What Is The Minimum Value Of The Objective Function, $C$, With The Given Constraints?Objective Function:$\[ C = 2x + 3y \\]Constraints:$\[ \begin{cases} 2x + 3y \leq 24 \\ 3x + Y \leq 15 \\ x \geq 2 \\ y \geq 3

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Introduction

In linear programming, the objective function is a mathematical expression that represents the goal of the problem. The constraints are the limitations or restrictions that the solution must satisfy. In this article, we will explore the minimum value of the objective function, $C$, with the given constraints. We will use the graphical method to visualize the feasible region and find the minimum value of the objective function.

Objective Function

The objective function is given by:

${ C = 2x + 3y }$

This function represents the cost or value of the solution. The goal is to minimize this function while satisfying the given constraints.

Constraints

The constraints are given by:

${ \begin{cases} 2x + 3y \leq 24 \\ 3x + y \leq 15 \\ x \geq 2 \\ y \geq 3 \end{cases} }$

These constraints represent the limitations or restrictions on the values of $x$ and $y$. The first constraint represents a linear inequality, while the second constraint represents another linear inequality. The third and fourth constraints represent non-negativity constraints.

Graphical Method

To visualize the feasible region, we will use the graphical method. We will plot the lines corresponding to the constraints and find the intersection points. The feasible region is the area that satisfies all the constraints.

Plotting the Constraints

First, we will plot the lines corresponding to the constraints.

• The line corresponding to the first constraint is $2x + 3y = 24$. We can plot this line by finding the x-intercept and y-intercept.
• The line corresponding to the second constraint is $3x + y = 15$. We can plot this line by finding the x-intercept and y-intercept.
• The line corresponding to the third constraint is $x = 2$. We can plot this line by finding the x-intercept.
• The line corresponding to the fourth constraint is $y = 3$. We can plot this line by finding the y-intercept.

Finding the Intersection Points

Next, we will find the intersection points of the lines.

• The intersection point of the first and second constraints is found by solving the system of equations: ${ \begin{cases} 2x + 3y = 24 \\ 3x + y = 15 \end{cases} }$ Solving this system of equations, we get $x = 3$ and $y = 5$.
• The intersection point of the second and third constraints is found by solving the system of equations: ${ \begin{cases} 3x + y = 15 \\ x = 2 \end{cases} }$ Solving this system of equations, we get $x = 2$ and $y = 9$.
• The intersection point of the third and fourth constraints is found by solving the system of equations: ${ \begin{cases} x = 2 \\ y = 3 \end{cases} }$ Solving this system of equations, we get $x = 2$ and $y = 3$.

Finding the Feasible Region

The feasible region is the area that satisfies all the constraints. We can find the feasible region by plotting the lines and finding the intersection points.

The feasible region is a polygon with vertices at $(2, 3)$, $(3, 5)$, and $(2, 9)$.

Finding the Minimum Value of the Objective Function

To find the minimum value of the objective function, we will use the graphical method. We will plot the lines corresponding to the objective function and find the intersection points.

Plotting the Objective Function

First, we will plot the lines corresponding to the objective function.

• The line corresponding to the objective function is $C = 2x + 3y$. We can plot this line by finding the x-intercept and y-intercept.

Finding the Intersection Points

Next, we will find the intersection points of the lines.

• The intersection point of the objective function and the first constraint is found by solving the system of equations: ${ \begin{cases} 2x + 3y = 24 \\ 2x + 3y = C \end{cases} }$ Solving this system of equations, we get $x = 4$ and $y = 4$.
• The intersection point of the objective function and the second constraint is found by solving the system of equations: ${ \begin{cases} 3x + y = 15 \\ 2x + 3y = C \end{cases} }$ Solving this system of equations, we get $x = 3$ and $y = 5$.
• The intersection point of the objective function and the third constraint is found by solving the system of equations: ${ \begin{cases} x = 2 \\ 2x + 3y = C \end{cases} }$ Solving this system of equations, we get $x = 2$ and $y = 3$.

Finding the Minimum Value of the Objective Function

The minimum value of the objective function is found by evaluating the objective function at the intersection points.

• The minimum value of the objective function is $C = 2(2) + 3(3) = 13$.

Conclusion

In this article, we explored the minimum value of the objective function, $C$, with the given constraints. We used the graphical method to visualize the feasible region and find the minimum value of the objective function. The minimum value of the objective function is $C = 13$.

Discussion

The graphical method is a powerful tool for solving linear programming problems. It allows us to visualize the feasible region and find the optimal solution. However, it can be time-consuming and difficult to use for large problems.

In this problem, we used the graphical method to find the minimum value of the objective function. We plotted the lines corresponding to the constraints and found the intersection points. We then evaluated the objective function at the intersection points to find the minimum value.

The minimum value of the objective function is $C = 13$. This is the optimal solution to the problem.

References

• [1] Chvatal, V. (1983). Linear Programming. W.H. Freeman and Company.
• [2] Dantzig, G. B. (1963). Linear Programming and Extensions. Princeton University Press.
• [3] Hillier, F. S., & Lieberman, G. J. (2015). Introduction to Operations Research. McGraw-Hill Education.

Keywords

• Linear programming
• Objective function
• Constraints
• Graphical method
• Feasible region
• Minimum value
• Optimization
  

Introduction

In our previous article, we explored the minimum value of the objective function, $C$, with the given constraints. We used the graphical method to visualize the feasible region and find the minimum value of the objective function. In this article, we will answer some frequently asked questions about linear programming and the minimum value of the objective function.

Q&A

Q1: What is linear programming?

A1: Linear programming is a method of optimization that involves finding the optimal solution to a problem by minimizing or maximizing a linear objective function subject to a set of linear constraints.

Q2: What is the objective function?

A2: The objective function is a mathematical expression that represents the goal of the problem. It is a linear function of the variables in the problem.

Q3: What are the constraints?

A3: The constraints are the limitations or restrictions that the solution must satisfy. They are linear inequalities or equalities that the variables in the problem must satisfy.

Q4: How do I find the feasible region?

A4: To find the feasible region, you need to plot the lines corresponding to the constraints and find the intersection points. The feasible region is the area that satisfies all the constraints.

Q5: How do I find the minimum value of the objective function?

A5: To find the minimum value of the objective function, you need to evaluate the objective function at the intersection points of the lines corresponding to the constraints. The minimum value of the objective function is the smallest value of the objective function in the feasible region.

Q6: What is the graphical method?

A6: The graphical method is a powerful tool for solving linear programming problems. It allows you to visualize the feasible region and find the optimal solution.

Q7: What are the advantages of the graphical method?

A7: The advantages of the graphical method include:

• It allows you to visualize the feasible region and find the optimal solution.
• It is a powerful tool for solving linear programming problems.
• It is easy to use and understand.

Q8: What are the disadvantages of the graphical method?

A8: The disadvantages of the graphical method include:

• It can be time-consuming and difficult to use for large problems.
• It requires a good understanding of linear programming and the graphical method.

Q9: How do I choose the best method for solving a linear programming problem?

A9: To choose the best method for solving a linear programming problem, you need to consider the following factors:

• The size of the problem.
• The complexity of the problem.
• The availability of computational resources.
• The level of expertise of the person solving the problem.

Q10: What are some common applications of linear programming?

A10: Some common applications of linear programming include:

• Resource allocation.
• Scheduling.
• Production planning.
• Inventory management.

Conclusion

In this article, we answered some frequently asked questions about linear programming and the minimum value of the objective function. We hope that this article has been helpful in understanding the concepts of linear programming and the graphical method.

Discussion

Linear programming is a powerful tool for solving optimization problems. It allows you to find the optimal solution to a problem by minimizing or maximizing a linear objective function subject to a set of linear constraints. The graphical method is a powerful tool for solving linear programming problems. It allows you to visualize the feasible region and find the optimal solution.

References

• [1] Chvatal, V. (1983). Linear Programming. W.H. Freeman and Company.
• [2] Dantzig, G. B. (1963). Linear Programming and Extensions. Princeton University Press.
• [3] Hillier, F. S., & Lieberman, G. J. (2015). Introduction to Operations Research. McGraw-Hill Education.

Keywords

• Linear programming
• Objective function
• Constraints
• Graphical method
• Feasible region
• Minimum value
• Optimization
• Resource allocation
• Scheduling
• Production planning
• Inventory management