• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
Page
  1. 1
    1
  2. 2
    2
  3. 3
    3
  4. 4
    4
  5. 5
    5
  6. 6
    6
  7. 7
    7
  8. 8
    8
  9. 9
    9
  10. 10
    10
  11. 11
    11
  12. 12
    12
  13. 13
    13
  14. 14
    14
  15. 15
    15
  16. 16
    16
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  • Level: GCSE
  • Subject: Maths
  • Word count: 4546

Producing a Box

Extracts from this document...

Introduction

image04.pngimage05.pngimage00.png

I was given a 12 by 12 square, and asked to reshape it into a ‘strawberry box’ type shape, to find the box with the largest volume.  Below, is a diagram of what I had to work with.

image01.png

To find out the volumes of my boxes, I will cut out a 1cm by 1cm square from each corner of my 12 by 12cm square and shape it into a box.  I will then find the volume of this box by multiplying the length by the breath by the height.  I will then repeat this method again, only I will be cutting off a 2cm by 2cm square, and a 3cm by 3cm square, from each corner, until I can no longer remove any more.  I will display these shapes on the following pages, and present the figures in a table after the last shape.

image06.png

Shape One

Volume = Length x Breath x Height

Length = 10cm

Breath = 10cm

Height = 1cm

Volume = 10 x 10 x 1

Volume = 100 cm³

Shape Two

Volume = Length x Breath x Height

Length = 8cm

Breath = 8cm

Height = 2cm

Volume = 8 x 8 x 2image06.png

Volume = 128 cm³

Shape Three

Volume = Length x Breath x Height

Length = 6cm

Breath = 6cm

Height = 3cm

Volume = 6 x 6 x 3

Volume = 108 cm³

Shape Four

image02.png

Volume = Length x Breath x Height

Length = 4cm

Breath = 4cm

Height = 4cm

Volume = 4 x 4 x 4image06.png

Volume = 64 cm³

Shape Five

Volume = Length x Breath x Height

Length = 2cm

Breath = 2cm

Height = 5cm

Volume = 2 x 2 x 5

Volume = 20 cm³

Shape

Length

Breath

Height

Volume (cm³)

1

10

10

1

100

2

8

8

2

128

3

6

6

3

108

4

4

4

4

64

5

2

2

5

20

By looking at this table, I can see that shape 2 has the largest volume.  To look at the possibility of there being a larger volume available, I am now going to place these figures in a graph on the next page.

image06.png

image13.png

After looking at the graph, I realise that there could be a larger volume available.  So I am now going to investigate this possibility.

Because shape 2 had the largest volume, and I had removed a

2cm by 2cm square from each corner to make shape 2.

...read more.

Middle

Length = 10 – 3.2

Length = 6.8                                                Length = Breath

Length = 6.8

Breath = 6.8

Height = 1.6

Volume = Length x Breath x Height

Volume = 6.8 x 6.8 x 1.6

Volume = 73.984 cm³

Shape 9

        Length = 10 – (Height + Height)

        Length = 10 – (1.7 + 1.7)

        Length = 10 – 3.4

        Length = 6.6                                                Length = Breath

Length = 6.6

Breath = 6.6

Height = 1.7

Volume = Length x Breath x Height

Volume = 6.6 x 6.6 x 1.7

Volume = 74.052 cm³

image06.png

Shape 10

Length = 10 – (Height + Height)

        Length = 10 – (1.8 + 1.8)

        Length = 10 – 3.6

        Length = 6.4                                                Length = Breath

Length = 6.4

Breath = 6.4

Height = 1.8

Volume = Length x Breath x Height

Volume = 6.4 x 6.4 x 1.8

Volume = 73.728 cm³

Shape

Length

Breath

Height

Volume (cm³)

5

6.2

6.2

1.9

73.036

7

7

7

1.5

73.5

8

6.8

6.8

1.6

73.984

9

6.6

6.6

1.7

74.052

10

6.4

6.4

1.8

73.728

By looking at my results like this, I can see that shape 9 is the largest possible box, you can make from a 10cm by 10cm cube.  Now, I want to check my previous formula against the 10 by 10cm cube.

image06.png

The Formula for Finding the Volume of A Strawberry Box

image08.png

                                                 x          x

                                        x                          x

                                        x                          x

                                                 x         x

Length = 10 – 2x

Breath = 10 – 2x

                Height = x

Volume = Length x Breath x Height

Volume = (10 – 2x)(10 – 2x) x x

image06.png

I will now check if my formula is correct

                        Let x = 3

Volume = (10 – 2x)(10 – 2x) x x

Volume = (10 – 6) x (10 – 6) x 3

Volume = 4 x 4 x 3

Volume = 48 cm³

By comparing my results here with Shape 3 on page 9, I can see that my formula has worked just fine.  However, I am going to try it out on yet another size of square (24cm by 24cm square) to be sure that it will work on sizes.

        I am going to workout the figures on the next few pages, and place them in a table at the end.

Shape 1

        Length = 24 – (Height + Height)

        Length = 24 – (1 + 1)

        Length = 24 – 2

        Length = 22                                                Length = Breath

Length = 22

Breath = 22

Height = 1

Volume = Length x Breath x Height

Volume = 22 x 22 x 1

Volume = 484 cm³

image06.png

Shape 2

        Length = 24 – (Height + Height)

        Length = 24 – (2 + 2)

        Length = 24 – 4

        Length = 20                                                Length = Breath

Length = 20

Breath = 20

Height = 2

Volume = Length x Breath x Height

Volume = 20 x 20 x 2

Volume = 800 cm³

Shape 3

        Length = 24 – (Height + Height)

        Length = 24 – (3 + 3)

        Length = 24 – 6

        Length = 18                                                Length = Breath

Length = 18

Breath = 18

Height = 3

Volume = Length x Breath x Height

Volume = 18 x 18 x 3

Volume = 972 cm³

image06.png

Shape 4

        Length = 24 – (Height + Height)

        Length = 24 – (4 + 4)

        Length = 24 – 8

        Length = 16                                                Length = Breath

Length = 16

Breath = 16

Height = 4

Volume = Length x Breath x Height

...read more.

Conclusion

Shape 1

Size of square removed = x = 1

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 2)(12 – 2) X 1

       Vol = 22  X  10  X  1

       Vol = 220cm³image06.png

Shape 2

Size of square removed = x = 2

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 4)(12 – 4) X 2

       Vol = 20  X  8  X  2

       Vol = 320cm³

Shape 3

Size of square removed = x = 3

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 6)(12 – 6) X 3

       Vol = 18  X  6  X  3

       Vol = 324cm³

Shape 4

Size of square removed = x = 4

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 8)(12 – 8) X 4

       Vol = 16  X  4  X  4

       Vol = 256cm³

image06.png

Shape 5

Size of square removed = x = 5

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 10)(12 – 10) X 5

       Vol = 14  X  2  X  5

       Vol = 140cm³

As before, I will place the results for finding the volume of a rectangular box with a net of size 12 by 24cm, in a table displayed below.

Shape

Length

Breadth

Height

Volume (cm³)

1

22

10

1

220

2

20

8

2

320

3

18

6

3

324

4

16

4

4

256

5

14

2

5

140

To make this information more presentable, I am going to put it into a graph below.

image10.png

image06.png

I have learned from working out the largest volume using the square, that the first table/graph doesn’t always show that largest volume of box.  This table and graph shows that the largest box has been found by removing a 2cm square from each corner of the original rectangle.  I am now going to check weather or not Shape 2 is the largest shape.  Checking this will be done by removing a 1.9cm square from each corner of the original rectangle to make Shape 6, and removing a 2.1cm square from each corner of the original rectangle to make Shape 7.

Shape 6

Size of square removed = x = 1.9

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 3.8)(12 – 3.8) X 1.9

       Vol = 20.2  X  8.2  X  1.9

                              Vol = 314.716cm³

Shape 7

Size of square removed = x = 2.1

Length    = 24 – 2x

Breadth  = 12 – 2x

Height     = x

Volume = (24 – 2x)(12 – 2x) X x

       Vol = (24 – 4.2)(12 – 4.2) X 2.1

       Vol = 19.8  X  7.8  X  2.1

                              Vol = 324.324cm³

I am now going to place these two new figures in a table along with Shape 2 so that I can compare my new results with my last ones.

Shape

Length

Breath

Height

Volume (cm³)

2

20

8

2

520

6

20.2

8.2

1.9

314.716

7

19.8

7.8

2.1

324.324

image06.pngimage11.png

image12.png

Barry                 McNickle

         12 E image06.png

...read more.

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

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Comparing length of words in newspapers essays

  1. Assesment of Reading Difficulties in Patient AM Following the Development of Vascular Dementia.

    more general - purpose cognitive mechanism exists and it is damage to this which results in alexia. They also argue that because reading is a relatively knew cognitive ability in evolutionary terms it makes sense for it to be "mediated by a neural substrate that sub serves other visuoperceptual functions

  2. How well can you estimate the length of an object?

    144 1.61 84 2.05 37 1.45 39 1.20 157 1.50 34 1.30 Whilst I was generating numbers, I came across a few repeats which I ignored and carried on. Also, as you can see on the table above, I came across an out-of-context number which I also replaced.

  1. To observe how, the length and amplitude of a pendulum string, affect the time ...

    Results: Length as a factor Length (cm) Time for 10 oscillations (seconds) Time for 1 oscillation (seconds) 10cm 3.40 0.34 20cm 5.35 0.535 30cm 7.80 0.78 40cm 8.89 0.889 50cm 10.47 1.047 60cm 12.6 1.26 Cross sectional area as a factor - 50cm (fixed length)

  2. This investigation looked to see whether the height on the shore would affect the ...

    Gibbula Umbilicalis are a good choice for this investigation as they are easy to find and identify due to their characteristic grey-green shell with purple zigzag pattern, and also the large quantity of them over the rocky shore. Planning Hypothesis I think there will a significant difference in the ratio

  1. The Open Box Problem

    to form this data into a graph, to help me see whether the highest number is a decimal or a whole number. I am now going to test whether or not the graph shown above is symmetrical by taking the numbers immediately above and below my highest point.

  2. Consumer responses to wine bottle back labels

    and wine consumption behaviour influence such opinions; and * to evaluate the significance that consumers attribute to back labels in influencing their purchasing behaviours. Research Design and Methodology Research was conducted among 56 students attending courses on wine and society in both South Australia and Western Australia in January and February 1999.

  1. "Broadsheet newspapers have a longer average word length than tabloid newspapers"

    table I produced two frequency polygons, one for the tabloids and one for the broadsheets data. These are the following observations I made:- o Both polygons, start with an almost equally steep ascent up until 3. This is the most dramatic difference in both graphs when the word length only increases by 1 letter.

  2. Open Box Problem.

    Length cut from each side ( x ) Length of Base ( l ) Area of Base ( a ) Volume ( v ) 1cm 18 324 324 2cm 16 256 512 3cm 14 196 588 4cm 12 144 576 5cm 10 100 500 6cm 8 64 384 7cm 6 36 252 8cm 4 16 128 9cm 2

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work