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 [18] . This optimization criterion consists in designing first a cylinder with a minimal area for a given volume in using Equations (2) and (4) and then redistributes the total volume and area that will be removed from de cylinder in order to get the desired shape of the bottle that we wish to manufacture. At the end of the process, the volume of the resultant bottle must be equal to that of the original cylinder, where the area of the resultant bottle could be less than that of the cylinder, as it is shown in theorem 2.1 below. Now, the unique proposition given in [19] tell us that the area of a cylinder of minimal area is among the area of a sphere and the area of a cube having all of them the same volume. But, the area of the bottle turns out to be less than that of the cylinder of minimal area, so we have the area of a bottle closer to the area of a sphere than the area of a cylinder of minimal area. Of course, the volumes of the bottle, cylinder of minimal area and sphere are the same. Since the sphere is considered as a geometric object with maximum volume and lesser area, the Criterion of Optimization by Compensation turns out to be a good method for optimizing.

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 (Figure 1(a)) in order to get the desired shape of the bottle, it must be removed a solid which is enclosed by an external surface, the surface of the cylinder, and internal surface, the surface of revolution that will give the shape to the bottle. See in the upper part of Figure 1(b) signed with the arrow. Calculate the difference of areas between the cylindrical surface and the surface of revolution that both enclose the solid that was removed. With this difference must be formed a cylinder without bases whose volume must be equal approximately to the volume that was removed (see Figure 1(b));

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 Figure 1(c)).

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

Figure 2 shows the scheme to be used in order to prove the theorem given above.

From scheme of Figure 2 we have:

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 Figure 2 and according to the Criterion of Optimization by Compensation. This let us find a = 15.1384 u . With this data, the area and volume of the solid of revolution that is generated when the region enclosed by the polygonal OABGJMN and x-axis of Figure 2, rotates around x-axis, are equal to A s = 360.684 u 2 and V s = 546.249 u 3 respectively.

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 Figure 3, which is based in the criterion of optimization by compensation seen in the xy-plane.

The mathematical model taking into account the Criterion of Optimization by Compensation is as follows: consider the schema (shown in Figure 3) corresponding to a general bottle with a straight section along its silhouette, where V i , i = 1 , n + k ^ , are volumes generated when the regions below the curves represented by the non-constant polynomials P i ( x ) , i = 1 , n + k ^ of certain

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 Figure 3

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 Figure 3). By Criterion of Optimization by Compensation

∑ 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 Figure 3, where the coefficients of these polynomials are unknowns. Thus, the Optimization problem can be formulated as follows: given the equation

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.

We 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 Figure 4. If this were the case, we must try to choose other points by varying slightly some points of the initial set of points until the oscillations have vanished. Polynomial oscillation could be present when a number of chosen points are not appropriated in order to get a Lagrange polynomial interpolation. Therefore, in order to get a desirable solution for the problem of optimization and choose a set of points in such a way that the poly

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 Figure 5. The characteristics of the bottle are: Type of the product that it contains: carbonated drink; Factory: Coca Cola company; volume of the bottle: 537.5 ml; weight of the empty bottle without its cap: 25.18 g; material from which the bottle is manufactured: Polyethylene terephthalate (PET); base of the bottle: petaloid base of five cavities.

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 Table 1:

Figure 6(a) shows for the real bottle, the interface of the computer program elaborated in the high-level programming language MATLAB. This computer program has five options: interpolate, volume, area, surface of revolution and optimize in order to perform the tasks. The calculations are performed by considering the bottle with a flat base. Adjustments are made in Input Data of each section. In this case, it was considered five sections ( S i , i = 1 , 5 ^ ) for the bottle. In Optimizer, the volume of the bottle to be optimized must be entered. As a result, in using Equations (2)-(4), we get optimum results for the cylinder, such as: radio, height and area. In Output Data of Figure 6(a) are shown the results of the volume and area of the bottle, according to the adjustments that were made in Input data of each section. When the optimization is being performed the value of the area varies until get approximately the value of the area obtained in Optimum results for the cylinder, keeping approximately constant the value of the volume in Optimizer (see Figure 6(a)). In using the computer program and with the data of the Table 1, the real bottle in three dimensions is shown in Figure 6(b). As, it can be seen from the results of the calculations in using the

computer program, the optimum area of the bottle according to the Criterion of Optimization by Compensation must be 365.96 cm² when the base of the bottle is flat.

It is worth to emphasize that, the volume of 551.63 cm³ in Output Data, of Figure 6(a), for the bottle exceeds 537.5 cm³ 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 cm³ approximately that have to be subtracted from 551.63 cm³. So we have 537.63 cm³ which is a good approximation of the volume of the real bottle. Similarly, the area 435.972 cm² 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 Figure 7, we find that the area of the bottle with petaloid base is 435.23 cm², 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 Table 2, being shown a set of them in Figure 8 by circles in sections S1, S2, S3, S4 and S5. The task is to determine by Lagrange interpolation the polynomials that pass through these points. The number x 4 in the x-axis, is the right extreme of the section S4 which is a straight line, and left extreme of section S5 represented by the polynomial P 4 ( x ) . This number must be determined by moving section S5 to the right or left according to the sign of h 1 , positive or negative in P 4 ( x − h 1 ) as it is shown in Figure 8. These polynomials are the starting points to get the final polynomials that determine the silhouette of the redesigned bottle and hence optimized. As it can be seen

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 Table 3, are expressed in the form of decimals in order to decrease distortions in the design of the bottle, optimization and during passing data to the CNC lathe. Greater the number of considered decimal places better will be the design. These criteria are taken into account in all the calculations that are made in all the process of optimization and must be presented in this way.

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 Table 2, x 4 instead of 9.71 and x 5 instead of 12.01 and a new Table with new values for x 4 and x 5 must be found.

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 Figure 8 we see that P 1 ( x ) = 1.1 , P 4 ( x ) = 4.4 , x 1 = 1.5 and r = 1.1 , h c = 8.81 , r c = 4.41 . In using the data of Table 3, Equation (67) can be written as,

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,

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 Figure 9). To solve the system, Equations (88), (89) and (91) are renamed as f 1 , f 2 and f 3 , where:

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

Figure 9 shows the SNLEs, where it is possible to see the entry functions f 1 , f 2 and f 3 ; Jacobian matrix f 1 x , f 2 x , f 3 x , f 1 y , f 2 y , f 3 y , f 1 z , f 2 z and f 3 z ; initial points x o , y o and z o as well as the approximations to roots.

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 Figure 9, are similar to values of Table 2; and P 4 ( x − h 1 ) = P 4 ( x + 0.0023 ) ≈ P 4 ( x ) , P 3 ( x ) , ⋯ , P 1 ( x ) similar to values of Table 3. With these results, and in using the Optimizer of Five Sections, we get the polynomials for the optimized redesigned bottle shown in Lagrange polynomial coefficients of Figure 10, and which are shown in Table 4.

Therefore, with the data of Table 2 where x 5 = 12.007 ≈ 12.01 and x 4 = 9.70766 ≈ 9.71 and following the same procedure that was made for the real bottle, the results for the redesigned bottle, after applying criterion of optimization by compensation, are shown in Figure 10.

In Output data of this figure, we can see that we can manufacture a bottle with an area of 363.588 cm² less than 365.96 cm² 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 [19] .

When considering a petaloid base for the bottle of Figure 10, and according to the design of a new petaloid base following similar calculations to that made

for the real bottle (see Figure 6), the area of the bottle is modified from 365.96 cm² to 366.17 m².

Table 4 shows the Lagrange polynomials, of each section, of the optimized redesigned bottle. We can see similar results to that given in Table 3, which means that the optimization criterion by compensation works very well.