Plastic Polymer Copolymer



An Illustration of Linear Programming -

Plastic Soul Company: An Illustration of Linear Programming - Graphical Solution

Plastic Soul Company manufactures two types of plastics: a functional polymer branded Standard Type 2 and a copolymer branded Vintage Type 1. It yearns to find the optimal mix of plastics to be produced that yields the maximum profit after taking into account the constraints present during the manufacturing process.

In determining which model to use to be able to solve the given problem, we should identify what type of problem is present. Here, the company has only one objective that is to maximize their profit. They want to find out which combination of products should be produced and how many of them. But along the manufacturing process there are several restrictions to be taken account for. The problem presents three types of constraints. The first restriction is that the machine producing the products can only handle up to a certain weight. The truck delivering the products also has this same kind of weight restriction. And lastly, there exist a short supply of flame-retarding agent needed for the production. With these restrictions come different courses of action that the management can choose from. They can decide how much they have to devote on each of the products. And the last thing is that the objective and constraints can be expressed as inequalities or in terms of linear equations. These equations are stated in detail in the next discussions. With all these properties that the problem has, it fits the requirements to be able to considered a linear programming problem (Hanna, Render & Stair, 2007). Furthermore the problem has two decision variables, for which the graphical solution of a linear programming model can be used (Hanna et al., 2007).

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Linear programming optimizes linear objective functions which, are subject to constraints. In lay man's term, linear programming is the process of searching for the best solution obtainable under certain circumstances. The word programming refers to planning: planning to yield desired outcome after considering the limitations. Linear programming can be applied in solving problems in business management: operation research. It is applicable to merchandising or manufacturing company. Normally, it is used to determine the optimal mix of products to be produced to either maximize profit given limited resources or minimize cost of production while maximizing available resources. Moreover, another purpose of linear programming is to properly allocate limited resources to prevent any surplus or wastes.

Objective Function and Constraints

Plastic Soul Company makes two types of plastics, the Standard Type 2 and Vintage Type 1, and they want to maximize their profit by knowing how much of each type they should produce. Knowing that the contribution from each truckload of Standard is $950 and the contribution of each truckload of Vintage is $1,200, we can derive the objective function, max z = 950x1 + 1200x2, where x1 is the Standard Type 2 while the Vintage Type 1 is the x2.

Other that the implicit constraint, x1, x2 ≥ 0, there are three other constraints found in this problem. Given that the machine can produce not more than 70 metric tons of any mix output of plastic per day and knowing that one truckload of Standard polymer weighs 1.4 metric tons and Vintage copolymer weighs 2.8 metric tons, we can derive our first constraint, 1.4x1 + 2.8x2 ≤ 70. It is also stated that the loading facility can handle up to 30 trucks per day so we can derive our second constraint, 1x1 + 1x2 ≤ 30, to determine how many of each type of plastics should be loaded if the facility could handle only up to 30 trucks per day. Also, it is stated that a truckload of the Standard polymer requires an input of 3 canisters of agent and 1 canister of agent for the vintage although the plant can only provide 65 canisters of agent per day, at most. So we can identify our third constraint which is 3x1 + 1x2 ≤ 65 to know how much of each type of plastic can maximize the 65 canisters of agent per day.

Solution Using QM for Windows

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To solve the problem given the variables and constraints, we can use the QM for Windows program. In QM for Windows, choose the Linear Programming in the Module tab and create a new file (see Figure 1). A dialogue box, as seen in Figure 2, will then appear for you to create the data set for Linear Programming. For the title, enter the title of the case which is the Plastic Soul Company. Then enter the corresponding number of Constraints and Variables, which is 3 and 2 respectively. Click on the objective identified in the problem which is to maximize profit then click OK.

Provide the name of the types of plastic in the first row by replacing the label Standard with Standard Type 2 and Vintage Type 1. Subsequently, replace the constraints in the first column with the label of the constraints in the problem, metric tons/day, trucks/day and canisters of flame retardant. Take note that the Maximize and RHS label would not be changed (see Figure 3).

Given the two variables, z = 950x1 + 1200x2, and 3 constraints, 1.4x1 + 2.8x2 ≤ 70 , 1x1 + 1x2 ≤ 30 and 3x1 + 1x1 ≤ 65, enter first the corresponding coefficient of the variables, 950 and 1,200, into maximize row. Input the coefficient of the constraints into their corresponding rows, for x1 in the Standard Type 2 column and x2 in the Vintage type 1 column then choose the sign for the constraint, <=, then place the values in the Right hand side of the constraint into the RHS column (see Figure 3).

After entering all the data, press F9 to solve the problem, and choose Cascade in the Window tab to see the tables such as the Linear Programming Results, Ranging, Solution List, Iterations and Graph of the problem cascaded in the Work Place (see Figure 4). As for Linear Programming, we would only need the Table for the Results (Table 1) and the Graph (Figure 5) of the problem.


The lines represent the constraints. They are the limitations or the maximum use of the resources. In this problem, the lines represent ≤ constraints. Therefore, the points beyond the lines are beyond the limitations of the limited resources that the firm has. For example, the line passing through the points (30,0) and (0,30) represents the constraint that the maximum trucks the loading facility could handle is 30 trucks. Any point beyond that line like (31,0) and (17,17) are not possible answers for they go beyond the limitations of the constraint, thus using more than what the firm could use given the limited resources. The same is true with the lines passing though points (50,0) and (0,25), and through points (21 2/3, 0) and (0,65).

The feasible region or the shaded region represents all the points or possible product combinations. All the points found inside the feasible region including the points on the lines or boundary of the feasible region is possible product combinations. All the points satisfy the constraints and do not go beyond the constraint lines thus not violating any constraint. These points in the feasible region are the firm's alternatives which they can choose from given limited resources.

The corner points of the feasible region represent the possible optimal solution or mix of products to be produced. Each corner point maximizes a certain constraint. In this problem, there are five corner points or vertices. The vertex (0,0) satisfies the constraint x1 , x2 ≤ 0. The vertex (0,25) maximizes the total metric tons to be produced which is represented by the constraint 1.4x1 + 2.8x2 ≤ 70. Producing only 25 truckloads of Vintage Type 1 and none for Standard Type 2 would give a total production of 70 metric tons of plastic daily. The vertex (21 2/3, 0) maximizes the use of the 65 canisters of flame-retarding agent available which is represented by the constraint 3x1 + 1x2 ≤ 65. Producing 21 and two-thirds truckload of Standard Type 2 would consume all 65 canisters of the agent. The vertex (17 ½,12 ½) maximizes the number of trucks to be handled by the loading facility per day which is represented by the constraint x1 + x2 ≤ 30. Producing this 17 and a half truckload of Standard Type 2 and 12 and a half truckload of Vintage Type 1 would require 30 trucks in total. The vertex (10,20) maximizes both the total production capacity and the number of trucks that could be handled daily which are represented by the constraints 1.4x1 + 2.8x2 ≤ 70 and x1 + x2 ≤ 30 respectively. Producing 10 truckloads of Standard Type 2 and 20 truckloads of Vintage Type 1 would give a total production of 70 metric tons of plastic and would require 30 trucks daily.

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Given that the vertices of the feasible region maximize the use of the resources, the firm must choose the mix that would yield the highest profit. In our computations, we have found out that the optimum mix is 10 truckloads of Standard Type 2 and 20 truckloads of Vintage Type 1. We substituted it in the objective function defined as MaxZ = 950x1 + 1,200x2 and obtained a profit of $33,500. However, there are 15 canisters of the flame-retarding agent that are unused.

Conclusion and Recommendation

In order to maximize profit, Plastic Soul Company should produce 10 trucks of functional polymer branded Standard Type 2 generating P9,500 in and 20 trucks of copolymer branded Vintage Type 1 generating P24,000 having a total profit of P33500. Using the given information, maximum of 30 trucks per day which can load either Standard polymer, Vintage copolymer or any mixture of the two products, we concluded that there are 15 canister of the agent unutilized.

Given constraints: 1.4X1 +2.8 X2 =70

3X1 + 1X2 = 65

1 X1 + 1 X2 = 30

Proof: 1.4(10) + 2.8(20) = 70

3(10) + 1(20) = 50

1(10) + 1(20) = 30

Therefore, it will be better for the company to only allot 50 canisters of the agent to produce the plastics given and transfer the excess canisters to other products.


Hanna, M. E., Render, B. M., & Stair, R. M. (2007). Quantitative Analysis for Management

Studies (9th ed.). Singapore: Pearson Hall.