• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month
  1. 1
  2. 2
  3. 3
  4. 4
  5. 5
  6. 6
  7. 7
  8. 8
  9. 9
  10. 10
  11. 11
  12. 12
  13. 13
  14. 14

Coding and Modelling - The tools used in my spreadsheets.

Extracts from this document...


Coding and Modellingimage00.pngimage01.png

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:



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


...read more.


                15 cm                                                         (15 cm – 2x)

                                       x cmimage11.png

                                        (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


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


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.

...read more.


2 – 4ac             hence x = - (-140) + or - √ (140)2 – 4 (300)(12)

               2a                                                     2(12)  

x = 140 + or - √ 19600 - 14400


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.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics 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

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    The Gradient Function

    5 star(s)

    nx^n-1 works for fractional powers, and not just when x = 1/2. y=x1/4 x y Second point x second point y gradient 3 1.316074 3.1 1.326906811 0.143409 3 1.316074 3.01 1.317169373 0.14528 3 1.316074 3.001 1.316183672 0.145471 4 1.414214 4.1 1.422970721 0.123786 4 1.414214 4.01 1.415096618 0.125045 4 1.414214 4.001

  2. Numerical integration can be described as set of algorithms for calculating the numerical value ...

    The above theory was vital in my investigation. Finding the value of M524288 under the curve is practically impossible, unless extrapolated estimates are used. This can be explained using table 1.0 below: Mn Values of M Difference Ratio of Differences M1 0.9582363067372210 0.0187590424355787 0.2634168677131520 M2 0.9394772643016420 0.0049414481996782 0.2550173797225570 M4 0.9345358161019640 0.0012601551719167 0.2514868156098100 M8 0.9332756609300470 0.0003169124113596 0.2503932714168220 M16 0.9329587485186880

  1. The Gradient Fraction

    If you round the values up and down to their lowest terms and highest terms, they will equal to the following: x=2: 7.82 (round up to 8) the 2 has been multiplied by 4. x=3: 12.41 (round down to 12)

  2. Functions Coursework - A2 Maths

    If x1 is an approximate root of the equation f(x)=0, the Newton-Raphson formula for finding a second approximation x2 from x1 is x2 = x1 - (f(x1)/(f /( x1)) In the case f(x)=x3-7x-4, the Newton-Raphson formula simplifies to: x2 = (2x13+4)/(3x12-7)

  1. Numerical Differentiation

    x y 0 0 1 -0.070373 2 -0.1784121 3 -0.4830918 4 5.0632911 5 0.5107252 Newton-Raphson Method xr+1 = xr - f (xr) f '(xr) y= 4ln|x| - x + 2 dy= -1 + 4 dx x0 1 x0 6 x1 0.666666667 x1 15.50111363 x2 0.724372086 x2 12.08113601 x3 0.727506177 x3

  2. Estimate a consumption function for the UK economy explaining the economic theory and statistical ...

    The function now looks as follows: ct = 0.7273 + 0.9403yt As mentioned earlier, consumers look to their expected lifetime income when deciding expenditure. This theory has been explored and modelled by a number of economists most notably Friedman with his Permanent Income Hypothesis (PIH).

  1. Mathematical Investigation

    Continuation from Part I-III: For the sake of a clear and easier understanding of the sine functions, the reflections of the sine functions will be explained here. Sine functions of y=-sin(x) and y=sin (-x) are investigated: Figure #6: Graphs of the sine functions of y=sin (-x)

  2. The open box problem

    is 1. I will now draw a graph that will also show the maximum amount for x and to verify that x is 1 and that the greatest volume possible is 16. The graph shows also shows that the maximum volume is about 16 and x is 1.

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