20 cm is the length and width of the square, therefore
20 = 2H + L
Rearranging this it becomes
L = 20 - 2H
The formula for the volume is
Width * Length * Height
Or
(20 - 2H) * (20 - 2H) * H
The maximum area it found when the cut out is 1/6, meaning that:
H = 20/6
If I substitute this into the equation for maximum volume then it will become
V = (L - 2L/6)(L - 2L/6) L/6
I multiply the 2 in each bracket with L/6, giving you
V = (L - 2L/6) (L - 2L/6) L/6
To multiply brackets out I'll separate them, and write the 2nd bracket out twice.
V = L (L - 2L/6) - 2L/6 (L - 2L/6) L/6
Then complete multiplication leaving
³V = (L² - 2L²/6 - 2L²/6 - 4L²/36) L/6
Then I multiply the whole thing by L/6, which gives me
V = L³/6 - 2L³/36 - 2L³/36 + 4L³/216
I put the whole equation over a common denominator to help me with simplification, meaning the whole thing becomes over 216. This means that it becomes:
V = 4L³ + 36L³ - 12L³ - 12L³
216
This can be simplified so you end up with:
40L³ - 24L³
216
Which then cancels out to give you
16L³
216
You then divide the top and bottom by 8, which is a factor of both, and you will end up with:
2L³
27
I am now going to continue my investigation by looking at the shape of rectangles. As there are too many combinations of lengths and widths of rectangles for me to possibly even begin to investigate I am going to investigate in the following manner.
I shall begin with a width of 20cm, and a length of 40cm, this is a ratio of 1:2, the length being twice as long as the width.
The square cut outs will rise in measurements of 1cm at a time, beginning with a cut out of 1 cm, and 9cm being the maximum for there to be a box left.
I will then investigate various other ratios.
I have again constructed a spreadsheet to assist with my workings out. Firstly I am looking at whole numbers, a beginning with a width of 20cm and a length of 40cm, and cut outs from 1 to 9cm.
Here are the results I got.
You should be able to see from this table, and the graph I have drawn that that the largest volume achieved is 1536cm², and this is obtained when the amount cut off from the corners measures 4cm². I can also see that to make my results more accurate, to more decimal places, I next need to look between 4 and 5cm² cut offs.
Below are the formulae I used for the above spreadsheet, which only differs slightly to the one that was used for the square. All that has changed is the length being 40, which is reflected below.
Next are my results from calculating the volume when the cut out area is measured to one decimal place.
I have also drawn a graph to illustrate my results.
From these results, where the largest volume of 1539.522cm² is achieved with a cut off of 4.2cm², I see that I should look between 4.2 and 4.3cm² as a cut off next to get even more accurate results which are measuring to 2 decimal places.
Below are my results showing the size of the cut off calculated to 2 decimal places.
I am going to finally calculate the size of the cut off to 3 decimal places for a very accurate result.
The table above and my graph tell me that the largest volume will lie somewhere between the cut off sizes of 4.22 and 4.23cm², so I will finally look in between these numbers for the cut off size to achieve my most accurate results.
Below is my final table of results, which shows the cut off measured to 3 decimal places. The largest volume is highlighted in bold.
I have drawn one more graph to show these results easily. You should be able to see from this table, and from the graph that the largest volume I have obtained is 1539.600701cm², and that this is got when the cut off area from each corner of the box is measuring 4.226cm².
When I divide the cut out which gives the maximum volume, 4.226 by 20 this gives me the proportion that should be cut off to give the maximum volume, which is 0.2113, almost a quarter.
The length of the rectangle is double the width (ratio 1:2) so length = 2L
Volume will be expressed as:
V = L * W * H
As square cut out is x.
Take it away from both sides of both length & width so it becomes
Length = 2L - 2x
Width = 2Lx
So the cut out is h = x
Expression for volume will be:
L * W * H
(2L - 2x) (L - 2x) * x
Multiplying it out becomes
(2L² - 2xL - 4xL + 4x ²) * x
Multiply X through and the equation will become
2L² (x) - 6xL (x) + 4x² (x)
So this can be
2L²X - 6X²L + 4X³
4X² - 6X²L + 2L²
I have fully worked out which cut off gives the maximum area with a ratio of 1:3, as I did above, but with a width of 20cm and a length of 60cm. I worked again to 3 decimal places. All of my results are shown below. I am not going to draw any further graphs, as they are all fairly similar.
When I divide the largest cut off, 4.515 by 20, this gives me the ratio that needs to be cut off each corner to achieve the largest volume. This is 0.2257, which is again almost a quarter.
Next I fully worked out the cut off giving the largest volume when the width: length ratio is 1:10. The tables of results are shown below.
Again I calculated the ratio that needs to be cut off the card to create the box with the largest volume. This is 0.24345, which is close to ¼ again.
The final volume I am going to work out is with a width: length ratio of 1:100.
Here are my results for this.
I worked out the proportion that needs to be cut off the box to give maximum volume; this was 0.24445, which is very close to ¼.
Ð stands for delta.
Firstly we should consider a graph of y = x² as shown below.
The line through X and Y has almost the correct gradient.
It's gradient is
Increase in y-coordinate from X to Y
Increase in x-coordinate from X to Y
You have to find an expression for , which represents
the gradient of the graph at the point X.
So
y = x²
y + Ðy = (x + Ðx)(x +Ðx)
Multiply out brackets.
y+ Ðy = x² + x Ðx + x Ðx + (Ðx) ²
Add like terms together.
y + Ðy = x² + 2xÐx + (Ðx)²
Now here the x² at the end is y in terms of x.
Ðy = x² + 2xÐx + (Ðx) ² - x²
Then you divide by Ðx, which gives you
Ðy 2xÐx - (Ðx) ²
Ðx = Ðx
Ðy
Ðx = 2x + Ðx
Now because delta (Ð) is so tiny that it is insignificant, we forget all about it, which leaves us with
Ðy/Ðx = 2x = 0
I can use calculus to help me complete my calculations to solve the problem, for proof through exhaustion.
I have drawn a graph showing the proportions that I have worked out. I can see that they tend towards ¼. This is the amount that should be cut off each corner to give the maximum volume possible.
Open Box Investigation
An open box is to be made from a sheet of card. Identical squares are to be cut out of the four corners of the card as shown below in the net. The card is then folded along the red lines to form the box.
The main aim is to determine the size of the square cut out which makes the volume of the box as large as possible for any given square sheet of card.
Net:
Investigation points:
- For any sized square sheet of card, investigate the size of the square cut out which makes an open box of the largest volume.
- For any sized rectangular sheet of card, investigate the size of the square cut out which makes an open box of the largest volume.
10cm Square Card
I am going to begin by investigating a 10×10cm square sheet of card. I shall, step by step increase the card size until I have found a formula to support this investigation.
First, I shall start by looking into whole numbers being cut out of the corners of the box. I will use the formula:
Volume = (Length – (2 × Cut Out)) × (Width – (2 × Cut Out)) × Height
to construct a spreadsheet. This will allow me to quickly and accurately calculate the volume of the box. Below are the results I got through this spreadsheet.
As you can see from the table above, the largest volume lies between the 1 and 2cm cut out. Therefore to make my results more accurate, I am going to investigate the volume of the box with the cut out to 1 decimal place.
Below is a table showing this.
From this set of results it shows me that to achieve the maximum volume; I will need to need to look at cut outs measuring to 2 decimal places between 1.6 and 1.7cm.
The table below shows the cut out measured to 2 decimal places.
Looking at the table above you can see that the maximum volume is achieved with the cut out measured to 2 decimal places, of 1.67cm. Therefore I have no reason to continue with this particular measurement and so shall go on to investigate the maximum possible volume of a 12×12cm sheet of card.
12cm Square Card
I am now going to further my investigation by looking into the maximum possible volume that can be achieved for a 12×12cm sheet of card.
Above are my initial results, looking into the cut out sizes of between 1 and 5cm. I did not need to exceed this as you can see from the results that the volumes begin to decrease at the 4cm cut out, so it is not necessary to continue as the volumes will only decrease further into negative numbers. As highlighted, the optimum volume occurs between the 2 and 3cm cut outs and so I will continue investigating to 1 decimal place.
From the table it is clear that the largest volume occurs when the cut out size is 2cm. For that reason I will not prolong this stage of my investigation and shall continue onto the next.
15cm Square Card
In order to gain enough evidence to support my investigation I now need to study the cut out sizes for a 15×15cm square sheet of card.
Below are my findings.
As shown in the table the maximum obtainable volume for this size of card lies between the 2 and 3cm cut out. Therefore, I will carry on to investigate to 1 decimal place.
These result show that there may be a chance that a higher volume could be achieved, so to get a more precise result I will look at the cut out sizes to 2 decimal places.
This table shows that that no higher volume can be achieved for a 15×15cm square of card than the 2.5cm cut out and so this concludes this section of the investigation.
Results
The table below shows the cut out measurements for each size of card that produce the maximum volume for an open box.
From my series of results I have realised a pattern that is consistent through out each stage of this investigation. I have found that if you divide that card size by the square cut out, the answer is always 6. Therefore, if you divide the card size by 6, the answer will give you the cut out size that provides the maximum volume for that size of square card.
Card Size ÷ 6 =Cut Out
Example; 10 ÷ 6 = 1.67
12 ÷ 6 = 2
15 ÷ 6 = 2.5
Prediction
With the formula that I have found I am now able to predict the outcome for a larger sized square sheet of card.
I predict that for an 18×18cm sheet of card the cut out size that will provide the largest possible volume, will be 3cm.
Card Size ÷ 6 = Cut Out
18 ÷ 6 = 3cm
To test my prediction I will construct a further spreadsheet to determine the cut out sizes for an 18×18cm.
This shows that the largest volume lies between the 3 and 4cm cut out and so, so far, proves that my prediction is correct. To obtain an even more accurate result, I shall take these results to 1 decimal place.
This table shows that the highest volume that can be achieved is with a 3cm cut out. This proves my prediction but as confirmation I intend to repeat this section, substituting the card size of 18cm to 72cm.
I predict that for a 72×72cm sheet of card the cut out size that will provide the largest possible volume, will be 48cm.
Card Size ÷ 6 = Cut Out
72 ÷ 6 = 48cm
To, again test my prediction I will construct another spreadsheet to establish the cut out sizes for a 72×72cm.
Again, this table shows the largest possible volume lying between 2 numbers so I will continue on investigating to 1 decimal place.
From this second trial of my prediction, I have again proved that it is correct. This second trial has therefore proved the accuracy of my prediction.
Justifying my Results
I appreciate that finding the maximum volume of an open box requires me to find the maximum volume of x, which occurs where the gradient is 0.
