• 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
  • Level: GCSE
  • Subject: Maths
  • Word count: 1845

The Open Box Problem

Extracts from this document...

Introduction

The Open Box Problem

Equations for a square

Total length of paper = x

Total length of cutout = a

Width of base of box = c = x – 2a

Area of base = c2

Area of sides = 4ac

Volume = ac2

Square piece of paper with dimensions 12 x 12.

Cut out

Length

Width

Area of base

Area of sides

Volume

1

10

10

100

40

100

2

8

8

64

64

128

2.1

7.8

7.8

60.84

65.52

127.764

2.2

7.6

7.6

57.76

66.88

127.072

2.3

7.4

7.4

54.76

68.08

125.948

2.4

7.2

7.2

51.84

69.12

124.416

2.5

7

7

49

70

122.5

2.6

6.8

6.8

46.24

70.72

120.224

2.7

6.6

6.6

43.56

71.28

117.612

2.8

6.4

6.4

40.96

71.68

114.688

2.9

6.2

6.2

38.44

71.92

111.476

3

6

6

36

72

108

3.1

5.8

5.8

33.64

71.92

104.284

3.2

5.6

5.6

31.36

71.68

100.352

3.3

5.4

5.4

29.16

71.28

96.228

3.4

5.2

5.2

27.04

70.72

91.936

3.5

5

5

25

70

87.5

3.6

4.8

4.8

23.04

69.12

82.944

3.7

4.6

4.6

21.16

68.08

78.292

3.8

4.4

4.4

19.36

66.88

73.568

3.9

4.2

4.2

17.64

65.52

68.796

4

4

4

16

64

64

5

2

2

4

40

20

A graph to show the comparison of the cutout to the volume of a 12 x 12 square.

image00.png

Square piece of paper with dimensions 16 x 16.

Cut out

Length

Width

Area of base

Area of sides

Volume

1

14

14

196

56

196

2

12

12

144

96

288

2.1

11.8

11.8

139.24

99.12

292.404

2.2

11.6

11.6

134.56

102.08

296.032

2.3

11.4

11.4

129.96

104.88

298.908

2.4

11.2

11.2

125.44

107.52

301.056

2.5

11

11

121

110

302.5

2.6

10.8

10.8

116.64

112.32

303.264

2.7

10.6

10.6

112.36

114.48

303.372

2.8

10.4

10.4

108.16

116.48

302.848

2.9

10.2

10.2

104.04

118.32

301.716

3

10

10

100

120

300

3.1

9.8

9.8

96.04

121.52

297.724

3.2

9.6

9.6

92.16

122.88

294.912

3.3

9.4

9.4

88.36

124.08

291.588

3.4

9.2

9.2

84.64

125.12

287.776

3.5

9

9

81

126

283.5

3.6

8.8

8.8

77.44

126.72

278.784

3.7

8.6

8.6

73.96

127.28

273.652

3.8

8.4

8.4

70.56

127.68

268.128

3.9

8.2

8.2

67.24

127.92

262.236

4

8

8

64

128

256

5

6

6

36

120

180

...read more.

Middle

I have predicted that for a 20 x 20 square piece of paper that the maximum volume I can get will be with a cutout of 3cm.

Square piece of paper with dimensions 20 x 20

Cut out

Length

Width

Area of base

Area of sides

Volume

1

18

18

324

72

324

2

16

16

256

128

512

2.1

15.8

15.8

249.64

132.72

524.244

2.2

15.6

15.6

243.36

137.28

535.392

2.3

15.4

15.4

237.16

141.68

545.468

2.4

15.2

15.2

231.04

145.92

554.496

2.5

15

15

225

150

562.5

2.6

14.8

14.8

219.04

153.92

569.504

2.7

14.6

14.6

213.16

157.68

575.532

2.8

14.4

14.4

207.36

161.28

580.608

2.9

14.2

14.2

201.64

164.72

584.756

3

14

14

196

168

588

3.1

13.8

13.8

190.44

171.12

590.364

3.2

13.6

13.6

184.96

174.08

591.872

3.3

13.4

13.4

179.56

176.88

592.548

3.4

13.2

13.2

174.24

179.52

592.416

3.5

13

13

169

182

591.5

3.6

12.8

12.8

163.84

184.32

589.824

3.7

12.6

12.6

158.76

186.48

587.412

3.8

12.4

12.4

153.76

188.48

584.288

3.9

12.2

12.2

148.84

190.32

580.476

4

12

12

144

192

576

5

10

10

100

200

500

A graph to show the comparison of the cutout to the volume of a 19 x 19 square.

image02.png

The formula for the maximum volume of the box for square shaped pieces of paper is:

x = side of square

/ = divide by

x/6 = side of square / 6

...read more.

Conclusion

="c1">56

108

168

4

12

2

24

112

96

A graph to show the comparison of the cutout to the volume of a 10 x 20 rectangle.

image03.png

Rectangle piece of paper with dimensions 10 x 30.

Cut out

Width

Length

Area of base

Area of sides

Volume

1

28

8

224

72

224

1.1

27.8

7.8

216.84

78.32

238.524

1.2

27.6

7.6

209.76

84.48

251.712

1.3

27.4

7.4

202.76

90.48

263.588

1.4

27.2

7.2

195.84

96.32

274.176

1.5

27

7

189

102

283.5

1.6

26.8

6.8

182.24

107.52

291.584

1.7

26.6

6.6

175.56

112.88

298.452

1.8

26.4

6.4

168.96

118.08

304.128

1.9

26.2

6.2

162.44

123.12

308.636

2

26

6

156

128

312

2.1

25.8

5.8

149.64

132.72

314.244

2.2

25.6

5.6

143.36

137.28

315.392

2.3

25.4

5.4

137.16

141.68

315.468

2.4

25.2

5.2

131.04

145.92

314.496

2.5

25

5

125

150

312.5

2.6

24.8

4.8

119.04

153.92

309.504

2.7

24.6

4.6

113.16

157.68

305.532

2.8

24.4

4.4

107.36

161.28

300.608

2.9

24.2

4.2

101.64

164.72

294.756

3

24

4

96

168

288

4

22

2

44

192

176

A graph to show the comparison of the cutout to the volume of a 10 x 30 rectangle.

image04.png

Rectangle piece of paper with dimensions 10 x 40.

Cut out

Width

Length

Area of base

Area of sides

Volume

1

38

8

304

92

304

1.1

37.8

7.8

294.84

100.32

324.32

1.2

37.6

7.6

285.76

108.48

342.91

1.3

37.4

7.4

276.76

116.48

359.79

1.4

37.2

7.2

267.84

124.32

374.98

1.5

37

7

259

132

388.5

1.6

36.8

6.8

250.24

139.52

400.38

1.7

36.6

6.6

241.56

146.88

410.65

1.8

36.4

6.4

232.96

154.08

419.33

1.9

36.2

6.2

224.44

161.12

426.44

2

36

6

216

168

432

2.1

35.8

5.8

207.64

174.72

436.04

2.2

35.6

5.6

199.36

181.28

438.59

2.3

35.4

5.4

191.16

187.68

439.67

2.4

35.2

5.2

183.04

193.92

439.3

2.5

35

5

175

200

437.5

2.6

34.8

4.8

167.04

205.92

434.3

2.7

34.6

4.6

159.16

211.68

429.73

2.8

34.4

4.4

151.36

217.28

423.81

2.9

34.2

4.2

143.64

222.72

416.56

3

34

4

136

228

408

4

32

2

64

272

256

A graph to show the comparison of the cutout to the volume of a 10 x 40 rectangle.

image05.png

...read more.

This student written piece of work is one of many that can be found in our GCSE Fencing Problem 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 Fencing Problem essays

  1. Fencing problem.

    100 Opposite = TAN 540 � 100 Opposite = 137.6m = Height. I shall now substitute the height into the formula below: Area of a triangle = 1/2 � Base � Height Area of a triangle = 1/2 � 200 � 137.6 = 13760m2 I have now found the area of one triangle.

  2. Fencing Problem

    This is shown in my investigation to find the largest area of a triangle with a perimeter of 1000 metres. For example; Quadrilaterals I am now going to investigate quadrilaterals. I will investigate; Squares, Rectangles, Parallelograms and Trapeziums. I hypothesize that the largest area for a quadrilateral shape with a

  1. Fencing Problem

    170 34527.16 333.44 48112.515 190 37401.54 334 48112.233 210 39982.81 335 48110.712 230 42253.70 336 48107.879 250 44194.17 337 48103.725 270 45780.73 350 47925.724 Observation From the results of the table, the area increases as the base increase. However, there reaches a point where the areas stops increasing and in fact starts to decrease.

  2. Fencing problem.

    Below is a sample table (it would occupy too much space if I placed all the results down); however, the full results will be shown in the graph. Table of Results Width (m) Length (m) Area (m�) 1 499 499 2 498 996 3 497 1491 4 496 1984 5

  1. Investigation of Open Ended Tubes.

    If two sides should be the same size for optimum area, we can calculate the optimum Length of the third one. First some limitations: 2a+b = 320 a+b>a 2a>b h� = a�-(a-x)� = (320-2a)�-x� this again shows that a>h and 320-2a>h.

  2. The Fencing Problem

    Area using base formula: Area = 500B - B� Area = 500 ( 250 - 250� = 62,500m� Area using height formula: Area = 500H - H� Area = 500 ( 250 - 250� = 62,500m� Test two: This next test will be on a rectangle, the rectangle base is

  1. fencing problem part 2/8

    This can only be achieved if the 'abcdCos2? ' part of the formula is a low as possible because this number is taken away from the number to be square rooted. Therefore abcdCos2 ? should equal zero. To make abcdCos2 ?

  2. Arceology paper

    To go along with these methods, as well as to map the entire area, aerial photography may be useful if the site is an open area. If the area were a woodland area, this technique would not work well and should be substituted with topographic survey.

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