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  • Published on: 14th September 2019
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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 second-hand 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 +4000a-1

dAda=4a-4000a-2

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

8a-3000a2=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 non-existent 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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