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• Level: GCSE
• Subject: Maths
• Word count: 3314

Open Box Problem.

Extracts from this document...

Introduction

Open Box Problem

Introduction

The open box problem is about an open box, which is to be made from a sheet of card. Identical sized squares are cut off the four corners as shown in the diagram below.

The main aim of this investigation is to find the size of the square cut which makes the volume of the box as large as possible for any given rectangular or square sheet of card.

Squares

I am going to begin the open box investigation by investigating a square with the width and length of 24cm. The formula that needs to be used to get the volume of a box is:

Volume = Length x Width x Height

If I am to use a square of side length 24cm, then I can calculate the side lengths minus the cut out squares using the following equation.

Volume = x(24-2x)(24-2x)

x = is the square that is to be cut out.

24 = is the size of the length and the width

For this square I will make a table to show the volume of different cuts. I will use whole numbers for the different cuts to find roughly where the maximum volume occurs.

24cm by 24cm Square

For this square I will make a table to show the volume of different cuts. I will use whole numbers for the different cuts to find roughly where the maximum volume occurs.

 Cut x L (24-2x) W (24-2x) Volume 1 22 22 484 2 20 20 800 3 18 18 972 4 16 16 1024 5 14 14 980 6 12 12 864 7 10 10 700 8 8 8 512 9 6 6 324 10 4 4 160 11 2 2 44 12 0 0 0

...read more.

Middle

21

0

0

0

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.

 Cut x L (48-2x) W (48-2x) Volume 1 46 46 2116 2 44 44 3872 3 42 42 5292 4 40 40 6400 5 38 38 7220 6 36 36 7776 7 34 34 8092 8 32 32 8192 9 30 30 8100 10 28 28 7840 11 26 26 7436 12 24 24 6912 13 22 22 6292 14 20 20 5600 15 18 18 4860 16 16 16 4096 17 14 14 3332 18 12 12 2592 19 10 10 1900 20 8 8 1280 21 6 6 756 22 4 4 352 23 2 2 92 24 0 0 0

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:

• 1:2
• 1:3
• 1:4

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

 Cut x L (10-2x) W (20-2x) Volume 1 8 18 144 2 6 16 192 3 4 14 168 4 2 12 96 5 0 10 0

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

 Cut x L (20-2x) W (40-2x) Volume 1 18 38 684 2 16 36 1152 3 14 34 1428 4 12 32 1536 5 10 30 1500 6 8 28 1344 7 6 26 1092 8 4 24 768 9 2 22 396 10 0 20 0

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.

 Cut x L (10,000-2x) W (20,000-2x) Volume 2000 6000 16000 192000000000 2113 5774 15774 192450087588

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

 Cut x L (20-2x) W (60-2x) Volume 1 18 58 1044 2 16 56 1792 3 14 54 2268 4 12 52 2496 5 10 50 2500 6 8 48 2304 7 6 46 1932 8 4 44 1408 9 2 42 756 10 0 40 0
...read more.

Conclusion

Volume = x(20-2x)(80-2x)

40cm:160cm

 Cut x L (40-2x) W (160-2x) Volume 1 38 158 6004 2 36 156 11232 3 34 154 15708 4 32 152 19456 5 30 150 22500 6 28 148 24864 7 26 146 26572 8 24 144 27648 9 22 142 28116 10 20 140 28000 11 18 138 27324 12 16 136 26112 13 14 134 24388 14 12 132 22176 15 10 130 19500 16 8 128 16384 17 6 126 12852 18 4 124 8928 19 2 122 4636 20 0 120 0

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

 Cut x L (100-2x) W (400-2x) Volume 23 54 354 439668 25 50 350 437500

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

 Cut x L (1000-2x) W (4000-2x) Volume 232 536 3536 439708672 250 500 3500 437500000

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.

...read more.

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