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  • Level: GCSE
  • Subject: Maths
  • Word count: 2973

Maths Investigative task on perimeter of a rectangle and volume of shapes

Extracts from this document...

Introduction

MINIMUM PERIMETER

  1. Your task is to find the value of dimension that minimizes the perimeter. By using technology, list all the possible values of length, width, and perimeter with area 1000m2.

Data attached at the back. Color code: Yellow shows the best value that matches the requirements, red matches with red and green matches with green.  Here is some sample data:

Length (m)

Width (m)

Perimeter (m)

Area (m2)

1

1000

2002

1000

2

500

1004

1000

3

333.333

672.667

1000

4

250

508

1000

5

200

410

1000

6

166.667

345.333

1000

7

142.857

299.714

1000

8

125

266

1000

9

111.111

240.222

1000

10

100

220

1000

11

90.909

203.818

1000

12

83.333

190.667

1000

13

76.923

179.846

1000

14

71.429

170.857

1000

15

66.667

163.333

1000

16

62.5

157

1000

17

58.824

151.647

1000

18

55.556

147.111

1000

19

52.632

143.263

1000

20

50

140

1000

21

47.619

137.238

1000

22

45.455

134.909

1000

23

43.478

132.957

1000

24

41.667

131.333

1000

25

40

130

1000

26

38.462

128.923

1000

27

37.037

128.074

1000

28

35.714

127.429

1000

29

34.483

126.966

1000

30

33.333

126.667

1000

31

32.258

126.516

1000

32

31.25

126.5

1000

33

30.303

126.606

1000

34

29.412

126.824

1000

35

28.571

127.143

1000

36

27.778

127.556

1000

37

27.027

128.054

1000

38

26.316

128.632

1000

39

25.641

129.282

1000

40

25

130

1000

41

24.39

130.78

1000

42

23.81

131.619

1000

43

23.256

132.512

1000

44

22.727

133.455

1000

45

22.222

134.444

1000

46

21.739

135.478

1000

The answer to this question is when the perimeter=126.5m, length=32 and width is 31.25.

  1. Make a formula and predict the kind of graph and test if the hypothesis holds on your findings.

For the formula of this question, I have used the standard equation for the perimeter of the rectangle.

image00.png

To get the equation in two variables, we will have to convert one of the variables in terms of another variable. To do this, I used the area formula:

image01.png

As the area is kept constant at 1000 in the investigation, the formula can be made into,

image02.png

From this we can transfer the variable length to isolate width, thus getting a new equation.

image03.png

Substituting this in the original perimeter equation,

image04.png

image05.png

image06.png

image07.png

To prove my formula right, I will make use of two examples.

X=5 (when length is 5m)

The perimeter will be:

image08.png

X=10 (when length is 10m)

The perimeter will be:

image09.png

As seen above, my formula functions perfectly.

According to the question, the graph should be made on length/width by perimeter. As the perimeter can only be zero when either the length or width is zero which is an impossible scenario, the graph will never touch the x-axis or y-axis as neither the perimeter nor the length or width will be zero. This is a trait of a reciprocal graph.

...read more.

Middle

50

14

3.571

35.143

50

15

3.333

36.667

50

16

3.125

38.25

50

17

2.941

39.882

50

18

2.778

41.556

50

19

2.632

43.263

50

20

2.5

45

50

21

2.381

46.762

50

22

2.273

48.545

50

23

2.174

50.348

50

24

2.083

52.167

50

25

2

54

50

26

1.923

55.846

50

27

1.852

57.704

50

28

1.786

59.571

50

29

1.724

61.448

50

30

1.667

63.333

50

31

1.613

65.226

50

32

1.563

67.125

50

33

1.515

69.03

50

34

1.471

70.941

50

35

1.429

72.857

50

36

1.389

74.778

50

37

1.351

76.703

50

38

1.316

78.632

50

39

1.282

80.564

50

40

1.25

82.5

50

41

1.22

84.439

50

42

1.19

86.381

50

43

1.163

88.326

50

44

1.136

90.273

50

45

1.111

92.222

50

46

1.087

94.174

50

47

1.064

96.128

50

48

1.042

98.083

50

49

1.02

100.041

50

50

1

102

50

As seen above, the minimum perimeter is achieved when the dimensions are 7m by 7.143m with a perimeter of 28.286m while maintaining an area of 50m2. Proving with the formula:image17.png

 which is 28.28image14.png

MAXIMIZING AREA

Anna has 40 m of fencing. She wishes to form a rectangular enclosure in which she will keep chickens. To help maximize the area she can enclose, she uses an existing fence. The 40 m of fencing forms the other three sides of the rectangle. Your task is to determine the rectangular shape, which encloses the maximum area of ground.

  1. Your task is to find the value of dimension that maximizes the area. By using technology, list all the possible values of length, width, and area with fencing 40 m.

The data is attached at the back. Color coding: Yellow is width, blue is area.

Width (m)

Length (m)

Perimeter (m)

Area (m2)

1

38

40

38

2

36

40

72

3

34

40

102

4

32

40

128

5

30

40

150

6

28

40

168

7

26

40

182

8

24

40

192

9

22

40

198

10

20

40

200

15

10

40

150

16

8

40

128

17

6

40

102

18

4

40

72

19

2

40

38

The data that satisfies the solution is most area which is 200m2 with the length=20 and breadth=10.

  1. Make a formula and predict the kind of graph and test if the hypothesis holds on your findings.

As this formula is about area, we will use the area formula:

image18.png

We will also use the perimeter formula, but in this only three sides are equal to the perimeter. image19.png

Isolating l,

image20.png

Substituting to the area equation,

image21.png

image22.png

(Where x is equal to the width or breadth)

To prove my formula, I will use 2 examples:

X=2

When x=2 the area is:

image22.png

image23.png

image24.png

X=4

When x=4 the area is:

image22.png

image25.png

image26.png

So, we can see above the formulas function properly.

We can also prove this equation by another method. As the graph is a parabola, it is a quadratic equation. By using the standard quadratic equation, we can find the equation:

Taking three points on the graph (1, 38), (2, 72) and (3, 102) and substituting to the general quadratic equation image27.png

,

image28.png

image29.png

image30.png

Solving them simultaneously,

1X4…..image31.png

…….1’

1’-2……image32.png

……..4

1X9….image33.png

……..1’’

1’’-3…….image34.png

…….5

4X3…….image35.png

So image36.png

. Substituting to 5, image37.png

. Substituting to 1. image38.png

So, image39.png

The graph of this equation is quadratic as visible from the equation and the data. And the data increases first, and then decreases, it displays a parabolic behavior. The equations too are in the power of 2, making it a quadratic curve or parabola. To prove this, here is the graph:image63.jpg

  1. What relationship do you observe?

This graph is a quadratic graph also known as a parabola. This graph is an upward parabola because the co-efficient of x2 is negative. At the beginning of the graph from 1 to 10, the graph is increasing and later on from 10 to 20 it is decreasing. This shows that this is a quadratic equation which is also visible from the equation and graph.  As there is no y-intercept in the graph, the value of c is 0. In this case we also observe that the area is the highest when the dimensions are close to that of a square.

  1. What degree of accuracy have you used in your findings? Why?

I have used one decimal place in my data for length and width and two for my area. This makes my result and my method quite accurate as there is a high degree of accuracy. I did this so that there is no repetition of any data, as said earlier, the data was quite close, so using a accuracy of 1 decimal place would give me repeated data, and thus cause a negative impact on the graph too. As I have made use of technology to calculate my results, they are even more accurate as there is very less room for error.  I have also done this for very accurate data which is better than the ones with lower decimal places, while giving me the same graph as any higher decimal places.

  1. Suggest any method that you may use for the improvement of your work.

Although my graph, findings and formulae I have used are very accurate, I could have made a few changes in my investigative. I could have used another graphing software instead of Excel, I would have gotten a much accurate graph than the one attached above. I could have also used it to confirm my graph and my findings. As I have used three methods to find my equation, my equation is very accurate. But forming another equation would also increase the validity of the results. I could have also used the graph to find the equation of the graph which would improve the accuracy of my equation.

  1. Janeth started her own summer business -putting on birthday parties for small children. Because all the children wanted to sit together, she had to place the card tables together into rectangles. Only one child could sit on each side of a card table. Her first party had eighteen children. How many tables did Janneth need to borrow with a maximum area?

For finding the general equation to find the maximum area, I just tried modifying the formula used for finding the minimum perimeter in terms of area. So the general equation is:

image40.png

As we can see below, the best dimensions that maximize the area are 5 tables by 4 tables where 20 tables are required. But if we don’t include the tables in the center, than only 14 tables are required, thus this is the best dimension. This also goes along with my hypothesis saying that the area is the highest when the dimensions are the closest to that of a square.

Length (m)

Width (m)

Perimeter  (m)

Area (m2)

1

8

18

8

2

7

18

14

3

6

18

18

4

5

18

20

5

4

18

20

6

3

18

18

7

2

18

14

8

1

18

8

Checking with general equation:

image41.png

But as tables cannot be in decimals, image42.png

Example 1: A child is building a model of a house. She has a limited perimeter of the base Styrofoam of 8m. Help her choose the dimensions of the Styrofoam to maximize the area.

Length (m)

Width (m)

Perimeter (m)

Area (m2)

0.1

3.9

8

0.39

0.2

3.8

8

0.76

0.3

3.7

8

1.11

0.4

3.6

8

1.44

0.5

3.5

8

1.75

0.6

3.4

8

2.04

0.7

3.3

8

2.31

0.8

3.2

8

2.56

0.9

3.1

8

2.79

1

3

8

3

1.1

2.9

8

3.19

1.2

2.8

8

3.36

1.3

2.7

8

3.51

1.4

2.6

8

3.64

1.5

2.5

8

3.75

1.6

2.4

8

3.84

1.7

2.3

8

3.91

1.8

2.2

8

3.96

1.9

2.1

8

3.99

2

2

8

4

2.1

1.9

8

3.99

2.2

1.8

8

3.96

2.3

1.7

8

3.91

2.4

1.6

8

3.84

2.5

1.5

8

3.75

2.6

1.4

8

3.64

2.7

1.3

8

3.51

2.8

1.2

8

3.36

2.9

1.1

8

3.19

3

1

8

3

3.1

0.9

8

2.79

3.2

0.8

8

2.56

3.3

0.7

8

2.31

3.4

0.6

8

2.04

3.5

0.5

8

1.75

3.6

0.4

8

1.44

3.7

0.3

8

1.11

3.8

0.2

8

0.76

3.9

0.1

8

0.39

 The maximum area is achieved when the Styrofoam is with the dimensions 2m by 2m with the area of 4m2. Checking with the formula: image43.png

Example 2: A company has to build a swimming pool in an apartment. They have a limited amount of concrete to build enough wall to cover 100m. Formulate a dimension that will help maximize the amount of area they can cover, so that a larger pool can be built.

Length (m)

Width (m)

Perimeter  (m)

Area (m2)

1

49

100

49

2

48

100

96

3

47

100

141

4

46

100

184

5

45

100

225

6

44

100

264

7

43

100

301

8

42

100

336

9

41

100

369

10

40

100

400

11

39

100

429

12

38

100

456

13

37

100

481

14

36

100

504

15

35

100

525

16

34

100

544

17

33

100

561

18

32

100

576

19

31

100

589

20

30

100

600

21

29

100

609

22

28

100

616

23

27

100

621

24

26

100

624

25

25

100

625

26

24

100

624

27

23

100

621

28

22

100

616

29

21

100

609

30

20

100

600

31

19

100

589

32

18

100

576

33

17

100

561

34

16

100

544

35

15

100

525

36

14

100

504

37

13

100

481

38

12

100

456

39

11

100

429

40

10

100

400

41

9

100

369

42

8

100

336

43

7

100

301

44

6

100

264

45

5

100

225

46

4

100

184

47

3

100

141

48

2

100

96

49

1

100

49

As seen above, the dimensions that maximize the area are 25m by 25m giving an area of 625m2. We c

...read more.

Conclusion

  1. Generalize and form this for any dimension of your choice, use your understanding to show situation where we can use the concept of maximizing the volume. Solve the example below and give two more examples.

A box must have base x cm by 2x cm. What are the dimensions of the box of maximum volume if its length + width + depth ≤ 140 cm?

Using the basic dimensions as 28cm by 22cm:

Length (cm)

Width (cm)

Depth (cm)

Volume (cm3)

0.1

0.2

139.7

2.794

0.2

0.4

139.4

11.152

0.3

0.6

139.1

25.038

0.4

0.8

138.8

44.416

0.5

1

138.5

69.25

0.6

1.2

138.2

99.504

0.7

1.4

137.9

135.142

0.8

1.6

137.6

176.128

0.9

1.8

137.3

222.426

1

2

137.0

274

1.1

2.2

136.7

330.814

1.2

2.4

136.4

392.832

1.3

2.6

136.1

460.018

1.4

2.8

135.8

532.336

1.5

3

135.5

609.75

1.6

3.2

135.2

692.224

1.7

3.4

134.9

779.722

1.8

3.6

134.6

872.208

1.9

3.8

134.3

969.646

2

4

134.0

1072

2.1

4.2

133.7

1179.234

2.2

4.4

133.4

1291.312

2.3

4.6

133.1

1408.198

2.4

4.8

132.8

1529.856

2.5

5

132.5

1656.25

2.6

5.2

132.2

1787.344

2.7

5.4

131.9

1923.102

2.8

5.6

131.6

2063.488

2.9

5.8

131.3

2208.466

3

6

131.0

2358

3.1

6.2

130.7

2512.054

3.2

6.4

130.4

2670.592

3.3

6.6

130.1

2833.578

3.4

6.8

129.8

3000.976

3.5

7

129.5

3172.75

3.6

7.2

129.2

3348.864

3.7

7.4

128.9

3529.282

3.8

7.6

128.6

3713.968

3.9

7.8

128.3

3902.886

4

8

128.0

4096

4.1

8.2

127.7

4293.274

4.2

8.4

127.4

4494.672

4.3

8.6

127.1

4700.158

4.4

8.8

126.8

4909.696

4.5

9

126.5

5123.25

4.6

9.2

126.2

5340.784

4.7

9.4

125.9

5562.262

4.8

9.6

125.6

5787.648

4.9

9.8

125.3

6016.906

5

10

125.0

6250

5.1

10.2

124.7

6486.894

5.2

10.4

124.4

6727.552

5.3

10.6

124.1

6971.938

5.4

10.8

123.8

7220.016

5.5

11

123.5

7471.75

5.6

11.2

123.2

7727.104

5.7

11.4

122.9

7986.042

5.8

11.6

122.6

8248.528

5.9

11.8

122.3

8514.526

6

12

122.0

8784

6.1

12.2

121.7

9056.914

6.2

12.4

121.4

9333.232

6.3

12.6

121.1

9612.918

6.4

12.8

120.8

9895.936

6.5

13

120.5

10182.25

6.6

13.2

120.2

10471.824

6.7

13.4

119.9

10764.622

6.8

13.6

119.6

11060.608

6.9

13.8

119.3

11359.746

7

14

119.0

11662

7.1

14.2

118.7

11967.334

7.2

14.4

118.4

12275.712

7.3

14.6

118.1

12587.098

7.4

14.8

117.8

12901.456

7.5

15

117.5

13218.75

7.6

15.2

117.2

13538.944

7.7

15.4

116.9

13862.002

7.8

15.6

116.6

14187.888

7.9

15.8

116.3

14516.566

8

16

116.0

14848

8.1

16.2

115.7

15182.154

8.2

16.4

115.4

15518.992

8.3

16.6

115.1

15858.478

8.4

16.8

114.8

16200.576

8.5

17

114.5

16545.25

8.6

17.2

114.2

16892.464

8.7

17.4

113.9

17242.182

8.8

17.6

113.6

17594.368

8.9

17.8

113.3

17948.986

9

18

113.0

18306

9.1

18.2

112.7

18665.374

9.2

18.4

112.4

19027.072

9.3

18.6

112.1

19391.058

9.4

18.8

111.8

19757.296

9.5

19

111.5

20125.75

9.6

19.2

111.2

20496.384

9.7

19.4

110.9

20869.162

9.8

19.6

110.6

21244.048

9.9

19.8

110.3

21621.006

10

20

110.0

22000

10.1

20.2

109.7

22380.994

10.2

20.4

109.4

22763.952

10.3

20.6

109.1

23148.838

10.4

20.8

108.8

23535.616

10.5

21

108.5

23924.25

10.6

21.2

108.2

24314.704

10.7

21.4

107.9

24706.942

10.8

21.6

107.6

25100.928

10.9

21.8

107.3

25496.626

11

22

107.0

25894

So the dimensions that give the maximum volume with the sheet of size 28cm by 22cm is with the dimensions 11cm by 22cm by 107 cm. I could have gone further, but as the size of the width of the sheet is on 22 cm, this is the maximum volume available.

Example 1: Packaging companies require maximizing their volume in order to decrease packaging cost and put more food in the box. Suggest the dimensions of the box with the most volume for a plastic sheet 20cm by 15cm.

Length (cm)

Width (cm)

Depth (cm)

Volume (cm3)

19

14

0.5

133

18

13

1

234

17

12

1.5

306

16

11

2

352

15

10

2.5

375

14

9

3

378

13

8

3.5

364

12

7

4

336

11

6

4.5

297

10

5

5

250

9

4

5.5

198

8

3

6

144

7

2

6.5

91

6

1

7

42

5.2

0.2

7.4

7.696

The maximum volume achievable for this dimension is 7.696cm3 with the dimensions 5.2cm by 0.2cm by 7.4cm.

Example 2: A bottle manufacturing company is making special cuboid shaped bottles. They have little budget to test the bottles on people, so they have to maximize the volume. Suggest the size of the square that should be cut off from a 50cm by 25cm to maximize the volume.

Length (cm)

Width (cm)

Depth (cm)

Volume (cm3)

49.6

24.6

0.2

244.032

49.2

24.2

0.4

476.256

48.8

23.8

0.6

696.864

48.4

23.4

0.8

906.048

48

23

1

1104

47.6

22.6

1.2

1290.912

47.2

22.2

1.4

1466.976

46.8

21.8

1.6

1632.384

46.4

21.4

1.8

1787.328

46

21

2

1932

45.6

20.6

2.2

2066.592

45.2

20.2

2.4

2191.296

44.8

19.8

2.6

2306.304

44.4

19.4

2.8

2411.808

44

19

3

2508

43.6

18.6

3.2

2595.072

43.2

18.2

3.4

2673.216

42.8

17.8

3.6

2742.624

42.4

17.4

3.8

2803.488

42

17

4

2856

41.6

16.6

4.2

2900.352

41.2

16.2

4.4

2936.736

40.8

15.8

4.6

2965.344

40.4

15.4

4.8

2986.368

40

15

5

3000

39.6

14.6

5.2

3006.432

39.2

14.2

5.4

3005.856

38.8

13.8

5.6

2998.464

38.4

13.4

5.8

2984.448

38

13

6

2964

37.6

12.6

6.2

2937.312

37.2

12.2

6.4

2904.576

36.8

11.8

6.6

2865.984

36.4

11.4

6.8

2821.728

...read more.

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