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In this article, the mathematical foundations of the so called Criterion of Optimization by Compensation for designing commercial bottles with a straight section along its silhouette and with lateral surfaces of revolution is presented. Such mathematical model uses as main tools, Lagrange polynomial interpolation and Newton’s Method for Nonlinear Systems being first necessary to formulate and demonstrate a theorem. It was redesigned and manufactured a bottle of a half-liter of Fanta soda of the well-known Coca Cola Company, which uses 18.86% more material that such criterion establishes. It was expected that the redesigned bottle use 4.91% more of material with respect to what is established by the Criterion of Optimization by Compensation. However, it was reported a 13% of mistake due to important limitations that must be overcome.

During the production stages of a bottle until being finally sent to the consumers, the manufacturers and merchants must confront a more exigent market and society every day. The bottle has to satisfy not only the necessity of containing, protecting, preserving, commercializing and distributing merchandise, if not also, of retraining, decrease of the ecological impacts and minimization of costs. Therefore, it is necessary to design appropriate bottles according to necessities above indicated, making this evident the necessity of generating and transmitting the knowledge of science and technology, in which the concept of optimization is of great importance. For instance, Fletcher [

Up to now, there has not been found studies about a clear mathematical model of an optimization method of a bottle with a lateral surface of revolution and a straight section along its silhouette. In the case of PET bottles, there are some studies about optimizing and redesigning the whole or part of the body of them by using different software programs. Masood and KeshavaMurthy [

On the other hand, for aluminum bottles, Han et al. [

Therefore, since most containers whose shapes are determined by surface of revolution and manufactured from plastic or metal material, the enterprises that decide to apply criterion of optimization on its manufacture, could try to reach the following achievements: reduction of rubbish, saving energy, decrease of the negative environmental impact and hence a friendly environmental image.

In this work, we report a mathematical model of the Criterion of Optimization by Compensation proposed by Reyna and Moore [

Such Criterion of Optimization by Compensation tells us that in order to design a bottle with a minimal total superficial area, it must first design a cylinder with a minimal total superficial area. From the cylinder, it must be get the shape of the bottle by removing certain solid parts of the cylinder. With the solid parts removed, it must be formed a cylinder whose volume is equal to the sum of the volumes of the solid parts removed from the initial cylinder. The lateral surface area of the formed cylinder must be equal to or less than the sum of the outer surface areas of the solid parts removed from the initial cylinder. This new formed cylinder has to fit the straight section of the bottle that is being designing. Since de cylinder has a minimal area, it is optimized, and at a beginning, the superficial area of the bottle is optimized too. Indeed, theorem 2.1 (see below) shows that the area of the bottle results to be less than that of the cylinder. However, this is only a descriptive fact and a mathematical modelling turns out to be necessary. Then, this work presents a mathematical modelling as the one required, according to the Criterion of Optimization by Compensation.

Flow diagram of the Optimization method for designing an optimized bottle of volume V. The starting point is a cylinder of minimal area of volume V. Following the Criterion of Optimization by Compensation according to steps 1, 2,…, s = 3 , k + 1 , where k is a finite number, the final silhouette of the optimized bottle can be found. From this, the three dimensional bottle is obtained.

To design a container with a straight cylindrical shape and of a certain volumetrically capacity in using a smaller amount of material, lead us to determine a minimum of the following function

A ( r ) = 2 π r 2 + 2 V r (1)

where V, A and r are the volume, total area and radius of the cylinder respectively.

Deriving the function A ( r ) with respect to r and equaling to zero we have that

4 π r − 2 V r 2 = 0

Solving this equation with respect to r

r = V 2 π 3 (2)

Which is a critical point for the function given in Equation (1)

But

d 2 A ( r ) d r 2 = 4 π + 4 V r 3 (3)

By replacing (2) in (3)

d 2 A ( V 2 π 3 ) d r 2 = 12 π

Which is major than zero. This means that the value of r given by Equation (2) corresponds to a minimum and which let us determine the height h of the cylinder in using V = π r 2 h , equal to

h = 4 V π 3 (4)

As it can be seen in this case, it is not difficult to get the values of r given by Equation (2) and h given by Equation (4) which let us minimize the area of a cylinder for a given value of its volume V. Furthermore Equation (3) tell us that

Equation (1) is always concave above because 4 π + 4 V r 3 > 0 , ∀ r > 0 . This

means that given by Equation (2) corresponds to a global minimum.

In the case of designing containers with a non-cylindrical shape as it can be the general case, of a bottle, its silhouette could result whimsical due to aesthetic considerations as many others, so it is impossible to write an Equation (function), as in the case of the cylinder discussed in Section 2.1. Therefore, in order to solve this apparent difficult, we will use the Criterion of Optimization by Compensation proposed by Reyna and Moore [

An outline of the Criterion of Optimization by Compensation, in order to minimize the area of a bottle, keeping a constant volume is as follow:

1) Know the volumetric capacity V of the bottle that will be optimized;

2) Optimize the total area of a cylinder whose volumetric capacity is V;

3) In each zone of the optimized cylinder where will be modified (

4) Fit the cylinder without bases, obtained in the step 3, in the cylindrical zone of the optimized cylinder. In this way, it is possible to obtain an optimized bottle with lateral surface of revolution (see

The theorem below is based on the Criterion of Optimization by Compensation. This theorem tell us that it is possible to deform a cylinder of volume V c and minimal area A c to a solid of revolution of volume V s and area A s such that V c = V s and A s < A c . This theorem is formulated in order to show in particular that the area of a bottle could be less than the corresponding minimal area of a cylinder, both enclosing the same volume, so that the optimization of the bottle is better.

Theorem 2.1 There exists a solid of revolution of the non-convex type whose surface is less than the surface of a right circular cylinder of minimal area, keeping both of them the same volume.

Proof

From scheme of

V c = ∑ i = 1 6 V i (5)

Due to the Criterion of Optimization by Compensation,

V 7 = ∑ i = 4 6 V i (6)

replacing Equation (6) in (5)

V c = ∑ i = 1 3 V i + V 7 (7)

We have as well that,

V c = π r c 2 h c (8)

V 1 = π r 2 x 0 (9)

V 2 = π ∫ x 0 x 1 [ P ( x ) ] 2 d x (10)

V 3 = π H 2 ( h c − x 1 ) (11)

V 7 = π ∫ h c a [ Q ( x ) ] 2 d x (12)

P ( x ) = α 1 x + α 2 (13)

Q ( x ) = β 1 x + β 2 (14)

By replacing Equation (13) in (10)

V 2 = π ∫ x 0 x 1 [ α 1 x + α 2 ] 2 d x

V 2 = π [ α 1 2 ( x 1 3 − x 0 3 ) 3 + α 1 α 2 ( x 1 2 − x 0 2 ) + α 2 2 ( x 1 − x 0 ) ] (15)

Then,

V 2 = V 2 ( α 1 , α 2 ) (16)

By replacing Equation (14) in (12),

V 7 = π ∫ h c a [ β 1 x + β 2 ] 2 d x

V 7 = π [ β 1 2 ( a 3 − h c 3 ) 3 + β 1 β 2 ( a 2 − h c 2 ) + β 2 2 ( a − h c ) ] (17)

Then,

V 7 = V 7 ( β 1 , β 2 , a ) (18)

By replacing Equations. (8), (9), (11), (16) and (18) in (7),

#Math_52# (19)

Let A c be such that,

A c = ∑ i = 0 5 A i (20)

Where,

A 0 = π r 2 (21)

A 1 = 2 π r x 0 (22)

A 2 = 2 π ∫ x 0 x 1 P ( x ) 1 + [ P ′ ( x ) ] 2 d x (23)

A 3 = 2 π H ( h c − x 1 ) (24)

A 4 = 2 π ∫ h c a Q ( x ) 1 + [ Q ′ ( x ) ] 2 d x (25)

A 5 = π R 2 (26)

A c = 2 π r c h c + 2 π r c 2 (27)

Deriving Equation (13),

P ′ ( x ) = α 1 (28)

Deriving Equation (14),

Q ′ ( x ) = β 1 (29)

Replacing Equations (13) and (28) in (23) and integrating,

A 2 = 2 π 1 + α 1 2 [ α 1 x 1 2 2 + α 2 x 1 − α 1 x 0 2 2 − α 2 x 0 ] (30)

Then,

A 2 = A 2 ( α 1 , α 2 ) (31)

Replacing Equations (14) and (29) in (25) and integrating,

A 4 = 2 π 1 + β 1 2 [ β 1 a 2 2 + β 2 a − β 1 h c 2 2 − β 2 h c ] (32)

Then,

A 4 = A 4 ( β 1 , β 2 , a ) (33)

Replacing Equations (21), (22), (24), (26), (31) and (33) in (20),

2 π r c h c + 2 π r c 2 − π r 2 − 2 π r x 0 − A 2 ( α 1 , α 2 ) − 2 π H ( h c − x 1 ) − A 4 ( β 1 , β 2 , a ) − π R 2 = 0 (34)

Now, the conditions that the polynomials P ( x ) and Q ( x ) must satisfy are:

P ( x 0 ) = r (35)

P ( x 1 ) = H (36)

Q ( h c ) = H (37)

Q ( a ) = R (38)

From Equations (35) and (13),

α 1 x 0 + α 2 − r = 0 (39)

From Equations (36) and (13),

α 1 x 1 + α 2 − H = 0 (40)

From Equations (37) and (14),

β 1 h c + β 2 − H = 0 (41)

From Equations (38) and (14),

β 1 a + β 2 − R = 0 (42)

Taking into account Equations (19), (34), (39), (40), (41) and (42) we get the following system of equations,

α 1 x 0 + α 2 − r = 0 (43)

α 1 x 1 + α 2 − H = 0 (44)

β 1 h c + β 2 − H = 0 (45)

β 1 a + β 2 − R = 0 (46)

#Math_83# (47)

2 π r c h c + 2 π r c 2 − π r 2 − 2 π r x 0 − A 2 ( α 1 , α 2 ) − 2 π H ( h c − x 1 ) − A 4 ( β 1 , β 2 , a ) − π R 2 = 0 (48)

Hence, we have six equations in five unknowns.

From Equations (43) and (44),

α 1 = H − r x 1 − x 0 (49)

α 2 = r x 1 − H x 0 x 1 − x 0 (50)

From Equations. (45) and (46),

β 1 = R − H a − h c (51)

β 2 = H a − R h c a − h c (52)

By replacing Equations (15) and (17) in (47),

π r c 2 h c − π r 2 x 0 − π [ α 1 2 ( x 1 3 − x 0 3 ) 3 + α 1 α 2 ( x 1 2 − x 0 2 ) + α 2 2 ( x 1 − x 0 ) ] − π H 2 ( h c − x 1 ) − π [ β 1 2 ( a 3 − h c 3 ) 3 + β 1 β 2 ( a 2 − h c 2 ) + β 2 2 ( a − h c ) ] = 0 (53)

By replacing Equations (49), (50), (51) and (52) in (53) and simplifying,

π r c 2 h c − π r 2 x 0 − π [ ( H − r x 1 − x 0 ) 2 ( x 1 3 − x 0 3 ) 3 + ( H − r ) ( r x 1 − H r 0 ) ( x 1 + x 0 x 1 − x 0 ) + ( r x 1 − H x 0 ) 2 x 1 − x 0 ] − π H 2 ( h c − x 1 ) − π [ ( R − H a − h c ) 2 ( a 3 − h c 3 ) 3 + ( R − H ) ( H a − R h c ) ( a + h c a − h c ) + ( H a − R h c ) 2 a − h c ] = 0 (54)

Equation (54) is an equation in the unknown parameter a.

By replacing Equations (30) and (32) in (48)

2 π r c h c + 2 π r c 2 − π r 2 − 2 π r x 0 − 2 π 1 + α 1 2 [ α 1 x 1 2 2 + α 2 x 1 − α 1 x 0 2 2 − α 2 x 0 ] − 2 π H ( h c − x 1 ) − 2 π 1 + β 1 2 [ β 1 a 2 2 + β 2 a − β 1 h c 2 2 − β 2 h c ] − π R 2 = 0 (55)

By replacing Equations (49), (50), (51) and (52) in (55),

2 π r c h c + 2 π r c 2 − π r 2 − 2 π r x 0 − 2 π 1 + ( H − r x 1 − x 0 ) 2 [ ( H − r x 1 − x 0 ) x 1 2 2 + ( r x 1 − H x 0 x 1 − x 0 ) x 1 − ( H − r x 1 − x 0 ) x 0 2 2 − ( r x 1 − H x 0 x 1 − x 0 ) x 0 ] − 2 π H ( h c − x 1 ) − 2 π 1 + ( R − H a − h c ) 2 × [ ( R − H a − h c ) a 2 2 + ( H a − R h c a − h c ) a − ( R − H a − h c ) h c 2 2 − ( H a − R h c a − h c ) h c ] − π R 2 = 0 (56)

Equation (56) as well as Equation (54), is an equation in the unknown parameter a.

Since we have a theorem of existence, we solve numerically Equation (54) in the unknown parameter a in using the testing data r = 1.2 u , r c = 4.43 u , h c = 8.86 u , x 0 = 2 u , H = 4.2 u , R = 2.7 u and x 1 = 4.7 u . Here, from Equa-

tions (2) and (4) r c = h c 2 and the value h c = 8.86 u is arbitrary. The rest of

values for r , x 0 , H and R are given heuristically by looking

The minimal area of the cylinder of revolution that is generated when the region enclosed by the rectangle ODIK rotates around the x-axis, is equal to A c = 369.921 u 2 , while its volume is equal to V c = 546.249 u 3 which is the same value for the solid of revolution. So, we have V c = V s and A s < A c which prove the theorem.

Equation (56) led us to a no wished solution since a = 15.5758 u increase the volume of the solid of revolution to V s = 562.862 u 3 such that the surface of minimal area of the cylinder is A c = A s = 369.921 u 2 . Say, we have A c = A s and V c < V s .

In this section, the mathematical model of the criterion of optimization by compensation for designing commercial bottles with lateral surfaces of revolution and a straight section along its silhouette is presented. The start point is the schema shown in

The mathematical model taking into account the Criterion of Optimization by Compensation is as follows: consider the schema (shown in

degree rotates around the x-axis. The value of k depends on the number of polynomials needed to complete the shape of the silhouette of the bottle out of the rectangle with vertices ( 0 , 0 ) , ( 0 , r c ) , ( h c , r c ) and ( h c , 0 ) . V i , 1 , n ^ are volumes generated when the regions above of the polynomials P i ( x ) , 1 , n ^ , rotates around the x-axis. V 0 and V 0 are volumes below and above the constant polynomial P ( x ) = P 0 ( x n ) . Then, from

V c = ∑ i = 0 n ( V i + V i ) (57)

where V c is the volume of the cylinder of minimal area with radius r c and height h c which is generated when the rectangle with vertices ( 0 , 0 ) , ( 0 , r c ) , ( h c , r c ) and ( h c , 0 ) rotates around the x-axis (see

∑ i = 0 n V i = π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + ∑ i = 1 k V n + i (58)

Putting Equation (58) in (57)

V c = ∑ i = 0 n V i + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + ∑ i = 1 k V n + i (59)

But,

V c = π r c 2 h c (60)

V 1 = π ∫ 0 x 1 [ P 1 ( x ) ] 2 d x (61)

V 2 = π ∫ x 1 x 2 [ P 2 ( x ) ] 2 d x (62)

⋮

V n = π ∫ x n − 1 x n [ P n ( x ) ] 2 d x (63)

V 0 = π [ P 0 ( x n ) ] 2 ( h c − x n ) (64)

V n + 1 = π ∫ x n + 1 x n + 2 [ P n + 1 ( x ) ] 2 d x V n + 2 = π ∫ x n + 2 x n + 3 [ P n + 2 ( x ) ] 2 d x (65)

⋮

V n + k = π ∫ x n + k x n + k + 1 [ P n + k ( x ) ] 2 d x (66)

Putting Equations (60)?(66) in (59), we have finally,

π r c 2 h c = π ∫ 0 x 1 [ P 1 ( x ) ] 2 d x + π ∑ i = 1 n − 1 ∫ x i x i + 1 [ P i + 1 ( x ) ] 2 d x + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + π ∑ i = 1 k ∫ x n + i x n + i + 1 [ P n + i ( x ) ] 2 d x (67)

Equation (67) is a fundamental equation derived from the Criterion of Optimization by Compensation.

Now let P 1 ( x ) = ∑ i = 0 m 1 a 1 i x i , P 2 ( x ) = ∑ i = 0 m 2 a 2 i x i , ⋯ , P n ( x ) = ∑ i = 0 m n a n i x i ,#Math_158#, P n + 2 ( x ) = ∑ i = 0 m n + 2 a n + 2 i x i , ⋯ , P n + k ( x ) = ∑ i = 0 m n + k a n + k i x i be,

the polynomials that are shown in

π r c 2 h c = π ∫ 0 x 1 [ P 1 ( x ) ] 2 d x + π ∑ i = 1 n − 1 ∫ x i x i + 1 [ P i + 1 ( x ) ] 2 d x + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + π ∑ i = 1 k ∫ x n + i x n + i + 1 [ P n + i ( x ) ] 2 d x (68)

Subject to the following conditions:

P 1 ( x 0 ) = r (69)

P 1 ( x 1 ) = P 2 ( x 1 ) = R 1 (70)

P 2 ( x 2 ) = P 3 ( x 2 ) = R 2 (71)

P n − 1 ( x n − 1 ) = P n ( x n − 1 ) = R n − 1

P n ( x n ) = P 0 ( x n ) = R n = R 0 (72)

P 0 ( x n ) = P 0 ( x n + 1 ) = P n + 1 ( x n + 1 − h 1 ) = R n = R 0 = R n + 1 (73)

P n + 1 ( x n + 2 − h 1 ) = P n + 2 ( x n + 2 − h 2 ) = R n + 2 (74)

P n + 2 ( x n + 3 − h 2 ) = P n + 3 ( x n + 3 − h 3 ) = R n + 3 (75)

P n + 3 ( x n + 4 − h 3 ) = P n + 4 ( x n + 4 − h 4 ) = R n + 4 (76)

⋮

P n + k − 1 ( x n + k − h k − 1 ) = P n + k ( x n + k − h k ) = R n + k (77)

P n + k ( x n + k + 1 − h k ) = R n + k + 1 (78)

P 1 ( t i ) = P 1 i , 0 ≤ t i ≤ x 1 , 0 ≤ i ≤ m 1 (79)

P 2 ( t i ) = P 2 i , x 1 ≤ t i ≤ x 2 , 0 ≤ i ≤ m 2 (80)

⋮

P n ( t i ) = P n i , x n − 1 ≤ t i ≤ x n , 0 ≤ i ≤ m n (81)

P n + 1 ( t i ) = P n + 1 i , x n + 1 ≤ t i ≤ x n + 2 , 0 ≤ i ≤ m n + 1 (82)

P n + 2 ( t i ) = P n + 2 i , x n + 2 ≤ t i ≤ x n + 3 , 0 ≤ i ≤ m n + 2 (83)

⋮

P n + k ( t i ) = P n + k i , x n + k ≤ t i ≤ x n + k + 1 , 0 ≤ i ≤ m n + k (84)

where R i , i = 1 , ⋯ , n + k in Equations (70)-(78) are constant ordinates given by the designer. On the other hand, in order to give the shape of the silhouette of the bottle, ( t i , P 1 i ) , 0 ≤ t i ≤ x 1 , 0 ≤ i ≤ m 1 ; ( t i , P 2 i ) , x 1 ≤ t i ≤ x 2 , 0 ≤ i ≤ m 2 ; ⋯ , ( t i , P n i ) , x n − 1 ≤ t i ≤ x n , 0 ≤ i ≤ m n are also given by the designer such that each of them determines a set of points through which the polynomials obtained by interpolation, P i ( t ) , i = 1 , n ^ , passes respectively. Similarly, the ( t i , P n + 1 i ) , x n + 1 ≤ t i ≤ x n + 2 , 0 ≤ i ≤ m n + 1 ; ( t i , P n + 2 i ) , x n + 2 ≤ t i ≤ x n + 3 , 0 ≤ i ≤ m n + 2 ; ⋯ , ( t i , P n + k i ) , x n + k ≤ t i ≤ x n + k + 1 , 0 ≤ i ≤ m n + k ; are also given by the designer, such that each of them determines a set of points through which the polynomials obtained by interpolation, P n + i ( t ) , i = 1 , k ^ , passes respectively. These polynomials are such that their corresponding polynomials by translation according to, P n + i ( t − h i ) , i = 1 , k ^ fit with the remaining part of the bottle in the interval x n + k ≤ t i ≤ x n + k + 1 , k ≥ 1 . The r and x i , 1 , n ^ , are constants given by the designer, while r c and h c are the radius and height of a cylinder of minimal area. We must find a 1 i , 0 ≤ i ≤ m 1 ; a 2 i , 0 ≤ i ≤ m 2 ; ⋯ ; a n i , 0 ≤ i ≤ m n ; a n + 1 i , 0 ≤ i ≤ m n + 1 ; a n + 2 i , 0 ≤ i ≤ m n + 2 ; ⋯ ; a n + k i , 0 ≤ i ≤ m n + k ; h i , 1 ≤ i ≤ k ; x n + i , 1 ≤ i ≤ k + 1 such that the area of the bottle of volume V c is less or equal than the area of a cylinder of minimal area according to the theorem given above.

Solving the Problem of OptimizationWe can solve the optimization problem in using Lagrange polynomial interpolation and Newton’s Method for Nonlinear Systems according to the following steps:

Step 1 Find the constants a 1 i , 0 ≤ i ≤ m 1 ; a 2 i , 0 ≤ i ≤ m 2 ; ⋯ ; a n i , 0 ≤ i ≤ m n ; a n + 1 i , 0 ≤ i ≤ m n + 1 ; a n + 2 i , 0 ≤ i ≤ m n + 2 ; ⋯ ; a n + k i , 0 ≤ i ≤ m n + k , in using Lagrange interpolation, say, by determining the coefficients of the Lagrange interpolating polynomial in each case respectively. These coefficients are the constants we are trying to find.

Step 2 The x n + 1 , x n + 2 and h 1 can be found by solving the system of Equations (85), (86) and (87) given below, in using the Newton’s Method.

∑ i = 1 n I i + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + π ∫ x n + 1 x n + 2 [ P n + 1 ( x − h 1 ) ] 2 d x = p 1 ( π r c 2 h c ) , 0 < p 1 < 1 (85)

where,

I 1 , I 2 , ⋯ , I n are integrals that correspond to the polynomials

P 1 ( x ) , P 2 ( x ) , ⋯ , P n ( x ) , and can be calculated as well as π [ P 0 ( x n ) ] 2 ( h c − x n ) .

P n + 1 ( x n + 1 − h 1 ) = R n = R 0 = R n + 1 (86)

P n + 1 ( x n + 2 − h 1 ) = R n + 2 (87)

Step 3 x n + 3 and1 can be found by solving the system of equations:

P n + 2 ( x n + 2 − h 2 ) = R n + 2

∑ i = 1 n I i + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + I n + 1 + π ∫ x n + 2 x n + 3 [ P n + 2 ( x − h 2 ) ] 2 d x = p 2 ( π r c 2 h c ) , 0 < p 1 < p 2 < 1

I n + 1 can be calculated in using the data calculated in step 2.

Step 4 x n + 4 and h 3 can be found by solving the system of equations:

P n + 3 ( x n + 3 − h 3 ) = R n + 3

∑ i = 1 n I i + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + I n + 1 + I n + 2 + π ∫ x n + 3 x n + 4 [ P n + 3 ( x − h 3 ) ] 2 d x = p 3 ( π r c 2 h c ) , 0 < p 1 < p 2 < p 3 < 1

I n + 2 can be calculated in using the data calculated in step 3.

Step k x n + k and h k − 1 can be found by solving the system of equations:

P n + k − 1 ( x n + k − 1 − h k − 1 ) = R n + k − 1

∑ i = 1 n I i + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + ∑ i = 1 k − 2 I n + i + π ∫ x n + k − 1 x n + k [ P n + k − 1 ( x − h k − 1 ) ] 2 d x = p k − 1 ( π r c 2 h c ) , 0 < p 1 < p 2 < p 3 < ⋯ < p k − 1 < 1

I n + k − 2 can be calculated in using the data calculated in step k − 1 .

Step k+1 x n + k + 1 and h k can be found by solving the system of equations:

P n + k ( x n + k − h k ) = R n + k

∑ i = 1 n I i + π [ P 0 ( x n ) ] 2 ( h c − x n ) + π ( P 0 ( x n ) ) 2 ( x n + 1 − h c ) + ∑ i = 1 k − 1 I n + i + π ∫ x n + k x n + k + 1 [ P n + k ( x − h k ) ] 2 d x = p k ( π r c 2 h c ) , 0 < p 1 < p 2 < p 3 < ⋯ < p k = 1

I n + k − 1 can be calculated in using the data calculated in step k.

The number of unknown coefficients of the polynomials depends on the number of points that were arbitrary chosen, in order to get a Lagrange polynomial that must describe the silhouette of the bottle. Hence, we must be careful when the number of points are chosen, since the polynomial in that region could present slight oscillations and not describing the shape of the silhouette of the bottle in that region, as shown in

nomials should be free of oscillations, we must use a computer program (optimizer of five sections).

As an application of the criterion by compensation, a bottle of Fanta soda of the Coca Cola Company was considered, as shown in

We first analyse the real bottle, in order to determine its superficial area mainly. In fact, after cutting along the bottle through the middle and putting half of it in a coordinate system, previously drawn on a millimetered paper, the x and y coordinates of the silhouette of the real bottle are shown in

Section 1 | Section 2 | Section 3 | Section 4 | Section 5 | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|

x | y | x | y | x | y | x | y | x | y | |||||

0.00 | 1.10 | 3.00 | 1.10 | 9.50 | 2.14 | 14.00 | 3.15 | 20.00 | 3.15 | |||||

3.00 | 1.10 | 3.50 | 1.50 | 10.00 | 2.20 | 20.00 | 3.15 | 20.50 | 3.25 | |||||

4.30 | 2.45 | 11.00 | 2.65 | 21.00 | 3.30 | |||||||||

5.05 | 2.85 | 12.00 | 3.15 | 21.50 | 3.32 | |||||||||

6.05 7.00 8.00 9.00 9.50 | 2.95 2.80 2.45 2.17 2.14 | 13.05 13.50 14.00 | 3.32 3.30 3.15 | 22.00 23.26 | 3.29 2.95 | |||||||||

computer program, the optimum area of the bottle according to the Criterion of Optimization by Compensation must be 365.96 cm^{2} when the base of the bottle is flat.

It is worth to emphasize that, the volume of 551.63 cm^{3} in Output Data, of ^{3} because the program calculates the volume of the bottle with flat base. By measuring the volume of the cavities of the petaloid base (base of the real bottle) in laboratory by using water for filling such cavities, it is found a total volume of 14 cm^{3} approximately that have to be subtracted from 551.63 cm^{3}. So we have 537.63 cm^{3} which is a good approximation of the volume of the real bottle. Similarly, the area 435.972 cm^{2} in Output Data for the bottle is due to fact that the program calculates the area of the bottle by considering it with flat base. By calculations made apart in the zone of the petaloid base in order to determine its area by approximation by triangles and circles as shown in ^{2}, so that the difference in areas between the bottle with flat base and petaloid base is not significant.

In order to redesign the bottle, it is necessary to give some points through which the silhouette of the redesigned bottle passes. These points are given by the designer and are shown in

Section 1 | Section 2 | Section 3 | Section 4 | Section 5 | ||||||
---|---|---|---|---|---|---|---|---|---|---|

x | y | x | y | x | y | x | y | x | y | |

0.00 | 1.10 | 1.50 | 1.10 | 5.50 | 3.50 | 8.30 | 4.40 | 9.71 | 4.40 | |

1.50 | 1.10 | 2.50 | 3.20 | 6.00 | 3.65 | 9.71 | 4.40 | 10.01 | 4.50 | |

3.50 | 4.00 | 6.50 | 4.10 | 11.01 | 4.50 | |||||

4.00 | 4.00 | 7.00 | 4.45 | 12.01 | 3.80 | |||||

5.00 | 3.60 | 7.50 | 4.55 | |||||||

5.50 | 3.50 | 8.00 | 4.50 | |||||||

8.30 | 4.40 | |||||||||

according to the development of the model in Section 2.3, in general the polynomials P n + i ( x ) , i = 1 , k ^ are susceptible of changing, depending if some of h 1 , h 2 , ⋯ , h k are zero or different of zero. In this particular case we have only P 4 ( x − h 1 ) from P 4 ( x ) and we will find that h 1 ≈ 0 , so that the Lagrange polynomials of the redesigned bottle will be equal to the Lagrange polynomials of the optimized redesigned bottle.

It is worth to emphasize that, the coefficients of the polynomials, in

In order to get the optimized redesigned bottle, rest to find the polynomial P 4 ( x − h 1 ) from P 4 ( x ) that fit correctly with the design of the silhouette of the bottle. To do this we put in

Equations (73) and (74) with R 4 = 4.4 , R 5 = 3.8 where n = 3 , and in using the polynomial P 4 ( x ) become,

P 4 ( x 5 − h 1 ) − 3.8 = 0 , from Equation (74),

− 0.0406912 ( x 5 − h 1 ) 3 + 0.99403 ( x 5 − h 1 ) 2 − 7.40008 ( x 5 − h 1 ) + 15.9863 = 0 (88)

P 4 ( x 4 − h 1 ) − 4.4 = 0 , from Equation (73),

− 0.0406912 ( x 4 − h 1 ) 3 + 0.99403 ( x 4 − h 1 ) 2 − 7.40008 ( x 4 − h 1 ) + 15.3863 = 0 (89)

By looking

531.798009333735 = π ∫ 1.5 5.5 [ P 2 ( x ) ] 2 d x + π ∫ 5.5 8.3 [ P 3 ( x ) ] 2 d x + π [ P 3 ( 8.3 ) ] 2 ( x 4 − 8.3 ) + π ∫ x 4 x 5 [ P 4 ( x − h 1 ) ] 2 d x

where P 3 ( 8.3 ) = P 4 ( x 4 ) = 4.4 .

By numerical integration of the first two integrals and taking into account that P 3 ( 8.3 ) = 4.4 , we have,

Section | Lagrange polynomials of the redesigned bottle |
---|---|

S1 | P 1 ( x ) = 1.1 |

S2 | P 2 ( x ) = 0.0034921 x 5 − 0.040476 x 4 + 0.21706 x 3 − 1.3089 x 2 + 5.7384 x − 5.1167 |

S3 | P 3 ( x ) = − 0.036829113 x 6 + 1.4915791 x 5 − 24.923167 x 4 + 219.65835 x 3 − 1075.681 x 2 + 2772.3003 x − 2931.6348 |

S4 | P 0 ( x ) = 4.4 (Straight section) |

S5 | P 4 ( x ) = − 0.0406912 x 3 + 0.99403 x 2 − 7.40008 x + 19.7863 |

225.332507809904 = π ( x 4 − 8.3 ) ( 4.4 ) 2 + π ∫ x 4 x 5 [ P 4 ( x − h 1 ) ] 2 d x

From which

#Math_303# (90)

Since

π ∫ x 4 x 5 [ P 4 ( x − h 1 ) ] 2 d x = π ( 0.00165577375744 ( x 5 − h 1 ) 7 7 − 0.080896547072 ( x 5 − h 1 ) 6 6 + 1.590331911492 ( x 5 − h 1 ) 5 5 − 16.32205962592 ( x 5 − h 1 ) 4 4 + 94.0975355844 ( x 5 − h 1 ) 3 3 − 292.840405808 ( x 5 − h 1 ) 2 2 + 391.49766769 x 5 ) − π ( 0.00165577375744 ( x 4 − h 1 ) 7 7 − 0.080896547072 ( x 4 − h 1 ) 6 6 + 1.590331911492 ( x 4 − h 1 ) 5 5 − 16.32205962592 ( x 4 − h 1 ) 4 4 + 94.0975355844 ( x 4 − h 1 ) 3 3 − 292.840405808 ( x 4 − h 1 ) 2 2 + 391.49766769 x 4 )

where integration was performed analytically, Equation (90) becomes finally,

60.8212337734984 x 4 + π ( 0.00165577375744 ( x 5 − h 1 ) 7 7 − 0.080896547072 ( x 5 − h 1 ) 6 6 + 1.590331911492 ( x 5 − h 1 ) 5 5 − 16.32205962592 ( x 5 − h 1 ) 4 4 + 94.0975355844 ( x 5 − h 1 ) 3 3 − 292.840405808 ( x 5 − h 1 ) 2 2 + 391.49766769 x 5 ) − π ( 0.00165577375744 ( x 4 − h 1 ) 7 7 − 0.080896547072 ( x 4 − h 1 ) 6 6 + 1.590331911492 ( x 4 − h 1 ) 5 5 − 16.32205962592 ( x 4 − h 1 ) 4 4 + 94.0975355844 ( x 4 − h 1 ) 3 3 − 292.840405808 ( x 4 − h 1 ) 2 2 + 391.49766769 x 4 ) − 730.148748129941 = 0 (91)

Now, must be solved the system of Equations (88), (89) and (91).

Since a system of three nonlinear equations in three unknowns has been obtained, must be solved numerically in using the Newton’s Method for Nonlinear Systems, whose interface Solver For System of Nonlinear Equations (SNLEs) shows us the results of the values for the unknowns after perform 5 iterations (see

f 1 = − 0.0406912 ( x 5 − h 1 ) 3 + 0.99403 ( x 5 − h 1 ) 2 − 7.40008 ( x 5 − h 1 ) + 15.9863

f 2 = − 0.0406912 ( x 4 − h 1 ) 3 + 0.99403 ( x 4 − h 1 ) 2 − 7.40008 ( x 4 − h 1 ) + 15.3863

f 3 = 60.8212337734984 x 4 + π ( 0.00165577375744 ( x 5 − h 1 ) 7 7 − 0.080896547072 ( x 5 − h 1 ) 6 6 + 1.590331911492 ( x 5 − h 1 ) 5 5 − 16.32205962592 ( x 5 − h 1 ) 4 4 + 94.0975355844 ( x 5 − h 1 ) 3 3 − 292.840405808 ( x 5 − h 1 ) 2 2 + 391.49766769 x 5 ) − π ( 0.00165577375744 ( x 4 − h 1 ) 7 7 − 0.080896547072 ( x 4 − h 1 ) 6 6 + 1.590331911492 ( x 4 − h 1 ) 5 5 − 16.32205962592 ( x 4 − h 1 ) 4 4 + 94.0975355844 ( x 4 − h 1 ) 3 3 − 292.840405808 ( x 4 − h 1 ) 2 2 + 391.49766769 x 4 ) − 730.148748129941

Additionally, we need the elements of the Jacobian matrix, which are:

f 1 x = − 0.1220736 ( x 5 − h 1 ) 2 + 1.98806 ( x 5 − h 1 ) − 7.40008

f 1 y = 0

f 1 z = 0.1220736 ( x 5 − h 1 ) 2 − 1.98806 ( x 5 − h 1 ) + 7.40008

f 2 x = 0

f 2 y = − 0.1220736 ( x 4 − h 1 ) 2 + 1.98806 ( x 4 − h 1 ) − 7.40008

f 2 z = 0.1220736 ( x 4 − h ) 2 − 1.98806 ( x 4 − h ) + 7.40008

f 3 x = π ( 0.00165577375744 ( x 5 − h 1 ) 6 − 0.080896547072 ( x 5 − h 1 ) 5 + 1.590331911492 ( x 5 − h 1 ) 4 − 16.32205962592 ( x 5 − h 1 ) 3 + 94.0975355844 ( x 5 − h 1 ) 2 − 292.840405808 ( x 5 − h 1 ) + 391.49766769 )

f 3 y = − π ( 0.00165577375744 ( x 4 − h 1 ) 6 − 0.080896547072 ( x 4 − h 1 ) 5 + 1.590331911492 ( x 4 − h 1 ) 4 − 16.32205962592 ( x 4 − h 1 ) 3 − 94.0975355844 ( x 4 − h 1 ) 2 − 292.840405808 ( x 4 − h 1 ) + 391.49766769 ) + 60.8212337734984

f 3 z = π ( − 0.00165577375744 ( x 5 − h 1 ) 6 + 0.080896547072 ( x 5 − h 1 ) 5 − 1.590331911492 ( x 5 − h 1 ) 4 + 16.32205962592 ( x 5 − h 1 ) 3 − 94.0975355844 ( x 5 − h 1 ) 2 + 292.840405808 ( x 5 − h 1 ) ) − π ( − 0.00165577375744 ( x 4 − h 1 ) 6 + 0.080896547072 ( x 4 − h 1 ) 5 − 1.590331911492 ( x 4 − h 1 ) 4 + 16.32205962592 ( x 4 − h 1 ) 3 − 94.0975355844 ( x 4 − h 1 ) 2 + 292.840405808 ( x 4 − h 1 ) )

In order to enter Equations (88), (89), (91) and the elements of the Jacobian matrix to the SNLEs, in Matlab language, we change to an equivalent form. Then, taking into account x = x 5 , y = x 4 and z = h 1 , we have

f 1 = 92501 ∗ z 12500 − 92501 ∗ x 12500 + 99403 ∗ ( x − z ) 2 100000 − 3179 ∗ ( x − z ) 3 78125 + 22498717100947801 1407374883553280

f 2 = 92501 ∗ z 12500 − 92501 ∗ y 12500 + 99403 ∗ ( y − z ) 2 100000 − 3179 ∗ ( y − z ) 3 78125 + 21654292170815833 1407374883553280

f 3 = 484 ∗ π y 25 − 31017640839698776609179259142809 ∗ π ∗ ( x − z ) 79228162514264337593543950336 − ( 31017640839698776609179259142809 ∗ π ∗ ( y − z ) ) 79228162514264337593543950336 − 515170290029661592553 ∗ π ∗ ( x − z ) 2 3518437208883200000 + 1379484460268183051814177 ∗ π ∗ ( x − z ) 3 43980465111040000000000 − ( 22432867934940286126331 ∗ π ∗ ( x − z ) 4 ) 5497558138880000000000 + ( 397582977873 ∗ π ∗ ( x − z ) 5 ) 1250000000000 − ( 316002137 ∗ π ∗ ( x − z ) 6 ) 23437500000 + ( 10106041 ∗ π ∗ ( x − z ) 7 ) 42724609375 + ( 515170290029661592553 ∗ π ∗ ( y − z ) 2 ) 3518437208883200000

− ( 1379484460268183051814177 ∗ π ∗ ( y − z ) 3 ) 43980465111040000000000 + ( 22432867934940286126331 ∗ π ∗ ( y − z ) 4 ) 5497558138880000000000 − ( 397582977873 ∗ π ∗ ( y − z ) 5 ) 1250000000000 + ( 316002137 ∗ π ∗ ( y − z ) 6 ) 23437500000 − ( 10106041 ∗ π ∗ ( y − z ) 7 ) 42724609375 − 6422456308599681 8796093022208

f 1 x = 99403 ∗ x 50000 − 99403 ∗ z 50000 − 9537 ∗ ( x − z ) 2 78125 − 92501 12500

f 1 y = 0

f 1 z = 99403 ∗ z 50000 − 99403 ∗ x 50000 − 9537 ∗ ( x − z ) 2 78125 − 92501 12500

f 2 x = 0

f 2 y = 99403 ∗ y 50000 − 99403 ∗ z 50000 − 9537 ∗ ( y − z ) 2 78125 − 92501 12500

f 2 z = 99403 ∗ z 50000 − 99403 ∗ y 50000 − 9537 ∗ ( y − z ) 2 78125 − 92501 12500

f 3 x = 31017640839698776609179259142809 ∗ π 79228162514264337593543950336 − 515170290029661592553 ∗ π ∗ ( 2 x − 2 z ) 3518437208883200000 + 4138453380804549155442531 ∗ π ∗ ( x − z ) 2 43980465111040000000000 − 22432867934940286126331 ∗ π ∗ ( x − z ) 3 1374389534720000000000 + 397582977873 ∗ π ∗ ( x − z ) 4 250000000000 − 316002137 ∗ π ∗ ( x − z ) 5 3906250000 + 10106041 ∗ π ∗ ( x − z ) 6 6103515625

f 3 y = 515170290029661592553 ∗ π ∗ ( 2 y − 2 z ) 3518437208883200000 − 737094590335565475834206206607601 ∗ π 1980704062856608439838598758400 − 4138453380804549155442531 ∗ π ∗ ( y − z ) 2 43980465111040000000000 + 22432867934940286126331 ∗ π ∗ ( y − z ) 3 1374389534720000000000 − 397582977873 ∗ π ∗ ( y − z ) 4 250000000000 + 316002137 ∗ π ∗ ( y − z ) 5 3906250000

− 10106041 ∗ π ∗ ( y − z ) 6 6103515625

f 3 z = 515170290029661592553 ∗ π ∗ ( 2 x − 2 z ) 3518437208883200000 − 4138453380804549155442531 ∗ π ∗ ( x − z ) 2 43980465111040000000000 + 22432867934940286126331 ∗ π ∗ ( x − z ) 3 1374389534720000000000 − 397582977873 ∗ π ∗ ( x − z ) 4 250000000000 + 316002137 ∗ π ∗ ( x − z ) 5 3906250000 − 10106041 ∗ π ∗ ( x − z ) 6 6103515625 − 515170290029661592553 ∗ π ∗ ( 2 y − 2 z ) 3518437208883200000 + 4138453380804549155442531 ∗ π ∗ ( y − z ) 2 43980465111040000000000 − 22432867934940286126331 ∗ π ∗ ( y − z ) 3 1374389534720000000000 + 397582977873 ∗ π ∗ ( y − z ) 4 250000000000 − 316002137 ∗ π ∗ ( y − z ) 5 3906250000 + 10106041 ∗ π ∗ ( y − z ) 6 6103515625

In Approximations to roots of the Solver for System of Nonlinear Equations, we can see that x 5 = 12.0077 ≈ 12.01 , x 4 = 9.70766 ≈ 9.71 and h 1 = − 0.00231716 ≈ − 0.0023 . The values of x 5 = 12.0077 ≈ 12.01 , x 4 = 9.70766 ≈ 9.71 , x 3 = 8.3 , x 2 = 5.5 and x 1 = 1.5 , in

Therefore, with the data of

In Output data of this figure, we can see that we can manufacture a bottle with an area of 363.588 cm^{2} less than 365.96 cm^{2} specified in optimum area of optimum results for the bottle, which is in accordance with theorem 2.1 and similar results reported by Reyna and Morales [

When considering a petaloid base for the bottle of

Section | Lagrange polynomials of the optimized redesigned bottle |
---|---|

S1 | P 1 ( x ) = 1.1 |

S2 | P 2 ( x ) = 0.0034921 x 5 − 0.040476 x 4 + 0.21706 x 3 − 1.3089 x 2 + 5.7384 x − 5.1167 |

S3 | P 3 ( x ) = − 0.036829113 x 6 + 1.4915791 x 5 − 24.923167 x 4 + 219.65835 x 3 − 1075.681 x 2 + 2772.3003 x − 2931.6348 |

S4 | P 0 ( x ) = 4.4 |

S5 | P 4 ( x ) = − 0.0406912 x 3 + 0.99403 x 2 − 7.40008 x + 19.7863 |

for the real bottle (see ^{2} to 366.17 m^{2}.

On the basis of the Criterion of Optimization by Compensation it was obtained a mathematical model that let us optimize (minimize) the superficial area of a bottle with a straight section along its silhouette. The minimal area that this mathematical model let us obtain is in the sense that it tends to the area of the sphere which is considered as geometric object with maximal volume and less area. Say, in general the area of the bottle is between the area of the sphere and the area of the cylinder. This is not difficult to show in using proposition presented in [

Results of the application of the Criterion of Optimization by Compensation for the case of the half-liter bottle of the Fanta soda, are summarized in ^{3} with a superficial area of 435.23 cm^{2} approximately. However, in using the Criterion of Optimization by Compensation for almost the same volume (537.63 cm^{3}), the area can be reduced to 366.17 cm^{2} (with petaloid base) so that it is necessary only 21.185 g of PET plastic keeping the original thickness of the wall of the bottle. It is worth to emphasize that 21.185 g was obtained using a rule of three with the data of

lation areaofthebottle − areaoftheoptimizedbottle areaoftheoptimizedbottle × 100 ) respecting to what

is established by the Criterion of Optimization by Compensation and, it has 69.06 cm^{2} more superficial area or it uses 3.995 g more of PET plastic. In indications such as * and ** are specified that the results 366.17 cm^{2} and 537.63 cm^{3} are

Real | Optimum | Deviation | Volume-real | |||||||
---|---|---|---|---|---|---|---|---|---|---|

area | weight | area | weight | area | weight | Water | Matlab | |||

(cm^{2}) | (g) | (cm^{2}) | (g) | (cm^{2}) | (g) | (cm^{3}) | (cm^{3}) | |||

435.23 | 25.18 | 366.17 | 21.185 | 69.06 | 3.995 | 537.5 | 537.63 | |||

** It was | ||||||||||

* with | * with | 18.86% | Subtracted | |||||||

non-flat | non-flat | 14 cm^{3 }due to | ||||||||

base | base | non-flat base | ||||||||

due to the petaloid base of the bottle. The difference in areas between the bottlewith flat base and petaloid base is not significant in the present work.

The cost that it is paid when the optimization is achieved in using the Criterion of Optimization by Compensation is that the height of the bottle diminishes and a widening happens. The view at scale of the real bottle and the redesigned bottle optimized in extreme is shown in

Since no necessarily, an optimization in extreme must be achieved, it was attempted to manufacture a bottle with a mistake of 4.91% considered arbitrarily by the designer, say, with an area of 383.916 cm^{2} (This value is obtained by using

the relation areaofthebottle − areaoftheoptimizedbottle areaoftheoptimizedbottle × 100 = 4.91 , where

the area of the optimized bottle is 365.96 cm^{2}). Such attempt of manufacturing the bottle was achieved with certain limitations such as:

1) It was impossible to find preforms of PET plastic of 23.25 g approximately, as it was required by calculations.

2) It was impossible to find preforms of PET plastic whose design must be according to the

length of the lip of the bottle as it is required in the design.

3) The mechanism of exportation of data, to the numerical control lathe distorted the data. So, volume and area of the bottle were slightly altered.

In ^{2}, while the area of the bottle in Optimum results is

Section 1 | Section 2 | Section 3 | Section 4 | Section 5 | Section 6 | Section 7 | |||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|

x | y | x | y | x | y | x | y | x | y | x | y | x | y |

0.0 | 1.10 | 2.70 | 1.25 | 4.30 | 3.00 | 6.60 | 3.35 | 8.40 | 3.385 | 9.75 | 3.915 | 13.20 | 3.915 |

2.7 | 1.25 | 4.00 | 2.68 | 5.30 | 3.62 | 7.30 | 2.985 | 8.90 | 3.88 | 13.2 | 3.915 | 14.00 | 4.00 |

4.30 | 3.00 | 5.90 | 3.65 | 7.80 | 3.00 | 9.75 | 3.915 | 15.00 | 3.96 | ||||

6.60 | 3.35 | 8.40 | 3.385 | 16.20 | 3.085 |

365.960 cm^{2} being in both cases when the base of the bottle is flat. This means that there is a mistake of 4.91% with respect to optimum results. As a consequence of inserting to the bottle a petaloid base plus the limitations stated above the mistake increase as it is indicated below.

The above last limitation (number 3) allowed us to obtain a bottle whose geometric design had a mistake of 13% (This value was obtained by direct measurement of the area of the mold of the bottle and by using the formula

areaofthemoldofthebottle − areaoftheoptimizedbottle areaoftheoptimizedbottle × 100 , where the area

of the optimized bottle is 365.96 cm^{2} according to Optimum results for the bottle of

Therefore, according to this reports, we can note that, in all cases the wall

Section | Lagrange polynomials of the manufactured bottle |
---|---|

S1 | P 1 ( x ) = 0.0 5 ^ x + 1.1 |

S2 | P 2 ( x ) = − 0.020833333333335 x 2 + 1.239583333333336 x − 1.945000000000007 |

S3 | P 3 ( x ) = − 0.005166029622552 x 3 − 0.276176540850457 x 2 + 3.629662267080676 x − 7.090307990922156 |

S4 | P 4 ( x ) = 0.053631553631554 x 3 − 0.704280904280836 x 2 + 1.489891774891476 x + 8.776333333333241 |

S5 | P 5 ( x ) = − 0.702832244008714 x 2 + 13.148997821350747 x − 57.474738562091602 |

S6 | P 0 ( x ) = 3.915 |

S7 | P 6 ( x ) = − 0.0773358585859 x 3 + 0.031823232323232 x 2 − 0.435284469696971 x + 2.018724999999995 |

thickness of the bottles was changed, and the weight of the bottle reduced in average~4.0 g, similarly to the found results in this work. However, in this work the wall thickness of the bottle have to remain unchanged.

In summary, we propose the mathematical foundations of the so-called Criterion of Optimization by Compensation and in using this, an optimized and redesigned half-liter bottle of Fanta soda of the well-known Coca Cola Company, was manufactured. This manufactured bottle was designed with a mistake of 4.91% with respect to what such criterion of optimization stablishes. However, it was reported a mistake of 13% in each manufactured bottle due to important technical limitations listed above that must be overcome. In general, in spite of such limitations a good bottle was obtained with such 13% of mistake, resulting the thickness of the wall of the bottle 75% of that of the real bottle because a preform of 22 g of PET plastic was used, instead of a preform of 23.25 g as it was required.

The authors would like to thank Universidad Nacional del Santa of Peru for its financial support in the development of the present research and the Brazilian Agency CNPq for its partial financial support. Process N^{o} 460733/2014?1.

Zegarra, L.B.R., Armas, L.E.G. Reyna, A.D., Vergara, J.A.L. and Obeso, F.A.V. (2017) Mathematical Model of the Criterion of Optimization by Compensation for Designing Commercial Bottles with Lateral Surfaces of Revolution and a Straight Section along Its Silhouette. Open Journal of Optimization, 6, 115-147. https://doi.org/10.4236/ojop.2017.63009