4 6
= 6
Therefore, according to this equation, the cut of x, which will give the open box its maximum volume, is 6cm. I will now construct a table to prove that this answer is right just like I did for the two squares I investigated before.
Volume = x(36-2x)(36-2x)
As you can see, the table above shows us that the maximum volume for this open box is when cut x is 6cm, which was the answer I predicted using the equation L/6.
The scatter diagram on the next page also shows us that the maximum value is 6. Notice that the shape of the scatter diagram is the same as the two scatter diagrams done for the other two squares.
42cm by 42cm Square
I will now investigate a 42cm by 42cm square and what is the size of cut x, which will give the maximum volume of the open box. Using the equation L/6 I will attempt to predict the value of x, which will give the maximum value of the open box:
Length/width of square 42
4 6
= 7
Therefore according to this equation, the cut of x, which will give the open box its maximum volume is 7cm.
Volume = x(42-2x)(42-2x)
As you can see the table above shows that the cut of x, which will give the maximum volume for the open box is 7. This proves my prediction right again using the L/6 formula.
The scatter diagram below also shows that when the cut of x is 7, the open box will have its highest volume. Notice yet again the shape of the scatter diagram is the same.
48cm by 48cm Square
I will now be investigating a 48cm by 48cm square. This is the last square that I will be investigating. I will use the formula L/6 again to predict the cut of x, which will give this open box its maximum volume:
Length/width of square 48
4 6
= 8
Therefore according to this equation, the value of the cut of x, which will give this open box its maximum volume, is 8cm.
Volume = x(48-2x)(48-2x)
The table above proves that the maximum volume of the open box is reached when the cut x 8cm.
The scatter diagram below also shows that the maximum volume of the open box is reached when the cut of x is 8cm. Notice yet again the shape of the graph is the same.
The reason I am went through squares with constant measurements (24cm, 30cm, 36cm, 42cm, 48cm) is because I am trying to prove that L/6 will equal the cut of x which will give the open box its maximum volume, and these numbers all easily divide by 6.
Rectangles
In this part of my coursework, I will be investigating the measurements of cut x to give a maximum volume for an open box made out of a rectangle. I will be investigating rectangles in the ratio of:
The reason I am investigating rectangles with ratios of length by width is to be systematic. For example I might investigate a rectangle in the ratio of 1:2 with the these measurements:
This therefore means to calculate the volume of its open box the formula is now:
Volume = x(20-2x)(40-2x)
Unlike before, the values at the beginning of the brackets aren’t the same. They are the measurements of the length and width, which is in the ratio of 1:2.
Rectangles in the ratio of 1:2
10cm:20cm
As you can see, the table above shows the volume of an open box for different measurements for cut x for a rectangle with the measurements of 10cm by 20cm. The table shows that the maximum volume is when the cut x is 2. Notice that this is 10/5, which is the length of the rectangle divided by 5. To calculate the volume for this particular open box, I used the formula that I stated above except I substituted the values in the brackets:
Volume = x(10-2x)(20-2x)
20cm:40cm
The table above shows that the maximum volume occurs when the cut of x is equal to 4. Notice that this is again 20/5 which is the length divided by 5. To calculate the volume for this open box the formula was:
Volume = x(20-2x)(40-2x)
100cm:200cm
I calculated the volume for this open box using this formula:
Volume = x(100-2x)(200-2x)
The reason I am doing numbers with such big ratios is because I want to test the fact that when the rectangle has a ratio of 1:2, the length/5 will then give the cut of x which will give that certain rectangle its maximum volume. I have constructed a table in an excel spreadsheet, just like the ones before to investigate the cut of x which will give this open box its maximum volume. I will not include the table for this, as it will take up to much space.
However for this ratio I have found out that this time, the value of x, which gives this open box its maximum volume is 21, which isn’t 100/5. The answer to 100/5 is 20. Although this is close, as the values gets larger the smaller the divisor seems to be. The divisor is between 4.7 and 4.8 for this ratio.
10,000cm:20,000cm
I calculated the volume for this particular open box using the formula:
Volume = x(10,000-2x)(20,000-2x)
As you can see, the ratio of the rectangle I have chosen is a very large one. This is again because I want to test what the divisor will turn out to be. As you can see from the table on the next page, there is a very significant difference between 10,000/5, which is 2000, and the value of x, which give this open box its maximum volume, which is 2113. The difference is 113, which is quite big.
As you can see I have only put 10,000/5 for cut x, which is 2000 and, 2113, which is the cut of x which give the open box its maximum volume. This is because the original table would have taken up a lot of space and is insignificant data.
As you would have seen in the table above, there is a big difference in the measurement of x, which gives the open box its maximum volume. This time the ratio of 10,000:20,000 is much bigger than 100:200. However, the divisor is still between 4.7 and 4.8. Therefore I have reason to believe that the limit is between 4.7 and 4.8 for rectangles in the ratio of 1:2.
Rectangles in the ratio of 1:3
20cm:60cm
As you can see, the table above shows the volume of an open box for different measurements for cut x for a rectangle with the measurements of 20cm by 60cm, which is in the ration of 1:3. The table shows that the maximum volume is when the cut x is 5. Notice that this is 20/4, which is the length of the rectangle divided by 4. To calculate the volume for this particular open box, I used this formula:
Volume = x(20-2x)(60-2x)
30cm:90cm
The table above shows that the cut of x, which will give this open box its maximum volume, is 7. However, the divisor this time is not 4. It seems to have increased to 4.2-4.3. The formula I used to find out the volume is:
Volume = x(30-2x)(90-2x)
100cm:300cm
The table above shows the cut of x that gives this open box its maximum, volume, which is 23cm, and the cut equalling to 100/4, which is 25cm. The divisor for this open box, which calculated the cut of x, which in turn gives this open box, its maximum volume is around 4.3. Notice that even thought the values of the same ratio gets larger, the divisor is staying in the same region. Therefore, I predict that 4.3-4.5 is the limit.
1000cm:3000cm
The table above shows the volume of a box with the length of 1000cm and the width of 3000cm. The first row, which is in bold, shows the value of x, which gives this particular open box its maximum volume. The last row, which is in bold shows 1000/4, which is the length divided by 4. Instead of being 4, the divisor is now around 4.42. Notice that even though the values have increased by ten times, the divisor is still in between 4.3-4.5. Therefore, I have reason to believe that the limit for rectangles in the ratio of 1:3 is around 4.3-4.5.00. The formula I used to calculate the volumes for different cuts is:
Volume = x(1000-2x)(3000-2x)
Rectangles in the ratio of 1:4
1:4 is the final ratio I will be looking at. However, I might take a brief look into rectangles in the ratio of 1:5 to prove the limit of the divisor where the answer give the cut of x which gives the open box its maximum volume.
20cm:80cm
The table above shows that the value of x, which gives this open box its maximum value, is 5. Notice that even though the ratio has gone form 1:3 to 1:4, the length (which is 20cm for both ratios) divided by 4 still gives the cut of x that gives this open box its maximum volume. The formula used to calculate the volume for different cuts is:
Volume = x(20-2x)(80-2x)
40cm:160cm
The table above shows that the value of x, which gives the maximum volume for this particular open box, is 9cm. However, this is not the length divided by 4, which is supposed to be 40/4, which is 10. Instead the divisor is now 4.4. The formula used to calculate the volume of the open box for different cuts is:
Volume = x(40-2x)(160-2x)
100cm:400cm
The table above shows that the cut of x, which gives the maximum volume, is 23cm. The divisor is between 4.3 and 4.4. Notice that the previous rectangle had the same divisor. Therefore, I believe that the limit is around 4.3-4.4. The formula used to calculate the volume for the different square cuts is:
Volume = x(100-2x)(400-2x)
1000cm:4000cm
In the table above, I have included the cut of x, which gives the open box its maximum volume and the cut of x, which is equal to length/4. As you can see from the table above, the cut of x, which gives this open box its maximum volume, is 232cm. Yet again the divisor is also between 4.3-4.4. Therefore, one can conclude that the limit of the divisor for the ratios of 1:3, and 1:4.
From that, I can predict that for any ratio (except 1:1 and 1:2), the length divided by 4-4.5 will give the square cut of x which will give the open box its maximum volume.
