SL Math IA Draft 1
How can differential calculus and optimisation be used to determine the minimum dimensions for differently shaped packagings of potato chips ?
Introduction:
Product packaging plays a salient role in influencing consumer decisions in purchasing a firm's products. This is because a product's appearance communicates many values such as the quality of the product and the company's values. Packaging is as important as the product itself as it serves as a crucial marketing and communicating tool. Since young, I have always been drawn to design my own utilities like my wardrobe cabinet and the packaging of the gifts that I hand out. Additionally, the thought of earning money on my own and what it takes to manage a business has been on my mind ever since I started engaging in secondhand businesses through a website called carousell.
Rationale
My inspiration for this investigation arose after showcasing special interests in designing and having to engage with small business dealings that sparked my interest in starting a business in the future. As such, I decided to explore the product packaging of food products that I am exposed to in day to day life since it was the most thought of commodity. Extensive research led me to find out that a lot of common household product packaging use excessive materials which causes difficulty when disposing them. It is said that the greater the number of materials used in packaging, the harder it is for recycling machines to separate them. I came across potato chips and decided to explore the math behind optimising the dimensions of packaging them.
Aim:
As a result, the main aim of my investigation is to design the product packaging for potato chips by using the least materials by minimising the surface area of the packagings.This is to ensure that lesser materials are being utilised in making the packaging and hence, reduce the negative impacts this has on the environment due to the aforementioned factors. In doing this, I will be employing differential calculus to analyze the dimensions of 3 different product packaging shapes and hence justify which option is the best.
Assumptions:
Assuming I am designing the product packaging of my potato chips brand, I would first have a few things in mind. Firstly, I would definitely be profit motivated but at the same time want the best for my customers. More importantly, assuming that my overall product is environmentally friendly, I would want to minimise the surface area of 3 different potato chip packaging, by employing optimization and calculus to consider the packaging design of the potato chips. With the aid of a fixed volume in differentiating the surface areas, I would also be able to determine the cost of producing each packaging. In order to account for the cost of materials, I shall assume that the packaging would cost $0.015/cm2, thus determining which type of packaging would give the smallest price. The volume I am assuming each packaging should take up is fixed to be 1000 cm3. Lastly, I shall assume that my potato chips are flat in shape so as to minimise space.
Measurements
In order to calculate the dimensions of the different packaging, I have to first create and examine three types of product structures for the potato chip packaging.
A box with a square base
A rectangular based box with ratio dimensions of 2 x 1
A cylindrical tubing with a circular cross sectional area
Let the variable SA denote surface area, V denote volume and C the total amount of money used to design these packaging. Let the variables for respective structures, a denote length , r denote radius and b denote height.
Potato Chip Packaging with a square base
The formula for volume of a square based box would be
V= length x base x height
= a2b
1000 = a2b
b=1000x2
The total surface area would be,
SA= 2a2+4ab
= 2a2+4a1000b2)
= 2a2+4000aa2
= 2a2+4000a
The graph below illustrates the function of the SA, 2a2+4000a
Since a represents the length, x > 0
The shape of the graph is one of a minimum.
Using differentiation , the minimum surface area obtained is as follows
SA=2a2+4000a
SA=2a2 +4000a1
dAda=4a4000a2
dAda=4a 4000a2
Since at dAda=0, the minimum value is obtained,
dAda=4a 4000a2
4a 4000a2=0
4a3=4000
a3=1000
a=10 cm
Therefore, the height of the chips box would also be 10 cm since
b =1000a2
= 1000102
= 10 cm
Total SA=2(10)2+400010
= 600cm2
Using the second derivative test,
d2Ada2 = 4+8000a3
At a =10 cm,
d2Ada2 = 4 +80001000
= 12
As d2Ada2 > 0, the point is a relative minimum.
Total cost of surface material = 0.015 600
= $9.00
Potato Chip Packaging with rectangular based box with ratio dimensions of 2 x 1
The volume of the rectangular box would be
V=2a2b
1000 = 2a2b
b= 500a2
Substituting this into the SA formula, the surface area will be given as
SA= 4a2+6a(500a2)
= 4a2+3000a
The graph of the function would thus be,
Once again, it can be seen that when SA =0, a minimum point is obtained for the graph.
The minimum point of a is determined by differentiating SA.
SA= 4a2+3000a
dAda=8a 3000a2
8a3000a2=0
8a = 3000a2
8a3=3000
a3 =375
a=3375
a=7.21 cm (3.sf)
Therefore, the height of the packaging would be
b =500a2
=500(7.21)2
= 9.61 cm (3.sf)
Total SA= 4(7.21)2+30007.21
= 624 cm2(3.sf)
Using the second derivative,
d2Ada2=8+6000a3
= 8 +60007.213
= 24.0
As d2Ada2 > 0, the point is a relative minimum.
Total Cost of Surface Material
= 624 0.015
= $ 9.36
Potato Chip Packaging with a cylindrical tubing with a circular cross sectional area
Volume = r2b
The height written in terms of volume is as follows:
V=r2b
b = 1000r2
The total surface area of a cylinder is
SA=2rb +2r2
= 2r(1000r2)+ 2r2
= 2000r+2r2
The graph of the surface area of a cylinder is illustrated below
The minimum shape of the graph is obtained when dAdr=0
The minimum point of x is determined by differentiating SA.
SA=2000r+2r2
dAdr=2000r2+4r
2000r2+4r=0
4r3=2000
r3=500
r = 5.42 cm (3.sf)
Therefore, the height of the packaging would be,
b=1000r2
=1000(5.42)2
= 10.8 cm (3.sf)
Total SA=20005.42+2(5.42)2
= 554 cm2(3.sf)
Using the second derivative,
d2Adx2=4000r3+4
= 4+40005.423
= 37.7 (3.sf)
As d2Adx2 > 0, the point is a relative minimum.
Total Cost of Surface Material = 0.015 554
= $8.31
For all three packaging structures, a local minimum value is obtained respectively and the value for a is a minimum value as confirmed by the diagram as well as the second derivative test.
Comparison of all 3 packaging models
Type of Packaging
Length/cm
Width/cm
Height/cm
Surface Area/ cm2
Total Cost of Surface Material /$
Square
10
10
10
600
9.00
Rectangular
14.44
7.21
9.61
624
9.36
Cylindrical

10.84
10.8
554
8.31
From the above, it seems like the cheapest option would be a cylindrical packaging which is rather favourable since it lowers the cost of production. However, this means that there would be the least space to fit my potato chips, which may hinder my customer's ability to dig out the chips from the bottom of the tubing. This may pose as an inconvenience to my customers, making them have an unpleasant experience when consuming the chips. Although the square packaging is more expensive, the surface area is slightly bigger and this gives rise to greater space capacity. I most likely would not consider the rectangular packaging because of the high material cost. Being profit motivated, it is unlikely for me to choose a packaging that adds additional cost to my product.
Extension
A possible extension is to utilise implicit differentiation to find the minimum value, using the case of the cylinder packaging. Let Vbe Volume, h be Height and rbe radius.
Using this volume formula, h may be determined as Vr2and hence substituted into the equation for surface area.
SA=2rh +2r2
ddrSA=2h +2rdhdr+4r
Using implicit differentiation on the formula for volume,
V=r2h
dVdr=dhdrr2+2rh
dhdrr2+2rh=0
r2dhdr=2rh
dhdr=2 hr
Substituting this formula into ddrSA ,
ddrSA = 2h + 2r (2hr )+ 4r
= 2h +4r
ddrSA = 0 when
h = 2r
V=(r2)(2r)
= 2r3
r = 3V2
By utilising the volume formula to find r in terms of V, we see that in the case h = 0,
SA = 2rh +2r2
= 2(3V2)2
= 2 (3V2)2
= 32v
When h=2r= 3V22
SA=2rh +2r2
= 2r(2r) +2r2
= 6r2
= 6(3V2)2
= 6(3v242)
= 354v2
The minimum surface area will occur at one of the critical points. The domain of this function will be (0,∞). However, the radius cannot be 0 because this would mean that the cylinder will be so tall that it becomes nonexistent and would not be able to store any of the chips. On the hand, r also cannot be ∞ because this would mean the cylinder would be too flat to have a height and once again unable to store the chips. Therefore, it would have to take place at the minimum point.
Therefore, as mentioned above, the solution in which h=2r has a minimum surface area where ddrSA =0.
Checking with a packaging volume of a 1000 cm3, the radius of the cylinder would be
r = 3V2
= 310002
= 5.42 cm (3.sf)
Therefore, the surface area would be given as,
SA=354(1000)2
= 554cm2 (3.sf)
Conclusion :
In deciding which product packaging shape to choose, I needed to know which is more important to me — consumer's preference and interest, amount of space available for the chips, or the lowest cost? If I want to maximise consumers preference and interest, I would likely consider the square boxed packaging because it gives the chips enough space and makes it easier to eat the chips, although it is the second most costly to make. The benefits of the square packaging largely takes concerns the the consumers and makes it easier to construct the packaging since the dimensions are the same. However,if I take the cylindrical packaging with the smallest surface area, I would pay the least amount but have a lack of space. This may pose as a problem because the chips may not be able to fill the tube. The rectangular packaging would be the appeal the least because of its high production cost. Since I am producing a packaging design for my product, I would look at how to minimise cost and maximise profits. Hence to make my decision, I will select the square packaging with a slightly larger area and the one with a higher cost as compared to the cylindrical packaging. Although the cost of $9.00 is not the lowest, it provides me with a more feasible packaging for my products and allows me to store my chips without having to worry about the lack of space. Moreover, as mentioned, it is important that I take care of my customers and look at how I package my products from their point of view as consumers.
Lastly, this IA has been a challenging but engaging experience as it allowed me to figure out how much thought is put into packaging products. Previously, I knew nothing about how producers make decisions on branding and packaging their products. In the past, I only knew the use of differential calculus in my math sums but having explored this IA, I was exposed to how people use differentiation in the real world. I believe this IA can be extended by the use of Implicit Differentiation. In my IA, I only focused on using explicit differentiation by setting a fixed variable. Implicit differentiation is an alternative method used to find the maximum or minimum value of a function and this may be an interesting avenue for further research on optimising product packaging by forming a new equation.
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