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• Level: GCSE
• Subject: Maths
• Word count: 1324

# The Open Box Problem

Extracts from this document...

Introduction

Introduction The Open Box Problem A square piece of card has four equal sized squares cut out from each of it's four corners and is folded to make an open top box. The problem is to find out which value of x (the side length of the cut out) will maximise the volume in the resulting box, which is equivalent to the height (x) * the width (the side length of the square - 2x) * the depth (the same value as the width). This problem can be expanded to finding the maximised volume for a box that is made from a rectangular piece of card. Methods of Finding the Solution to the Problem - Trial and Error Approach There are two main ways of finding the solution to the problem. The first is a trial and error approach. For square pieces of card, a table is constructed showing the cut out side length and the resulting volume. In the example, the square being investigated has a side length of 20 cm. The length of the cut out is increased until the resulting volume goes down. ...read more.

Middle

3.33 6.01 25 x 25 4.17 6 30 x 30 5 6 35 x 35 5.83 6 Analysis It seems that the optimal cut out length is a sixth of the length of the original square. Testing this hypothesis, the optimal cut outs for 40 x 40, 45 x 45 and 50 x 50 should be 6.67, 7.5 and 8.33 consecutively. These results show that the predicted numbers result in a larger volume than those cut outs bigger or smaller by 0.01: Analysing Results for Rectangles Introduction Unlike finding the optimal cut out from squares, rectangles will be a lot harder as the optimal cut relationship will be with two variables; length and width, rather than just one. One approach to this problem is looking at the optical cuts of rectangles whose sides follow a general rule, for example the length to width ratio could be 1 : 2 or 3 : 5. For now, I shall analyse the results for this method. Results for 1 : 2 Rectangles Width Height Cut Out Volume Width / Cut Out Height / Cut Out 5 10 1.06 24.06 4.74 9.47 6 12 1.27 41.57 4.74 9.47 ...read more.

Conclusion

3.64 886.55 4.95 8.24 21 35 4.25 1,407.81 4.94 8.24 Results for 4 : 5 Rectangles Width Height Cut Out Volume Width / Cut Out Height / Cut Out 8 10 1.47 52.51 5.44 6.8 12 15 2.2 177.23 5.43 6.79 16 20 2.95 420.11 5.43 6.79 20 25 3.68 820.53 5.43 6.79 24 30 4.41 1,417.87 5.43 6.79 28 35 5.15 2,251.53 5.43 6.79 Analysing the Results Collecting averages for the different ratios, this table shows them, as well as the results for the squares (which are the same as rectangles with a ratio of 1 : 1. As the two ratios are proportionate to the ratios of the sides of the rectangles, I have included a column showing height / cut out ratio divided by the ratio of height side. Ratio Width / Cut Out Height / Cut Out Base 1 : 1 6 6 6 1 : 2 4.74 9.47 4.74 1 : 3 4.43 13.3 4.43 2 : 3 5.11 7.65 2.55 1 : 5 4.24 21.18 4.24 2 : 5 4.55 11.36 2.27 3 : 5 4.95 8.24 1.65 4 : 5 5.43 6.79 1.36 ?? ?? ?? ?? ...read more.

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# Related GCSE Open Box Problem essays

1. ## Investigation: The open box problem.

�x1.8 V = (12-(2x1.9)) �x1.9 V = (12-(2x2.1)) �x2.1 V = 127.008 V = 127.756 V = 127.764 X = 2.2 X = 2.3 X = 2.4 V = (12-(2x2.2)) �x2.2 V = (12-(2x2.3)) �x2.3 V = (12-(2x2.4)) �x2.4 V = 127.072 V = 125.948 V = 124.416 X = 2.5 V = (12-(2x2.5))

2. ## Tbe Open Box Problem

6 x = 25 � V175 6 x = 25 + 13.22875656 = 6.371459427 6 x = 25 - 13.22875656 = 1.961873907 or 1.962 (3.d.p) 6 x must be 1.962 because otherwise 6.371 (3.d.p) is over twice the length. I shall prove that this formula is correct. Cut off (cm)

1. ## THE OPEN BOX PROBLEM

As there are too many combinations of lengths and widths of rectangles for me to possibly even begin to investigate I am going to investigate two different rectangles, with the ratio between the length and width as 2:1 and 3:1.

2. ## Open box Problem.

(2W - X -X )*(W - X - X)*X(cm) 1.21 41.49864 1.22 41.52099 1.23 41.53907 1.24 41.5529 1.25 41.5625 1.26 41.5679 1.27 41.56913 1.28 41.56621 1.29 41.55916 1.3 41.548 From the results we can see, 1.27cm is the maximum value of X.

1. ## Open Box Problem

73.9375 1.751 6.498 6.498 1.751 73.934231 1.752 6.496 6.496 1.752 73.93092403 1.753 6.494 6.494 1.753 73.92757911 1.754 6.492 6.492 1.754 73.92419626 1.755 6.49 6.49 1.755 73.9207755 1.76 6.48 6.48 1.76 73.903104 1.77 6.46 6.46 1.77 73.864932 1.8 6.4 6.4 1.8 73.728 2 6 6 2 72 3 4 4 3

2. ## The Open Box Problem

________________________________________________________ Rectangular piece of card with dimensions 5 x 15 Small Side Volume Length Width Height 1 39 13 3 1 2 22 11 1 2 1.5 36 12 2 1.5 1.4 37.576 12.2 2.2 1.4 1.3 38.688 12.4 2.4 1.3 1.2 39.312 12.6 2.6 1.2 1.1 39.424 12.8 2.8

1. ## Maths Courseowrk - Open Box

These are the values when l=10 X L V=x(10-2x) 2 1 10 64 1.5 10 73.5 2 10 72 2.5 10 62.5 3 10 48 3.5 10 31.5 4 10 16 4.5 10 4.5 1.75 10 73.9375 Here is a graph when l=10 These are the values when l=20 X L v=x(20-2x)

2. ## The Open Box Problem

591.5 4 576 4.5 544.5 5 500 I then homed in between 3 and 3.5cm for 'x': size of cut out 'x' (cm) volume (cm�) 3 588 3.1 590.364 3.2 591.872 3.3 592.548 3.4 592.416 3.5 591.5 I then homed in again with values of 'x' to two decimal places: size of cut out 'x' (cm)

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