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Area & Volume Exploration – Component proportional changes

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Introduction

Area & Volume Exploration – Component proportional changes

Question 2:          Suppose that you are required to design an open steel tray from a sheet of steel that measures 120cm by 80cm by cutting squares from each corner, and folding to form a tray. What size should the squares be cut from each corner of the sheet so that the maximum volume is obtained for the tray.

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Height = x cm

Width  = 80cm - 2x cm

Length = 120cm – 2 x cm

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The task is to find the optimal size of the squares that need to be cut out of the corners in order to find the maximum obtainable Volume inside the tray once it has been folded. To work this out there are a number of different methods for doing this. One method is Trial and Improvement.

Trial and Improvement

 Height

(Value of xcm)

2 xcm

Length (cm)

(120cm – 2 x cm)

Width (cm)

(80cm - 2x cm)

Volume (cm3)

(120x  - 2 x 2)×(80 - 2x)

+/-

5

10

110

70

38’500

7.5

15

105

65

51’187.5

+

10

20

100

60

60’000

+

12.5

25

95

55

65’312.5

+

=

15

30

90

50

67’500

+

17.5

35

85

45

66’937.5

-

...read more.

Middle

+

15

30

90

50

67’500

+

16

32

88

48

67’584

+

=

17

34

86

46

67’252

-

The Volume starts to decrease after the value of 16 for x. So we know that the Maximum volume lies between 15 and 17. We can now continue the trial and improvement table working only between 15 and 17.

Height

(Value of xcm)

2 xcm

Length (cm)

(120cm – 2 x cm)

Width (cm)

(80cm - 2x cm)

Volume (cm3)

(120x  - 2 x 2)×(80 - 2x)

+/-

15

30

90

50

67’500

15.25

30.5

89.5

49.5

67’561.3125

+

15.5

31

89

49

67’595

+

15.65

31.3

88.7

48.7

67’603.1485

+

=

15.75

31.5

88.5

48.5

67’602.9375

-

The table shows that the volume is at its greatest at 15.65 and decreases after this value. For increased accuracy  we can continue the trial and improvement table working between the x values of 15.50 and 15.8.

Height

(Value of xcm)

2 xcm

Length (cm)

(120cm – 2 x cm)

Width (cm)

(80cm - 2x cm)

Volume (cm3)

(120x  - 2 x 2)×(80 - 2x)

+/-

15.500

31

89

49

67’595

15.525

31.05

88.95

48.95

67’597.44131

+

15.550

31.1

88.9

48.9

67’599.1155

+

15.575

31.15

88.85

48.85

67’600.52294

+

15.600

31.2

88.8

48.8

67’601.664

+

15.625

31.25

88.75

48.75

67’02.53.906

+

15.650

31.3

88.7

48.7

67’603.1485

+

15.675

31.35

88.65

48.65

67’603.49269

+

15.700

31.4

88.6

48.6

67’603.572

+ =

15.725

31.45

88.55

48.55

67’603.38681

-

The optimal value of xto create a maximum volumehas now been narrowed down to between 15.675 and 15.725.

...read more.

Conclusion

2. The maximum possible volume of the tray is 67’603.57731cm3

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There are other methods that could be used to determine the maximum volume and the optimal value of X.One method could be differential calculus. This is applying the rules of differentiation to the object in question. This would be a quicker and more accurate method to determine the maximum volume and vale of X.

This investigation could be useful in engineering projects where a product must be produced with the maximum efficiency. This could mean being produced so that the volume inside is at its maximum possible and the wastage of material is at its least. Differentiation can also be used to determine dimensions of building sites or fields etc.

This exercises a very useful task that is widely used throughout many types of engineering and construction.

...read more.

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