Soft-beverage is something that we all have liked since we were young. Since childhood, I have always been interested in the work done by my father. Being a bottle manufacturer for soft-beverage companies, he always had to search for bottles with maximum volume and minimum manufacturing cost. It is important to know how the companies work to produce containers having the maximum volume with the least production cost as this has a huge impact on the other factors involved in manufacturing. Decrease in production cost can also increase income and can further help the economy. I would thus, embark upon deeper investigation of factors affecting cost of bottle manufacturing. I subsequently learned about how the surface area of a product like a bottle can affect its production cost. Hence from an early age, I was encouraged to extrapolate beyond my academic scope to know about the factors affecting manufacturing cost and when I learnt about calculus in school and the concept of optimization, I realized that I could indeed find a bottle shape that would incur minimum manufacturing cost. According to me, education should have a perfect balance between practicality and theories. Being an inquirer and a student of IB, I have always learnt to apply theoretical knowledge in real life in order to achieve a practical goal. I have also been inspired by &nbsp;the renowned beverage company &ndash; PepsiCo &ndash; and have always wanted to know how they optimize their products. I have seen several videos and read several articles to know more about this topic. However, I have not been able to find the exact reasoning behind this so, this has motivated me to reach the aim of this exploration.

My investigation aims at optimization of soft-beverage bottle such that it occupies maximum volume but incurs minimum manufacturing cost.

Soft-beverage

A carbonated or non-carbonated beverage generally obtained from fruit concentrates, is termed as a soft-beverage. Soft beverages are packed in various containers ranging from Plastic jars and PET bottles to cardboard tetra packs.

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Bottles

Bottles are containers used to store liquid contents which are generally enclosed by a lid and have a narrow neck with a wide base. Bottles might be made of glass, polyethylene terephthalate (PET) or plastic. However, in this exploration, three PET soft-beverage bottles made by PepsiCo have been taken and considered. To find bottle shape with maximum capacity but minimum production cost, three bottles of identical volume (=750 mL) have been taken. We thus, refer to these distinct shapes as Type 1, Type 2, Type 3, respectively.

type 1 pepsico bottle.svg

type 2 pepsico.svg

type 3.svg

Exploration Methodology

In this investigation, three bottles of equal volume were taken to analyze differences arising between surface areas due to the bottle constitution and shape. First the images were placed in Desmos software, functions&nbsp;marked representing different parts of the bottle&rsquo;s surface, and finally the area under curve integrated using a calculator software such that it yields area of figure during revolution about X-axis. Finally the summation of areas was scaled by relating units assumed in place of actual height (in cm). This helped in the proposal of a final model that incurs minimum production cost as it has minimum area but offers the same volume as the others. This ideal bottle was then mapped on Desmos and its area was found out resulting in the justification of the model proposed.

Using Desmos web application, a function was drawn for each of the surface of the soft-beverage bottles in parts and accordingly integrated within proper limits to obtain surface area.

Firstly, the image of a bottle is placed in the Desmos software and the surface of the bottle is traced and the equation of the curve is obtained. Then the curved part is assumed to be infinitesimally divided into smaller and smaller parts such that the curved edge appears to be comparatively straight. However whenever the figure is rotated about the X-axis, each of these infinitesimally small parts form a circle and the summation of the circumferences of the circles yields the required area. Each of the radii, can be obtained from the integration of the region below the infinitesimal parts.

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\(\text{Surface area }= \sum\limits_{i=1}n^2\pi\phi(\lambda_i^{**})\Delta\lambda\sqrt{1+\phi'(\lambda_i^*)^2}\)

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Where&nbsp;&lambda;i* and &lambda;i** are in the interval

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\(\biggl[{λ_{i-1}},λ_{i}\biggl{]}\)

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\(∴ Surface area = \displaystyle\int\limits^{v}_{\mu}(2πϕ(\lambda)\sqrt{(1+ϕ'(\lambda)^2)}\)

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In the calculations, the working formula used is thus --

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\(Α = 2π\displaystyle\int\limits^{v}_{\mu}ϕ(\lambda)\sqrt{1+[ϕ'(\lambda)]^2} dλ\)&hellip;equation (1)

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Where A is the surface area of the figure formed by rotation of function f(x) about the X-axis.

parts of type 1 bottle shaded with numbers.svg

Part 1 -

The function demarcating Part 1 of the bottle, is given by -

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y = 0.07x + 0.88

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(\displaystyle\int\limits^{-3.99}_{-4.7}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y=\frac{d}{dx}(0.07x+0.88) = 0.07\)

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\(Α = 2π \displaystyle\int\limits^{-3.99}_{-4.7} (0.07x +0.88)\sqrt{1+[0.07]^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 0. 4099

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&Alpha; = 2. 575 sq. units

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The solution is given below -

Part 2 -

The function demarcating Part 2 of the bottle, is given by -

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y = 0.5

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(Α = 2π \displaystyle\int\limits^{-3.7}_{-3.9}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y' = \frac{d}{dx}(0.5)=0\)

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\(∴ Α = 2π \displaystyle\int\limits^{-3.7}_{-3.9}0.5\sqrt{1+(0)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 0.1

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&Alpha; = 0. 6283 sq. units

The solution is given below -

calculation of surface area of part 1.svg

calculation of surface area of part 2.svg

Part 3 -

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The function demarcating Part 3 of the bottle, is given by -

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y = - 0. 23 x3 - 2. 33 x2 - 7. 09 x - 5. 43

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(A = 2π\displaystyle\int\limits^{-2.04}_{-3.7}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y'=\frac{d}{dx}(-0.23x^3-2.33x^2-7.09x-5.43)=-0.69x^{2}-4.66x - 7.09\)

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\(A = 2π \displaystyle\int\limits^{-2.04}_{-3.7}(- 0. 23x3 - 2. 33x2 - 7. 09x - 5. 43)\sqrt{1+(-0.\,69x^2-4.\,66x-7.\,09)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 2. 0266

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&Alpha; = 12. 7335sq. units

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The solution is given below -

calculation of surface area of part 3.svg

Part 4 -

The function demarcating Part 4 of the bottle, is given by -

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y = 1.3

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(Α = 2π\displaystyle\int\limits^{0.5}_{-2.04}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y' = \frac{d}{dx} (1.3) = 0\)

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\(Α = 2π  \displaystyle\int\limits^{0.5}_{-2.04}1.3\sqrt{1+(0)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 3. 302

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&Alpha; = 20. 747 sq. units

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The solution is given below -

calculation of surface area of part-four.svg

Part 5 -

The function demarcating Part 5 of the bottle, is given by -

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y = - 0. 3x + 1. 45

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(Α = 2π\displaystyle\int\limits^{1.25}_{0.5}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y'=\frac{d}{dx} (- 0.3x + 1. 45) = - 0. 3\)

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\(Α = 2π \displaystyle\int\limits^{1.25}_{0.5}( - 0.3x + 1.45)\sqrt{1+( - 0.3)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 0. 9299

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&Alpha; = 5. 8427 sq. units

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The solution is given below -

calculation of surface area of part 5.svg

Part 6 -

The function demarcating Part 6 of the bottle, is given by -

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y = 0. 12x + 0. 92

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(A = 2π \ \displaystyle\int\limits^{2}_{1.25}​y​​ \sqrt{1+[y']^2}dx\)

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Here,

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\(y' = \frac{d}{dx}(0.12x + 0.92) = 0.12\)

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\(A = 2π \ \displaystyle\int\limits^{2}_{1.25}(0. 12x + 0. 92)\sqrt{1+(0.12)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 0. 8422

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&Alpha; = 5. 2917 sq. units

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The solution is given below -

calculation of surface area of part 6.svg

Part 7 -

The function demarcating Part 7 of the bottle, is given by -

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y = - 0. 1x + 1. 36

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Integrating the function as per equation (1), within the appropriate limits, we get -

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\(A = 2π \displaystyle\int\limits^{2.89}_{2}y\sqrt{1+[y']^2}dx\)

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Here,

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\(y'=\frac{d}{dx} (- 0. 1x + 1. 36) = - 0. 1\)

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\(A = 2π\displaystyle\int\limits^{2.89}_{2}(-0.1x+1.36)\sqrt{1+(-0.1)^2}dx\)

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The above integral has been solved using a calculator and the result thus obtained is -

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&Alpha; = 2&pi; &times; 0. 9977

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&Alpha; = 6. 2687 sq. units

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The solution is given below -

calculation of surface area of part 7.svg

Optimization Of Soft-Beverage Bottle Such That It Occupies Maximum Volume But Incurs Minimum Manufacturing Cost.

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Ilan Grapel

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