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With the growing trials at eliminating extra packaging and minimizing the cost of production, of the ubiquitous variances in package designs, perhaps a cylindrical one is the most relevant. Moreover, I was born and brought up in a family of packaging designers. My grandfather owns a company that manufactures bottles and cylindrical packages for water bottles, soft drinks, paper holders, and so on. I thus enquired him about the factors he keeps in mind while manufacturing his products. He said that two equivalent factors are indispensable when it comes to the manufacture of cylindrical products and bottles- first the cost and second the aesthetic sense. Furthermore, bottles as cylindrical packages are more commonly available in terms of production and sale. I thus embarked upon investigating to identify and optimize the bottle shape that will have minimum cost and subsequently minimum surface Area, without losing its aesthetic sense. Other factors have not been considered as the size and volume depending upon the requirement and the material cannot be changed without incurring a loss of quality which is unacceptable. I thus, for the sake of convenience, considered three bottles of the identical volume of popular designs available in the market, retailed by the reverent Pepsi Co. , and thus found out the optimized shape by considering the surface Area to be minimum. I hence stated my research topic as follows: Investigating optimized dimensions for cylindrical packages in a manner such that containers’ environmental costs do not outweigh their aesthetic value.

Bottles are containers for collecting and storing beverages, liquids, or similar items, usually enclosed by a cap. They can be glass, PET, plastics, paper cardboard, etc. But, in this investigation, 3 PET beverage bottles retailed by PepsiCo have been taken into account and considered. To optimize bottle shape with minimum cost, three bottles of identical volume have been taken, each being 750 mL. These bottles are henceforth

In this exploration, firstly image of a bottle has been placed in the application Desmos and equations for the surfaces of the bottles were obtained. Then the Area under curve around the surface of the bottle was founded by calculation using integrating within proper limits. The summation of partial Areas gave the total surface area of the bottle. These values were scaled to the real-life prototypes by the unitary method and compared to identify the appropriate optimized shape.

the Desmos, was used for drawing a function that was needed (in parts) for every surface of the bottles and was integrated based on proper limits to find out the surface area.

Initially, on the Desmos software, the bottle’s image was positioned and the curve’s equation was found after tracing the bottle’s surface. Next, it was assumed that the curved part would be divided infinitesimally into smaller portions till the curved part can appear to be straight. Yet, when the figure is revolved about the X-axis, each of these infinitesimally minute parts form a circle and the addition of the perimeters of the circles along the height, gives the required total surface area. Each of the respective radii, can be found by the integration of the area below the minute parts.

Surface area =\(\sum\limits_{i=1}n^2πς\biggl(\varrho_i^{**}\biggl) \Delta \varrho \sqrt{1+ ς^{'}\Big(\varrho_i ^*\Big)^2}\)

Where i^{*} and ϱ_{i}^{**} are in the interval \(\biggl[ϱ_{i-1},ϱ_i\biggl]\)

∴ Surface area = \(\displaystyle\int\limits^{η}_{ζ}\{2πς(\varrho)\sqrt{\{1+ς^{'}(\varrho)^2\}}\)

For calculations the following formula has been used:

\(AREA = 2π\displaystyle\int\limits^{η}_{ζ}{ζ}(\varrho)\sqrt{1+\biggl[ς^{'}(\varrho)\biggl]^2}\)dϱ…equation {1}

Where we find surface area of the figure shaped by revolution of function f{ψ} about the X-axis.

**Part 1**

Part 1 of the Bottle is bounded by the function

ϖ = 0.071 ψ + 0.878

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{-4.7}^{-3.99}\)ϖ\(\sqrt{1+[ϖ']^2} \)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\)(0. 071 ψ + 0. 878) = 0. 07

AREA = 2π \(\displaystyle\int\limits_{-4.7}^{-3.99}\{0 .07ψ+0. 88\}\sqrt{1+[0. 07]}^2\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 0. 409854

AREA = 2. 575 sq. units

Below is the calculation and solution

**Part 2**

Part 2 of the Bottle is bounded by the function

ϖ = 0. 501

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{-3.9}^{-3.7}\)ϖ\(\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\)(0. 501) = 0

∴ AREA = 2π \(\displaystyle\int\limits_{-3.9}^{-3.7}\)0.5 \(\sqrt{1+\{0\}^2}\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 0. 1

AREA = 0. 628319 sq. units

Below is the calculation and solution

**Part 3**

Part 3 of the Bottle is bounded by the function

ϖ = - 0. 231ψ^{3} - 2. 332ψ^{2} - 7. 089ψ - 5. 4302

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{-3.7}^{-2.04}\)ϖ \(\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\)\(\biggl(\)= - 0. 2313ψ^{3} - 2. 3322ψ^{2} - 7. 089ψ - 5. 4302\(\biggl)\)= - 0. 69ψ^{2} - 4. 66ψ - 7. 09

AREA = 2π \(\displaystyle\int\limits_{-3.7}^{-2.04}\)\(\{-0.23ψ^3-2.33ψ^2-2.33ψ^2-7.09ψ-5.43\}\sqrt{1+\{-0.69ψ^2-4.66ψ-7.09\}^2} \)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 2. 0266

AREA = 12. 7333sq. units

Below is the calculation and solution

**Part 4**

Part 4 of the Bottle is bounded by the function

ϖ = 1. 3001

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{-2.04}^{0.5}ϖ\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\)(1. 3001) = 0

AREA = 2π \(\displaystyle\int\limits_{-2.04}^{0.5}1.3\sqrt{1+\{0\}^2}\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 3. 302

AREA = 20. 747 sq. units

Below is the calculation and solution

**Part 5**

Part 5 of the Bottle is bounded by the function

ϖ = - 0. 302ψ + 1. 449

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{\,0.5}^{1.25}ϖ\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\) (- 0. 302ψ + 1. 449) = - 0. 3

AREA = 2π\(\displaystyle\int\limits_{\,\,0.5}^{1.25}\{-0.3ψ+1.45\}\sqrt{1+\{-0.3\}^2}\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 0. 92984

AREA = 5. 8427 sq. units

Below is the calculation and solution

**Part 6**

Part 6 of the Bottle is bounded by the function

ϖ = 0. 119ψ + 0. 918

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π \(\displaystyle\int\limits_{\,1.25}^{2}ϖ\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\)(0. 119ψ + 0. 918) = 0. 12

AREA = 2π\(\displaystyle\int\limits_{\,1.\,25}^{2}\{0 .\,12ψ\,+\,0.\,92\}\sqrt{1+\{0.\,12\}^2}\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 0. 842249

AREA = 5. 2917 sq. units

Below is the calculation and solution

**Part 7**

Part 7 of the Bottle is bounded by the function

ϖ = - 0. 099ψ + 1. 356

Within appropriate limits, the function is integrated as per equation (1)

AREA = 2π\(\displaystyle\int\limits_{2}^{2.89}ϖ\sqrt{1+[ϖ']^2}\)dψ

Here,

ϖ^{'} = \(\frac{d}{dψ}\) (- 0. 099ψ + 1. 356) = - 0. 1

AREA = 2π \(\displaystyle\int\limits_{2}^{2.89}\{-0.1ψ+1.36\}\sqrt{1+\{-0.1\}^2}\)dψ

The result obtained after integrating the above using a calculator, is as follows

AREA = 2π × 0. 997747

AREA = 6. 2687 sq. units

Below is the calculation and solution