | 6.676 | 1.662 | 740736 |

1.663 | 6.674 | 6.674 | 1.663 | 74.0738 |

1.664 | 6.672 | 6.672 | 1.664 | 74.0739 |

1.665 | 6.67 | 6.67 | 1.665 | 74.074 |

1.666 | 6.668 | 6.668 | 1.666 | 74.074065 |

1.667 | 6.666 | 6.666 | 1.667 | 74.074072 |

1.668 | 6.664 | 6.664 | 1.668 | 74.074 |

1.669 | 6.662 | 6.662 | 1.669 | 74.074 |

The highest is 1.667. I will show the graph to show the change in volume for a 10 by 10 box.

Middle

10.02

2.49

249.997

2.5

10

10

2.5

250

2.51

9.98

9.98

2.51

249.997

2.52

9.96

9.96

2.52

249.988

2.53

9.94

9.94

2.53

249.973

2.54

9.92

9.92

2.54

249.952

2.55

9.9

9.9

2.55

249.926

2.56

9.88

9.88

2.56

249.893

2.57

9.86

9.86

2.57

249.854

2.58

9.84

9.84

2.58

249.81

2.59

9.82

9.82

2.59

249.76

2.6

9.8

9.8

2.6

249.704

The maximum value of x is 2.5cm. This shows my prediction was correct. This is that a square box has to be divided by six to find the cut out size in which would give the maximum volume.

To help find the highest value of x to make the largest volume, I will use algebra, with aid from textbook sources. I will need to divide the cut out square by the original length of the box.

L= Length of box

Y= Length of box- 2x

10= 2x+y

x = 10/6 or L/6

I will substitute these into simple equations:

V = (L – 2L/6)(L – 2L/6) L/6

I multiplied the 2 in each bracket by L/6 which gives me:

V = (L – 2L/6)(L- 2L/6)L/6

This makes

V = L (L- 2L/6) – “L/6 (L – 2L/6) L/6

This is:

V = (L² - 2L²/6 – 2L²/6 – 4L²)L/6

I will multiply the lot by L/6

V = L³/6 – 2L³/36 – 2L³/36 + 4L³/216

Now I have put the denominator as 216

V = (4L³ + 36L³ - 12L³ - 12L³)/216

This makes

40L³ - 24L³ / 216

This is simplified as

V = 2L³/ 27

I will now test this using different size squares.

- 10cm by 10cm = 2x10³/27 = 74.074074
- 15cm by 15cm = 2x15³/27 = 250

This proves that the equation is correct.

Also there is another way to find out the highest volume shown below. I will show this formula using a 10 by 10 square.

V = x(10-2x)(10-2x) This is minus 2x because of the lengths of the square cut-outs.

= x(100-20x-20x+4x²)

= 4x³-40x²+100x

Using the gradient function I can work this out. The gradient function is this g = anx n-1.

This makes

12x²-80x+100 = 0

Using ax²+bx+c = 0

x = - b +/- √b²-4ac

2a

x = 80 +/- √-80² - (4x12x100)

2x12

x = 80 +/- √6400 – 4800

24

x = 80 +/- 40

24

If it were minus then this is the result- 1.667

If it were plus then this is the result- 5 This would be impossible as this would make up the length of the square so minus is correct.

I am now going to investigate the open box with a rectangle. I will begin this part of my investigation with a general formula for the volume of the rectangle.

The volume of an open box can be expressed like this:

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

This is simplified as:

V = x (4x² + LW – 2Wx – 2Lx)

V = 4x³ + 2Lx² - 2Wx² + LWx

V = x(LW) + 4x³ - x²(2L + 2W)

I can now differentiate When dv/dx = 0

dv/dx = LW + 12x² - 2(2L + 2W)

0 = 12x² - (4L + 4W) + LW

This is now a quadratic equation, we can use this formula:

x = -b ± √-(4L + 4W)² - 4 x 12 x LW

2 x 12

This is simplified as:

x = 4(L + W) ± √-(4L + 4W) – 48LW

24

Multiplying the two brackets together:

x = 4(L + W) ± √16L² + 16W2 – 16LW

24

Factorising:

x = 4(L + W) ± √16 (W² + L² - LW)

24

x = 4(L + W) ± 4√(W² + L² - LW)

24

I can simplify by dividing by the common factor of 4:

x = L + W ± √W² + L² - LW

6

I cannot simplify anymore, but using this I am able to find the value of x which would give me the largest volume of the rectangular box. I shall make predictions to some rectangles and prove them. Firstly I will find the largest volume of the box if the rectangle was 10cm by 15cm.

Firstly my prediction:

x = 10 + 15 ± √15² + 10² - 10x15

6

x = 25 ± √225 + 100 -150

6

x = 25 ± √175

6

x = 25 + 13.22875656 = 6.371459427

6

x = 25 – 13.22875656 = 1.961873907 or 1.962 (3.d.p)

6

x must be 1.962 because otherwise 6.371 (3.d.p) is over twice the length. I shall prove that this formula is correct.

Cut off (cm) | Width (cm) | Length (cm) | Height (cm) | Volume (cm²) |

1 | 13 | 8 | 1 | 104 |

2 | 11 | 6 | 2 | 132 |

3 | 9 | 4 | 3 | 108 |

4 | 7 | 2 | 4 | 56 |

Using the graph I can see that the highest is between 1 and 3.

Into decimal points as it is more accurate between 1and 2.

Cut off (cm) | Width (cm) | Length (cm) | Height (cm) | Volume (cm²) |

1.1 | 12.8 | 7.8 | 1.1 | 109.824 |

1.2 | 12.6 | 7.6 | 1.2 | 114.912 |

1.3 | 12.4 | 7.4 | 1.3 | 119.288 |

1.4 | 12.2 | 7.2 | 1.4 | 122.976 |

1.5 | 12 | 7 | 1.5 | 126 |

1.6 | 11.8 | 6.8 | 1.6 | 128.384 |

1.7 | 11.6 | 6.6 | 1.7 | 130.152 |

1.8 | 11.4 | 6.4 | 1.8 | 131.328 |

1.9 | 11.2 | 6.2 | 1.9 | 131.936 |

2 | 11 | 6 | 2 | 132 |

2.1 | 10.8 | 5.8 | 2.1 | 131.544 |

The highest is between 1.9 and 2 so I will go into 2 decimal places.

Cut off (cm) | Width (cm) | Length (cm) | Height (cm) | Volume (cm²) |

1.91 | 11.18 | 6.18 | 1.91 | 131.966484 |

1.92 | 11.16 | 6.16 | 1.92 | 131.991552 |

1.93 | 11.14 | 6.14 | 1.93 | 132.011228 |

1.94 | 11.12 | 6.12 | 1.94 | 132.025536 |

1.95 | 11.1 | 6.1 | 1.95 | 132.0345 |

1.96 | 11.08 | 6.08 | 1.96 | 132.038144 |

1.97 | 11.06 | 6.06 | 1.97 | 132.036492 |

1.98 | 11.04 | 6.04 | 1.98 | 132.029568 |

1.99 | 11.02 | 6.02 | 1.99 | 132.017396 |

Conclusion

Height (cm)

Volume (cm²)

4.221

31.558

11.558

4.221

1539.598623

4.222

31.556

11.556

4.222

1539.599316

4.223

31.554

11.554

4.223

1539.59987

4.224

31.552

11.552

4.224

1539.600286

4.225

31.55

11.55

4.225

1539.600563

4.226

31.548

11.548

4.226

1539.600701

4.227

31.546

11.546

4.227

1539.6007

4.228

31.544

11.544

4.228

1539.600561

4.229

31.542

11.542

4.229

1539.600284

This shows that 4.226cm needs to be cut off to give the highest volume for a box of a 20 by 40cm rectangle. This proves that my formula is correct.

To show that the formula; x = L + W ± √W² + L² - LW

6

Is correct by the formula obtained earlier:

x = -b ± √ b² - 4ac

2a

Also the formula v = x (L-2x)(W-2x)

Using the rectangle 10 by 15:

v = x (10-2x)(15-2x)

v = x (4x² - 50x + 150)

v = 4x³ - 50x² + 150x

Now differentiate:

dv/dx = 4x³ - 50x² + 150x

When dv/dx = 0

0 = 12x² - 100x + 150

Simplified by the common factor of four:

0 = 3x² - 25x + 37.5

This is now ready for the formula:

x = 25 ±√25²-4x3x37.5

2x3

x = 25 ±√175

6

x = 25 + 13.22875656= 6.371459427

6

x = 25 – 13.22875656 = 1.961873907

6

1.962 (3.d.p) is correct as 6.371 (3.d.p) would be too large. This shows that my formula is correct. I shall show this with the 20 by 40 rectangle:

v = (20-2x)(40-2x)

v = (4x²-120x+800)

v = 4x³ - 120x² + 800x

Now differentiate:

dv/dx = 4x³ - 120x² + 800x

When dv/dx = 0

0 = 12x² - 240x + 800

Divide by the common factor of four:

0 = 3x² - 60 + 200

Now I can use the formula:

x = 60 ±√60² - 4x3x200

2x3

x = 60±√3600 – 4x3x200

6

x = 60 ±/1200

6

x = 60 + 34.64101615 = 15.77350269

6

x = 60 - 34.64101615 = 4.226497308

6

4.226 (3.d.p) is correct because 15.774 (3.d.p) is too large to make the box.

These results confirm that the formula is correct:

x = L + W ± √W² + L² - LW

6

This formula shows the size of the cut off which gives the highest possible volume for a rectangular 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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