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

Extracts from this document...

Introduction

MINIMUM PERIMETER

- 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.

- 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.

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:

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

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

Substituting this in the original perimeter equation,

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

X=5 (when length is 5m)

The perimeter will be:

X=10 (when length is 10m)

The perimeter will be:

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.

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:

which is 28.28

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.

- 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.

- 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:

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

Isolating l,

Substituting to the area equation,

(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:

X=4

When x=4 the area is:

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

,

Solving them simultaneously,

1X4…..

…….1’

1’-2……

……..4

1X9….

……..1’’

1’’-3…….

…….5

4X3…….

So

. Substituting to 5,

. Substituting to 1.

So,

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:

- 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.

- 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.

- 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.

- 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:

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:

But as tables cannot be in decimals,

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:

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

Conclusion

- 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 |

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