# The Open Box Problem.

Extracts from this document...

Introduction

In this investigation, I will be investigating the same as I did in the 'Second Investigation'. This will be the maximum volume, which can be made from a certain size rectangle piece of card, with different size sections cut from their corners. The type of cuboids I will be using are opened topped boxes. I will keep one side the same all of the time.

The size sections that I will be cutting from the rectangle piece of card will all be the same size. The section sizes will go up to the maximum length possible for the piece of card.

During this investigation, I will not account for the ‘tabs’, which would normally be needed to hold the box sides together.

I still predict that to obtain the largest volume from a certain size, Rectangle, piece of card; the length of the section cut from the corner, needs to equal a sixth of side A. Side B can be anything. The only difference between this one and the last experiment will be that I will have only one variable, which I hope will give a rule to find the size of cut out needed to leave the maximum volume cuboid.

X = 1/6

Middle

1.8

4.8

5.184

0.61

0.61

1.78

4.78

5.1901

0.62

0.62

1.76

4.76

5.1941

0.63

0.63

1.74

4.74

5.1960

0.64

0.64

1.72

4.72

5.1958

0.65

0.65

1.70

4.70

5.1935

0.66

0.66

1.68

4.68

5.1892

0.67

0.67

1.66

4.66

5.1829

0.68

0.68

1.64

4.64

5.1745

0.69

0.69

1.62

4.62

5.1642

0.7

0.7

1.6

4.6

5.152

0.8

0.8

1.4

4.4

4.928

0.9

0.9

1.2

4.2

4.536

1.0

1.0

1.0

4.0

4.000

1.1

1.1

0.8

3.8

3.344

1.2

1.2

0.6

3.6

2.592

1.3

1.3

0.4

3.4

1.768

1.4

1.4

0.2

3.2

0.896

1.5

7cm by 3cm, piece of card

Length of the section (cm) | Height of the section (cm) | Depth of the section (cm) | Width of the section (cm) | Volume of the cubeoid (cm3) |

0.1 | 0.1 | 2.8 | 6.8 | 1.904 |

0.2 | 0.2 | 2.6 | 6.6 | 3.432 |

0.3 | 0.3 | 2.4 | 6.4 | 4.608 |

0.4 | 0.4 | 2.2 | 6.2 | 5.456 |

0.5 | 0.5 | 2.0 | 6.0 | 6.000 |

0.6 | 0.6 | 1.8 | 5.8 | 6.264 |

0.61 | 0.61 | 1.78 | 5.78 | 6.276 |

0.62 | 0.62 | 1.76 | 5.76 | 6.2853 |

0.63 | 0.63 | 1.74 | 5.74 | 6.2922 |

0.64 | 0.64 | 1.72 | 5.72 | 6.2966 |

0.65 | 0.65 | 1.70 | 5.70 | 6.2985 |

0.66 | 0.66 | 1.68 | 5.68 | 6.2980 |

0.67 | 0.67 | 1.66 | 5.66 | 6.2951 |

0.68 | 0.68 | 1.64 | 5.64 | 6.2897 |

0.69 | 0.69 | 1.62 | 5.62 | 6.2820 |

0.7 | 0.7 | 1.6 | 5.6 | 6.272 |

0.71 | 0.71 | 1.58 | 5.58 | 6.2596 |

0.8 | 0.8 | 1.4 | 5.4 | 6.048 |

0.9 | 0.9 | 1.2 | 5.2 | 5.616 |

1.0 | 1.0 | 1.0 | 5.0 | 5.000 |

1.1 | 1.1 | 0.8 | 4.8 | 4.224 |

1.2 | 1.2 | 0.6 | 4.6 | 3.312 |

1.3 | 1.3 | 0.4 | 4.4 | 2.288 |

1.4 | 1.4 | 0.2 | 4.2 | 1.176 |

1.5 |

8cm by 3cm, piece of card

Length of the section (cm) | Height of the section (cm) |

Conclusion

dy/dx = 12x2-4Ax-4Bx+AB

= 12x2-(4A+4B)x+AB

To work out the minimum or maximum values, the greadient needs to be "0".

To find x, you need to use = -b ± \/(b2-4ac)

_________________

2a

Substitute as follows:

a = 12

b = -4(A+B)

c = AB

Therefore, x = +4(A+B) ± \/(-4(A+B)2-(4 . 12 . AB)

_______________________________________________

2 . 12

= 4(A+B) ± \/16(A2+2AB+B2)-48AB

____________________________________________

24

= 4(A+B) ± \/16A2+32AB+16B2-48AB

_______________________________________________

24

= 4(A+B) ± \/16A2-16AB+16B2

________________________________________

24

= 4(A+B) ± 4 . \/A2-AB+B2

________________________________________

24

= A+B ± \/A2-AB+B2

__________________________

6

= 1/6 [ A+B ± \/A2-AB+B2 ]

- = 1/6 [ A+B ± \/A2-AB+B2 ]

This is a general rule for any rectangle (including squares). Squares also have their own formula, as A = B :

x = 1/6 [ A+A ± \/A2-AA+A2 ]

= 1/6 [ 2A ± \/A2 ]

= 1/6 [ 2A ± A ]

= 1/6 [ 3A ] or 1/6 [ A ]

= 1/2 A or 1/6 A

= MIN MAX

Therefore, the maximum volume is when x = 1/6 A

Conclusion

This proves by calculus the result that I obtained in the First Open Box problem. By substituting different ratios of sides I can also prove a relationship for any rectangle. For example I could use B = 2A, or B = 8A, or B = nA and obtain a result as follows :

x = 1/6 [ A+nA ± \/(A2 - A.nA + (nA)2) ]

= 1/6 [ (1+n)A ± \/(A2 - nA2 + n2A2) ]

= 1/6 [ (1+n)A ± \/A2 (1-n+n2) ]

= 1/6 [ (1+n)A ± A\/(1-n)2 ]

= 1/6 A [ 1+n ± (1-n) ]

= 1/6 A [ 2 ] or 1/6 A [ 2n ]

= MIN MAX

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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