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 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.
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. 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 \(\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\)\(\biggl(\)= - 0. 2313ψ3 - 2. 3322ψ2 - 7. 089ψ - 5. 4302\(\biggl)\)= - 0. 69ψ2 - 4. 66ψ - 7. 09
AREA = 2π \(\displaystyle\int\limits_{-3.7}^{-2.04}\)\(\{-0.23ψ^3-2.33ψ^2-2.33ψ^2-7.09ψ-5.43\}\sqrt{1+\{-0.69ψ^2-4.66ψ-7.09\}^2} \)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:
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\(\displaystyle\int\limits_{\,\,2.89}^{3.8}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\)(0. 258ψ + 0. 309) = 0. 26
AREA = 2π \(\displaystyle\int\limits_{\,\,2.89}^{3.8}\{0.26ψ+0.31\}\sqrt{1+\{0.26\}^2}\)dψ
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:
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\(\displaystyle\int\limits_{\,\,3.8}^{4.8}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\biggl(-\,0.511ψ^2+4.072ψ-6.79\biggl)=-1.02ψ+4.07\)
AREA = 2π \(\displaystyle\int\limits_{\,\,3.8}^{4.8}\{-\,0.51ψ^2+4.07ψ-6.8\}\sqrt{1+\{-1.02ψ+4.07\}^2}\)dψ
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:
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.
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 \(\displaystyle\int\limits_{-4.34}^{-3.58}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\) (0. 028ψ + 0. 651) = 0. 03
AREA = 2π \(\displaystyle\int\limits_{-4.34}^{-3.58}\{0.03ψ+0.65\}\sqrt{1+\{0.03\}^2}\)dψ
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:
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\(\displaystyle\int\limits_{-3.45}^{-3.3}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\{0.449\}=0\)
AREA = 2π \(\displaystyle\int\limits_{-3.45}^{-3.3}\{0.45\}\sqrt{1+\{0\}^2}\)dψ
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:
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\(\displaystyle\int\limits_{-3.3}^{-0.62}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ'\(\frac{d}{dψ}\biggl(\)- 0. 231ψ2 - 0. 689ψ + 0. 722\(\biggl)\)= - 0. 46ψ - 0. 69
AREA = 2π \(\displaystyle\int\limits_{-3.3}^{-0.62}\{-\,0.23ψ^2-0.69ψ+0.72\}\sqrt{1+\{-\,0.46ψ - 0.69\}^2}\)dψ
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:
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 \(\displaystyle\int\limits_{-0.62}^{2.15}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\)(1.078) = 0
AREA = 2π\(\displaystyle\int\limits_{-0.62}^{2.15}1.08\sqrt{1+\{0\}^2}\)dψ
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:
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 \(\displaystyle\int\limits_{\,\,2.15}^{4.57}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\biggl(\)- 0. 221ψ2 + 1. 399ψ - 0. 92\(\biggl)\)= - 0. 44ψ + 1. 4
AREA = 2π \(\displaystyle\int\limits_{\,\,2.15}^{4.57}\{-\,0.22ψ^2+1.4ψ-0.9\}\sqrt{1+\{-\,0.44ψ+1.4\}^2}\)dψ
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:
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
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\(\displaystyle\int\limits_{-4.42}^{-3.78}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\)(0. 049ψ + 0. 738) = 0.05
AREA = 2π \(\displaystyle\int\limits_{-4.42}^{-3.78}\{0.05ψ\,+\,0.74\}\sqrt{1+\{0.05\}^2}\)dψ
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:
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:
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\(\displaystyle\int\limits_{-3.48}^{-1.85}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
\(c\frac{d}{d\psi}\biggl(-0.367\psi^2-1.489\psi-0.256\biggl)=-0.74\psi-1.49\)
AREA = 2π\(\displaystyle\int\limits_{-3.48}^{-1.85}\{-0.37\psi^2-1.49\psi-0.26\}\sqrt{1+\{-0.74\psi-1.49\}^2}\)dψ
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:
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\(\displaystyle\int\limits_{-1.85}^{0.4}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{dψ}\)(1.201) = 0
AREA = 2π \(\displaystyle\int\limits_{-1.85}^{0.4}1.2\sqrt{1+\{0\}^2}\)dψ
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:
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\(\displaystyle\int\limits_{,0.4}^{3}ϖ\sqrt{1+[ϖ']^2}\)dψ
Here,
ϖ' = \(\frac{d}{d\psi}\biggl(\)0.0045Ψ7 - 0.089Ψ6 + 0.565 - 2.021Ψ4 + 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:
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.
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:
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:
Obtaining surface area using Desmos:
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\(\displaystyle\int\limits_{-4.28}^{-3.37}0.37\sqrt{1+\{0\}^2}\)dψ
AREA = 2.1156 sq units
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 \(\displaystyle\int\limits_{-3.37}^{-1.96}\{0.57ψ\,+\,2.3\}\sqrt{1+\{0.57\}^2}\)dψ
AREA = 7.964 sq units
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 \(\displaystyle\int\limits_{-1.96}^{4.27}1.18\sqrt{1+\{0\}^2}\)dψ
AREA = 46.1902 sq units
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.
This investigation shows that the optimized bottle for achieveing minimum production cost, is a cylindrical one with no curved edge.
Strength
Limitations