Research article

Industrial optimization using three-factor Cobb-Douglas production function of non-linear programming with application

  • Received: 22 July 2023 Revised: 09 October 2023 Accepted: 16 October 2023 Published: 06 November 2023
  • MSC : 90C30, 90C90

  • This paper is about the effectiveness of the Cobb-Douglas (C-D) production function in industrial optimization, estimating the number of factors used in the production process of the water industry, for instance, capital and human labor. Moreover, we have modeled a nonlinear optimization problem for a local water industry using two and three factors of production. For this purpose, we have taken into account the Cobb-Douglas production function with different production factors using the Lagrange multiplier method with the ordinary least squares method. In the course of the solution, a linear function is used to calculate the cost function, and the C-D production function is used to calculate the production function. The Lagrange multiplier method with the ordinary least squares method is then used to solve the constrained optimization problem for the product of production. Furthermore, we compared the outcomes from both examples of two- and three-factor C-D production functions in order to validate the Lagrange multiplier method for the C-D production function. Moreover, the three-factor C-D production function is solved by the Lagrange multiplier method with the ordinary least squares method, which provides optimal results as compared to previous studies in literature. The validity of the proposed methodology is explained by using the products of a local production industry in Pakistan.

    Citation: Shakoor Muhammad, Fazal Hanan, Sayyar Ali Shah, Aihua Yuan, Wahab Khan, Hua Sun. Industrial optimization using three-factor Cobb-Douglas production function of non-linear programming with application[J]. AIMS Mathematics, 2023, 8(12): 29956-29974. doi: 10.3934/math.20231532

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  • This paper is about the effectiveness of the Cobb-Douglas (C-D) production function in industrial optimization, estimating the number of factors used in the production process of the water industry, for instance, capital and human labor. Moreover, we have modeled a nonlinear optimization problem for a local water industry using two and three factors of production. For this purpose, we have taken into account the Cobb-Douglas production function with different production factors using the Lagrange multiplier method with the ordinary least squares method. In the course of the solution, a linear function is used to calculate the cost function, and the C-D production function is used to calculate the production function. The Lagrange multiplier method with the ordinary least squares method is then used to solve the constrained optimization problem for the product of production. Furthermore, we compared the outcomes from both examples of two- and three-factor C-D production functions in order to validate the Lagrange multiplier method for the C-D production function. Moreover, the three-factor C-D production function is solved by the Lagrange multiplier method with the ordinary least squares method, which provides optimal results as compared to previous studies in literature. The validity of the proposed methodology is explained by using the products of a local production industry in Pakistan.



    Optimization is a useful technique for determining the optimal solution to a problem. In other words, optimization is the problem of choosing suitable inputs under given circumstances in order to get the best possible output. For instance, optimization can be used in production models to adjust different inputs and make them more effective in order to get the best output for the production of a particular industry [1]. In this scenario, once we have modeled a problem, it can be solved using the available optimization techniques to find the optimal solution [2].

    An optimization problem usually consists of three ingredients: an objective function, a set of constraints and a number of decision variables. There are two types of optimization problems: constrained optimization problems and unconstrained optimization problems. Constrained optimization problems have restriction(s) on the objective function, while unconstrained optimization problems have no restriction on the objective function [3,4].

    If at least one of the objective functions or the constraint function is nonlinear, then the problem is known as a nonlinear optimization problem. There are many techniques that have been used for the maximization or minimization of nonlinear optimization problems. Sometimes, a problem cannot be modeled correctly using linear programming; therefore, one can use nonlinear programming approaches [5,6] to model the problem. For the constrained optimization problem under consideration in this paper, the objective function, known as the cost function, is linear while the constraint function, known as the Cobb-Douglas (C-D) production function, is nonlinear. In the course of solution, the constrained optimization problem is converted into an unconstrained optimization problem and then solved by the Lagrange multiplier method using the ordinary least squares approach.

    There have been many problems in literature in which the C-D production function has been used. A mixed-integer linear programming (MILP) model was established for the optimization of production scheduling [7]. A two-factor C-D production function is carried out with the aim of picking the suitable C-D production model for calculating the production process of the selected manufacturing industries in Bangladesh [8]. By combining the allometric scaling concept, which is used to estimate the parameters of the C-D function, with the application in transportation problems in China, a novel algorithm for creating geographical C-D models is developed [9].

    For the improvement of different factors in the Polish metallurgical industry, the power regression C-D function was used with the aim of developing a number of production factors, for instance, net production, production sold and volume of steel production [10]. The proper management of a country's resources is an important issue for its economic development. Along similar lines, the optimization of water management for three industries that rely on water demand prediction, subject to a number of ambiguities, is handled by the use of the C-D production function [11]. A two-factor production function was used to model sustainable economic development with the goal of labor production in relation to the commodity production system's capital-labor ratio [12]. An economic model has been presented by the application of proper Inada conditions to the C-D production function, which converges to or diverges from per capita product and the steady state of capital [13]. Moreover, a two-factor C-D production function was presented in which the effects of labor force and capital on agricultural heritage systems are carried out to maximize profit as well as the sustainability of agricultural heritage through the application of a two-factor C-D production function in which they examine the impacts of major factors on agricultural productivity [14]. An algorithmic or analytical procedure was conducted to handle the issue of optimal utilization of resources towards a feasible and profitable model via the C-D production function [15]. An application of the C-D production function model was used to find the role of land in urban economic growth [16]. A general oligopolistic market equilibrium with nonlinear programming was considered, in which each firm's factor contributes to the system, and then solved by tensor variational inequality [17,18].

    It becomes necessary in a real-world optimization problem to adopt a set of nonlinear terms in a mathematical model in order to get particular operational features of the decision-making problem. On the other hand, when nonlinear terms exist in the course of a solution, they add to the computational complexity of the problem. For this purpose, the researchers have developed proper transformation as well as linearization methods for the optimization problems that consist of nonlinear terms.

    In this paper, a special type of production function, which was founded by Cobb and Douglas in 1928 and is known as the C-D production function, is under consideration. This function is based on empirical studies that have been applied to the economy for optimal production [19]. This function supplies a number of different inputs to the problem and, as a result, produces a unique output for the problem. These inputs may be two or more than two in number, depending on the factors of production used in the industry. Also, the function having more than two factors is known as the generalized C-D production function [20,21]. There have been many results in literature in which two-factor C-D function was presented for production. This work presents the optimal solution technique of the C-D function for three factors of production using the Lagrange multiplier method and the ordinary least squares method, with applications. Moreover, Nervole's C-D production function with three inputs using the ordinary least squares method has been presented in [22]. Furthermore, this work also developed an environment to transform and linearize an optimization problem with nonlinear objectives or nonlinear constraints by using existing techniques [23].

    The proposed work is summarized as follows:

    (1) To develop a model in order to transform as well as linearize the nonlinear terms.

    (2) To choose a suitable model [1] as an optimization problem and to solve it using a novel methodology known as C-D production function using the Lagrange multiplier method with the ordinary least squares method.

    (3) To find whether the best possible solution to the problem is obtained or not [2].

    (4) To apply the C-D production function using the Lagrange multiplier method with the ordinary least squares method for the solution of the two-factor C-D production function as well as for the three-factor C-D production function.

    (5) To apply the case study to industrial optimization in order to minimize costs more efficiently.

    First we solved the two-factor C-D production function [24] using the Lagrange multiplier method with the ordinary least squares method. Second we used three factors of the C-D production function with the same technique. The obtained results show that cost minimization in cases of three factors of production is more efficient as compared to cost minimization in cases of two factors of production.

    The problem under consideration depends on three basic factors:

    ● Cost of the company for manufacturing production.

    ● Quantity of production.

    ● Income from the sales of production according to market prices.

    For instance, the role of the water industry is to filter water in different purifying tanks, then add chemicals, pack them properly, and then sell them in the market. In this task, the main objective of the industry is to minimize costs or maximize production. There have been many approaches in literature in which the total cost of an industry is presented as a linear function [24,25].

    K=ni=1qixi+Rs=qTx+Rs (2.1)

    where n represents production factors.

    x=[x1x2...xn]

    is a vector of production factors.

    q=[q1q2...qn]

    is the vector of prices of production factors.

    The following relationships present the production volume [26]:

    y=p(x)=p(x1,x2...xn). (2.2)

    The income of the company from production sales is given by:

    S(y)=bTy (2.3)

    where,

    y=[y1y2...yn]

    is a vector of the quantity of produced goods.

    b=[b1b2...bn]

    is the vector prices of produced goods.

    The profit of the company is the difference between its income and its cost of production.

    Z(x,y)=S(y)K(x). (2.4)

    The main focus of this paper is to minimize the cost at a specific level of production and, consequently, to maximize production with more factors of production.

    The optimization problem originating from our data analysis consists of a linear function and a nonlinear function. We have used the linear function as a cost function, while the nonlinear C-D production function is used as production output. Let us consider a linear cost function as an objective function and the nonlinear C-D production function as a constrained function. Then, this becomes a constrained optimization problem with an equality constraint. For the sake of simplicity, we convert this constrained optimization problem to an unconstrained optimization problem by using the Lagrange multiplier method, and then we solve this constrained optimization problem for stationary points. Let us start with a function Z of two independent variables that is subject to an equality constraint function g. The objective and constraint function are given as follows:

    minz(x,y).

    Subject to:

    g(x,y)=0.

    We have, using the Lagrange multiplier,

    L(x,y,λ)=z(x,y)+λg(x,y). (2.5)

    In order to solve the problem, we must first determine the values of x, y, and λ. Note that we consider the cost function to be an objective function and the C-D production function as a constraint function, and solve it for different factors of production. Furthermore, we extend this work as an application to optimize production factors in a specific industry.

    The production function for three inputs is as follows:

    P=AXa1Ya2Za3. (2.6)

    After linearizing, we get

    lnP=lnA+a1lnX+a2lnY+a3lnZ. (2.7)

    After this, we will use least squares linear regression to find out the structural parameters, which are A,a1,a2,a3. The corresponding cost functions is given by

    C(X,Y,Z)=α1+α2X+α3Y+α4Z. (2.8)

    Along similar lines as Eq (2.5), we can get the corresponding Lagrange function for three-factor C-D production function.

    In other words, the overall methodology can also be summarized as follows:

    (1) First, model an objective and constraint function using two inputs for the given industry.

    (2) Transform a constrained optimization problem into an unconstrained optimization problem.

    (3) Solve the unconstrained problem using the Lagrange multiplier method with ordinary least squares.

    (4) Repeat the above procedure for three factors of production for the given industry.

    (5) Compare the outcomes in both cases.

    Given input prices, a cost function shows how much it costs to produce various output levels. In the course of solving such problems, one or both productions and factors of production may be stated by using their values. It is good practice to present the products of an industry in proper units that have a number of production components. In a similar way, human labor can be calculated when needed. When the availability of aggregated data is smaller, it can be measured with headcount or work time. When a higher level of aggregation is available, then the value of human labor seems more suitable. The most challenging task is the quantitative description of the capital used in the industry. In the majority of analyses, it becomes challenging due to the use of a number of factors of production. The greater a company's asset base, the less it is associated with high productivity [27]. In the following, we analyzed a local water industry with a two-factor C-D production function and a three-factor C-D production function using the Lagrange multiplier method with the ordinary least squares method and compared their results.

    For this problem, the data is taken from a local water industry Abysin Water Industry, which is one of the local registered branches of Chemtronics Water Services in Lahore, Pakistan. First, the problem is solved for two factors, and then it is generalized. In this case, the cost function consists of two factors (labor and capital) of production, which is given by:

    C(X,Y)=α1+α2X+α3Y (3.1)

    where,

    C(X, Y) represents the cost of the industry in rupees,

    α1 represents the fixed cost of the industry,

    α2 represents the unit price of labor per hour,

    α3 represents the unit price of capital per kg,

    X represents the number of labor hours,

    Y represents the amount of capital in kg.

    All the prices in the industry are in Pakistani currency (the rupee). The raw materials used as capital are taken in kilograms, i.e., the unit for capital is kg.

    Here,

    α1=75000,α2=1400,α3=100.

    Putting these values in Eq (3.1), we get,

    C(X,Y)=75000+1400X+100Y. (3.2)

    For production, we use the C-D production function, which is given by:

    P=AXaYb (3.3)

    where

    P represents the amount of production in liters.

    After Eq (3.3) we have,

    lnP=lnA+alnX+blnY. (3.4)

    The analyzed data for the water industry with two inputs, i.e., human labor and capital, and production as an output for the year 2022 is given in Table 1.

    Table 1.  Inputs and output values of the water industry with two inputs.
    Months Production(P) Labors(X) Capital(Y) Ln(P) Ln(X) Ln(Y)
    in litres in hours in kg
    Jan. 36,000 300 900 10.49127 5.703782 6.802395
    Feb. 34,000 272 833 10.43412 5.605802 6.725034
    March 36,700 315 924 10.51053 5.752573 6.828712
    April 37,000 330 942 10.51867 5.799093 6.848005
    May 35,700 278.5 874 10.48291 5.629418 6.77308
    June 36,400 321 922 10.50232 5.771441 6.826545
    July 33,600 269 860 10.42228 5.594711 6.756932
    Aug. 35,500 308 950 10.47729 5.7301 6.856462
    Sep. 36,875 338 957 10.51529 5.823046 6.863803
    Oct. 37,000 321 914 10.51867 5.771441 6.817831
    Nov. 33,600 266 875 10.42228 5.583496 6.774224
    Dec. 36,500 336 890 10.50507 5.817111 6.791221

     | Show Table
    DownLoad: CSV

    We have used ordinary least squares regression in order to find the structural parameters of the given C-D production function. The data analysis has been done using Microsoft Excel, which is given in Regressions 1 and 2 for two and three inputs, respectively.

    Figure Regression 1.  Linear regression for two inputs.
    Figure Regression 2.  Linear regression for three inputs.

    Structural parameters for two inputs are as follow:

    A=3212.468,a=0.3568,b=0.0542,
    L(X,Y,λ)=C(X,Y)+λH(X,Y), (3.5)

    where,

    H(X,Y)=360003212.468X0.3568Y0.0542.

    Using Eq (3.2) and H(x, y) in Eq (3.5), we have,

    L=75000+1400X+100Y+λ(360003212.468X0.3568Y0.0542). (3.6)

    Now taking partial derivatives of L with respect to X,Y and λ respectively and equating to zero, i.e.,

    LX=0,

    we get,

    14001146.2085λX0.6432Y0.0542=0,

    which implies that

    λ=1.2214X0.6432Y0.0542. (3.7)

    Now, when

    LY=0,

    we get

    100174.1157λX0.3568Y0.9458=0,

    which implies

    λ=100X0.3568Y0.9458. (3.8)

    Similarly, when

    Lλ=0,

    we get,

    360003212.468X0.3568Y0.0542=0. (3.9)

    Comparing (3.7) and (3.8),

    1.2214X0.6432Y0.0542=100X0.3568Y0.9458,

    we get

    Y=2.1266X. (3.10)

    Putting Eq (3.10) in Eq (3.9), we have

    360003212.468X0.3568(2.1266X)0.0542=0.

    Solving for X, we have

    X=323.7554. (3.11)

    From (3.10), we have

    Y=688.4982, (3.12)

    and using the values of X and Y in Eq (3.2), we have,

    C(X,Y)=597107. (3.13)

    Table 2 compares the actual cost with the estimated cost.

    Table 2.  Actual and estimated cost values for water industry using two inputs.
    Months Estimated cost Actual cost Error (R) Square of the error (R2)
    in rupees in rupees
    Jan. 585,000 608,000 23000 529000000
    Feb. 539,100 570,000 30900 954810000
    March 608,400 588,000 -20400 416160000
    April 631,200 608,000 -23200 538240000
    May 552,300 598,000 45700 2088490000
    June 616,600 621,000 4400 19360000
    July 537,600 574,000 36400 1324960000
    Aug. 601,200 564,000 -37200 1383840000
    Sep. 643,900 622,000 -21900 479610000
    Oct. 615,800 598,000 -17800 316840000
    Nov. 534,900 582,000 47100 2218410000
    Dec. 634,400 632,000 -2400 5760000
    R2 10275480000
    Mean square error 856290000

     | Show Table
    DownLoad: CSV

    Figure 1 represents the actual and theoretical costs of the industry. For example, in January 2022, we can see that the actual cost was greater than the estimated cost, but from the data analysis, our calculated cost is less than both the actual and estimated cost. Similarly, if we compare the cost from the table with our calculations, we can see the difference. It means that there is a sufficient difference in both costs, meaning that costs are minimized to a great extent.

    Figure 1.  Estimated cost versus actual cost of the industry using two inputs.

    In this case, the cost function consists of three factors: capital, chemicals, and labor. The cost function is given by:

    C(X,Y,Z)=α1+α2X+α3Y+α4Z (3.14)

    where

    α1 represents the fixed cost of the industry,

    α2 represents the unit price of labor per hour,

    α3 represents the unit price of chemicals per kg,

    α4 represents the unit price of capital per kg,

    X represents the number of labor hours,

    Y represents the amount of chemicals in kg,

    Z represents the amount of capitals in kg,

    where

    α1=75000,α2=1250,α3=2000,α4=100.

    Putting the above values in Eq (3.14), we get

    C(X,Y,Z)=75000+1250X+2000Y+100Z. (3.15)

    Now, the production function, as in [26], is given by

    P=AXa1Ya2Za3 (3.16)

    where

    P represents amount of production in liters.

    After linearizing, we get

    lnP=lnA+a1lnX+a3lnY+a3lnZ. (3.17)

    A data analysis of the water industry consisting of three inputs for the year 2022 is given in Table 3. The three inputs are human labor, capital, and chemicals, respectively.

    Table 3.  Inputs and output values of the water industry with three inputs.
    Months Labors(X) Chemicals(Y) Capitals(Z) Production(P) Ln(X) Ln(Y) Ln(Z) Ln(P)
    in hours in kg in kg in litres
    Jan. 240 60 900 36,000 5.4806 4.0943 6.8023 10.4912
    Feb. 220 52 833 34,000 5.3936 3.9512 6.7250 10.4341
    March 252 63 924 36,700 5.5294 4.1431 6.8287 10.5105
    April 264 66 942 37,000 5.5759 4.1896 6.8480 10.5186
    May 222 56.5 874 35,700 5.4026 4.0342 6.773 10.4829
    June 260 61 922 36,400 5.5606 4.1108 6.8265 10.5023
    July 215 54 860 33,600 5.3706 3.9889 6.7569 10.4222
    Aug. 250 58 950 35,500 5.5214 4.0604 6.8564 10.4772
    Sep. 268 70 957 36,875 5.5909 4.2484 6.8638 10.5152
    Oct. 256 65 914 37,000 5.5451 4.1743 6.8178 10.5186
    Nov. 212 54 875 33,600 5.3565 3.9889 6.7742 10.4222
    Dec. 262 74 890 36,500 5.5683 4.3040 6.7912 10.5050

     | Show Table
    DownLoad: CSV

    This data is collected from the water industry for the year 2022, and all results are calculated on a monthly basis.

    From these analyses, we have,

    A=3938.188,a1=0.2944,a2=0.0673,a3=0.0457.

    Putting these values in (3.16), we get

    P=3938.188X0.2944Y0.0673Z0.0457. (3.18)

    Using the Lagrange multipliers method, we have,

    L(X,Y,Z,λ)=C(X,Y,Z)+λH(X,Y,Z), (3.19)

    where

    H(X,Y,Z)=360003938.188X0.2944Y0.0673Z0.0457. (3.20)

    Putting Eqs (3.15) and (3.20) in Eq (3.19), we obtain

    L=75000+1250X+2000Y+100Z+λ(360003938.188X0.2944Y0.0673Z0.0457). (3.21)

    Now, taking partial derivatives of L with respect to X, Y, Z and λ and setting all partial derivatives equal to zero, we have,

    LX=0.

    we then get

    12501159.4925λX0.7056Y0.0673Z0.0457=0.

    Solving the above equation for λ, we get

    λ=12501159.4925X0.7056Y0.0673Z0.0457. (3.22)

    also setting

    LY=0,

    we get

    2000265.04λX0.2944Y0.9327Z0.0457=0.

    Solving above equation for λ, we get

    λ=7.5460X0.2944Y0.9327Z0.0457. (3.23)

    Finally setting

    LZ=0,

    we get

    100179.9751λX0.2944Y0.0673Z0.9543=0.

    Solving above equation for λ, we get

    λ=100179.9751X0.2944Y0.0673Z0.9543. (3.24)

    Similarly, when

    Lλ=0,

    we get

    360003938.188X0.2944Y0.0673Z0.0457=0. (3.25)

    Comparing (3.22) and (3.23), we get

    12501159.4925X0.7056Y0.0673Z0.0457=7.5460X0.2944Y0.9327Z0.0457

    which implies

    X=6.9996Y. (3.26)

    Now, comparing (3.23) and (3.24), we get

    7.5460X0.2944Y0.9327Z0.0457=100179.9751X0.2944Y0.0673Z0.9543,

    which implies

    Z=13.5809Y. (3.27)

    Using (3.25) and (3.26) in (3.27)

    (6.9996Y)0.2944Y0.0673(13.5809Y)0.0457=9.1412,

    yields,

    Y0.4074=4.5755.

    Thus,

    Y=41.7932. (3.28)

    Putting (3.28) in (3.26) and (3.27) respectively, we get

    X=292.5356,Z=567.5892, (3.29)

    and using the values of X, Y and Z in Eq (3.15), we have

    C(X,Y,Z)=581014. (3.30)

    Table 4 presents actual and estimated costs on a monthly basis for the water industry for the year 2022.

    Table 4.  Actual and estimated cost values for the water industry using three inputs.
    Months Estimated cost Actual cost Error (R) Square of the error (R2)
    in rupees in rupees
    Jan. 585,000 608,000 23000 529000000
    Feb. 537,300 570,000 32700 1069290000
    March 608,400 588,000 -20400 416160000
    April 631,200 608,000 -23200 538240000
    May 552,900 598,000 45100 2034010000
    June 614,200 621,000 6800 46240000
    July 537,750 574,000 36250 1314062500
    Aug. 598,500 564,000 -34500 1190250000
    Sep. 645,700 622,000 -23700 561690000
    Oct. 616,400 598,000 -18400 338560000
    Nov. 535,500 582,000 46500 2162250000
    Dec. 639,500 632,000 -7500 56250000
    R2 10256002500
    Mean square error 854666875

     | Show Table
    DownLoad: CSV

    Figure 2 represents the actual and estimated cost of the industry for the three-factors C-D production function.

    Figure 2.  Estimated cost versus actual cost for the industry for three inputs.

    We have solved the C-D production function with two and three factors of production using the Lagrange multiplier method with the ordinary least squares method. Moreover, this is an optimal solution approach for the C-D production function with three factors of production using the Lagrange multiplier method with the ordinary least squares method. Despite the fact that the proposed approach is a different solution technique as compared to the existing solution techniques in the literature, we still compared the general features of the proposed methodology for the water industry with different factors to Nervole's approach. In the following, we compared the presented solution approach with Nervole's approach [22]. In Table 5, we have differentiated Nerlove's C-D function and the C-D production function with the Lagrange multiplier method.

    Table 5.  Comparative analysis between Nerlove's C-D function and the C-D production function with the Lagrange multiplier method.
    No. Nerlove's C-D cost function C-D production function
    1 Nerlove used the C-D cost model We used the C-D production function
    2 Nerlove used the function to estimate the cost We used the given function as an output
    3 Nerlove approach needs much algebra for evaluation It has a simple implementation
    4 Computational complexity is too much Having less computational complexity

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    In addition, the paired t-test is used to determine if there is a significant difference between the means of two related data sets as well as to find the mean square error from the findings of both the two- and three-factor C-D production functions. In the case of the findings of the two-factor C-D function, the outcomes of a paired t-test on two data sets result in a P-value of 0.547 and degrees of freedom (df) of 11. The paired t-test compares the means of the two variables to determine if there is a significant difference between them. The null hypothesis is that there is no significant difference between the means of the two variables. This means that there is no significant difference between the means of the two variables at the 5% level of significance. This showed that the P-value is greater than the critical value of 0.05, indicating that we cannot reject the null hypothesis.

    In the case of the findings from the three-factor C-D function, the findings of a paired t-test result in a P-value of 0.559 and degrees of freedom (df) of 11. Moreover, the findings from the three-factor C-D using a paired t-test were also conducted, which also showed that there is no significant difference between the means of the two variables with the same 5% level of significance.

    Based on the findings of the paired t-test, we conclude that there is no significant difference between the means of the two related variables from the findings of two and three factors in the C-D production functions. This shows that the C-D production function plays a key role in the production problem when using the Lagrange multiplier method with the ordinary least squares method. Furthermore, we have worked on the cost comparison of two and three factors in the C-D production function. In the case of two factors of production, the cost value is 597107 per unit, while in the case of three factors of production, the cost value is 581014 per unit. In both cases, our calculated cost is less than the actual cost of the industry. Besides, the cost calculation for the three-factor C-D production function is less than that of the two-factor C-D production function. Clearly, we can see the differences in Tables 69.

    Table 6.  Data set for the two-factor C-D production function.
    Paired samples statistics (Pair 1)
    Mean N Std. deviation Std. error mean
    Estimated cost 591700.0000 12 40670.96351 11740.69586
    Actual cost 597083.3333 12 21997.76159 6350.20679
    Paired samples correlations (Pair 1)
    N Correlation Sig.
    Estimated cost and Actual cost 12 0.690 0.013

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    Table 7.  Paired sample test.
    Paird differences
    Mean Std. derivation Std. eror mean 95s% confidence interval of the difference t df Sig. (2-tailed)
    Lower Upper
    Pair1 estimated cost-actual cost 5383.33333 30041.96560 8672.36846 -24471.08762 13704.42096 -0.621 11 0.547

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    Table 8.  Data set for the three-factor C-D production function.
    Paired samples statistics (Pair 1)
    Mean N Std. deviation Std. error mean
    Estimated cost 591862.5000 12 41320.23016 11928.12300
    Actual cost 597083.3333 12 21997.76159 6350.20679
    Paired samples correlations (Pair 1)
    N Correlation Sig.
    Estimated cost and Actual cost 12 0.709 0.010

     | Show Table
    DownLoad: CSV
    Table 9.  Paired sample test.
    Paird differences
    Mean Std. derivation Std. error mean 95s% Confidence interval of the difference t df Sig. (2-tailed)
    Lower Upper
    Pair1 estimated cost-Actual cost 5220.83333 30043.78213 8672.89265 -24309.74179 13866.07513 -0.602 11 0.559

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    In this paper, the optimal solution developed by C-D is carried out with different production factors. From these analyses, we concluded that the C-D production function plays a key role in the production problem when using the Lagrange multiplier method with the ordinary least squares method. Moreover, we solved the constrained optimization problem with a two-factor and three-factor C-D production function using the Lagrange multiplier with the ordinary least squares method. In the case of two factors of production, the cost value is 597107 per unit, while in the case of three factors of production, the cost value is 581014 per unit. This showed that, with more production factors in the C-D production function, the cost value is minimized to a high extent. This validates that the C-D production function with more factors using the Lagrange multiplier is more effective than previous approaches in literature. Moreover, the presented solution methodology is compared to Nervole's C-D production function. This means that the individual expression of each factor as an input has a key role in obtaining the best optimized results. Moreover, we optimized the overall cost of the water industry using the three-factor C-D production function as an application of C-D production using the Lagrange multiplier method with the ordinary least squares method.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    We are thankful to the National Natural Science Foundation of China (under research grant no. 22150410332) and the start-up foundation for the introduction of talent at Jiangsu University of Science and Technology, China.

    The authors declare that they have no competing interest.



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