The Open Box Problem
An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below.(fig 1)
FIG 1
The card is then folded along the dotted lines to make the box.
The main aim is to determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card.
I am going to begin by investigating a square with a side length of 20 cm. Using this side length, the maximum whole number I can cut off each corner is 9cm, as otherwise I would not have any box left.
I am going to begin by looking into whole numbers being cut out of the box corners.
The formula that needs to be used to get the volume of a box is: ( * = multiply)
Volume = Length * Width * Height
If I am to use a square of side length 20cm, then I can calculate the side lengths minus the cut out squares using the following equation.
Volume = Length -- (2 * Cut Out) * Width -- (2 * Cut Out) * Height
Using a square, both the length & the width are equal. I am using a length/width of 20cm. I am going to call the cut out ""x."" Therefore the equation can be changed to:
Volume = 20 -- (2x) * 20 -- (2x) * Height
If I were using a cut out of length 1cm, the equation for this would be as follows:
Volume = 20 -- (2 * 1) * 20 -- (2 * 1) * 1
So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:
(20 -- 2) * (20 -- 2) * 1
8 * 18 * 1
= 324cm³
I used these formulae to construct a table, which would allow me to quickly and accurately calculate the volume of the box.
Cut out (cm²)
formula
volume (cm³)
Volume = 20 -- (2*1) * 20 -- (2*1) * 1
324
2
Volume = 20 -- (2*2) * 20 -- (2*2) * 2
512
3
Volume = 20 -- (2*3) * 20 -- (2*3) * 3
588
4
Volume = 20 -- (2*4) * 20 -- (2*4) * 4
576
5
Volume = 20 -- (2*5) * 20 -- (2*5) * 5
500
6
Volume = 20 -- (2*6) * 20 -- (2*6) * 6
384
7
Volume = 20 -- (2*7) * 20 -- (2*7) * 7
252
8
Volume = 20 -- (2*8) * 20 -- (2*8) * 8
28
9
Volume = 20 -- (2*9) * 20 -- (2*9) * 9
36
As you can see by the table above, the largest volume is achieved when an area of 3cm² is cut off each corner of the box. I have also drawn a graph to show my results. (See below fig 2)By looking at this graph, and my table of results, I can see that to achieve the maximum volume I will need to look at cut outs of between 3 and 4 cm². This is because the volume the second highest volume is 4cm² cutout, so the highest has got to be between 3 and 4 cm²
FIG 2
To try to make my results more accurate, I am going to investigate the volume of the box with the cut out to more than 1 decimal place.
Below is a table showing the cut out to 1 decimal place, with the largest volume achieved highlighted in bold.
Cut out (cm²)
formula
Volume (cm³)
3.1
Volume = 20 -- (2*3.1) * 20 -- (2*3.1) * 3.1
590.364
3.2
Volume = 20 -- (2*3.2) * 20 -- (2*3.2) * 3.2
591.872
3.3
Volume = 20 -- (2*3.3) * 20 -- (2*3.3) * 3.3
592.548
3.4
Volume = 20 -- (2*3.4) * 20 -- (2*3.4) * 3.4
592.416
3.5
Volume = 20 -- (2*3.5) * 20 -- (2*3.5) * 3.5
591.5
3.6
Volume = 20 -- (2*3.6) * 20 -- (2*3.6) * 3.6
589.824
3.7
...
This is a preview of the whole essay
3.1
Volume = 20 -- (2*3.1) * 20 -- (2*3.1) * 3.1
590.364
3.2
Volume = 20 -- (2*3.2) * 20 -- (2*3.2) * 3.2
591.872
3.3
Volume = 20 -- (2*3.3) * 20 -- (2*3.3) * 3.3
592.548
3.4
Volume = 20 -- (2*3.4) * 20 -- (2*3.4) * 3.4
592.416
3.5
Volume = 20 -- (2*3.5) * 20 -- (2*3.5) * 3.5
591.5
3.6
Volume = 20 -- (2*3.6) * 20 -- (2*3.6) * 3.6
589.824
3.7
Volume = 20 -- (2*3.7) * 20 -- (2*3.7) * 3.7
587.412
3.8
Volume = 20 -- (2*3.8) * 20 -- (2*3.8) * 3.8
584.288
3.9
Volume = 20 -- (2*3.9) * 20 -- (2*3.9) * 3.9
580.476
As you can see by this table, the largest volume is arrived at when the corners cut off the box are 3.3cm². I have also drawn another graph to illustrate these results. This graph shows me that to get even more accurate results, to 2 decimal places, I am going to need to look at cut offs measuring between 3.3 and 3.4 cm. Again this is because 3.4 is the next highest volume, meaning that the best has got to fit between these two points
The table below shows the cut off measured to 2 decimal places.
Cut out (cm²)
formula
Volume (cm³)
3.31
Volume = 20 -- (2*3.31) * 20 -- (2*3.31) * 3.31
592.570764
3.32
Volume = 20 -- (2*3.32) * 20 -- (2*3.32) * 3.32
592.585472
3.33
Volume = 20 -- (2*3.33) * 20 -- (2*3.33) * 3.33
592.592148
3.34
Volume = 20 -- (2*3.34) * 20 -- (2*3.34) * 3.34
592.590186
3.35
Volume = 20 -- (2*3.35) * 20 -- (2*3.35) * 3.35
592.5815
3.36
Volume = 20 -- (2*3.36) * 20 -- (2*3.36) * 3.36
592.564224
3.37
Volume = 20 -- (2*3.37) * 20 -- (2*3.37) * 3.37
592.539012
3.38
Volume = 20 -- (2*3.38) * 20 -- (2*3.38) * 3.38
592.505888
3.39
Volume = 20 -- (2*3.39) * 20 -- (2*3.39) * 3.39
592.464876
Looking at the table you should be able to see again the largest volume in bold, is with a cut out of 3.33cm².
If I wish to work out the proportion of the box that needs to be cut away to obtain the maximum area, I need to divide 3.33 by 20. In doing this I get an answer of 0.1665, or a proportion 1/6.
You can find a general formula allowing you to achieve maximum area by using algebraic expressions for the height length and width.
20 cm is the length and width of the square, therefore
20 = 2H + L
Rearranging this it becomes
L = 20 -- 2H
The formula for the volume is
Width * Length * Height
Or
(20 - 2H) * (20 -- 2H) * H
The maximum area it found when the cut out is 1/6, meaning that:
H = 20/6
If I substitute this into the equation for maximum volume then it will become
V = (L -- 2L/6)(L -- 2L/6) L/6
I multiply the 2 in each bracket with L/6, giving you
V = (L -- 2L/6) (L -- 2L/6) L/6
To multiply brackets out I'll separate them, and write the 2nd bracket out twice.
V = L (L -- 2L/6) -- 2L/6 (L -- 2L/6) L/6
Then complete multiplication leaving
V = (L² - 2L²/6 -- 2L²/6 -- 4L²/36) L/6
Then I multiply the whole thing by L/6, which gives me
V = L³/6 -- 2L³/36 -- 2L³/36 + 4L³/216
I put the whole equation over a common denominator to help me with simplification, meaning the whole thing becomes over 216. This means that it becomes:
V = 4L³ + 36L³ - 12L³ - 12L³/216
This can be simplified so you end up with:
40L³ - 24L³/216
Which then cancels out to give you
6L³/216
You then divide the top and bottom by 8, which is a factor of both, and you will end up with:
2L³/27
I am now going to continue my investigation by looking at the shape of rectangles. 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 in the following manner.
I shall begin with a width of 20cm, and a length of 40cm, this is a ratio of 1:2, the length being twice as long as the width.
FIG 3
The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 9cm being the maximum for there to be a box left.
I will then investigate various other ratios.
I have again constructed a table to assist with my workings out. Firstly I am looking at whole numbers, a beginning with a width of 20cm and a length of 40cm, and cut outs from 1 to 9cm.
Here are the results I got.
cut out (cm²)
Formula
Volume (cm³)
Volume = 40 -- (2*1) * 20 -- (2*1) * 1
684
2
Volume = 40 -- (2*2) * 20 -- (2*2) * 2
152
3
Volume = 40 -- (2*3) * 20 -- (2*3) * 3
428
4
Volume = 40 -- (2*4) * 20 -- (2*4) * 4
536
5
Volume = 40 -- (2*5) * 20 -- (2*5) * 5
500
6
Volume = 40 -- (2*6) * 20 -- (2*6) * 6
344
7
Volume = 40 -- (2*7) * 20 -- (2*7) * 7
092
8
Volume = 40 -- (2*8) * 20 -- (2*8) * 8
768
9
Volume = 40 -- (2*9) * 20 -- (2*9) * 9
396
You should be able to see from this table, that the largest volume achieved is 1536cm², and this is obtained when the amount cut off from the corners measures 4cm². I can also see that to make my results more accurate, to more decimal places, I next need to look between 4 and 5cm² cut offs.
Next are my results from calculating the volume when the cut out area is measured to one decimal place.
Cut out (cm²)
Formula
Volume (cm³)
4.1
Volume = 40 -- (2*4.1) * 20 -- (2*4.1) * 4.1
538.484
4.2
Volume = 40 -- (2*4.2) * 20 -- (2*4.2) * 4.2
539.552
4.3
Volume = 40 -- (2*4.3) * 20 -- (2*4.3) * 4.3
539.228
4.4
Volume = 40 -- (2*4.4) * 20 -- (2*4.4) * 4.4
537.536
4.5
Volume = 40 -- (2*4.5) * 20 -- (2*4.5) * 4.5
534.5
4.6
Volume = 40 -- (2*4.6) * 20 -- (2*4.6) * 4.6
530.144
4.7
Volume = 40 -- (2*4.7) * 20 -- (2*4.7) * 4.7
524.492
4.8
Volume = 40 -- (2*4.8) * 20 -- (2*4.8) * 4.8
517.568
4.9
Volume = 40 -- (2*4.9) * 20 -- (2*4.9) * 4.9
509.396
I have also drawn a graph to illustrate my results.
FIG 4
From these results, where the largest volume of 1539.522cm² is achieved with a cut off of 4.2cm², I see that I should look between 4.2 and 4.3cm² as a cut off next to get even more accurate results which are measuring to 2 decimal places.
Below are my results showing the size of the cut off calculated to 2 decimal places.
Cut out (cm²)
Formula
Volume(cm³)
4.21
Volume = 40 -- (2*4.21) * 20 -- (2*4.21) * 4.21
539.581844
4.22
Volume = 40 -- (2*4.22) * 20 -- (2*4.22) * 4.22
539.597792
4.23
Volume = 40 -- (2*4.23) * 20 -- (2*4.23) * 4.23
539.599868
4.24
Volume = 40 -- (2*4.24) * 20 -- (2*4.24) * 4.24
539.588096
4.25
Volume = 40 -- (2*4.25) * 20 -- (2*4.25) * 4.25
539.5625
4.26
Volume = 40 -- (2*4.26) * 20 -- (2*4.26) * 4.26
539.523104
4.27
Volume = 40 -- (2*4.27) * 20 -- (2*4.27) * 4.27
539.469932
4.28
Volume = 40 -- (2*4.28) * 20 -- (2*4.28) * 4.28
539.403008
4.29
Volume = 40 -- (2*4.29) * 20 -- (2*4.29) * 4.29
539.322356
The table above you can tell that the largest volume occurs, to two decimal places, when 4.23cm² is cut off.
I have drawn one more graph to show these results easily. You should be able to see from this that the largest volume I have obtained is 1539.599868 cm², and that this is got when the cut off area from each corner of the box is measuring 4.23cm².
FIG 5
When I divide the cut out which gives the maximum volume, 4.23 by 20 this gives me the proportion that should be cut off to give the maximum volume, which is 0.2115, which is close to 1/4.
The length of the rectangle is double the width (ratio 1:2) so length = 2L
Volume will be expressed as:
V = L * W * H
As square cut out is x.
Take it away from both sides of both length & width so it becomes
Length = 2L - 2x
Width = 2Lx
So the cut out is h = x
Expression for volume will be:
L * W * H
(2L - 2x) (L - 2x) * x
Multiplying it out becomes
(2L² - 2xL - 4xL + 4x ²) * x
Multiply X through and the equation will become
2L² (x) - 6xL (x) + 4x² (x)
So this can be
2L²X - 6X²L + 4X³
4X² - 6X²L + 2L²
I have fully worked out which cut off gives the maximum area with a ratio of 1:3, as I did above, but with a width of 20cm and a length of 60cm.
Next I have fully worked out the cut off giving the largest volume when the width: length ratio is 1:10. I have used length 100 and width 10cm. The tables of results are shown below.
Cut out (cm²)
Formula
Volume(cm³)
Volume = 100 -- (2*1) * 10 -- (2*1) * 1
784
2
Volume = 100 -- (2*2) * 10 -- (2*2) * 2
152
3
Volume = 100 -- (2*3) * 10 -- (2*3) * 3
128
4
Volume = 100 -- (2*4) * 10 -- (2*4) * 4
736
Cut out (cm²)
Formula
Volume(cm³)
2.1
Volume = 100 -- (2*2.1) * 10 -- (2*2.1) * 2.1
166.844
2.2
Volume = 100 -- (2*2.2) * 10 -- (2*2.2) * 2.2
177.792
2.3
Volume = 100 -- (2*2.3) * 10 -- (2*2.3) * 2.3
184.868
2.4
Volume = 100 -- (2*2.4) * 10 -- (2*2.4) * 2.4
188.096
2.5
Volume = 100 -- (2*2.5) * 10 -- (2*2.5) * 2.5
187.5
2.6
Volume = 100 -- (2*2.6) * 10 -- (2*2.6) * 2.6
183.104
Cut out (cm²)
Formula
Volume(cm³)
2.41
Volume = 100 -- (2*2.41) * 10 -- (2*2.41) * 2.41
188.208084
2.42
Volume = 100 -- (2*2.42) * 10 -- (2*2.42) * 2.42
188.281952
2.43
Volume = 100 -- (2*2.43) * 10 -- (2*2.43) * 2.43
188.317628
2.44
Volume = 100 -- (2*2.44) * 10 -- (2*2.44) * 2.44
188.315136
2.45
Volume = 100 -- (2*2.45) * 10 -- (2*2.45) * 2.45
188.2745
Again I calculated the ratio that needs to be cut off the card to create the box with the largest volume. This is 0.2434, which is close to 1/4 again.
The final volume I am going to work out is with a width: length ratio of 1:100. I will use 1000 length and 10 width
Here are my results for this.
Cut out (cm²)
Formula
Volume(cm³)
Volume = 1000 -- (2*1) * 10 -- (2*1) * 1
7984
2
Volume = 1000 -- (2*2) * 10 -- (2*2) * 2
1952
3
Volume = 1000 -- (2*3) * 10 -- (2*3) * 3
1928
4
Volume = 1000 -- (2*4) * 10 -- (2*4) * 4
7936
Cut out (cm²)
Formula
Volume(cm³)
2.1
Volume = 1000 -- (2*2.1) * 10 -- (2*2.1) * 2.1
2128.44
2.2
Volume = 1000 -- (2*2.2) * 10 -- (2*2.2) * 2.2
2265.792
2.3
Volume = 1000 -- (2*2.3) * 10 -- (2*2.3) * 2.3
2362.868
2.4
Volume = 1000 -- (2*2.4) * 10 -- (2*2.4) * 2.4
2420.096
2.5
Volume = 1000 -- (2*2.5) * 10 -- (2*2.5) * 2.5
2437.5
2.6
Volume = 1000 -- (2*2.6) * 10 -- (2*2.6) * 2.6
2415.104
I worked out the proportion that needs to be cut off the box to give maximum volume; this was 0.2444, which is very close to 1/4.
Ð stands for delta.
Firstly we should consider a graph of y = x² as shown below.
The line through X and Y has almost the correct gradient. It''s gradient is
Increase in y-coordinate from X to Y
Increase in x-coordinate from X to Y
You have to find an expression for , which represents the gradient of the graph at the point X.
So
y = x²
y + Ðy = (x + Ðx)(x +Ðx)
Multiply out brackets.
y+ Ðy = x² + x Ðx + x Ðx + (Ðx) ²
Add like terms together.
y + Ðy = x² + 2xÐx + (Ðx)²
Now here the x² at the end is y in terms of x.
Ðy = x² + 2xÐx + (Ðx) ² - x²
Then you divide by ÐÐx, which gives you
Ðy 2xÐx -- (Ðx) ²
Ðx = Ðx
Ðy
Ðx = 2x + Ðx
Now because delta (Ð) is so tiny that it is insignificant, we forget all about it, which leaves us with
Ðy/Ðx = 2x = 0
I can use calculators to help me complete my calculations to solve the problem, for proof through exhaustion.
I have drawn a graph showing the proportions that I have worked out. I can see that they tend towards 1/4. This is the amount that should be cut off each corner to give the maximum volume possible.
FIG 6
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Max Box Sam Norton
