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• Level: GCSE
• Subject: Maths
• Word count: 1383

# The Open Box Problem.

Extracts from this document...

Introduction

Chloe Bell

Mathematics Course Work

The Open Box Problem

I have to find the maximum volume that can be possibly obtained from a piece of rectangular

card with an area of 240cm2. The card must be cut and folded to form an open box.

In order to determine the best possible shaped cuboid to maximise the volume, it is necessary

to consider a large range of dimensioned rectangles with an area of

240cm2.  I have tabulated these in table 1.  In my initial investigation I am using whole numbers

in multiples of 1cm.

Table of Rectangles which can be constructed from a 240cm2 piece of Card

240cm2

Length (cm)

240

1

120

2

80

3

60

4

48

5

40

6

30

8

24

10

20

12

16

15 *

15

16 *

Table 1

* Note: The table has been stopped at 15 and 16 as it starts to repeat its self

In order to construct an open box it is necessary to cut out squares from the corners of the

end of the card to allow the box to be formed properly.  The size of these squares determines

the height of the box.  I need to do this for different sized squares at the corners of each

rectangle to determine the largest volume.

Middle

15

270

Table 3

From the results in Table 3 a pattern is emerging which indicates that the volume increases as

the shape becomes more square i.e. as the length and breadth become the same.  To test this,

I have decided to calculate the volume for a perfect square at varying heights.

Perfect Square

Fig 4

I have generated a second formula to aid calculating volume, which takes account of Length &

cm = 15.492cm

Volume = L x B x H

Volume = (L - 2H) 2 x H

Table 4 shows the volumes of varying cuboids with equal lengths, breadths with different

heights. It is not necessary to carry on with calculations below 7.492cm as the volume starts to

decrease.  Full table of results can be found in appendices 2 and a graphically representation

of table 4 can be found in appendices 5

Length (cm)

Height (cm)

Volume (cm3)

13.492

13.492

1

182.032

11.492

11.492

2

264.129

9.493

9.493

3

270.290

7.492

7.492

4

224.516

Table 4

The largest volume obtainable falls between 2 and 4cm height.  I suspect this is not the largest

possible volume since I have only dealt with whole numbers (integers) so far.  I have therefore

carried out a 2 point decimal search in 0.01cm increments.

Conclusion

differentiated to support my findings.  Fig 6 shows the dimensions that I will consider.

Fig 6

Card = a * b = A        or  = a         and         Box = xyh

Volume = xyh

V = (a - 2h) (b - 2h) h

V = abh - 2h2(a + b) + 4h3

V = Ah - 2h2 ( + b) + 4h3

To maximize volume differentiate with respect to b (Variable in the equation) assuming h is

kept constant

(Volume)         =  {A- 4h ( + b) + 12h2}

= 4hA - 4h

b2

For Maximum Volume set equal to zero (Gradient at the top of the graph is 0, represents

maximum volume)

4hA - 4h = 0

b2

4hA - 4h = 0          (Divide both sides by 4h)   = A - 1 = 0   or    A=b2

b2                                                        b2

Therefore b=

Since A = ab then a =  =  =  = b  (A square piece of paper gives best result)

Therefore Substitute finding into Volume = abh - 2h2(a + b) + 4h3

Therefore

V = Ah - 2h2 (2 ) + 4h3

V = 4h3 - 4 h2 + Ah

Differentiate with respect to h to find best height, assuming a constant value of A,

(Vol.) = 12h2 - 8 h + A

Set to zero for best h

12h2 - 8 h + A = 0

h =

2*12

h =   *

=

24

2  *

6

h =      or        (Impossible to construct a box from this)

6                   2

Therefore

Length of sides         a = b =         and         Height  =

4

This student written piece of work is one of many that can be found in our GCSE Height and Weight of Pupils and other Mayfield High School investigations section.

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