The open box problem
An open box is to be made from a sheet of card. Identical squares are to be cut off at the corners so that the card can be folded into the open box. The diagram below shows the sheet of card and the four corners, which are to be cut off.
There are two objectives I shall be investigating:
. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given square sheet of card.
2. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given rectangular piece of card.
First I shall investigate objective 1 as I think it will be easier to do.
Objective 1: The square sheet of card.
There are 2 ways to solve this problem: I can use algebra, or I can use trial and improvement. I think I will start by using trial and improvement; I will construct a series of tables and graphs and see if I can find any patterns. If so, then I will be able to come up with a hypothesis, which I can then test to see if I can solve this first objective. Then I will attempt to solve the same problem using algebra.
Method 1: Trial and improvement
To do this I need to make up some dimensions and then apply them to the square. I will choose the dimensions 6x6 to start with. I will call the length of the square cuttings x. The second drawing shows the open box with the dimensions included.
x
x
6cm
6-2x
6cm 6-2x x
If we look at the second drawing we can see that to get the two lengths we have to take away 2x because of the two square cuttings on each side. If we then multiply these sides together and then by x to get the volume we end up with V= x(6-2x)^2.
I will now construct a table to show a range of values for x and the volume of the open box, using the equation.
X
0.5
.0
.5
2.0
2.5
3
V (volume)
2.5
6
3.5
8
2.5
0
We can see here that the maximum volume for the dimensions 6x6 lies between 0.5 and 1.5; so I will now draw another table that focuses in more between the numbers 0.5 and 1.5.
X
0.5
0.6
0.7
0.8
0.9
.0
.1
.2
.3
.4
.5
V
2.5
3.824
4.812
5.488
5.876
6
5.884
5.552
5.028
4.336
3.5
After looking at this table I have concluded that the maximum volume for the open box dimensions of 6x6 is 16, and x (that makes the volume at it's maximum is) is 1.
I will now draw a graph that will also show the maximum amount for x and to verify that x is 1 and that the greatest volume possible is 16.
The graph shows also shows that the maximum volume is about 16 and x is 1. To verify this I am going to draw another graph to show that this is correct.
So we can see here the graph definitely shows that the largest volume possible is 16 and that the length of the square cutting for this would be 1. The graph is symmetrical.
Now that I have found the maximum volume possible for 6x6, I will now investigate more squares by now finding the maximum volume possible for a square with lengths 12x12, then I will see if I can see a pattern.
X
X
12cm
12-2x
X
12cm 12-2x
Here I use the same equation as before except that I replace 6 with 12 so the equation would be V=x(12-2x)^2. So I will now draw tables and a graph to show the maximum volume and the value of x.
X
0.5
.0
.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
V
60.5
00
21.5
28
22.5
08
87.5
64
40.5
20
5.5
0
Here is another table focusing in between 1.5 and 2.5 as the maximum volume clearly lies between those two values.
X
.5
.6
.7
.8
.9
2.0
2.1
2.2
2.3
2.4
2.5
V
21.5
23.904
25.732
27.008
27.756
28
27.764
27.072
25.948
24.416
22.5
From this table we can see that the maximum volume is 128 and the x is 2. To again prove this I will now construct two tables, one to show the graph of the equation, and another showing the same graph but closer up to see exactly what the maximum ...
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X
.5
.6
.7
.8
.9
2.0
2.1
2.2
2.3
2.4
2.5
V
21.5
23.904
25.732
27.008
27.756
28
27.764
27.072
25.948
24.416
22.5
From this table we can see that the maximum volume is 128 and the x is 2. To again prove this I will now construct two tables, one to show the graph of the equation, and another showing the same graph but closer up to see exactly what the maximum volume and x is.
From this graph we can guess that the maximum volume lies between 125 and 135 so I will now construct another graph to zoom in more on the line.
The graph shows that the value of x is definitely 2, and that the maximum volume is about 128, to verify this the box on the left of the graph says what the value of y is (volume) at x when x is 2; and it says that y is 128. The graph is symmetrical.
So I have found out the maximum volume for a square with lengths of 12x12. I will now try to find the maximum volume for a square of lengths 15 and then draw a table and hopefully find a pattern.
x
x
12
15-2x
x
15-2x
12
Again I use the same equation as before but this time I replace 12 with 15. So the equation will be V= x(15-2x)^2 . I will now construct tables and two graphs to again show the maximum volume and x.
X
0.5
.0
.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
V
98
69
216
242
250
243
224
96
62
25
88
54
26
7
0
The table shows that the maximum volume lies between 2.0 and 3.0 so I will construct another table to give a closer up view of this.
X
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
V
242
244.944
247.192
248.768
249.696
250
249.704
248.832
247.408
245.456
243
This table now suggests that the maximum volume is 250 and that x is 2.5. To prove this again I will construct two graphs.
Here it looks like the maximum volume of the square 15x15 is 250 but I am unsure of x; so I will construct another graph to show it closer up.
We can clearly see on this graph that x is 2.5 and that the maximum volume is 250. This graph is also symmetrical.
Now that I tested have 3 different squares I will now draw a table to show the volumes and values of x for the squares together to find a pattern.
Dimensions
Maximum Volume
X (that allows volume to be its maximum)
6 x 6
6
2 x 12
28
2
5 x 15
250
2.5
Looking at this table we can see that x appears to be one sixth of the lengths of the square. This seems to be the pattern that could work on all squares. To prove that this is not just a coincidence, I will now test my hypothesis on another square, one with dimensions of 24x24. If the value of x turns out to be 4 (one sixth of 24) then I will know that this is right.
x
x
24
24-2x
x
24-2x
24
Once again I will use the same equation but change the 15 to a 24 this time so the equation will be V=x(24-2x)^2. Again I will construct tables and graphs to find out the maximum volume and the value of x.
X
0.5
.0
.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
V
264.5
484
661.5
800
902.5
972
011.5
024
012.5
980
929.5
864
786.5
X
7.0
7.5
8.0
8.5
9.0
9.5
0.0
0.5
1.0
1.5
2.0
V
700
607.5
512
416.5
324
237.5
00
94.5
44
1.5
0
This table shows that the maximum volume lies between 3.5 and 4.5 so I will now draw another table that will take a closer look at it.
X
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
V
011.5
016.1
019.572
022.048
023.516
024
023.524
022.112
019.788
016.576
012.5
Clearly here it seems that x is 4 and that the maximum volume is 1024. So I think that my hypothesis is right in saying that x will always be one sixth of the length. But just to know for certain I will now create another two graphs to show this.
Here we can see that the volume is a little over a thousand but we can be pretty sure that x is 4. I will now get an even closer look by zooming into the graph more.
It is too hard to see the volume but the box to the side says that when x is 4, the volume is 1024 so we can be certain this and of my hypothesis.
So by creating these tables and graphs I have gradually progressed into discovering a hypothesis. And by testing out this hypothesis on one final square, it proves that my hypothesis is correct. So x =1/6L or x=L/6, and V=x(l-2x)^2.
Method 2: Using Algebra
Although I have already solved the squares problem, I will now attempt to solve it using algebra. To do this I will use the equation to find the volume and merge this with some other sort of expression to get 1 equation, which I will then solve to find the volume.
So I know that V=x(L-2x)^2 from the previous method, but just to be sure it is right, I will test it out using numbers instead of the symbols.
Lets use the square 6x6 so x will be 1 and V will be 16.
V=x(L-2x)^2
6=1(6-2*1)(6-2*1)
6=1(36-12-12+4)
6=36-12-12+4)
6=16
So that proves the equation is correct.
As well as this I already know that x=L/6, so if I now substitute this into the original equation, I should be able to sole that to find the volume of any square; lets see if it works.
Substitute x=L/6 into V=x(l-2x)(l-2x)
So V=L/6(L-2L/6)(L-2L/6)
V=L/6(L-L/3)(L-L/3)
V=L/6(L^2-L^2/3-L^2+L^2/9)
V=L/6(L^2-2L^2/3+L^2/9)
V=L/6(L^2-5L^2/9)
V=L/6(4L^2/9)
V=2/27L^3
To check this is correct, lets try this out with the same square (6x6).
So V=2/27L^3
6=2/27(6)^3
6=2/27(216)
6=16
So by two methods I have found the maximum volume for different squares, found the values of x and come up with an equation that works for any square.
I will now move on to the rectangles problem.
Objective 2: The rectangular piece of card
Again there are two ways I can solve this problem, there is trial and improvement and the algebra way. I will start with the trial and improvement method, by drawing graphs and tables, and then spotting any patterns. If so, then I can come up with a hypothesis.
Method 1: Trial and improvement
The rectangular problem is a lot more complex then the square problem because the lengths and widths can be anything (not the same as in the square), so I think I will investigate this by choosing different ratios of length to width and see what I can come up with. I will start with a ratio of length to width 2:1 with dimensions of 6x3.
x
x
3
6-2x x
3-2x
6
So if I multiply the length by width by height I get the equation. V=x(L-2x)(W-2x). And if I replace the dimensions with L and W I get the equation V=x(6-2x)(3-2x).
I will now create a table showing the different values of x and the volumes of this rectangle.
X
0
0.5
.5
V
0
5
4
0
We can see here that the maximum amount of x is between 0 and 1 so I will now draw another table showing that focuses in more on the values between 0 and 1.
X
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
.0
V
.624
2.912
3.888
4.576
5
5.184
5.152
4.928
4.536
4
Here we can see that the maximum volume is at about 0.6 but it could go either way, so to get a better look at it I am going to draw a graph to show this.
I cannot get an exact number for the volume and value of x (because the actual that x is around 0.63 and the maximum volume is around 5.2 (at the bottom it says the co-ordinates of where the cursor was hovered at). The graph is not symmetrical.
I will now choose another set of dimensions (8x4) still for the ratio 2:1 and see if I spot a pattern.
x
x
4 8-2x
x
4-2x
8
If I now swap the dimensions for this rectangle with the equation I used before I get V=x(8-2x)(4-2x). Using this equation I will again draw graphs and tables to find out the maximum volume and the value for x.
X
0.5
.0
.5
2.0
v
0.5
2
7.5
0
Here we can see the maximum volume lies between 0.5 and 1.5 so I will now draw another table focusing in on the values to try to find out the maximum volume.
X
0.5
0.6
0.7
0.8
0.9
.0
.1
.2
.3
.4
.5
v
0.5
1.424
2.012
2.288
2.276
2
1.484
0.752
9.828
8.736
7.5
This table shows the maximum volume at about 12.288, and the value of x at 0.8, although again it could go either way because the actual answer will be too many decimal places. I will now show this further by producing a graph.
The graph tells me that the maximum volume is at about 12.325 and the value of x is at 0.8427 (at the bottom it says the point at which the cursor was hovered). The graph is no symmetrical.
At the moment I do not think I can see a pattern between these values so I will choose another set of dimensions (12x6).
x
x
6
12-2x x
12 6-2x
Again if I swap the dimensions in the equation then I get the equation
V=x(12-2x)(6-2x). I will use this equation to draw tables and graphs to show the maxim volume of the open box and the value of x.
X
0.5
.0
.5
2.0
2.5
3.0
V
40
40.5
32
30
7.5
0
This table shows that the maximum volume is between 0.5 and 1.5. I will now draw another table to show this closer up so that I will be able to see the maximum volume more clearly.
X
.1
.2
.3
.4
.5
.6
.7
.8
.9
2.0
V
40.964
41.472
41.548
41.216
40.5
39.424
38.012
36.288
34.278
32
This table shows that the maximum volume is about 41.548 and the value for x for this is 1.4. To check this I will now draw a graph, which will hopefully show in more detail the maximum volume and value for x.
The graph shows that the maximum volume is about 41.64 and x is 1.264 ( at the bottom it says the co-ordinates of where the cursor was hovered). This is not the exact answer because the actual answer but is many decimal places long. I will now look to see if I can spot any patterns.
Through random investigation and experimenting I have worked out the width of each rectangle divided by the value of x for that rectangle and come up with this table.
Dimensions of Rectangle
Value for x
Width/x
6x3
0.6342
4.73
8x4
0.8454
4.73
2x8
.2679
4.73
As you can see in the table the width/x for each of these rectangles comes out to be 4.73. It is not exactly 4.73, but around that because of the many decimal places of x in each rectangle.
I will now test this out on 1 more case for ratio 2:1 and then move on to 3:1, and maybe see if the pattern continues. I will try this for the dimensions 10x5.
x
x
5
10-2x
x
5-2x
10
If I swap the dimensions in the equation I get the equation V=x(10-2x)(5-2x). I will now draw table sand graphs to find out the volume and the value for x.
X
0.5
.5
2
2.5
V
8
24
21
2
0
We can see here that the maximum volume is between 0.5 and 1.5 so I will draw another table focusing in on the data.
X
0.5
0.6
0.7
0.8
0.8
.0
.1
.2
.3
.4
.5
V
8
20.064
21.672
22.848
23.616
24
24.024
23.712
23.088
22.176
21
I will now use a graph as well to show the volume and value of x and see if my hypothesis works.
The graph shows that the maximum volume is about 24.07 and the value is about 1.056.
If I now work out x by my hypothesis (5/1.056), it should come out to 4.73.
5/1.056=4.73
This shows that my hypothesis works but this may only be for this ratio; so now I will move on to the ratio of length to width 3:1.
For the ratio 3:1 I will start with the dimensions 9x3.
x
x
3
9-2x
x
9 3-2x
If I put the dimensions 9 and 3 into the original equation it comes out to be
V=x(9-2x)(3-2x). Using this equation I will create tables and graphs to find out the maximum volume and value of x, and to see if the pattern relates to the ratio 2:1.
X
0.5
.5
V
8
7
0
This table shows that the maximum volume lies between 0.5 and 1.5 so I need to draw another table to show this closer up and to find the maximum volume and value for x.
X
.0
.1
.2
.3
.4
.5
.6
.7
.8
2.9
2.0
v
8
8.424
8.512
8.288
7.778
7
5.954
4.752
3.328
.736
0
The table shows that the maximum volume is about 8.512 and the value for x is about 1.2. I will now draw a graph to show this in more detail and try to get a more precise result.
The graph shows that the maximum volume is about 8.535 and the value for x is about 0.678 (the bottom says the co-ordinates of where the cursor is hovering over).
This is not the exact answer because the actual answer has many decimal places but it is close. This graph is not symmetrical.
I will now investigate with the dimensions 12:4 and then see if I can continue the pattern.
x
x
4
12-2x
x
12 12-2x
I will replace the dimensions in the equation with the dimensions 12 and 4 to get the equation V=x(12-2x)(4-2x). I will now use this equation to draw a table and a graph to work out the maximum volume and the value of x.
X
0.5
.0
.5
2.0
V
6.5
20
3.5
0
This table shows that the maximum volume is in between 0.5 and 1.5 so now I will draw another table that focuses in on the table to show more detailed results.
X
0.5
0.6
0.7
0.8
0.9
.0
.1
.2
.3
.4
.5
V
6.5
8.144
9.292
9.968
20.196
20
9.404
8.432
7.108
5.456
3.5
This table shows that the maximum volume is about 20.196 and I will now draw a graph to prove these results.
This graph shows that the maximum volume is about 20.208 and the value of x is about 0.901 (the bottom says the co-ordinates of the point where the cursor hovers). The graph is not symmetrical.
Now that I have investigated two of the rectangles for 3:1 I will now see if they are linked with the ratio 2¨in rectangles. With the 2:1 rectangles I found out that the width/x = 4.7 and I will now see if that works with the rectangles of ratio 3:1.
Dimensions
Value for x
Width/x
9x3
0.677
4.43
2x4
0.903
4.43
As you can see the table is different because the width/x =4.43 instead of 4,73 for the ratio 2:1. But this means that width/x =4.43 is the pattern for rectangles of ratio 3:1. This gives me the idea that for each ratio of rectangles, they will have their own pattern. But I have not yet found a link between all rectangle sof any ratio; this is why I have decided to stop this method of trial and improvement, a si believe if I go any further I will not find the relationship between all rectangles; so I will now move on to method 2, using algebra.
Method 2: Using Algebra
The second method involves using algebra. To do this I will use the equation before and attempt to solve it. Somewhere in this equation I will need to use something called the gradient function. The gradient function is used to find the derivative in an equation. The gradient function is nx(n-1). I will give an example now as to how apply the gradient function in an equation. If I had the expression 4x^3, I would multiply the gradient (4) by the power (2) to get 8, then the power has 1 taken away from it. So 4x^3 would become 12x^2.
I will now attempt to use algebra to find an equation for x.
I already know the equation from before as V=x(L-2x)(W-2x)
So V=x(L-2x)(W-2x)
V= xLW-2x^2(L+W)+4x^3
Now I will apply the gradient function, nx(n-1).
V=LW-4x(L+W)+12x^2
V=12x^2-4x(L+W)+LW
Now you can see that a quadratic equation has formed, because there is a an expression with x^2, plus an expression with x, plus another expression with no x. V is replaced with 0 because when the volume is at it's max, the gradient (x) will be 0 (because the line will be at it's peak). The 0 is then replaced with x. I solve this quadratic equation using the formula -B+-sqrt(B^2-4AC)
2A
A=12 B=-4(L+W)
C=LW
x=(4(L+W) +- sqrt((4(L+W)^2 - 48LW))/(24
24
x=(4(L+W) +- sqrt(16L^2 + 16W^2 - 16LW))/24
24
x=(L + W +- sqrt(L^2 + W^2 - LW))/6
6
x=L+W+-sqrt(L^2+W^2-LW)
6
The final equation uses the minus instead of a plus in the equation because if you used the minus the equation would not work.
So x=L+W-sqrt(L^2+W^2-LW)
6
So I have now found an equation that satisfies all rectangles. So by using trial and improvement and algebra I have investigated the two problems involving squares and rectangles and have found ways to find out the maximum volume, and the value of x.
