Optimization of soft-beverage bottle such that it occupies maximum volume but incurs minimum manufacturing cost.

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  the renowned beverage company – PepsiCo – 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.

 

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.

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 marked representing different parts of the bottle’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.

 

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

 

Where λ_i^* and λ_i^** are in the interval

 

\(\biggl[{λ_{i-1}},λ_{i}\biggl{]}\)

 

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

 

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

 

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

 

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

Part 1 -

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

 

y = 0.07x + 0.88

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 0. 4099

 

Α = 2. 575 sq. units

 

The solution is given below -

Part 2 -

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

 

y = 0.5

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 0.1

 

Α = 0. 6283 sq. units

The solution is given below -

Part 3 -

 

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

 

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

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 2. 0266

 

Α = 12. 7335sq. units

 

The solution is given below -

Part 4 -

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

 

y = 1.3

 

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

 

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

 

Here,

 

\(y' = \frac{d}{dx} (1.3) = 0\)

 

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

 

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

 

Α = 2π × 3. 302

 

Α = 20. 747 sq. units

 

The solution is given below -

Part 5 -

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

 

y = - 0. 3x + 1. 45

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 0. 9299

 

Α = 5. 8427 sq. units

 

The solution is given below -

Part 6 -

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

 

y = 0. 12x + 0. 92

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 0. 8422

 

Α = 5. 2917 sq. units

 

The solution is given below -

Part 7 -

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

 

y = - 0. 1x + 1. 36

 

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

 

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

 

Here,

 

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

 

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

 

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

 

Α = 2π × 0. 9977

 

Α = 6. 2687 sq. units

 

The solution is given below -

Part 8 -

The function demarcating Part 8 of the bottle, is given by --

 

y = 0. 26x + 0. 31

 

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

 

\(A = 2π \ \displaystyle\int\limits^{3.8}_{2.89}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0.26x\,+0.31=)=0.26\)

 

\(A = 2π  \ \displaystyle\int\limits^{3.8}_{2.89}(0.26x \,+0.31)\sqrt{1+(0.26)^2}dx\)

 

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

 

Α = 2π × 1. 1092

 

Α = 6. 9693 sq. units

 

The solution is given below -

Part 9 -

The function demarcating Part 9 of the bottle, is given by -

 

y = - 0. 51x² + 4. 07x - 6. 8

 

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

 

\(A = 2π\ \displaystyle\int\limits^{4.8}_{3.8}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(-0.51x^2\,+4.07x\,-6.8)= -1.02x + 4.07\)

 

\(A = 2π \ \displaystyle\int\limits^{4.8}_{3.8}(-0.51x^2+4.07x-6.8)\sqrt{1+(-1.02x\,+4.07)^2}dx\)

 

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

 

Α = 2π× 1. 3249

 

Α = 8. 3246 sq. units

 

The solution is given below -

Thus, total surface area of the Type 1 bottle will be: 69.382 sq.units

 

In Desmos, the height of the bottle has been considered to be 9.6 units. But in truth the bottle is about 21.2 cm high. Thus, by unitary method, the actual surface area becomes: 153.219 cm².

Part 1 -

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

 

y = 0. 03x + 0. 65

 

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

 

\(A = 2π\displaystyle\int\limits^{-3.58}_{-4.34}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0. 03x + 0. 65) = 0.03\)

 

\(A = 2π\displaystyle\int\limits^{-3.58}_{-4.34}(0.03x+0.65)\sqrt{1+(0.03)^2}dx\)

 

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

 

Α = 2π × 0. 4039

 

Α = 2. 5378 sq. units

 

The solution is given below -

Part 2 -

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

 

y = 0. 45

 

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

 

\(A = 2π\displaystyle\int\limits^{-3.3}_{-3.45}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0.45) = 0\)

 

\(A = 2π\displaystyle\int\limits^{-3.3}_{-3.45}(0.45)\sqrt{1+(0)^2}dx\)

 

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

 

Α = 2π × 0. 0675

 

Α = 0. 4241 sq. units

 

The solution is given below -

Part 3 -

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

 

y = - 0. 23x² - 0. 69x + 0. 72

 

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

 

\(A = 2π\displaystyle\int\limits^{-0.62}_{-3.3}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(-0.23x^3-0.69x+0.72)=-0.\,46x-0.69\)

 

\(A = 2π \displaystyle\int\limits^{-0.62}_{-3.3}(-0.23x^2-0.69x+0.72)\sqrt{1+(-0.46x-0.69)^2}dx\)

 

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

 

Α = 2π × 2. 994

 

Α = 18. 8119 sq. units

 

The solution is given below -

Part 4 -

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

 

y = 1. 08

 

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

 

\(A = 2π \displaystyle\int\limits^{2.15}_{-0.62}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(1.08)=0\)

 

\(A = 2π\displaystyle\int\limits^{2.15}_{-0.62}1.08\sqrt{1+(0)^2}dx\)

 

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

 

Α = 2π × 2. 9916

 

Α = 18. 7968 sq. units

 

The solution is given below -

Part 5 -

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

 

y = - 0. 22x² + 1. 4x - 0. 9

 

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

 

\(A = 2π \ \displaystyle\int\limits^{4.57}_{2.15}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(-0.22x^2+1.4x-0.9)=-0.44x+1.4)\)

 

\(A = 2π \ \displaystyle\int\limits^{4.57}_{2.15}(-0.22x^2+1.4x-0.9)\sqrt{1+(-0.44x+1.4)^2}dx\)

 

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

 

Α = 2π × 3. 0644

 

Α = 19. 2542 sq. units

 

The solution is given below -

Thus, total surface area of the Type 2 bottle will be: 59.825 sq.units

 

In Desmos, the height of the bottle has been considered to be 8.9 units. But in truth the bottle is about 19.6 cm high. Thus, by unitary method, the actual surface area becomes: 131.749 cm².

Part 1 -

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

 

y = 0. 05x + 0. 74

 

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

 

\(A = 2π\displaystyle\int\limits^{-3.78}_{-4.42}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0.05x+0.74)=0.05\)

 

\(A = 2π\displaystyle\int\limits^{-3.78}_{-4.42}(0.05x+0.74)\sqrt{1+(0.05)^2}dx\)

 

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

 

Α = 2π × 0. 3428

 

Α = 2. 1539 sq. units

 

The solution is given below -

Part 2 -

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

 

y = 0. 47

 

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

 

\(A = 2π\displaystyle\int\limits^{-3.48}_{-3.68}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0.47) = 0\)

 

\(A = 2π\displaystyle\int\limits^{-3.48}_{-3.68}0.47\sqrt{1+[0]^2}dx\)

 

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

 

Α = 2π × 0. 094

 

Α = 0. 5906 sq. units

 

The solution is given below -

Part 3 -

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

 

y = - 0. 37x² - 1. 49x - 0. 26

 

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

 

\(A = 2π\displaystyle\int\limits^{-1.85}_{-3.48}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(-0.37x^2-1.49x-0.26)=-0.74x-1.49\)

 

\(A = 2π\displaystyle\int\limits^{-1.85}_{-3.48}(-0.37x^2-1.49x-0.26)\sqrt{1+(-0.74x-1.49)^2}dx\)

 

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

 

Α = 2π × 1. 8265

 

Α = 11. 4762 sq. units

 

The solution is given below -

Part 4 -

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

 

y = 1. 2

 

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

 

 

\(A = 2π \displaystyle\int\limits^{0.4}_{-1.85}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(1.2)=0\)

 

\(A = 2π\displaystyle\int\limits^{0.4}_{-1.85}1.2\sqrt{1+[0]^2}dx\)

 

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

 

Α = 2π × 2. 7

 

Α = 16. 9646 sq. units

 

The solution is given below -

Part 5 -

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

 

y = 0. 005x ⁷ - 0. 09x ⁶ + 0. 6x ⁵ - 2. 02x ⁴+ 3. 6 1x ³ - 3. 26x ² + 1. 1x + 1. 1

 

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

 

\(A = 2π\ \displaystyle\int\limits^{3}_{0.4}y\sqrt{1+[y']^2}dx\)

 

Here,

 

\(y'=\frac{d}{dx}(0. 005x7 - 0. 09x6 + 0. 6x5 - 2. 02x4 + 3. 61x3 - 3. 26x2 +1. 1x + 1. 1) = 0. 035x6 - 0. 54x5 +3\)

 

\(A = 2π \ \displaystyle\int\limits^{3}_{0.4}(0. 005x7 - 0. 09x6 + 0. 6x5 - 2. 02x4 + 3. 61x3 - 3. 26x2 + 1. 1x + 1.1)\)

 

\(\sqrt{1+(0.035x^6 - 0.54x^5 + 3x^4 - 8. 08x^3 + 10.83x^2-6.5^2x+1.1)2}\)

 

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

 

Α = 2π× 2. 4212

 

Α = 15. 2128 sq. units

 

The solution is given below -

Thus, total surface area of the Type 2 bottle will be: 46.399 sq.units

 

In Desmos, the height of the bottle has been considered to be 8.5 units. But in truth the bottle is about 25 cm high. Thus, by unitary method, the actual surface area becomes: 136.468 cm².

It has been observed that more straight the bottle, lesser is the curved surface area but the height of the bottle significantly affects the total surface area after scaling. The Type 2 bottle has minimum area and Type 1 bottle has maximum area. So the cost of manufacturing a Type 2 bottle would be lesser compared to the Type 1 bottle. Thus for an average 750 ml bottle, by PepsiCo, the data obtained are 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:

Calculating surface area using Desmos as above,

Part 1 -

Equation of Part 1 obtained is: y = 0.37 with limits - 4.28 to -3.37

 

Thus,

 

\(  y' =\frac{d}{dx}0.37 = 0\)

 

\(A = 2π\displaystyle\int\limits^{-3.37}_{-4.28}0.37\sqrt{1+(0)^2}dx\)

 

Α = 2. 1156 sq units

Part 2 -

Equation of Part 2 obtained is: y = 0. 57x + 2. 3 with limits - 3. 37 to -1. 96

 

Thus,

 

\(y' = \frac{d}{dx}(0. 57x + 2. 3) = 0. 57\)

 

\(A = 2π \displaystyle\int\limits^{-1.96}_{-3.37}(0. 57x + 2. 3)\sqrt{1+(0.57)^2}dx\)

 

Α = 7. 964 sq units

Part 3:

Equation of Part 1 obtained is: y = 1. 18 with limits - 1. 96 to 4. 27

 

Thus, y' = \(\frac{d}{dx}\)(1.18) = 0

 

A = 2π \(\displaystyle\int\limits^{4.27}_{-1.96}1.18\sqrt{1+(0)^2}dx\)

 

Α = 46. 1902 sq units

Thus, total surface area of the Type 2 bottle will be: 56.2698 sq.units

 

In Desmos, the height of the bottle has been considered to be 8.55 units. But it was considered that the bottle is about 20 cm high. Thus, by unitary method, the actual surface area becomes: 131.625 cm².

 

This value is only slightly greater than the minimum area in the above three cases. So, the cost of manufacturing will be the least.

Optimization of soft-beverage bottle such that it occupies maximum volume but incurs minimum manufacturing cost.

 

This exploration has revealed that the ideal shape of the bottle which has maximum volume and minimum production cost is a straight cylindrical bottle a part of which converges towards the mouth so that the beverage can be accessed.

• It has been found that surface area of the ideal bottle is 131.625 cm², slightly greater than the ideal case among the three types (131.749 cm²).
• From the exploration it has been found that the surface area of Type 1 bottle is 153.219 cm² and the volume is 750 mL.
• It has further been found that the surface area of Type 2 bottle is 131.749 cm² and volume is 750 mL.
• Moreover, the surface area of Type 3 bottle is 136.468 cm² and the volume is 750 mL.
• It has been observed that Type 1 bottle has the maximum surface area incurring maximum manufacturing cost.
• Furthermore, it has been seen that Type 2 model has the minimum surface area, incurring minimum manufacturing cost
• It has also been found from the exploration, that a reduction in height i.e. till maximum height of 20 cm and increase in the straightness of the bottle decrease the surface area.
• Thus, it has been found that the ideal bottle should have a surface area of 131.625 cm² and the volume of nearly 750 mL.

Strength

• Integration method had been used in order to reduce error in approximation and to discover the deviation in the surface area over a small instant in X-axis.
• The same methodology had been used to understand the surface area of all the three types of bottles so that it increases the commonality among them and makes the exploration more reliable.
• An ideal new design had been proposed which has the minimum production cost at same capacity.
• The calculations were done using the TI-NSPIRE CX II T CAS Calculator which is permitted in the IB Diploma Programe.

 

Limitations

• As the images of bottles have been taken from the internet it may be possible that there is an error in their measurements i.e., their height and width.
• The shape of the ideal bottle is just a prototype and may have to be developed in the future.

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