# Coding and Modelling - The tools used in my spreadsheets.

Coding and Modelling

The tools used in my spreadsheets:

• ## Cell outline - to make my grid standout from the page

• Different font styles – to make the column headings standout from the rest of the text
• A centre justification tool – in order to make my spreadsheets look professional
• The graphing feature – enabling me to draw up graphs from my data
• Various mathematical calculations such as multiplying cells and the what if function – allowing me to save time and fully investigate my project

This is seen as an inappropriate graph because it only shows how the height affects the volume of the box when the question asks you how the cutout size affects the volume of a square of side 24cm:

Results:

For a square of size 24cm, the largest volume of box that can be made is 1024cm3, using a square cutout of 4cm.

For a square of size 120cm, the largest volume of box that can be made is 128000cm3, using a square cutout of 20 cm.

## Final Conclusion

In both cases, it would appear that a trend emerges.

## In each case the optimum cutout size is 1/6 of the size of the square sheet

i.e.: A square that is 24cm long has an optimum cutout size of 4cm and a maximum volume of 1024cm3

A square that is 120cm long has an optimum cutout size of 20cm and a maximum volume of 128000cm3

### Proof

A more mathematical approach can also be used to accurately determine the optimum cutout size, using a combination of quadratics and calculus to achieve a general result.

The volume of the box (y) is calculated by:

length * width * height

=(w - 2x)*(w - 2x)*x (where w is the length of the box)

=x(4x2 - 4xw + w2)

=4x3 - 4x2w + w2x

(plotting this equation against y will display volume against    cutout size so long as 0<x<w)

The maximum volume of box is achieved at the peak of the curve, i.e. when the gradient of the curve is equal to 0.

Applying Calculus and differentiating the formula can find the point where the graph has a gradient of 0.

Essentially differentiating is done by taking an equation:

xn and applying the principle nx(n-1)

Therefore if:

y=4x3 - 4x2w + w2x          =>dy=12x2 - 8wx + w2

dx

resulting in a quadratic equation, which can be solved by:

x=  -b + or -        √b2 – 4ac

2a                                        hence x= 8w √64w2 - 48w2

24

= 8w + or - 4w

24

= 4w

24

= w

6

The solution holds true with the trend noticed in the final conclusion.

In order to expand my investigation further, I will now proceed to investigate the largest square cutout size possible, which will enable me to have the biggest volume for a rectangular box of various lengths:

24cm

x cm

15cm                                                    (15cm – 2x)

x cm

(24cm -2x)

Question 2a) For a rectangle with a length of 24cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):

= Length * Width * Height

= (24 – 2x) * (15 – 2x) * x

= 360x – 78x2 + 4x3

Essentially differentiating is done by taking an equation:

xn and applying the principle nx(n-1)

...