Maths Courseowrk - Open Box

V=2(10-4)2

So the volume of the open box would be 144.

Tables

For the tables displayed below I have given the x value then the l value then used the formula to calculate the volume. At first I was going to work out the volume by hand and using a calculator however in the end I used excel which is a much more efficient method and is very quick and relatively simple to use. Using excel also eradicates any human errors that could have occurred through typing the wrong number in the calculator. Using excel will create more time for me to concentrate on the more challenging areas of this coursework. Using excel I could round off to an appropriate degree with a click of a mouse this would give me detailed and accurate answers that could easily be entered into a graph on the computer.

For the first table I used simple numbers to start and then developed the best cut out size to 5dp after the first table I realised that I would just use the graphs I would make to find the answer, this would save time and effort. However in the tables I have gone into 1 decimal place at least this will give me good values to plot a graph off. For the first table I realised that the highest values were when x was between 0.5 and 1 then I would use values in this number and see what the highest values were again until eventually I had an answer with a degree of accuracy with 5 decimal points or 6 significant figures.

Graphs

I complimented the tables with cubic graphs using the information I had for each of my value for x. (e.g. x=10) The graphs I have created are shown under each table. For this investigation I decided to use a program called Autograph which is available for me at school and at home and as we have been taught how to use it I thought this would be another way of analysing my results. This also came in useful as I had already all the values on the computer if needed on excel. Along with this I decided that autograph would be more accurate that by drawing the graphs by hand as I could have easily make a mistake and I am not as the smoothest person to draw a curve. For my graphs below the volume is shown along the y-axis and the cut out size along the x-axis hence the equation of the line is y=x(l-2x)2. The value of a depended on the square I was using so for the first table the value for l was 5. So to get a graph I only needed to enter the l value. I had to use y= and not v= as stated in the original formula as in autograph you could only use x= or y= and x was already in my formula so I had to use y=

On the next page is the table of values for when the l=5

X

L

v=x(5-2x) 2

0.5

5

8

5

9

.5

5

6

2

5

2

0.75

5

9.1875

0.8

5

9.248

0.85

5

9.2565

0.9

5

9.216

0.95

5

9.1295

0.831

5

9.259204764

0.832

5

9.259241472

0.833

5

9.259258148

0.834

5

9.259254816

0.835

5

9.2592315

0.836

5

9.259188224

0.837

5

9.259125012

0.838

5

9.259041888

0.839

5

9.258938876

0.8331

5

9.259258715

0.8332

5

9.259259081

0.8333

5

9.259259248

0.8334

5

9.259259215

0.8335

5

9.259258982

0.8336

5

9.259258548

0.8337

5

9.259257915

0.8338

5

9.259257082

0.8339

5

9.259256049

0.83331

5

9.259259254

0.83332

5

9.259259257

0.83333

5

9.259259259

0.83334

5

9.259259259

0.83335

5

9.259259256

0.83336

5

9.259259252

0.83337

5

9.259259246

0.83338

5

9.259259237

0.83339

5

9.259259227

This table of value is for a square with sides of 5cm in length. I originally went up in 0.5cm to get an accurate measure of where the highest volume will be. I then went up in 0.05 where I thought the largest possible volume will be. Then I continued although way until I got an answer containing five decimal points. So to get the highest volume the cut out size to 5 decimal places is 0.83333

This is the graph showing when l=5 where the highest volume for the open box is shown by the middle arrow and this gives a quite detailed look at what size of cut out will give the highest volume. The point we will look for is the first maximum which gives us the highest volume. The possible values for the cut out are between the points (0,0) and the point where the line passes downwards through the x axis.

These are the values when l=10

X

L

V=x(10-2x) 2

0

64

.5

0

73.5

2

0

72

2.5

0

62.5

3

0

48

3.5

0

31.5

4

0

6

4.5

0

4.5

.75

0

73.9375

Here is a graph when l=10

These are the values when l=20

X

L

v=x(20-2x) 2

0.5

20

80.5

20

324

.5

20

433.5

2

20

512

2.5

20

562.5

3

20

588

3.5

20

591.5

4

20

576

4.5

20

544.5

5

20

500

5.5

20

445.5

6

20

384

6.5

20

318.5

7

20

252

7.5

20

87.5

8

20

28

8.5

20

76.5

9

20

36

9.5

20

9.5

Here is a graph when l=20

These are the values when l=40

X

L

v=x(40-2x) 2

0.5

40

760.5

40

444

.5

40

2053.5

2

40

2592

2.5

40

3062.5

3

40

3468

3.5

40

3811.5

4

40

4096

4.5

40

4324.5

5

40

4500

5.5

40

4625.5

6

40

4704

6.5

40

4738.5

7

40

4732

7.5

40

4687.5

8

40

4608

8.5

40

4496.5

9

40

4356

9.5

40

4189.5

0

40

4000

0.5

40

3790.5

1

40

3564

1.5

40

3323.5

2

40

3072

2.5

40

2812.5

3

40

2548

3.5

40

2281.5

4

40

2016

4.5

40

754.5

5

40

500

5.5

40

255.5

6

40

024

6.5

40

808.5

7

40

612

7.5

40

437.5

8

40

288

8.5

40

66.5

9

40

76

9.5

40

9.5

Here is a graph when l=40

By looking at this graphs I realised they were all very similar and I decided to put them in a table with a/x to see what results I got. For this table I have decided to not round off my x value as this would make my other values less accurate. But for the last two columns I have left them to 1 significant figure to prevent complexity.

X

L

v=x(l-2x)2

l/x

0.83333

5

9

6

.66666

0

74

6

3.33333

20

523

6

6.66666

40

4741

6

In the above table I have placed the values of X which are the size of cut out that will give the greatest possible volume for the length of the box next to it. So when X=0.83333 the length of the box is 5. As you can see there is an obvious relationship between the L and X values as when L is divided by X the answer comes to roughly 6.

I have discovered that the relationship for the highest value of an open box is X=L/6. I gained two values where L/2 and L/6.I released it could not be L/2 when comparing this with my graphs turning points. The relationship is cemented by the writing on the table on the previous page as well. Here is my working to show how I mathematical discovered this relationship.

v=y(l-2y)2

v=y(l-2y)(l-2y)

v=(yl-2y2)(l-2y)

v=yl2 - 2ly2 - 2ly2 + 4y3

v=4x3 -4y2l + yl2

I then differentiated the x values and this is my results:

dy/dx = 12y2 - 8yl + l2

I then made a quadratic equation out of this:

(6y - l)(2y-l) = 0

This gave me the following answers:

6y - l = 0 or 2y - l = 0

6y = l 2y = l

y = l/6 y = l/2

y=l/6

I will prove this now by putting the formula to use. With l=6 which is my first tested table and in which I went to 5 decimal places.

Y=5/6

Y=0.833333333 recurring

This proves that the relationship for the formula is correct.

Graphs