As you can see by the table above, the largest volume is achieved when an area of 4cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 5 cm².
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. We can see that the maximum box area is made from the cut size of 4cms. Now I will try between 3-5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 4.2cm² is cut off each corner of the box. I have also drawn a graph to show my results.
- Evaluation for open box – squares
There is a connected between the x cut-off and the square size. If you look at the 1 d.p results you can see that the results (which are the highest volume) are 1/6 of there square size.
- x 6 = 19.8 to 2 significant figures that is 20
6.7 x 6 = 40.2 to 2 significant figures that is 40
4.2 x 6 = 25.2 to 2 significant figures that is 25
Therefore if X x 6 = l you can rearrange to give:
X = L
6
This formula can give you the largest volume for a square open box. We can use this formula to work out X, and then we can substitute X into the volume formula X(L-2X)2.
Formula in working:
Size of square = 20
X = L
6
X = 20
6
X = 3.3
Then Substitute X into X(L-2X)2 to get the volume,
3.3(20-2 x 3.3)2
=592.548
Therefore we can work out the nth rule
Nth rule
n = L
6
V = n(L-2n)2
Putting the nth rule into action
Find the biggest volume for size square 32
Rule
n = L
6
V = n(L-2n)2
X = 32
6
X = 5.3
V = 5.3(32-2 x 5.3)2
V = 2427.188
Proving that this is correct
As you can see from table, the nth is correct. Using the rule we found out that X = 5.3 and V = 2427.188.
From the table we see that X does equal 5.3 and V does equal 2427.188, therefore the nth rule is correct and can be used to find out the largest volume for a square open box, for any give number.
1. Evidence
To obtain evidence I will be used a series of methods:
Part 2, rectangle
I am now going to continue my investigation by looking at the shape of rectangles.
I am going investigate 3 different ratios for the rectangular open box. With each ratio I’m going consider 3 sizes.
Once I have obtained all information on the 3 ratios, I will look for patterns and try to formulate a rule to work out the largest volume for an open box square. The ratios that I will be using are:
- L=2W Ratio: 1:2
- L=3W Ratio: 1:3
- L=4W Ratio: 1:4
The formula that needs to be used to get the volume of a box is:
Volume = Length x Width x Height
The width of the rectangle is double the length (ratio 1:2) so width = 2W
As square cut out is X, the height will become X. Therefore the formula for volume will be:
V = X(l-2 X)(2w-2 X)
V = Volume
l = Length
X = Height and cut off
W = Width
I shall begin with a width of 10cm, and a length of 20cm, this is a ratio of 1:2, the width being twice as long as the length.
The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 4cm being the maximum for there to be a box left.
I’m to going cut out of length 1cm, the equation for this would be as follows:
Volume = 1(10-2 x1)(2x20-2x1)
So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:
1(10-2 x1)(2x20-2x1)
1 x 8 x 38
= 304³
I used the formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 4cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 2cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 2 and 3cm².
In the table below I have shown the formulae I have inputted into the spreadsheet to attain the above results.
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. We can see that the maximum box area is made from the cut size of 2cms. Now I will try between 2-3cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 2.3cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 15x30cm. The maximum whole number I can cut off each corner is 6cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(2w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 6cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
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. 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 4cm².
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. We can see that the maximum box area is made from the cut size of 3cms. Now I will try between 3-4cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 3.5cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 20x40cm. The maximum whole number I can cut off each corner is 9cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(2w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 9cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 5cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 4 and 5.5cm².
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. We can see that the maximum box area is made from the cut size of 5cms. Now I will try between 4-5.5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 4.6cm² is cut off each corner of the box. I have also drawn a graph to show my results.
In the table below is the records from results for l=2w ratio 1:2.
To work out the proportion between the size of the rectangle and cut-off we divide:
10/2.4 = 4.17
15/3.5 = 4.29
20/4.6 =4.35
And then we find the average,
4.17 + 4.29 + 4.35
3
=4.27
Therefore if you multiply any of the 1 d.p results by 4.27 you will get a result which is close to the size.
2.4 x 4.27 = 10.248 to 2 significant figures this is 10
3.5 x 4.27 = 14.945 to 2 significant figures this is 15
4.6 x 4.27 = 19.642 to 2 significant figures this is 20
Rearrange this rule you can get the formula to work out the highest volume for a rectangular open box at the ratio 1:2:
X = L
4.27
So the rule to work out X for the ratio 1:2 is X = L
4.27
Now that we know the formula to get X we can substitute X in the formula V = X(l-2 X)(2w-2 X) to get the volume.
Now will start to investigate rectangles where L=3W.
I will begin with a width of 10cm, and a length of 30cm, this is a ratio of 1:3, the width being 3 times as long as the length.
The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 4cm being the maximum for there to be a box left.
I’m to going cut out of length 1cm, the equation for this would be as follows:
Volume = 1(10-2 x1)(3x30-2x1)
So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:
1(10-2 x1)(3x30-2x1)
1 x 8 x 88
= 704cm³
I used the formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 4cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 2cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 2 and 3cm².
In the table below I have shown the formulae I have inputted into the spreadsheet to attain the above results.
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. We can see that the maximum box area is made from the cut size of 2cms. Now I will try between 2-3cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 2.4cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 15x45cm. The maximum whole number I can cut off each corner is 6cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(3w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 6cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 4cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 4.5cm².
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. We can see that the maximum box area is made from the cut size of 4cms. Now I will try between 3-4.5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 3.6cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 20x60cm. The maximum whole number I can cut off each corner is 9cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(3w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 9cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 5cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 4 and 5.5cm².
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. We can see that the maximum box area is made from the cut size of 5cms. Now I will try between 4-5.5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 4.9cm² is cut off each corner of the box. I have also drawn a graph to show my results.
In the table below is the records from results for l=2w ratio 1:2.
To work out the proportion between the size of the rectangle and cut-off we divide:
10/2.4 = 4.17
15/3.6 = 4.17
20/4.9s =4.08
And then we find the average,
4.17 + 4.17 + 4.08
3
=4.14
Therefore if you multiply any of the 1 d.p results by 4.27 you will get a result which is close to the size.
2.4 x 4.14 = 9.936 to 2 significant figures this is 10
3.6 x 4.14 = 14.904 to 2 significant figures this is 15
4.9 x 4.14 = 20.286 to 2 significant figures this is 20
Rearrange this rule you can get the formula to work out the highest volume for a rectangular open box at the ratio 1:3:
X = L
4.14
So the rule to work out X for the ratio 1:3 is X = L
4.14
Now that we know the formula to get X we can substitute X in the formula V = X(l-2 X)(2w-2 X) to get the volume.
Now will start to investigate rectangles where L=4W.
I will begin with a width of 10cm, and a length of 40cm, this is a ratio of 1:4, the width being 4 times as long as the length.
The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 4cm being the maximum for there to be a box left.
I’m to going cut out of length 1cm, the equation for this would be as follows:
Volume = 1(10-2 x1)(4x40-2x1)
So we can work out through this method that the volume of a box with corners of 1cm² cut out would be:
1(10-2 x1)(4x40-2x1)
1 x 8 x 158
= 1264cm³
I used the formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 4cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 2cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 2 and 3cm².
In the table below I have shown the formulae I have inputted into the spreadsheet to attain the above results.
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. We can see that the maximum box area is made from the cut size of 2cms. Now I will try between 2-3cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 2.5cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 15x60cm. The maximum whole number I can cut off each corner is 6cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(4w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 6cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 4cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 4.5cm².
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. We can see that the maximum box area is made from the cut size of 4cms. Now I will try between 3-4.5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 3.7cm² is cut off each corner of the box. I have also drawn a graph to show my results.
Now I’m going to investigate a rectangle with a side length of 20x80cm. The maximum whole number I can cut off each corner is 9cm, otherwise I would not have any box left. I am going to start off by looking at whole numbers.
Using the formula:
V = X(l-2 X)(3w-2 X)
I used these formulae to construct a spreadsheet, which would allow me to quickly and accurately calculate the volume of the box for cut-off size 1cm2 to 9cm2. Below are the results I got through this spreadsheet.
By using a table all results are shown in a clear way and is easier to look for a pattern.
As you can see by the table above, the largest volume is achieved when an area of 5cm² is cut off each corner of the box. I have also drawn a graph to show my results. 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 4 and 5.5cm².
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. We can see that the maximum box area is made from the cut size of 5cms. Now I will try between 4-5.5cms
Below is a table showing the cut out to 1 decimal place, with the largest area achieved highlighted in red.
As you can see by the table above, the largest volume is achieved when an area of 4.9cm² is cut off each corner of the box. I have also drawn a graph to show my results.
In the table below is the records from results for l=4w ratio 1:4.
To work out the proportion between the size of the rectangle and cut-off we divide:
10/2.5 = 4
15/3.7 = 4.05
20/4.9 =4.08
And then we find the average,
4 + 4.05 + 4.08
3
=4.04
Therefore if you multiply any of the 1 d.p results by 4.04 you will get a result which is close to the size.
2.5 x 4.04 = 10.1 to 2 significant figures this is 10
3.7 x 4.04 = 14.948 to 2 significant figures this is 15
4.9 x 4.04 = 19.796 to 2 significant figures this is 20
Rearrange this rule you can get the formula to work out the highest volume for a rectangular open box at the ratio 1:4:
X = L
4.04
So the rule to work out X for the ratio 1:4 is X = L
4.04
Now that we know the formula to get X we can substitute X in the formula V = X(l-2 X)(2w-2 X) to get the volume.
- Evaluation for open box – Rectangles
X=L
4
Is the limit. There is no smaller denominator
We can use this to calculate the cut-off size then substitute it in the volume equation.
Nth rule
L:nW
X = L
4
V= X(l-2X)(nW-2X)
Putting the nth rule into action
Find the biggest volume for size square 32
Rule
n = L
6
V = n(L-2n)2
X = 32
6
X = 5.3
V = 5.3(32-2 x 5.3)2
V = 2427.188
Proving that this is correct
As you can see from table, the nth is correct. Using the rule we found out that X = 5.3 and V = 2427.188.
From the table we see that X does equal 5.3 and V does equal 2427.188, therefore the nth rule is correct and can be used to find out the largest volume for a square open box, for any give number.