Maths Studies Project:

What is the most efficient method to decrease the surface area of an octagonal prism, whilst keeping the volume constant?

Maths Coursework
Title
What is the most efficient method to decrease the surface area of anElizabeth Shaw chocolate box, whilst keeping the volume constant?
Introduction- Aim
I want to find out the smallest possible surface area of the octagonal Elizabeth Shaw chocolate box. A standard box has a volume of Xcm³, and has a surface area of Xcm². The volume of the box will have to remain the same throughout the entirety of the experiment.

This is an example of the Elizabeth Shaw chocolate box, as you can see; it is a regular polygon with sides of 6.5cm length and a height of 3cm with a radius of 7.85cm. Minimising the surface area of the box would have a number of advantages; it would save costs of producing the packaging as well as producing fewer pollutants that would derive from producing the packaging, reducing our carbon footprint.
My aim is to use the 3 mathematical techniques- 1. Trial and improvement (using excel)
This will entail finding the minimum surface area by increasing the number of decimal places each time, for x value 2. Autograph (for PC)
I will insert the optimisation equation into the program which will construct a graph, where the lowest point (where the gradient is 0) will identify the minimum surface area obtainable. 3. Algebraic Differentiation
This is where I will
These will be compared and contrasted to show what the right minimisation of surface area is.
Hypothesis
I am assuming that the Autograph will be the most accurate method; the graph will give all possible values for the surface area of the octagonal prism. And the computer will find the minimum point where the gradient equals zero, much better that any method using human participation, which would lead to mistakes.
Getting the Original Volume of the Prism
The volume of an octagonal prism can be determined by the height times the surface area of the octagon which can be found by finding the area of one of the 8 triangles multiplying it by 8.
Area of the triangle
6.5

45° Area: base x height / 2 Height of triangle= 3.25/tan (22.5) = 7.846194078 =7.8462cm (4 s.f.)
6.5 x 7.846194078 / 2 = 25.50013075 =25.5cm² (4 s.f.)

Area of Octagon
25.5cm2
25.5cm2
25.5cm2
25.5cm2
25.5cm2
25.5cm2
25.5cm2
25.5cm2 Area of octagon: area of triangle x 8 Triangle area: 25.5cm²
25.50013075 x 8 = 204.001064 cm²

Total volume of octagon
Surface area of the octagon x height of the octagon
204.001046 x 3 = 612.0031381cm³ - this is to be kept constant throughout the experiment
Getting the equation
X
H
T
To get the equation that would be put into excel I needed to change all the values to X. Changing T to be in terms of x T = X2tan22.5 Tan (22.5) = 0.4142135624 T= X20.4142135624 T= X0.8284271247

Changing H to be in terms of X
The volume of the octagon is the surface area of the face multiplied by the height (H)
= 612.0031381= 0.5×8×x× x0.8284271247* H
(Rearranged formula)
H= 612.0031381×0.82842712474x²
H =126.75X²
Surface area= 2(area of octagon) + 8XH
= 2 (0.5×8 ×x×x0.8284271247) + 8x (126.75x²) = 2 (4x²0.8284271247) + 8(126.75x) =8x²0.8284271247+ 8×126.75x
Surface area equation = 9.6569x² + 1014x

Method 1: Using excel to find minimal surface area using trial and error with different values of X
The equation was placed into the program and the values for the calculated surface area automatically came out.
Surface area values with x= 3 – 6.5cm 3 | These values show that 3.5 to 4 cm show the smallest surface areas, if you go further and another decimal point between the two values, you get more accurate results

424.9121 | 3.5 | 408.0113107 | 4 | 408.0104 | 4.5 | 420.8855583 | 5 | 444.2225 | 5.5 | 476.4848614 | 6 | 516.6484 | 6.5 | 564.004025 |

Surface area values with x= 3 – 4cm 3 | 424.9121 | 3.1 | These values show that 3.7 to 3.8 cm show the smallest surface areas, if you go further and another decimal point between the two values, you get more accurate results

419.8996 | 3.2 | 415.7617 | 3.3 | 412.4364 | 3.4 | 409.8691 | 3.5 | 408.0113 | 3.6 | 406.8201 | 3.7 | 406.257 | 3.8 | 406.2877 | 3.9 | 406.8814 | 4 | 408.0104 |

Surface area values with x= 3.7 – 3.8cm 3.7 | 406.257015054 | 3.71 | These values show that 3.74 to 3.8 cm show the smallest surface areas, if you go further and another decimal point between the two values, you get more accurate results

406.233901171 | 3.72 | 406.216690121 | 3.73 | 406.205349962 | 3.74 | 406.199849092 | 3.75 | 406.200156250 | 3.76 | 406.206240504 | 3.77 | 406.218071251 | 3.78 | 406.235618214 | 3.79 | 406.258851432 | 3.8 | 406.2877413 |
Surface area values with x= 3.74 – 3.75cm 3.74 | 406.199849092 | 3.741 | These values show that 3.744 to 3.745 cm show the smallest surface areas, if you go further and another decimal point between the two values, you get more accurate results

406.199618960 | 3.742 | 406.199446876 | 3.743 | 406.199332811 | 3.744 | 406.199276732 | 3.745 | 406.199278609 | 3.746 | 406.199338411 | 3.747 | 406.199456107 | 3.748 | 406.199631666 | 3.749 | 406.199865057 | 3.75 | 406.200156250 |

Surface area values with x= 3.744 – 3.745cm 3.744 | 406.199276732 | 3.7441 | 406.199274312 | 3.7442 | These values show that 3.74 to 3.8 cm show the smallest surface areas, if you go further and another decimal point between the two values, you get more accurate results

406.199272472 | 3.7443 | 406.199271211 | 3.7444 | 406.199270530 | 3.7445 | 406.199270428 | 3.7446 | 406.199270905 | 3.7447 | 406.199271962 | 3.7448 | 406.199273598 | 3.7449 | 406.199275814 | 3.745 | 406.199278609 |

Surface area values with x= 3.7444 – 3.7445cm 3.7444 | 406.199270530 | 3.74441 | 406.199270493 | 3.74442 | This set of results shows that at with x at 5 decimal places, the smallest surface area is between 406.199270399 and 406.199270397, with x valuing at 3.74446 to 3.74447 3.74446 | 3.74447 |

406.199270463 | 3.74443 | 406.199270438 | 3.74444 | 406.199270419 | 3.74445 | 406.199270406 | 3.74446 | 406.199270399 | 3.74447 | 406.199270397 | 3.74448 | 406.199270402 | 3.74449 | 406.199270412 | 3.7445 | 406.199270428 |

Method 2 – Using Autograph

The diagram above is a graphical representation of what the values of x could be. Where the blue line goes through the x axis, is the minimum surface area for the octagonal prism (the bottom of the red line)

This is a zoomed in version of the graph, it shows that x is between 3.744 and 7.745

Again, this is a zoomed version of the graph above; it shows that x values between 3.74446 and 3.74447 at roughly around 3.744467 as the eye can see.

Method 3 - calculus
Differentiation
Surface area equation - = 9.6569x² + 1014x
This can be re-written as -
= 9.6569x²+ 1014x¯¹
This makes it easier to differentiate. To differentiate you need to multiply the initial co-efficient by the value of the power number, and then take away the value of 1 from that power number.
So when differentiated, the first part of the surface area equation looks like- d(A)dx= 19.3137x
The second part looks like - d(A)dx = -1014x¯²or 1014x²
Nowthe entirety of the equation has been differentiated - d(A)dx=19.3137x-1014x¯2 The value for x can be found my making d(A)dx = 0
0= 19.3137x-1014x¯2
This can be rearranged as -
1014x²=19.3137x
Then you multiply both sides by x² and divide then divide both sides by 19.3137 so that the x is on one side and the values are on the other, this would result in -
101419.3137=x³
The value of x can be found by cube rooting the fraction -
3101419.3137
= 3.74447 (rounded to 5 significant figures)

When the equation is differentiated again and x is placed in the end equation, the minimum or maximum dimensions? can be found, it depends whether x is above or below zero, if above, than it is theminimum, if below than it is themaximum. This can be show below – d2Adx2=19.3137+2028x¯³ (or2028x³)
When x is placed into the equation
19.3137+20283.74447³=57.941
As it is above zero, we know that x is theminimum surface area that it could be.

Validity
Evaluating method 1
Trial and error is not a quick and simple process, and a lot can go wrong. You have to be able with computers and input the correct equation without mislabelling one value, if not all your results could come out wrong. The results were accurate in the end though, this is because excel can go to number of decimal places, which means that the results are more accurate than the autograph method. The results are valid, as long as the optimisation equation is correct, the results should be right.
The only issue was that it did take a long time to do and there are other methods that would take less time and still produce an accurate result. Excel also heavily relies on human participation, so a lot can go wrong at each stage, for example I put in one wrong digit as the 1 decimal point in the equation I used in the program, and as I used the same equation for each phase of the excel, every single result was in fact, wrong to a certain extent. So I had to re-do the whole of the excel method for scratch.
Evaluating method 2
Using Autograph is another effective way of finding the value of x, this is because it give all of the possible values of x, as shown by a curve going through the x-axis, unfortunately, the program does not give the exact figure of x as it crosses the axis so the result is very approximate. It is a quick way of finding x as all you have to do is input the optimisation equation and press ‘enter’. It is a good way to visualise the results.
The problems with Autograph were mainly to do with not being able to measure where x crosses the x-axis, and you do need to zoom in a number of times to see the line with more decimal points.
Evaluating method 3
The method of differentiation is very accurate and needs only a little amount of time to do. With the first stage of differentiation you get the only value of x that shows the perfect value, and with the second stage showing you whether it is in fact the minimum surface area value.
The only limitation of differentiation is that there is a factor of human error possible, in completing the process of differentiation wrong.
Overall evaluation
All of the methods above are valid, mathematically proven methods. The results from each of them match, showing that they are in fact valid. They might only be valid to a certain extent as the original equation was done to only a certain number of decimal places. So the results are only valid to a certain number of decimal places

Conclusion
The aim of the project is to find the minimum surface area for an octagonal prism chocolate box, but still keep the same volume.
The is a strategy often used by corporations in the world, by reducing the surface area of a product, they are effectively reducing the business costs, by reducing the costs of production and transport, as the can make more of the product with the same amount of materials and not add to any atmospheric pollution and reduce transport, as more of them as they take up less space. And as the box is an adequate shape it can also be re-used.
The most effective method to find the best value of x, which would lead to the smallest possible surface area, was the differentiation, which produced a single, exact solution, which had could be read to a large number of decimal places, showing that it was a very accurate. With the only limitation being that the may be some human error, which was counter-acted by the fact that the result was the nearly the same as the results achieved in the other techniques.
The original equation was the surface area –
Surface area= 2(area of octagon) + 8XH
=8×0.5 ×Base ×height of triangle(base2tan22.5) ×8(Height of Prism×base)
=9.6569x² + 1014x
If we substitute x into the equation –
9.6569×3.74447² + 10143.74447
Surface area = 406.19927 | Original value(cm) | Changed value(cm) | x | 6.5 | 3.74447 | t | 7.8462 | 4.51998 | h | 3 | 9.03998 | SA | 564.004 | 406.19927 | (original values were found by reversing the equations used to change them into terms of x)

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[ 1 ]. http://www.google.co.uk/search?gcx=c&q=elizabeth+shaw+chocolate&safe=active&um=1&ie=UTF-8&tbm=isch&source=og&sa=N&hl=en&tab=wi&biw=1366&bih=667
[ 2 ]. REFERENCE TO TEXTBOOK FROM DXO