The Open Box Problem

The highest volume when the cut out size is to two decimal places is when 'x' is 1.67. On getting this number I tried working out how the cut out size related to the length of one side of the square piece of card.

I divided 10 by 1.67 and got the answer 5.988024. This answer is approximately 6 meaning that the cut out size is almost exactly a sixth of one length of the square piece of card. I realised that if I had homed in further I may have got even closer to the answer of one sixth.

I could not be sure that the answer of a sixth is correct as I had only tried it on one sized piece of card so to get proof I needed to investigate other different sized square pieces of card.

I investigated the ideal cut out size for a square piece of card 20 x 20.

I got these results:

size of cut out 'x' (cm)

volume (cm³)

324

.5

433.5

2

512

2.5

562.5

3

588

3.5

591.5

4

576

4.5

544.5

5

500

I then homed in between 3 and 3.5cm for 'x':

size of cut out 'x' (cm)

volume (cm³)

3

588

3.1

590.364

3.2

591.872

3.3

592.548

3.4

592.416

3.5

591.5

I then homed in again with values of 'x' to two decimal places:

size of cut out 'x' (cm)

volume (cm³)

3.31

592.570764

3.32

592.585472

3.33

592.592148

3.34

592.590816

3.35

592.5815

I found the optimum cut out size to get the largest volume is 3.33 when 'x' is investigated to two decimal places. One sixth of 20 is 3.33 to two decimal places so this backs up my theory that the optimum cut out size is one sixth of the length of one side of the square piece of card.

For square piece of card, use cut out size 1/6 of length to get largest volume

The graphs created both:

* Increase, reach a peak and then decrease.

* Peak at 1/6 of the length of one side of the square sheet of card

* Have the same 'arc' shape.

Algebra

The formula for the volume of the open box with the cut out size being 'x' is:

V=x(L-2x)(L-2x)

Simplifies:

V=x(L²-2xL-2xL+4x²)

V=xL²-4xL²+4x³

Gradient function:

V=L²-8xL+12x²

12x²-8xL+L²=0

12x²-6xL-2xL+L²=0

6x(2x-L)-L(2x-L)=0

(2x-L)(6x-L)=0

x=L/2 or x=L/6

(We know the gradient for the graphs at their peak must be 0 as they are horizontal)

To work out the maximum volume for a square piece of card substituting 'x' with 1/6L:

V=(1/6L)L²-4(1/6L)²L+4(1/6L)²

V=1/6L³-4/36L³+4/216L³

V=1/6L³-1/9L³+1/54L³

V=9/54L³-6/54L³+1/54L³

V=4/54L³

V=2/27L³

Maximum volume=2/27L³

This formula can be used to work out the maximum volume of any open box from a square shaped piece of card without needing to work out anything to do with the cut out size.

Question 2

I them moved on to question 2, which was to investigate the correct cut out size to satisfy the largest possible volume of the open box on any sized rectangular piece of card.

I had already found the cut out size for a rectangle of ratio 1:1 in the first question, as a square is also a type of rectangle. I first tried the formula for ratio 1:2 rectangles:

To work out the volume of the open box from a rectangular piece of card the formula is: V=(KW-2X)(W-2X)xX

I investigated the ideal cut out size for a rectangle 16 x 8:

size of cut out 'x' (cm)

volume (cm³)

84

.5

97.5

2

96

2.5

82.5

3

60

3.5

31.5

4

0

I then homed in between 1.5 and 2cm for 'X':

size of cut out 'x' (cm)

volume (cm³)

.5

97.5

.6

98.304

.7

98.532

.8

98.208

.9

97.356

2

96

I then homed in once more with values of 'X' to two decimal places:

size of cut out 'x' (cm)

volume (cm³)

.6

98.304

.61

98.352324

.62

98.394912

.63

98.431788

.64

98.462976

.65

98.4885

.66

98.508384

.67

98.522652

.68

98.531328

.69

98.534436

.7

98.532

The highest volume is 98.534436 when 'X' is to two decimal places being 1.69.

The highest volume when the cut out size is to two decimal places is when 'x' is 1.69. On getting this number I tried working out how the cut out size related to the width of the rectangular piece of card. I divided 8 by 1.69 and got the answer 4.733727811. This answer is close to a fifth.

I could not be sure that the answer of 4.733727811 is correct for all 1:2 rectangles so to get proof I needed to investigate other different sized rectangular pieces of card with ratio 1:2.

I then investigated it 18 x 9,

I got these results:

size of cut out 'x' (cm)

volume (cm³)

12

.5

35

2

40

2.5

30

3

08

3.5

77

4

40

I then homed in between 1.5 and 2cm for 'X':

size of cut out 'x' (cm)

volume (cm³)

.5

35

.6

37.344

.7

38.992

.8

39.968

.9

40.296

2

40

I then homed in to two decimal places:

size of cut out 'x' (cm)

volume (cm³)

.88

40.281088

.89

40.291676

.9

40.296

.91

40.294084

.92

40.285952

.93

40.271628

.94

40.251136

.95

40.2245

The graph and my results indicate that the highest volume is 140.296 when 'X' is to two decimal places. 'X' is 1.90. I then divided 9 by 1.9 and got the answer 4.736842105. This is very similar to the answer 4.733727811 that I got for the 16x8 rectangle so the answer is backed up. This number is close to being a fifth of the width of the rectangle.

I will now continue my investigation with rectangles of ratio 1:3. I predict the answer will be less than 1:2. So far 1:1 was 1/6 and 1:2 was around 1/5.

Here are my results for rectangular sheet 18 x 6:

size of cut out 'x' (cm)

volume (cm³)

64

.5

67.5

2

56

2.5

32.5

3

0

I then homed in between 1 and 2cm for 'x':

size of cut out 'x' (cm)

volume (cm³)

.1

66.044

.2

67.392

.3

68.068

.4

68.096

.5

67.5

.6

66.304

.7

64.532

.8

62.208

.9

59.356

2

56

I then homed in between 1.3 and 1.4cm for 'x':

size of cut out 'x' (cm)

volume (cm³)

.31

68.099564

.32

68.124672

.33

68.143348

.34

68.155616

.35

68.1615

.36

68.161024

.37

68.154212

.38

68.141088

.39

68.121676

.4

68.096

The graph and my results indicate that the highest volume is 68.1615 when 'X' is to two decimal places. 'X' is 1.35. I then divided 6 by 1.35 and got the answer 4.4444 reoccurring. I expected the number to be close to 1/4 but it was not as close as I had expected it to be. I then tried another 1:3 rectangle to see if I got the same results.

I investigated a rectangle sheet of dimensions 24x8:

size of cut out 'x' (cm)

volume (cm³)

32

.5

57.5

2

60

2.5

42.5

3

08

3.5

59.5

4

0

I then homed in between 1.5 and 2cm for 'x':

size of cut out 'x' (cm)

volume (cm³)

.5

57.5

.6

59.744

.7

61.092

.8

61.568

.9

61.196

2

60

I then homed in again between 1.8 and 1.9cm for 'x':

size of cut out 'x' (cm)

volume (cm³)

.8

61.568

.81

61.568564

.82

61.560672

.83

61.544348

.84

61.519616

.85

61.4865

.86

61.445024

.87

61.395212

.88

61.337088

.89

61.270676

.9

61.196

The highest volume for this rectangular sheet of card is 161.568564cm³ when 'X' is to two decimal places being 1.81cm. I then divided 8 by 1.81 and got the answer 4.419889503. This is very similar to the answer for rectangular sheet 18x6 so this is proof that for rectangles with ratio 1:3 the cut out size is about 1/4 when rounded down from 4.4....

I now have results for rectangles with 3 different ratios and I have created this table:

ratio

cut out size if width=1

fraction

:1

0.16666

1/6

:2

0.21111

about 1/5

:3

0.22666

about 1/4

I investigated some formulae to find the cut out size easily with less calculation.

This rectangular problem can be solved algebraically:

Area=(KW-2X)(W-2X)

Volume=(KW-2X)(W-2X)xX

=(KW²-2KWX-2XW-4X²)xX

=KW²X-2KWX²-2X²W-4X³

Gradient=KW²-4KWX-4XW+12X²

0=12X²-4WX(K+L)+KW²

A=12

B=-4W(K+1)

C=KW²

General Formula:

X=-B²± B²-4AC

2A

X=4W(K+1) ± -(4W(K+1))²-48KW²

24

X=4W(K+1) ± 16W²(K+1)(K+1)-48KW²

24

X=4W(K+1) ± 16W²(K²+2K+1)-48KW²

24

X=4W(K+1) ± 16K²W²+32KW²+16W²-48KW²

24

X=4W(K+1) ± 16K²W²-16KW²+16W²

24

X=4W(K+1) ± 16W²(K²-K+1)

24

X=W(K+1) ± 4W K²-K+1

24

X=W(K+1) ± W K²-K+1

6

X=cut out size

W=Width

K=Integer that is multiplied by 'W' to get the length.

This formula can be used to calculate the maximum cut out section of any size rectangular sheet of card, in order to obtain the maximum volume.