In the realm of linear programming, a fundamental task involves optimizing an objective function subject to a set of constraints. This optimization often entails finding the minimum or maximum value of the objective function within a feasible region defined by the constraints. In this article, we delve into a specific problem: determining the minimum value of the objective function C = 3x + 2y subject to the constraints x + y ≥ 4, x ≥ 0, and y ≥ 0. This is a classic example of a linear programming problem that can be solved using graphical methods or algebraic techniques. Understanding how to solve such problems is crucial in various fields, including economics, engineering, and operations research, where resource allocation and optimization are paramount.
To effectively address this problem, we will first explore the graphical representation of the constraints. By plotting the inequalities on a coordinate plane, we can identify the feasible region, which represents the set of all points (x, y) that satisfy all the given constraints simultaneously. This region will be a polygon, and the optimal solution (the minimum value of C) will occur at one of the vertices (corner points) of this polygon. Once the feasible region is established, we can evaluate the objective function C = 3x + 2y at each vertex. The vertex that yields the smallest value for C will be the solution to our problem. This approach not only provides the minimum value but also identifies the specific values of x and y that achieve this minimum. Furthermore, we will discuss the implications and interpretations of the solution within the context of real-world applications, illustrating how such optimization problems can be applied to practical scenarios.
Before diving into the solution, it's crucial to understand the components of the problem. We are given an objective function, C = 3x + 2y, which we aim to minimize. This function represents a linear relationship between two variables, x and y, and our goal is to find the smallest possible value of C while adhering to certain restrictions. These restrictions are known as constraints, and in this case, we have three constraints: x + y ≥ 4, x ≥ 0, and y ≥ 0. The constraint x + y ≥ 4 specifies a minimum requirement for the sum of x and y, while the constraints x ≥ 0 and y ≥ 0 restrict the values of x and y to be non-negative. These non-negativity constraints are common in real-world problems, as they often represent physical quantities that cannot be negative, such as the amount of resources used or the number of items produced.
The interplay between the objective function and the constraints defines the feasible region, which is the set of all points (x, y) that satisfy all the constraints simultaneously. The feasible region is a crucial concept in linear programming because the optimal solution—the point that minimizes or maximizes the objective function—must lie within this region. Graphically, the feasible region is represented by the intersection of the regions defined by the individual constraints. In this problem, the constraint x + y ≥ 4 corresponds to the region above the line x + y = 4, while the constraints x ≥ 0 and y ≥ 0 restrict the solution to the first quadrant of the coordinate plane. The feasible region, therefore, is the unbounded region in the first quadrant that lies above the line x + y = 4. Identifying this region is the first step towards solving the problem, as it narrows down the set of potential solutions to a manageable area.
To solve this linear programming problem, the first step involves graphically representing the constraints. Each constraint can be plotted as a line on the coordinate plane, and the feasible region is the area that satisfies all constraints simultaneously. Let's break down each constraint:
- x + y ≥ 4: To graph this inequality, we first treat it as an equation: x + y = 4. We can find two points on this line to plot it. For example, when x = 0, y = 4, and when y = 0, x = 4. So, the line passes through the points (0, 4) and (4, 0). Since the inequality is x + y ≥ 4, we are interested in the region above the line, including the line itself.
- x ≥ 0: This constraint represents all points to the right of the y-axis, including the y-axis.
- y ≥ 0: Similarly, this constraint represents all points above the x-axis, including the x-axis.
The feasible region is the area where all three constraints are satisfied. In this case, it is the unbounded region in the first quadrant (where both x and y are non-negative) that lies above the line x + y = 4. This region extends infinitely upwards and to the right. The corners of this feasible region are crucial because the minimum value of the objective function will occur at one of these corner points. For this particular problem, the corner points are (4, 0) and (0, 4). Understanding the graphical representation is essential as it provides a visual context for the problem and helps in identifying the potential solutions.
Having graphed the constraints, the next crucial step is to identify the corner points of the feasible region. These points are the vertices of the polygon formed by the intersection of the constraint lines. In linear programming, the optimal solution (either the minimum or maximum value of the objective function) always occurs at one of these corner points. Therefore, accurately identifying these points is essential for finding the solution.
In our problem, the feasible region is defined by the constraints x + y ≥ 4, x ≥ 0, and y ≥ 0. This region is an unbounded area in the first quadrant lying above the line x + y = 4. The corner points are the points where the constraint lines intersect. We have three lines to consider: x + y = 4, x = 0, and y = 0.
- Intersection of x + y = 4 and x = 0: Substituting x = 0 into the equation x + y = 4, we get 0 + y = 4, which gives us y = 4. So, the intersection point is (0, 4).
- Intersection of x + y = 4 and y = 0: Substituting y = 0 into the equation x + y = 4, we get x + 0 = 4, which gives us x = 4. So, the intersection point is (4, 0).
Thus, the corner points of the feasible region are (0, 4) and (4, 0). These are the only two corner points in this case, as the feasible region is unbounded and does not form a closed polygon. Now that we have identified these corner points, we can evaluate the objective function at each of them to determine the minimum value of C.
Once the corner points of the feasible region are identified, the next step is to evaluate the objective function at each of these points. This process involves substituting the coordinates of each corner point into the objective function and calculating the resulting value. The objective function in our problem is C = 3x + 2y, and we have identified two corner points: (0, 4) and (4, 0). By evaluating C at these points, we can determine which one yields the minimum value.
- At corner point (0, 4): Substituting x = 0 and y = 4 into the objective function, we get C = 3(0) + 2(4) = 0 + 8 = 8.
- At corner point (4, 0): Substituting x = 4 and y = 0 into the objective function, we get C = 3(4) + 2(0) = 12 + 0 = 12.
Comparing the values of C at these two corner points, we find that the minimum value of C is 8, which occurs at the point (0, 4). This indicates that the optimal solution to the problem is achieved when x = 0 and y = 4. This systematic approach of evaluating the objective function at each corner point is a fundamental principle of linear programming, ensuring that we find the global minimum or maximum within the feasible region. In the next section, we will present the final answer and summarize the solution process.
After evaluating the objective function C = 3x + 2y at the corner points (0, 4) and (4, 0), we found that:
- At (0, 4), C = 8
- At (4, 0), C = 12
Comparing these values, it is evident that the minimum value of the objective function C is 8. This minimum value occurs when x = 0 and y = 4. Therefore, the solution to the linear programming problem is that the minimum value of C is 8, and it is achieved at the point (0, 4). This means that under the given constraints, the smallest possible value of the expression 3x + 2y is 8, and this occurs when x is set to 0 and y is set to 4.
This solution highlights the power of linear programming in optimization problems. By systematically analyzing the constraints and the objective function, we can determine the optimal allocation of resources or variables to achieve the desired outcome. In this case, we have successfully identified the minimum value of the objective function, providing a clear and concise answer to the problem. This method is applicable to a wide range of scenarios where minimizing costs or maximizing profits is the primary goal.
In summary, we have successfully determined the minimum value of the objective function C = 3x + 2y subject to the constraints x + y ≥ 4, x ≥ 0, and y ≥ 0. By graphically representing the constraints, we identified the feasible region and its corner points (0, 4) and (4, 0). Evaluating the objective function at these corner points, we found that the minimum value of C is 8, which occurs at the point (0, 4). This solution demonstrates the effectiveness of linear programming techniques in solving optimization problems.
The process we followed involved several key steps:
- Understanding the problem: We clearly defined the objective function and the constraints.
- Graphical representation: We plotted the constraints on a coordinate plane to visualize the feasible region.
- Identifying corner points: We determined the vertices of the feasible region, which are the potential optimal solutions.
- Evaluating the objective function: We substituted the coordinates of the corner points into the objective function to find the minimum value.
This systematic approach ensures that we arrive at the correct solution in linear programming problems. The minimum value of the objective function C = 3x + 2y is 8. This solution underscores the practical applicability of linear programming in various fields, where optimization is crucial for decision-making and resource allocation. Understanding these concepts and techniques is essential for anyone involved in fields such as economics, engineering, and operations research.
C = 8