This is a graph I produced of the program Autograph. I have made it bold so that it is easier to interpret and I have also added the maximum value in a pink text box and the equation of the graph in a blue text box.
From the graph I can see that for a 10 by 10 square the cut out, to 3 decimal place, which gives the maximum volume is 3.333.
30 by 30 square
I have now found out that in a 10 by 10 square the cut out which gives the largest volume is between 5 and 6 but more towards 5 since 5 gives a larger volume. To obtain a more accurate result I will zoom in the shaded region.
After zooming in I have found out that a value between 5.1 and 5.2 for the cut out gives the highest volume. I have not found a pattern in the 30 by 30 square cut outs like I did with the 20 by 20 square. I have given the volume to 3 decimal place since it is a sensible degree of accuracy although the cut out is only to 1 decimal place therefore I will zoom in further to obtain a more accurate result.
I will zoom in again to increase the degree of accuracy in finding out what length of cut out gives the largest volume.
This is a graph I produced of the program Autograph. Like the previous graphs I have made it bold so that it is easier to interpret and I have also added the maximum value in a pink text box and the equation of the graph in a blue text box.
After choosing 3 different cut outs for the square I constructed a general table.
After analyzing the results I discovered a relationship between the length of the cut out and the size of the square and was able to derive a general expression. For an n by n square the cut out which maximises the volume is, where n is the length of the square.
I am now going to investigate if the general formula would work with a square with any size cut out. I am going to pick a random number such as 180 by 180.
60 by 60 square
The value of the cut out which produces the highest volume is 10. Since this is at the end of the table I extended the values to see if the volume would keep getting bigger.
Looking at the table you can see that the values do not keep increasing as the length of the cut out increase.
I zoomed in on the 10 to 11 region to obtain a more accurate result for the cut out.
Since the volume at 10 has reduced when I zoomed in I will not zoom in any further. I will now check if the value of the cut out will work with the general formulae. = 6. I can now confirm that the general formula works with squares of any size.
Task 2
For this part we are asked to determine for any sized rectangular sheet of card, investigate the size of the cut out square which makes an open box of the largest volume.
I decided to group the rectangles together in ratios. I constructed 3 graphs for each ratio. The ratios are 2: 1, 3: 1, 4: 1, and 5: 1.
2:1
These are the 3 cut out sizes I used for the 2:1 ratio rectangle.
After deciding on the sizes I put them into an equation. For example for the 40:20 cut out the equation would be: y = x(40–2x)(20–2x), where y is the volume. I have put y = because you cannot form a table by saying v = is Autograph.
Again like the previous task I have made the graph bold so that it is easier to interpret and I have also added the maximum value in a pink text box and the equation of the graph in a blue text box.
This is the graph for the 40:20 rectangle.
This is the graph for the 60:30 rectangle.
This is the graph for the 60:30 rectangle.
I will now try to find out a general rule for the 2:1 ratio rectangles. I have constructed a table for this. To find a general formula that maximises the volume, I have to first find a pattern between the results. The length of cut out which maximises volume is the x value on the maximum point on the graph.
After analysing the table of results I discovered that dividing the length of the cut out by the smallest value of the rectangle equals the same final answer for each size in the 2:1 ratio of rectangles. For example:
Therefore for an n by 2n rectangle the cut out which maximises the volume is 0.2113n.
These are the 3 cut out sizes I used for the 3:1 ratio rectangle.
This is the graph for the 60:20 rectangle.
This is the graph for the 90:30 rectangle.
This is the graph for the 120:40 rectangle.
I will now try to find out a general rule for the 3:1 ratio rectangles. I have constructed a table for this. To find a general formula that maximises the volume, I have to first find a pattern between the results. The length of cut out which maximises volume is the x value on the maximum point on the graph.
After analysing the table of results I discovered that dividing the length of the cut out by the smallest value of the rectangle equals the same final answer for each size in the 3:1 ratio of rectangles. For example:
Therefore for an n by 3n rectangle the cut out which maximises the volume is 0.2257n.
These are the 3 cut out sizes I used for the 4:1 ratio rectangle.
This is the graph for the 80:20 rectangle.
This is the graph for the 120:30 rectangle.
This is the graph for the 160:40 rectangle.
I will now try to find out a general rule for the 4:1 ratio rectangles. I have constructed a table for this. To find a general formula that maximises the volume, I have to first find a pattern between the results. The length of cut out which maximises volume is the x value on the maximum point on the graph.
After analysing the table of results I discovered that dividing the length of the cut out by the smallest value of the rectangle equals the same final answer for each size in the 4:1 ratio of rectangles. For example:
Therefore for an n by 4n rectangle the cut out which maximises the volume is 0.2324n.
These are the 3 cut out sizes I used for the 5:1 ratio rectangle.
This is the graph for the 100:20 rectangle.
This is the graph for the 150:30 rectangle.
This is the graph for the 200:40 rectangle.
I will now try to find out a general rule for the 5:1 ratio rectangles. I have constructed a table for this. To find a general formula that maximises the volume, I have to first find a pattern between the results. The length of cut out which maximises volume is the x value on the maximum point on the graph.
After analysing the table of results I discovered that dividing the length of the cut out by the smallest value of the rectangle equals the same final answer for each size in the 5:1 ratio of rectangles. For example: