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• Level: GCSE
• Subject: Maths
• Word count: 2540

# Boxed In.

Extracts from this document...

Introduction

Boxed In Coursework

Paul Mitchell

Mathematics Coursework

Mathematics Coursework

“Boxed In”

For my coursework I have been given a sheet of metal measuring 20cm x 20cm.  I have been asked to produce a box without a lid.  By cutting one centimetre squares from the corners of the sheet, I will be able to fold the corners up and create the box.

I am going to investigate further by cutting 2cm – 9cm squares off the corner of each box, and look at the results to find the biggest volume.

• 2cm Squares:

Volume = 162 x 2 = 512cm3

• 3cm Squares:

Volume = 142 x 3 = 588cm3

• 4cm Squares:

Volume = 122 x 4 = 576cm3

• 5cm Squares:

Volume = 102 x 5 = 500cm3

• 6cm Squares:

Middle

Volume = 82 x 6 = 384cm3

• 7cm Squares:

• 8cm Squares:

Volume = 42 x 8 = 126cm3

• 9cm Squares:

Volume = 22 x 9 = 36cm3

From my results I can see that the largest volume is 588cm3 from the cut out of 3cm.

I will now plot my results onto a graph.

From my graph, I can see that the maximum volume is at its highest between the cut out of 3 and 4cm.

I will now use trial and improvement method to work out the maximum volume.  I will work between the cut outs 3 and 4 centimetres.

To get the Length/Breath of the box; to work out the volume in my trial and improvement table; I will double the cut out number and subtract it from 20.

E.g. Cut out = 3.1

20 – (3.1 + 3.1) = Length/Breath

I will then take the length and square it (breath is the same as the length) and multiply it by the height; which is the size of the cut out.

E.g. 13.82 x 3.1 = 590.36 (the total volume of the box)

Therefore the whole calculation, given that (n = cut out      and      L = length) is…

Step 1: 20-(n + n) = L

Step 2: L2 x n = Volume

• Trial and Improvement to one decimal place:

3.1                             20 – (3.1 + 3.1) = 13.8

(13.8 x 13.8) x 3.1                          590.36 cm3

1. 20 – (3.2 + 3.2) = 13.6

(13.6 x 13.6) x 3.2                          591.87 cm3

1. 20 – (3.3 + 3.3) = 13.4

(13.4 x 13.4) x 3.3                          595.55 cm3

1. 20 – (3.4 + 3.4) = 13.2

(13.2 x 13.2) x 3.4                          592.42 cm3

1. 20 – (3.5 + 3.5) = 13

(13 x 13) x 3.5                                  591.5 cm3

From this table I can see that the cut out of 3.3cm gives the maximum volume.

I will now investigate further, and work out a cut out of between 3 and 4cm to two decimal places.

• Trial and Improvement to two decimal places:

1. 20 – (3.31 + 3.31) = 13.38

(13.38 x 13.38) x 3.31                       592.570 cm3

1. 20 – (3.32 + 3.32) = 13.36

(13.36 x 13.36) x 3.32                       592.585 cm3

1. 20 – (3.33 + 3.33) = 13.34

(13.34 x 13.34) x 3.33                       592.592 cm3

1. 20 – (3.34 + 3.34) = 13.32

(13.

Conclusion

I will now look at the 2.8cm cut out to 2 decimal places:

• Trial & Improvement to 2 decimal places:

I can see from my trial and improvement, that a cut out of 2.83cm yields a box with the largest size.  I will now create a formula for the box; ‘Y’ x ‘Z’ x ‘X’ and see what value I get for ‘x’.

I will now create my formula for the largest volume of the rectangle.

V = (20 – 2z) (15 – 2z) z

## V = (300 – 40z – 30z + 4z2) z                      V = (4z2 – 70z + 300)z

V = 4z3 – 70z2 + 300z

I will now differentiate the above equation to find the dv

dx

= 4z × 33-1 – 70z × 22-1 + 300z × 11-1                            = 12z2 – 140z + 300

At the turning point, the gradient is zero, so dy = 0. So substituting into my formula, gives:                                    dx

12z2 – 140z + 300 = 0                                                  3z2 – 35z + 75 = 0

• I will now use the ‘Quadratic Formula’ to get 2 values for ‘x’

ax2 + bx + c

V = 3z2 – 35z + 75                       (a = 3, b = -35 & C = 75)

## V = 8.83    or    V = 2.83

This proves that the largest volume for the rectangular box would come from a cut out of 2.83cm from each corner.  Now that I have the equation: (20 – 2z)(15 – 2z)z  I can substitute ‘x’ and ‘y’ into it to get a general formula for the biggest volume.

V = (x – 2z) (y – 2z) z

V = (4z2 – 2zx – 2zy - xy) z

V = 4z3 – 2z2x – 2z2y – xyz

V = 4z3 – 2z2(x + y) + 2xy

V = 4z3 – (x + y) 2z2 + xyz

= 4 × 33-1 – (x + y) 2 × 22-1 + xy × 1z1-1

= 12z2 – (x + y) 4z + xyz0

= 12z2 – (x + y) 4z + xy

Maths Coursework                Paul Mitchell

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