• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
• Level: GCSE
• Subject: Maths
• Word count: 1344

# Open box problem

Extracts from this document...

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

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.

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.

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.

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.

This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Number Stairs, Grids and Sequences essays

1. ## Mathematics Coursework: problem solving tasks

3 star(s)

Formulas I have inserted a column labeled terms. 1 represents the first arrangement of tiles, 2 represents the second arrangement of tiles, and so on. n represents the nth term. The common difference is the 2nd term - 1st term.

2. ## Open Box Problem.

The scatter diagram on the next page also shows us that the maximum value is 6. Notice that the shape of the scatter diagram is the same as the two scatter diagrams done for the other two squares.

1. ## Investigate the size of the cut out square, from any square sheet of card, ...

Now, I shall do some calculations to check for any other connections. 864 cm cubed divided by 18cm = 48 - No connection 18cm divided by 864cm cubed = 0.02083333333 - No connection 18cm

2. ## Mathematical Coursework: 3-step stairs

term 20. Thus making it's time consuming. After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes. The criterion of why I haven't explained what B stands for is because I haven't found it yet.

1. ## The Open Box Problem

� = V Is also relevant for the rectangle but A is split into W and L so the formula becomes: X (W - 2X) (L - 2X) = V As no immediate connection between X and V is obvious other means of calculation are necessary.

2. ## the Open Box Problem

My aim is to find the maximum volume. To solve x, I have to use the quadratic formula, since this quadratic equation cannot be factorised. a =12 b =-160 c=400 x = -b +- Vb�-4ac 2a x = 160+- V160�-4*12*400 2*12 x = 160+- V25600-19200 24 x = 160+- V6400

1. ## First Problem,The Open Box Problem

0.2 0.2 0.056 1.5 4cm by 4cm, piece of square card Length of the section (cm) Height of the section (cm) Depth of the section (cm) Width of the section (cm) Volume of the cube (cm3) 0.1 0.1 3.8 3.8 1.444 0.2 0.2 3.6 3.6 2.592 0.3 0.3 3.4 3.4

2. ## Investigate Borders - a fencing problem.

prediction was 28 which is the correct answer, for the number of squares needed for border number 6, and which also proves that the formula is in working order. Diagram of Borders of square: 4x1 Table of results for Borders of square: 4x1 Formula to find the number of squares

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to