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.

                15 cm                                                         (15 cm – 2x)

                                       x cm

                                        (23 cm –2x)

Question 2b)Similarly for a rectangle with a length of 23cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):

= Length * Width * Height

= (23 – 2x) * (15 – 2x) * x

= 345x – 76x2 + 4x3

Essentially differentiating is done by taking an equation:

xn and applying the principle nx(n-1)

Therefore if: y = 345x – 76x2 + 4x3                =>dy= 345 – 152x + 12x2

                                                                            dx

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

x =  -b + or -        √b2 – 4ac             hence x = - (-152) + or - √ (152)2 – 4 (345)(12)

               2a                                                     2(12)  

x = 152 + or - √ 23104 - 16560

                            24

x = 9.704cm (4s.f) and 2.963cm (4s.f)

Clearly the cutout size cannot be 9.704cm (4s.f) since the rectangular box is only 15cm wide. The only other possible cutout dimension would be 2.963cm (4s.f).

Instead of creating a new spreadsheet, I can change values within an old spreadsheet to suit the new condition I want.

For example in the above spreadsheet, I predict that if I change cell B2 to 23 cell B5 and the rest of the column going downwards will change subtracting 1 each time from the value because of the formula =B2-2*A5.

               2a                                                     2(12)  

x = 140 + or - √ 19600 - 14400

                            24

x = 8.838(4s.f) and 2.829cm (4s.f)

I have investigated five different rectangles, the results of which can be seen on the spreadsheet below:

Final Conclusion:

As you can see from the graph and the spreadsheet, my final prediction was right. As the length and width values get closer together, the optimum cutout size gets smaller because the volume size decreases.

I feel that my results have been very accurate because not only did I solve various formulae using differentiation but I also checked the results by constructing charts and graphs which showed me the optimum values. This has led to valid conclusions especially for the square Max Box where I achieved a general formula. I know that my other conclusion is valid because I have tested out the theory by doing lots of examples.

I feel that my results were also relevant because once I had found out that the cutout size affects the volume of the box, I was able to continue further in order to see whether a pattern emerged when investigating how the size of the length affected the cutout size when the width of the box remained constantly at 15cm. Unfortunately, I was unable to find a general formula for the rectangle Max Boxes.