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• Level: GCSE
• Subject: Maths
• Word count: 2763

# Maths Coursework

Extracts from this document...

Introduction

Maths Coursework

For this maths coursework, I will be investigating the volume of different sized open boxes. I will look at the different sizes of the squares to see which gives the biggest volume. I am going to be using both square and rectangular sheets of card for this task.

Here is a diagram to demonstrate what I will be doing and illustrate the layout of the sheet of card. The parts labelled ‘x’ will be the squares that I will cut out which its height and sizes will be identical as the other cut out squares.

x                                                x

x                                                           x

x                                                           x

x                                                 x

I will now illustrate a diagram of the box after it has been assembled. The sections labelled x is the height of the box.

To work out the volume of the box, I will use the following formula:

Length x Width x Height

For the first part of my coursework, I will be looking at square pieces of card. I will be investigating square pieces of card and study the relationships between the different sized square cut outs. I am going to look for the square cut out which gives the highest volume overall out of all the rectangles.

Middle

360

6

16

2

192

3.5

21

7

514.5

3.6

20.8

6.8

509.184

3.4

21.2

7.2

518.976

3.3

21.4

7.4

522.588

3.2

21.6

7.6

525.312

3.1

21.8

7.8

527.124

3*

22

8

528

2.9*

22.2

8.2

527.916

2.95*

22.1

8.1

528.0795

The square cut out which gave the biggest volume using the formula L x W x H was 2.95 x 2.95.

Secondly, I will look at the rectangle 15cm x 30cm.

 Size of cut out square(Height x) Length(30-2x) Width(15-2x) Volume (cm³)L x W x H 1 28 13 364 2 26 11 572 3* 24 9 648 4* 22 7 616 5 20 5 500 6 18 3 324 7 16 1 112 3.5 23 8 644 3.6 22.8 7.8 640.224 3.4 23.2 8.2 646.816 3.3 23.4 8.4 648.648 3.2* 23.6 8.6 649.472 3.1* 23.8 8.8 649.264 3.15* 23.7 8.7 649.4985

The cut out square which gave the largest volume was 3.15 x 3.15 by using the formula L x W H.

Last of all, I am going to look at 16cm x 32cm.

 Size of cut out square(Height x) Length(32-2x) Width(16-2x) Volume (cm³)L x W x H 1 30 14 420 2 28 12 672 3* 26 10 780 4* 24 8 768 5 22 6 660 6 20 4 480 7 18 2 252 3.5 25 9 787.5 3.6 24.8 8.8 785.664 3.4* 25.2 9.2 788.256 3.3* 25.4 9.4 787.908 3.35 25.3 9.3 788.2215

From using the formula L x W x H to calculate, the square cut out which gave the highest volume was 3.4 x 3.4.

Results Table - Maximum volume- ratio 1:2.

 Size of rectangle Size of cut out square (x) Proportion of rectangle (x÷length) 14 x 28 2.95 0.105357142   0.105 15 x 30 3.15 0.105   0.105 16 x 32 3.4 0.10625   0.105

Looking at my results, I can tell that the best square to cut the corners from is almost 105/1000 of the rectangle’s length.

30cm x 60cm

I predict that the cut out square will be 6.3 x 6.3.

 Size of cut out square (Height x) Length(60-2x) Width(30-2x) Volume (cm³)L x W x H 6.2 47.6 17.6 5194.112 6.3* 47.4 17.4 5195.988 6.4* 47.2 17.2 5195.776 6.35* 47.3 17.3 5196.1415

As I can see from my results table, the cut out square is 6.3 x 6.3; therefore, my prediction was correct.

Conclusion

Y=4x³-240x²+2000x

X    20          -2x

100   2000      -200x

-2x    -40x        4x²

Equation of the curve

= 12x²-480x+2000

a=12

b=-480

c=2000

x=-b± √ b²-4ac

2a

x= 480± √(480)²-4(12)(2000)

2(12)

x= 480± √ 134400

24

So x will equal:

x = 480+√134400                        or                   = 480- √134400

24                                                                24

x = (480+366.6060556)                                    (480-366.6060556)

24                                                             24

=32.27525232                                            =4.724747683

The size of the corner which gives the biggest volume is 4.724747683 or 4.7 (to 2 significant figures); it is unlikely to cut a corner of size 32.3cm from a rectangle which is 20cm x 100cm.

Differentiation for the ratio 1:10

The rectangle I will look at is 8cm x 80cm.

Length= 80-2x

Width= 8-2x

Height = x

Therefore the volume is V = (80 – 2x)(8-2x)x

V = (80-2x)(8x-2x²)

V = 640x – 160x² - 16x² - 4x³

V = 4x³ - 176x² + 640x

Equation of the curve:

dx

dv = 12x² - 352x + 640

a = 12

b = -352

c = 640

x = -b±b²-4ac

2a

x = 352±√(-352)²-4(12)(640)

2(12)

x = 27.38585602           or          = 1.947477314

Differentiation for 1:20

The rectangle I will look at is 5:100

Length = 100-2x

Width = 5-2x

Height = x

So the volume is V = (100-2x)(5-2x)x

V = (100-2x)(5x-2x²)

V = 500x – 200x² - 10x² + 4x³

V = 4x³ - 210x² + 500x

Equation of the curve:

dx

dv = 12x³ - 420x + 500

a= 12

b= -420

c= 500

x = -b±√b²-4ac

2a

x = 420±√-420²-4(12)(500)

2(12)

x = 33.76601775           or                 1.233982253

Conclusion

I am now going to collect my results. I will look at the proportion of both length and width, which gives the maximum volume.

Length:

 W:L Proportion (x÷l) 1:1 1/16 1:2 1/10 1:3 75/1000 1:4 0.0581 1:5 0.047 1:10 0.0243 1:20 0.0123

Width:

 W:L Proportion (x÷l) 1:1 0.16 1:2 0.211 1:3 0.225 1:4 0.232 1:5 0.236 1:10 0.243 1:20 0.246

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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## Here's what a teacher thought of this essay

3 star(s)

***
This is a well structured investigation. It uses high level mathematics to appropriately determine the relationship between length and volume. To improve this the mathematics should be described in more detail and linked to the desired
investigation outcomes. Strengths and improvements have been suggested throughout.

Marked by teacher Cornelia Bruce 18/07/2013

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