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Open box problem

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Introduction

Junaid Iqbal 11B        Mathematics Coursework

The Open Box Problem

Aim: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.

Part 1

I will be investigating the size of the cut out square which makes an open box of the largest volume for any sized square sheet of card.

The volume for any sized square sheet of card will be gained by using the following formula.

  Volume = height × area of base

    V = x × ((L - 2x) × (W - 2x))

X = height of cut out square

L= length of square

W = width of square

I am going to use the following formula to work out the volume of an 8cm by 8cm square sheet of card. I am going to keep the size of card the same while changing the height of the cut out square.

V = x × ((8 - 2x) × (8 - 2x))

SQUARE

Length (L)cm

Width (W)cm

Height (X)cm

Volume cm3 (2dp)

8

8

0

0.00000

8

8

0.25

14.06250

8

8

0.5

24.50000

8

8

0.75

31.68750

8

8

1

36.00000

8

8

1.25

37.81250

8

8

1.5

37.50000

8

8

1.75

35.43750

8

8

2

32.00000

8

8

2.25

27.56250

8

8

2.5

22.50000

8

8

2.75

17.18750

8

8

3

12.00000

8

8

3.25

7.31250

8

8

3.5

3.50000

8

8

3.75

0.93750

8

8

4

0.00000

...read more.

Middle

8

8

1.4

37.85600

8

8

1.45

37.71450

As it can clearly be seen from the table above that biggest volume has found to be 37.92cm3 not 37.81cm3 as shown in the previous table. This isn’t the highest possible volume; therefore I am now going to use calculus to find the maximum possible volume.  

V = x × ((8 - 2x) × (8 - 2x))

V = x × (64 - 16x - 16x + 4x2)

V = x × (64 - 32x + 4x2)

V = 64x - 32x2 + 4x3

If we rearrange this, it gives us:

4x3 - 32x2 + 64x

Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum.

dv   = 12x2 - 64x + 64

dx  

       =  12x2 - 64x + 64

        3

       =   3x2 – 16x + 16

The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve.

image00.png

Using the quadratic formula -b±√b2 – 4ac   :

                                                        2a

       =  16±√162 - 4(3) (16)

         2(3)

       =   16±√256 - 192

                  6

       =   16±√64

                     6                  

       =   16 + 8OR          =  16 - 8

        6             6

       =    4                                 =   4

3

The first value is out of range because it’s greater than half the width of the box.

...read more.

Conclusion

V = x × ((10 - 2x) × (8 - 2x))

V = x × (80 - 20x - 16x + 4x2)

V = x × (80 - 36x + 4x2)

V = 80x - 36x2 + 4x3

If we rearrange this, it gives us:

4x3 - 36x2 + 80x

Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum.

dv   = 12x2 - 72x + 80

dx  

       = 12x2 - 72x + 80

        4

       = 3x2 – 18x + 20

The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve.

image01.png

 Using the quadratic formula  -b±√b2 - 4ac    :

                                                          2a

       =  18±√182 - 4(3) (20)

         2(3)

       =   18±√324 - 240

                  6

       =   18±√80

                     6                  

       =   18 +√80OR          =  18 -√80

        6                 6

       =    4.49 (2dp)                      =   1.51 (2dp)

The first value is out of range because it’s greater than half the width of the box. So the second value must give the maximum height for the volume.

Now I am going to substitute it back into the equation.

1.51 × ((10 - 2 × (1.51) × (8 - 2 × (1.51)))

= 1.51 × ((6.98) × (4.98))

= 1.51 × 34.77

= 52.50 (2dp)

This is the highest possible volume for a 10 by 8cm rectangular sheet of card. This value could have also been found using the first method, but it would have taken lot more time.

...read more.

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