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Mathematics AA HL
Mathematics AA HL
Sample Internal Assessment
Sample Internal Assessment

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Table of content
Rationale
Background information
Process of calculation
Data Analysis
New design
Conclusion
Evaluation
Bibliography

Investigating optimized dimensions for cylindrical packages in a manner such that containers’ environmental costs do not outweigh their aesthetic value

Investigating optimized dimensions for cylindrical packages in a manner such that containers’ environmental costs do not outweigh their aesthetic value Reading Time
11 mins Read
Investigating optimized dimensions for cylindrical packages in a manner such that containers’ environmental costs do not outweigh their aesthetic value Word Count
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Table of content

Rationale

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.

Background information

Bottles

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

Figure 1 - Type 1 Bottle
Figure 1 - Type 1 Bottle
Figure 2 - Type 2 Bottle
Figure 2 - Type 2 Bottle
Figure 3 - Type 3 Bottle
Figure 3 - Type 3 Bottle

Exploration methodology

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.

Process of calculation

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.

Derivation of formula

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 ϱi** 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.

Type 1

Figure 4 - Type 1 Bottle With Obtained Functions
Figure 4 - Type 1 Bottle With Obtained Functions

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\(\sqrt{1+[ϖ']^2} \)

 

Here,

 

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

 

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

 

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:

Figure 5 - Calculation Of Integral
Figure 5 - Calculation Of Integral

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\(\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)(0. 501) = 0

 

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

 

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:

Figure 6 - Calculation Of Integral
Figure 6 - Calculation Of Integral

Part 3:

 

Part 3 of the Bottle is bounded by the function:

 

ϖ = - 0. 231ψ3 - 2. 332ψ2 - 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 \(\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)\(\biggl(\)= - 0. 2313ψ3 - 2. 3322ψ2 - 7. 089ψ - 5. 4302\(\biggl)\)= - 0. 69ψ2 - 4. 66ψ - 7. 09

 

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

 

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:

Figure 7 - Calculation Of Integral
Figure 7 - Calculation Of Integral

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

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)(1. 3001) = 0

 

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

 

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:

Figure 8 - Calculation Of Integral
Figure 8 - Calculation Of Integral

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:

Figure 9 - Calculation Of Integral
Figure 9 - Calculation Of Integral

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

 

Here,

 

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

 

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

 

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:

Figure 10 - Calculation Of Integral
Figure 10 - Calculation Of Integral

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

 

Here,

 

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

 

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

 

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:

Figure 11 - Calculation Of Integral
Figure 11 - Calculation Of Integral

Part 8:

 

Part 8 of the Bottle is bounded by the function:

 

ϖ = 0. 258ψ + 0. 309

 

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

 

AREA = 2π\(\displaystyle\int\limits_{\,\,2.89}^{3.8}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)(0. 258ψ + 0. 309) = 0. 26

 

AREA = 2π \(\displaystyle\int\limits_{\,\,2.89}^{3.8}\{0.26ψ+0.31\}\sqrt{1+\{0.26\}^2}\)

 

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

 

AREA = 2π × 1. 10922

 

AREA = 6. 9693 sq. units

 

Below is the calculation and solution:

Figure 12 - Calculation Of Integral
Figure 12 - Calculation Of Integral

Part 9:

 

Part 9 of the Bottle is bounded by the function:

 

ϖ = - 0. 511ψ2 + 4. 072ψ - 6. 79

 

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

 

AREA = 2π\(\displaystyle\int\limits_{\,\,3.8}^{4.8}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\biggl(-\,0.511ψ^2+4.072ψ-6.79\biggl)=-1.02ψ+4.07\)

 

AREA = 2π \(\displaystyle\int\limits_{\,\,3.8}^{4.8}\{-\,0.51ψ^2+4.07ψ-6.8\}\sqrt{1+\{-1.02ψ+4.07\}^2}\)

 

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

 

AREA = 2π × 1. 3249

 

AREA = 8. 3246 sq. units

 

Below is the calculation and solution:

Figure 13 - Calculation Of Integral
Figure 13 - Calculation Of Integral

69.382 sq.units is the total surface area of Type 1 Bottle.

 

9.6 units was the assumed height of bottle in Desmos. However in reality, 21.2 is the height of the actual bottle. By the unitary method, we thus write the actual total Surface Area as: 153.219 cm2.

Type 2

Figure 14 - Type 2 Bottle With Obtained Functions
Figure 14 - Type 2 Bottle With Obtained Functions

Part 1:

 

Part 1 of the Bottle is bounded by the function:

 

ϖ = 0. 028ψ + 0. 651

 

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

 

AREA = 2π \(\displaystyle\int\limits_{-4.34}^{-3.58}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ\(\frac{d}{dψ}\) (0. 028ψ + 0. 651) = 0. 03

 

 

AREA = 2π \(\displaystyle\int\limits_{-4.34}^{-3.58}\{0.03ψ+0.65\}\sqrt{1+\{0.03\}^2}\)

 

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

 

AREA = 2π×0. 403894

 

AREA = 2. 5378 sq. units

 

Below is the calculation and solution:

Figure 15 - Calculation Of Integral
Figure 15 - Calculation Of Integral

Part 2:

 

Part 2 of the Bottle is bounded by the function:

 

ϖ = 0. 449

 

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

 

AREA = 2π\(\displaystyle\int\limits_{-3.45}^{-3.3}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\{0.449\}=0\)

 

AREA = 2π \(\displaystyle\int\limits_{-3.45}^{-3.3}\{0.45\}\sqrt{1+\{0\}^2}\)

 

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

 

AREA = 2π × 0. 0675

 

AREA = 0. 4241 sq. units

 

Below is the calculation and solution:

Figure 16 - Calculation Of Integral
Figure 16 - Calculation Of Integral

Part 3:

 

Part 3 of the Bottle is bounded by the function:

 

ϖ = - 0. 231ψ2 - 0. 689ψ + 0. 722

 

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

 

AREA = 2π\(\displaystyle\int\limits_{-3.3}^{-0.62}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ'\(\frac{d}{dψ}\biggl(\)- 0. 231ψ2 - 0. 689ψ + 0. 722\(\biggl)\)= - 0. 46ψ - 0. 69

 

AREA = 2π \(\displaystyle\int\limits_{-3.3}^{-0.62}\{-\,0.23ψ^2-0.69ψ+0.72\}\sqrt{1+\{-\,0.46ψ - 0.69\}^2}\)

 

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

 

AREA = 2π × 2. 99395

 

AREA = 18. 8119 sq. units

 

Below is the calculation and solution:

Figure 17 - Calculation Of Integral
Figure 17 - Calculation Of Integral

Part 4:

 

Part 4 of the Bottle is bounded by the function:

 

ϖ = 1. 078

 

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

 

AREA = 2π \(\displaystyle\int\limits_{-0.62}^{2.15}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)(1.078) = 0

 

AREA = 2π\(\displaystyle\int\limits_{-0.62}^{2.15}1.08\sqrt{1+\{0\}^2}\)

 

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

 

AREA = 2π × 2. 9916 

 

AREA = 18. 7968 sq. units

 

Below is the calculation and solution:

Figure 18 - Calculation Of Integral
Figure 18 - Calculation Of Integral

Part 5:

 

Part 5 of the Bottle is bounded by the function:

 

ϖ = -0. 2212 + 1. 399ψ - 0. 92

 

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

 

AREA = 2π \(\displaystyle\int\limits_{\,\,2.15}^{4.57}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\biggl(\)- 0. 221ψ2 + 1. 399ψ - 0. 92\(\biggl)\)= - 0. 44ψ + 1. 4

 

AREA = 2π \(\displaystyle\int\limits_{\,\,2.15}^{4.57}\{-\,0.22ψ^2+1.4ψ-0.9\}\sqrt{1+\{-\,0.44ψ+1.4\}^2}\)

 

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

 

AREA = 2π × 3. 06435

 

AREA = 19. 2542 sq. units

 

Below is the calculation and solution:

Figure 19 - Calculation Of Integral
Figure 19 - Calculation Of Integral

59.825 sq.units is the total surface area of Type 1 Bottle.

 

8.9 units was the assumed height of bottle in Desmos. However in reality, 19.6 is the height of the actual bottle. By the unitary method, we thus write the actual total Surface Area as: 131.749 cm2

Type 3

Figure 20 - Type 3 Bottle With Obtained Functions
Figure 20 - Type 3 Bottle With Obtained Functions

Part 1:

 

Part 1 of the Bottle is bounded by the function:

 

ϖ = 0. 049ψ + 0. 738

 

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

 

AREA = 2π\(\displaystyle\int\limits_{-4.42}^{-3.78}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{dψ}\)(0. 049ψ + 0. 738) = 0.05

 

AREA = 2π \(\displaystyle\int\limits_{-4.42}^{-3.78}\{0.05ψ\,+\,0.74\}\sqrt{1+\{0.05\}^2}\)

 

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

 

AREA = 2π × 0. 342828

 

AREA = 2.1539 sq. units

 

Below is the calculation and solution:

Figure 21 - Calculation Of Integral
Figure 21 - Calculation Of Integral

Part 2:

Part 2 of the Bottle is bounded by the function:

 

ϖ = 0.467

 

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

 

AREA = 2π\(\displaystyle\int\limits^{-3.48}_{-3.68}ϖ\sqrt{1[ϖ']^2}d\psi\)

 

Here,

 

ϖ' = \(\frac{d}{d\psi}\)(0.467) = 0

 

 

AREA = 2π\(\displaystyle\int\limits^{-3.48}_{-3.68}0.47\sqrt{1+\{0\}^2}d\psi\)

 

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

 

AREA = 2π × 0. 094

 

AREA = 0.5906 sq. units

 

Below is the calculation and solution:

Figure 22 - Calculation Of Integral
Figure 22 - Calculation Of Integral

Part 3:

Part 3 of the Bottle is bounded by the function:

 

ϖ = - 0.3672ψ2 - 1.489ψ - 0.256

 

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

 

AREA = 2π\(\displaystyle\int\limits_{-3.48}^{-1.85}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

\(c\frac{d}{d\psi}\biggl(-0.367\psi^2-1.489\psi-0.256\biggl)=-0.74\psi-1.49\)

 

 AREA = 2π\(\displaystyle\int\limits_{-3.48}^{-1.85}\{-0.37\psi^2-1.49\psi-0.26\}\sqrt{1+\{-0.74\psi-1.49\}^2}\)

 

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

 

AREA=2π × 1.82648

 

AREA=11.4762 sq. units

 

Below is the calculation and solution:

Figure 23 - Calculation Of Integral
Figure 23 - Calculation Of Integral

Part 4:

Part 1 of the Bottle is bounded by the function:

 

ϖ = 1.201

 

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

 

 AREA = 2π\(\displaystyle\int\limits_{-1.85}^{0.4}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

 ϖ' = \(\frac{d}{dψ}\)(1.201) = 0

 

 AREA = 2π \(\displaystyle\int\limits_{-1.85}^{0.4}1.2\sqrt{1+\{0\}^2}\)

 

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

 

AREA = 2π × 2.7

 

AREA = 16.9646 sq. units

 

Below is the calculation and solution:

Figure 24 - Calculation Of Integral
Figure 24 - Calculation Of Integral

Part 5:

Part 5 of the Bottle is bounded by the function:

 

ϖ = 0.0045Ψ7 - 0.089Ψ6 + 0.56Ψ5 - 2.021Ψ4 + 3.612Ψ3 - 3.256Ψ2 + 1.11ψ + 1.09

 

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

 

 AREA = 2π\(\displaystyle\int\limits_{,0.4}^{3}ϖ\sqrt{1+[ϖ']^2}\)

 

Here,

 

ϖ' = \(\frac{d}{d\psi}\biggl(\)0.0045Ψ- 0.089Ψ6 + 0.565 - 2.021Ψ+ 3.612Ψ3 - 3.256Ψ2 + 1.11ψ + 1.09\(\biggl)\)= 0.035

 

 AREA = 2π \(\displaystyle\int\limits_{,0.4}^{3}\{0.005\psi^7-0.09\psi^6+0.6\psi^5-0.02\psi^4-3.26\psi^2+1.1\psi+1.1\}\sqrt{1+\{0.035}\)

 

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

 

AREA = 2π × 2.42119

 

AREA = 15.2128 sq. units

 

Below is the calculation and solution:

Figure 25 - Calculation Of Integral
Figure 25 - Calculation Of Integral

46.399 sq.units is the total surface area of Type 1 Bottle.

 

8.5 units was the assumed height of bottle in Desmos. However in reality, 25cm is the height of the actual bottle. By the unitary method, we thus write the actual total Surface Area as: 136.468 cm2.

Data Analysis

From the above calculations, it is clearly conspicuous that greater the height, more the surface area and lesser the curve on its body, lesser the surface area. From these two points, we can clearly decide that the cost of manufacturing a Type 2 bottle is significantly less compared to a Type 1 or even Type 3 bottle. The obtained are presented as follows:

Sl No.
Type of Bottle
Height {in cm}
Surface Area {in cm2}
1
Type 1
21.2
153.219
2
Type 2
19.6
131.749
3
Type 3
25
136.468
Figure 26 - Table On The Obtained Are Presented As Follows:

New design

It has been noted that for an ideal bottle of optimum volume {750 mL}, the height needs to be minimum {within 20cm} and the surface needs to be straight. For meeting volume requirements after decreasing height, the base can be broadened without incurring major increase in manufacturing cost. Moreover, the surface is proposed to be flat without curves or ridges. Thus, the design proposed is given below:

Figure 27 - Proposed Design Prototype
Figure 27 - Proposed Design Prototype

Obtaining surface area using Desmos:

Figure 28 - Finding Functions For Edges
Figure 28 - Finding Functions For Edges

Part 1:

Obtained equation for Part 1 is: ϖ = 0.37 with limits -4.28 to -3.37

 

Thus, ϖ' = \(\frac{d}{dψ}\)(0.37) = 0

 

 AREA = 2π\(\displaystyle\int\limits_{-4.28}^{-3.37}0.37\sqrt{1+\{0\}^2}\)

 

AREA = 2.1156 sq units

Figure 29 - Calculation Of Integral
Figure 29 - Calculation Of Integral

Part 2:

Obtained equation for Part 1 is: ϖ = 0.57ψ + 2.3 with limits -3.37 to -1.96

 

Thus, ϖ' = \(\frac{d}{dψ}\)(0.57ψ + 2.3) = 0.57

 

 AREA = 2π \(\displaystyle\int\limits_{-3.37}^{-1.96}\{0.57ψ\,+\,2.3\}\sqrt{1+\{0.57\}^2}\)

 

AREA = 7.964 sq units

Figure 30 - Calculation Of Integral
Figure 30 - Calculation Of Integral

Part 3:

Obtained equation for Part 1 is: ϖ = 1.18 with limits -1.96 to 4.27

Thus, ϖ' = \(\frac{d}{dψ}\)(1.18) = 0

 

 AREA = 2π \(\displaystyle\int\limits_{-1.96}^{4.27}1.18\sqrt{1+\{0\}^2}\)

 

AREA = 46.1902 sq units

Figure 31 - Calculation Of Integral
Figure 31 - Calculation Of Integral

56.2698 sq.units is the total surface area of Type 1 Bottle.

 

8.55 units was the assumed height of bottle in Desmos. However in reality, 20 cm is the assumed height of the actual bottle. By the unitary method, we thus write the actual total Surface Area as: 131.625 cm2

 

This value is approximately similar to the one obtained for Type 1 bottle. So we conclude that the cost of manufacturing this cylindrical bottle is minimum.

Conclusion

This investigation shows that the optimized bottle for achieveing minimum production cost, is a cylindrical one with no curved edge.

  • Ideal bottle has the surface area 131.625 cm2, which is minutely greater than the best case among the three taken into consideration {131.749 cm2}.
  • Type 1 bottle has the surface area 153.219 cm2 and volume 750 mL.
  • Type 2 bottle has the surface area 131.749 cm2 and volume 750 mL.
  • Type 3 bottle has the surface area 136.468 cm2 and volume 750 mL.
  • Maximum manufacturing cost and maximum surface area is found in Type 1 bottle.
  • Minimum manufacturing cost and minimum surface area is found in Type 2 bottle.
  • Reduction in height to at most 20 cm and decrease in curveness of edge, reduces the surface area of the cylindrical packaging.
  • Ideal prototype bottle has surface area 131.625 cm2 and volume nearly 750 mL.

Evaluation

Strength

  • The usage of method of integration, reduces error in approximation of area.
  • The application of identical methodology to each of the test cases, ensures global similarity in the examples under consideration, increasing the commonality and coherence.
  • The analysis was used to further propose a prototype package that would incur minimum cost of production.
  • The calculator TI-NSPIRE CX II T CAS was used for calculations, which is permitted in the IB curriculum.

 

Limitations

  • Since the data pertaining to bottles has been collected from varied sources, there is doubt regarding the factual correctness of empirical data.
  • The prototype suggested may require future modification and development.

Bibliography