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  • Level: GCSE
  • Subject: Maths
  • Word count: 3128

THE OPEN BOX PROBLEM

Extracts from this document...

Introduction

THE OPEN BOX PROBLEM

An open box is to be made from a sheet of card. Identical squares are cut off the four corners of the card, as shown below.

image00.pngimage02.pngimage01.pngimage00.pngimage01.png

image01.pngimage01.png

image01.pngimage01.pngimage01.pngimage01.png

image02.png

The card is then folded along the dotted lines to make the box.

The main aim is to determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card, but first I am going to experiment with a square to make it easier for me to investigate rectangles.  

I am going to begin by investigating a square with a side length of 10 cm. Using this side length, the maximum whole number I can cut off each corner is 4.9cm, as otherwise I would not have any box left.

I am going to begin by looking into going up in 0.1cm from 0cm being the cut out of the box corners.

The formula that needs to be used to get the volume of a box is:

Volume = Length * Width * Height

If I am to use a square of side length 10cm, then I can calculate the side lengths minus the cut out squares using the following equation.

Volume = Length - (2 * Cut Out) * Width - (2 * Cut Out) * Height

Using a square, both the length & the width are equal. I am using a length/width of 10cm. I am going to call the cut out "x." Therefore the equation can be changed to:

Volume = 10 - (2x) * 10 – (2x)

...read more.

Middle

2.45

249.925

2.46

10.08

10.08

2.46

249.952

2.47

10.06

10.06

2.47

249.973

2.48

10.04

10.04

2.48

249.988

2.49

10.02

10.02

2.49

249.997

2.50

10.00

10.00

2.50

250.000

2.51

9.98

9.98

2.51

249.997

As you can see from the table above the proportion of the box that needs to be cut away to obtain the maximum cut out size is 1/6 as I found out the maximum cut-out size is 2.5cm. I also made a graph to prove that 2.5cm is the maximum cut out size.

image09.png

To prove that 1/6 of the size of the box gets the maximum cut out size again, I tried to prove it with a box with measurements of 20cm by 20cm. So I worked out 1/6 of 20 which gave me 3.333 so in the spreadsheet I used numbers between 3 and 3.6 for the cut out square.

LEGNTH AND WIDTH OF THE CUT-OUT SIZE

LEGNTH

WIDTH

HEIGHT

VOLUME

3.00

14.00

14.00

3.00

588.000

3.10

13.80

13.80

3.10

590.364

3.20

13.60

13.60

3.20

591.872

3.30

13.40

13.40

3.30

592.548

3.40

13.20

13.20

3.40

592.416

3.50

13.00

13.00

3.50

591.500

3.60

12.80

12.80

3.60

589.824

As you can see the maximum cut out size is 3.30, which is around the answer, I worked out. So this definitely proves that 1/6 of the length or width of the box gives me the maximum cut out size of the box. I also made a graph to show that the maximum cut out size is around 3.30cm.

image10.png

Other than using Microsoft excel to show the maximum cut out size and volume you can also prove it through algebra. Using algebra I can prove the maximum volume. I already know that a 1/6 of the length or width (which is the same, as in a square all the sides are equal) gives me the maximum cut out size so I substituted this into the formula.

Volume= (x-2h) (x-2h) * h

(H=cut out size and x=the measurement of the side of the square)

Substitute (h=x/6)

Volume= (x-2x/6) (x-2x/6) * x/6

Multiply out

Volume= (x²-2x²/6-2x²/6+4x²/36)*x/6

Factorise

Volume= x³/6-2x³/36-2x³/36+4x³/216

Volume= 36x³-12x³-12x³+4x³

                         216image03.png

Volume= 40x³-24x³image04.png

216              

Volume= 16*x³image05.png

216

Volume= 2*x³ image06.png

27        

To prove this is correct I will use this formula to work out the maximum volume for a square with dimensions of 10cm by 10cm.

2*10³

           = 74.074 image05.png

  27

This is correct because this was the maximum volume I achieved when I calculated it in the spreadsheets.

I can also prove the maximum cut out size by using the gradient function. The gradient function also you to the gradient at any point on a line graph. To calculate the gradient function if y=a*x        (n*a) x   image07.png

For example: 3x²= (2*3) x

                  = 6x

This formula will give you the gradient when it equals 0. This can find the maximum cut out size as when the gradient=0 this will give the highest point in the graph as shown below:  

I will use this function to prove the maximum cut out size for a 10cm by 10cm square is 1.67cm:

Volume= (x-2h) (x-2h) * h

=  (X²-4xh+4h²) * h

              =  Xh²-4xh²+4h³

Substitute (h=10)

= 100x-40x+4x³  

Use gradient function

                0=100-80x+12x²                                                      (divide by 2) 0= 50-20x+6x²

                0= 25-20x+3x²

Solve by using common formulae

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 two different rectangles, with the ratio between the length and width as 2:1 and 3:1.

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.

This is the formula I put into the spreadsheet:

(2w-2x) (w-2x) x  

(w= the width of the rectangle and x= the cut out size)

Below are the results I got through this spreadsheet:

Card Size: 10cm by 20cm- TO FIND EXACT CUT OUT SIZE

Width

Length

Cut out size

Volume

6.00

16.00

2.00

192.000

5.98

15.98

2.01

192.076

5.96

15.96

2.02

192.146

5.94

15.94

2.03

192.208

5.92

15.92

2.04

192.263

5.90

15.90

2.05

192.311

5.88

15.88

2.06

192.351

5.86

15.86

2.07

192.385

5.84

15.84

2.08

192.412

5.82

15.82

2.09

192.431

5.80

15.80

2.10

192.444

5.78

15.78

2.11

192.450

5.76

15.76

2.12

192.449

5.74

15.74

2.13

192.440

5.72

15.72

2.14

192.425

5.70

15.70

2.15

192.404

5.68

15.68

2.16

192.375

5.66

15.66

2.17

192.339

5.64

15.64

2.18

192.297

5.62

15.62

2.19

192.248

5.60

15.60

2.20

192.192

...read more.

Conclusion

Optimum cut out size For Box Size 10cm by 30cmimage13.png

This graph shows that the maximum cut out size is around 2.26cm.

If I wish to work out the proportion of the box that needs to be cut away to obtain the maximum cut out size, I need to divide 2.26cm by 10. In doing this I get an answer of 0.226, or a proportion of 1/4.42. To see if this is correct I decided to look at different size square, 20cm by 60cm. I worked out a 1/4.42 of 20, which was 4.52cm so I made a spreadsheet to work out the maximum cut out size and volume of a 20cm by 60cm square. I looked between 4.45cm and 4.55cm for the maximum cut out size, as I knew the cut out size should be around that figure.

Card Size: 20cm by 60cm

Length

Width

Cut out size

Volume

51.20

11.20

4.40

2,523.136

51.18

11.18

4.41

2,523.368

51.16

11.16

4.42

2,523.580

51.14

11.14

4.43

2,523.769

51.12

11.12

4.44

2,523.938

51.10

11.10

4.45

2,524.085

51.08

11.08

4.46

2,524.210

51.06

11.06

4.47

2,524.314

51.04

11.04

4.48

2,524.398

51.02

11.02

4.49

2,524.459

51.00

11.00

4.50

2,524.500

50.98

10.98

4.51

2,524.519

50.96

10.96

4.52

2,524.518

50.94

10.94

4.53

2,524.495

50.92

10.92

4.54

2,524.451

50.90

10.90

4.55

2,524.386

50.88

10.88

4.56

2,524.299

50.86

10.86

4.57

2,524.192

50.84

10.84

4.58

2,524.064

50.82

10.82

4.59

2,523.914

50.80

10.80

4.60

2,523.744

As you can see the maximum cut out size is 4.51cm, which is around the answer, I worked out. So this definitely proves that 1/4.73 of the width of the box gives me the maximum cut out size of the box. I also made a graph to show that the maximum cut out size is around 4.51cm.

Optimum cut out size For Box Size 20cm by 60cm

image14.png

This graph shows that the maximum cut out size is around 4.51cm.

I need to prove through algebra how to get the maximum volume for the rectangle with a ratio of 3:1. I have already worked out that 1/4.42 of the width of the box gives me the maximum cut out size of the box, so I have to substitute this into the volume formula.

...read more.

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