I noticed that when I found out the maximum size volumes for both squares and rectangles that the bigger the square the bigger the volume. I also noticed that to find out the biggest volume the rectangle or square needed to be as equal in sides as possible to work out the maximum volume and cut out size.
I expected that the maximum volumes would become greater as the square or rectangle became bigger and that the cut out length would not be very big because it would make the box size smaller it the cut out length was bigger.
If you look at my tables and graphs you will notice that the more decimal places I add onto the equation the greater the size that the maximum volume will be. Again, if you look at the tables and graphs relations you will notice that the graphs to show the maximum cut out length will be much more accurate.
I began to look for a rule when I have sufficient measurements of rectangles in the ratio 1:2 to plot the graph to find the optimum cut out length. I plotted the graph by using the size of the rectangles as follows:
‘Y’ axis: I used the size of the rectangles to show the size of each rectangle volumes against,
‘X’ axis: I used the size of the cut out length to relate it to the size of the rectangle.
The formula to find the volume of a box is:
Size 1 = a
Size 2 = b
Cut-out = c
Length = a – 2c
Width = b – 2c
Height = c
Volume = Length x Width x Height
Volume = (a – 2c) x (b – 2c) x c
Volume = ab – 2ac -2ab + 4c
When searching for a rule for the different ratios of rectangles I noticed that each ratio had different formulae for which I had to discover each one by either trial and error or by plotting a graph as shown below.
When using excel now I know how to input formulas to make easier equations and to show them with a plotted graph with a trend line and the formula with it. This makes life a lot easier because otherwise I would have had to use trial and error. This would take a lot more time and would be a lot more hassle if I only had a short time to be working out such equations or formulas.
I think that the formula =x/6 for squares is that because there are 6 sides on a square and if you work out the average of each side on a square and divide it by 6 (because of 6 sides) it might add up to the cut out length. Although this is only a theory and I have not put it into practise.
I have found that the best method of working out the optimum cut out length and the maximum volume is to use a graph with a line of best fit, and using the gradient of this line it is possible to work out the optimum cut out length. The gradient of the line is different for each ratio, so a different graph must be plotted for each. I will explain how I plotted the graphs and how they are used to show the maximum volume for each ratio. This graph shows the gradient for rectangles in the ratio 1:2
Once the gradient has been found we can use the first number (in the case of rectangles in the ratio 1:2 it is 0.2113) to find the optimum cut out length of any rectangle in the same ratio e.g.
If I wanted to find the optimum cut out length of a rectangle with sides 15x30 I will multiply the smaller of the sides by 0.2113 to find the optimum cut out length =
15 x 0.2113 = 3.1695
OR
If I wanted to find the optimum cut out length of a rectangle with sides 15x30 I will multiply the larger of the sides by 0.10565 to find the optimum cut out length =
30 x 0.10565 = 3.1695
This means that 3.1695 is the cut out length that would give the maximum volume for a 15x30 rectangle.
However in order to use this method you first must be able to work out the maximum volume of at least 2 rectangles in the same ratio to be able to plot the graph. This is essential because you cannot plot a graph for one measurement. Finally, the more the measurements that you make for the ratio of 1:2 the more accurate that the graph is going to be.
If you look on my formulas page you will notice that I have all the equations for the ratios 1:2, 1:3 and 1:4 along with the squares formula which I have added above.
If I had more time I would create more squares and make it to 4 decimal places. This would increase the chance that my maximum volume would be bigger and be more precise to the maximum cut out length. I would also test more ratios and try to find a link between greater ratios a find a formula to make the ratios link to each other.