L= Original Length
Y= Original Length- 2x
10= 2x+y or y=10-2x
X=10/6 or L/6
To make an equation I am going to substitute these into a simple equation.
V = (L – 2L/6)(L – 2L/6) L/6
Then I multiplied the 2 in each bracket with L/6 which gives-
V = (L – 2L/6) (L – 2L/6) L/6
Then I will multiply brackets out by separating them. This gives-
V = L (L – 2L/6) – 2L/6 (L – 2L/6) L/6
Now I will complete the multiplication, which gives –
³V = (L² - 2L²/6 – 2L²/6 – 4L²/36) L/6
Now I will multiply everything by L/6/
V = L³/6 – 2L³/36 – 2L³/36 + 4L³/216
To make the equation simpler I will simplify the equation by putting everything over 216 because It is a common denominator.Now it looks like this-
V = 4L³ + 36L³ - 12L³ - 12L³
216
This can now be simplified to give me-
40L³ - 24L³
216
This can now be cancelled down to give me-
16L³
216
Both 16 and 216 can divide by 8 to give me this-
2L³
27
Next I will investigate Rectangles I will use a ratio of L= 2W
So if the
L =10 the W will = 5
L= 12 the W will =6
L=14 the W will =7
L=16 the W will = 8
The first one I will try will be the rectangle with a width of 5 and Length of 10.
As we can tell you can only use a cut size of 1 or 2 to give a volume. Because there is such little data to use, I am going to move on to a Rectangle with a Length of 20 and a width of 10.
As we can tell we can now go up to using a cut size of 4 cm’s. Now I am going to try and find a larger volume by trying between 2 and 3 cm’s because that is where the largest volume is at the moment to see what cut size will give the largest volume.
Now I am going to try between 2.1 and 2.2 to try and find the maximum volume.
Now I will try between 2.11 and 2.12 to try and find the maximum volume
From the graph we can tell that the line touches 192.45 twice. This shows that the maximum for a 20cm by 10 cm rectangle is 192.4503 roughly.
To see I will try between 2.111 and 2.112
As you can tell the volume does not change between 2.1113 and 2.1121. Therefore I shall not go any smaller. At the moment I cannot see anything, which may form a pattern so I will try another Rectangle.
Now I shall try a 40cm by 20cm rectangle.
The maximum volume is between 4cm and 5cm.
Now I will try from 4.0 to 5.0.
This shows that the maximum Volume is at 1540 cm3.
Now I shall try between 4.2 and 4.4.
This shows where the Volume reaches its maximum.
I still cannot see a pattern so I shall try a 50 by 25 rectangle.
As we can see the maximum volume is between 5 & 6.
Now I will go up in 1 decimal place.
This shows that the Maximum volume is between 3000 and 3020. Now I will try between 5.2 and 5.3 with a difference of two decimal places.
Now I will try between 5.28 and 5.3 to try and find the maximum volume. I will use three decimal places.
Now I will draw a graph to try and make the results easier to read.
Now we can see that the maximum volume is between 5.282 and 5.284. Now I will try between 5.282 and 5.284.
As we can tell from this there is no variation on the volume.
Therefore the maximum is 5.283 because this is the middle value.
When I divide the cut out which gives the largest volume by 20 this gives the proportion, which gives the maximum volume. The cut out is 5.283 when this is divided by 20 we get – 0.26415. This is just over a quarter.
The length of the rectangle is double the width (ratio of 1:2). This means that the length = 2L.
Therefore we can express the volume as V=L x W x H.
Because a square cut out is x we can take it away from both sides of the length and the width.
This means the formula becomes-
Length = 2L - 2x
Width = 2Lx
Because the cut out forms the height it is H. So the cut out is H=x
So we can now write the volume as-
L x W x H
(2L - 2x) (L - 2x) x X
When you multiply out the brackets you get-
(2L² - 2xL - 4xL + 4x ²) x X
Now I will try a rectangle with the ratio of 1:3. I will use a width of 20 and a length of 60.
Here we can see that a cut size of 5 gives the maximum volume. This may be because 5 is quarter of 20.
Now I shall try between 4 and 6 with one decimal place.
Now we can tell that the maximum cut is between 4.5 and 4.6, so I shall try two decimal places.
Now I will draw a graph to try and make the results easier to read.
I can tell that the maximum volume occurs with a cut size between 4.51 and 4.52. Now I shall try to three decimal places.
When I divide the maximum cut size of 4.515 by 20 this gives me the ratio that needs to be cut off each corner to make the largest volume. This is 0.2257, which is again almost a quarter.
Now I will try a rectangle with a ratio of 1:10 to see if it as a proportion of a quarter. I will use a width of 10 and a length of 100 because these numbers are easier to process. I predict that the maximum volume will be roughly 2.5 because 10/4 = 2.5. (a quarter)
From this data we can tell that the maximum volume is between 2 and 3. Now I shall use one decimal place to see which cut size will give a larger volume.
From this new data we can tell that the maximum volume is between 2.4 and 2.5. Now I shall use two decimal places to see what cut size will give the largest volume.
The largest volume is made from a cut size of 2.43. Now I shall try 3 decimal places between 2.43 and 2.44.
As we can tell between 2.433 and 2.436 the volume is the same.
Now I shall draw a graph to show my results-
Now I shall try between 2.433 and 2.436 to see if I can get a more accurate result.
Now we can tell that the cut size makes no difference at all to the volume, therefore I shall not bother to do any more.
2.435/10 = 0.2435, which again makes a proportion of nearly a ¼.
Now I can give a rule to show that to make the maximum volume of a rectangle you must divide the width by 4. This will give the cut size needed to make the maximum volume.
If I understood calculus it could be used to prove my rule.