French Holidays
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Investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them.
Extracts from this document...
Introduction
Introduction
The aim of this coursework is to investigate the volume of an open box constructed by one piece of rectangular card that has all four corners having had squares cut out of them. Firstly I will be studying the volume whilst changing the side of one length of the cut out square and the size of the original rectangle card. After I have investigated this relationship I will try to find out the formula for finding the cut size to get the largest volume for any specified original card size.
Square card size
I am going to begin by investigating a square card because this will give me a basic formula which I can elaborate on. I will start with a round number of 20cm for the length. This means that the maximum cut out square length I can cut out will be 9cm else I will have no box left.
The formula for the volume of any box is as follows:
Below, there is a diagram to explain where all these figures come from.
I have also included in this diagram the labels c, x and y, these show the cut out size and the original length and width of the card, I will now need to show the values of the width, length and height in terms of c, x and y.
Middle
47.2
5195.77600000000
30
6.33
17.34
47.34
5196.14254800000
30
6.34
17.32
47.32
5196.15241600000
30
6.35
17.3
47.3
5196.14150000000
30
6.339
17.322
47.322
5196.152364876
30
6.34
17.32
47.32
5196.152416000
30
6.341
17.318
47.318
5196.152259284
30
6.3396
17.3208
47.3208
5196.15242049254
30
6.3397
17.3206
47.3206
5196.15242248709
30
6.3398
17.3204
47.3204
5196.15242240317
The value of c for the highest volume is 6.3397 only up to 4 decimal places.
x | c | x/c |
20 | 4.2265 | 4.732048 |
30 | 6.3397 | 4.732085 |
I noticed whilst evaluating the results that if you divide x by c the values are very similar:
This is a very strong relationship; I can therefore predict that if I construct a similar table for another number, if you divide that number by 4.732, it will give its largest volume. I will use number 17: 17/4.732=3.59 therefore I predict that the largest volume will be made with a cut size of approx 3.59.
x=17
x | c | x-2c | 2x-2c | c(x-2c)(2x-2c) |
17 | 3 | 11 | 28 | 924 |
17 | 4 | 9 | 26 | 936 |
17 | 5 | 7 | 24 | 840 |
17 | 3.5 | 10 | 27 | 945 |
17 | 3.6 | 9.8 | 26.8 | 945.504 |
17 | 3.7 | 9.6 | 26.6 | 944.832 |
17 | 3.58 | 9.84 | 26.84 | 945.498048 |
17 | 3.59 | 9.82 | 26.82 | 945.506916 |
17 | 3.6 | 9.8 | 26.8 | 945.504000 |
17 | 3.592 | 9.816 | 26.816 | 945.507274752 |
17 | 3.593 | 9.814 | 26.814 | 945.507277428 |
17 | 3.594 | 9.812 | 26.812 | 945.507162336 |
17 | 3.5924 | 9.8152 | 26.8152 | 945.50728995610 |
17 | 3.5925 | 9.815 | 26.815 | 945.50729081250 |
17 | 3.5926 | 9.8148 | 26.8148 | 945.50729049110 |
The value of c for the highest volume is 3.5925 only up to 4 decimal places.
x | c | x/c |
20 | 4.2265 | 4.732048 |
30 | 6.3397 | 4.732085 |
17 | 3.5925 | 4.732081 |
Average = | 4.732071 |
My prediction proved to be correct and I have added a third row to the data and the average of the three results gives me a formula of the cut size for maximum volume:
This I can include in the formula for area:
Replace c with formula:
Multiply out the brackets:
Multiply by x/4.732071:
Put formula over a common denominator:
Simplify:
This then is the formula for the maximum volume of any open box constructed from a rectangle that’s length is twice as large as its width.
Ratio 1:3
The following formulae shows the values of the width, length and height in terms of c, x and y for the ratio of 1:3 and this time I know that the length is exactly three times the width.
Therefore I can replace these sub formulae into the first formula.
I have constructed a table in excel where I can input the data for the cut size and the original width of the card and it will calculate the volume, I will use it to find the largest volume through the cut size.
x=20
x | c | x-2c | 3x-2c | c(x-2c)(3x-2c) |
20 | 4 | 12 | 52 | 2496 |
20 | 5 | 10 | 50 | 2500 |
20 | 6 | 8 | 48 | 2304 |
20 | 4.4 | 11.2 | 51.2 | 2523.136 |
20 | 4.5 | 11 | 51 | 2524.500 |
20 | 4.6 | 10.8 | 50.8 | 2523.744 |
20 | 4.50 | 11 | 51 | 2524.500000 |
20 | 4.51 | 10.98 | 50.98 | 2524.519404 |
20 | 4.52 | 10.96 | 50.96 | 2524.517632 |
20 | 4.513 | 10.974 | 50.974 | 2524.521094788 |
20 | 4.514 | 10.972 | 50.972 | 2524.521234976 |
20 | 4.515 | 10.97 | 50.97 | 2524.521163500 |
20 | 4.5141 | 10.9718 | 50.9718 | 2524.5212373529 |
20 | 4.5142 | 10.9716 | 50.9716 | 2524.5212376132 |
20 | 4.5143 | 10.9714 | 50.9714 | 2524.5212357568 |
20 | 4.51415 | 10.9717 | 50.9717 | 2524.52123774759 |
20 | 4.51416 | 10.97168 | 50.97168 | 2524.52123776304 |
20 | 4.51417 | 10.97166 | 50.97166 | 2524.52123775731 |
To achieve even more accurate results I managed to go to fifth decimal place where the value for c was, yet even this is not totally accurate as there will be a much larger number of decimals in the c column to get the maximum number for V.
x | c | x/c |
20 | 4.51416 | 4.430503128 |
Conclusion
c
x-2c
2x-2c
c(x-2c)(2x-2c)
30
6
18
78
8424
30
7
16
76
8512
30
8
14
74
8288
30
6.7
16.6
76.6
8519.452
30
6.8
16.4
76.4
8520.128
30
6.9
16.2
76.2
8517.636
30
6.76
16.48
76.48
8520.2391
30
6.77
16.46
76.46
8520.2589
30
6.78
16.44
76.44
8520.2470
30
6.77
16.46
76.46
8520.2589320
30
6.771
16.458
76.458
8520.2591680
30
6.772
16.456
76.456
8520.2590866
30
6.7711
16.4578
76.4578
8520.2591741857
30
6.7712
16.4576
76.4576
8520.2591771525
30
6.7713
16.4574
76.4574
8520.2591769444
30
6.77123
16.45754
76.45754
8520.25917742344
30
6.77124
16.45752
76.45752
8520.25917745025
30
6.77125
16.4575
76.4575
8520.25917744531
30
6.771242
16.457516
76.457516
8520.25917745180
30
6.771243
16.457514
76.457514
8520.25917745210
30
6.771244
16.457512
76.457512
8520.25917745208
30
6.7712433
16.4575134
76.4575134
8520.25917745213
30
6.7712434
16.4575132
76.4575132
8520.25917745214
30
6.7712435
16.4575130
76.4575130
8520.25917745213
The value of c for the highest volume is 6.7712435 only up to 7 decimal places.
x | c | x/c |
20 | 4.51416 | 4.430503128 |
30 | 6.7712434 | 4.430500903 |
Proving my prediction correct, the value of x divided by c the values are very similar:
As before, this is a very strong relationship; I can therefore predict that if I construct a similar table for another number, if you divide that number by 4.4305009, it will give its largest volume. I will use number 17:
17/4.4305009=3.837036 therefore I predict that the largest volume will be made with a cut size of approx 3.837036.
x=17
x | c | x-2c | 2x-2c | c(x-2c)(2x-2c) |
17 | 3 | 11 | 45 | 1485 |
17 | 4 | 9 | 43 | 1548 |
17 | 5 | 7 | 41 | 1435 |
17 | 3.7 | 9.6 | 43.6 | 1548.672 |
17 | 3.8 | 9.4 | 43.4 | 1550.248 |
17 | 3.9 | 9.2 | 43.2 | 1550.016 |
17 | 3.83 | 9.34 | 43.34 | 1550.3671 |
17 | 3.84 | 9.32 | 43.32 | 1550.3708 |
17 | 3.85 | 9.3 | 43.3 | 1550.3565 |
17 | 3.836 | 9.328 | 43.328 | 1550.3715082 |
17 | 3.837 | 9.326 | 43.326 | 1550.3716050 |
17 | 3.838 | 9.324 | 43.324 | 1550.3715219 |
17 | 3.8369 | 9.3262 | 43.3262 | 1550.3716034296 |
17 | 3.837 | 9.326 | 43.326 | 1550.3716050120 |
17 | 3.8371 | 9.3258 | 43.3258 | 1550.3716047952 |
17 | 3.83703 | 9.32594 | 43.32594 | 1550.37160513588 |
17 | 3.83704 | 9.32592 | 43.32592 | 1550.37160514119 |
17 | 3.83705 | 9.3259 | 43.3259 | 1550.37160512851 |
17 | 3.837037 | 9.325926 | 43.325926 | 1550.37160514149 |
17 | 3.837038 | 9.325924 | 43.325924 | 1550.37160514157 |
17 | 3.837039 | 9.325922 | 43.325922 | 1550.37160514147 |
17 | 3.8370378 | 9.3259244 | 43.3259244 | 1550.37160514157000 |
17 | 3.8370379 | 9.3259242 | 43.3259242 | 1550.37160514157000 |
17 | 3.837038 | 9.3259240 | 43.3259240 | 1550.37160514157000 |
The value of c for the highest volume is 3.8370379 up to 7 decimal places.
x | c | x/c |
20 | 3.8370379 | 4.430500934 |
30 | 4.5141600 | 4.430503128 |
17 | 6.7712434 | 4.430500903 |
Average = | 4.430501655 |
Revised Formulae
My prediction proved to be correct and I have added a third row to the data and the average of the three results gives me a formula of the cut size for maximum volume:
This I can include in the formula for area:
Replace c with formula:
Multiply out the brackets:
Multiply by x/4.732071:
Put formula over a common denominator:
Simplify:
I now have the formulae for both the ratios 1:2 and 1:3, with examples for each.
This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.
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