Consider The Following Linear Programming Problem

Consider The Following Linear Programmin8x12yst1x3y 62x2y 8

Consider the following linear programming problem: Minimize 8X + 12Y subject to the constraints:

- 1X + 3Y ≥ 62

- 2X + 2Y ≥ 86

- 2X + 2Y ≥ 12

- X, Y ≥ 0

Use the graphical solution procedure to find the optimal solution. Determine the value of the objective function at this optimal point, specified as (X, Y) = (b) Assume that the objective function

Paper For Above instruction

The linear programming problem presented seeks to minimize the objective function 8X + 12Y subject to a set of inequality constraints. This form of optimization problem is classical in operations research and can be effectively solved graphically when involving only two decision variables, such as X and Y in this case. Graphical methods provide intuitive visualizations of feasible regions and optimal points, allowing for straightforward identification of optimal solutions, especially in problems with two variables.

Defining the Constraints and Feasible Region

The constraints are given as:

1. \(X + 3Y \geq 62\)

2. \(2X + 2Y \geq 86\)

3. \(2X + 2Y \geq 12\)

4. \(X, Y \geq 0\)

Before proceeding, it’s important to note that the constraints involving the same left-hand side but different right-hand sides need careful interpretation. The constraints involving \(2X + 2Y \geq 86\) and \(\geq 12\) are redundant because the larger right-hand side dominates the feasibility of the smaller one. Specifically, the constraint \(2X + 2Y \geq 86\) is more restrictive than \(2X + 2Y \geq 12\), which is always satisfied when the former is satisfied. Consequently, the latter can be considered redundant within the feasible region, although including all constraints explicitly can be instructive for comprehensiveness.

The graphical approach involves plotting the boundary lines of each constraint and identifying the feasible region defined by the inequalities:

- For \(X + 3Y \geq 62\), the boundary line is \(X + 3Y = 62\). The feasible region lies on or above this line.

- For \(2X + 2Y \geq 86\), the boundary line is \(2X + 2Y = 86\). The feasible region lies on or above this line.

- For the trivial lower bounds \(X \geq 0, Y \geq 0\), the feasible region is in the first quadrant.

Plotting the Constraints

Let's analyze the boundary lines:

- \(X + 3Y = 62\) passes through points:

- \(X = 0 \Rightarrow 3Y = 62 \Rightarrow Y \approx 20.67\)

- \(Y = 0 \Rightarrow X = 62\)

- \(2X + 2Y = 86\) simplifies to \(X + Y = 43\):

- \(X = 0 \Rightarrow Y = 43\)

- \(Y = 0 \Rightarrow X = 43\)

Given the inequalities:

- The feasible region is on or above line \(X + 3Y = 62\).

- On or above \(X + Y = 43\).

- In the first quadrant.

The most restrictive constraints at various points define the feasible region. To identify the feasible area, find the intersection point(s) of these boundary lines.

Solving for Intersection Points

- Intersect \(X + 3Y = 62\) and \(X + Y = 43\):

From \(X + Y = 43 \Rightarrow X = 43 - Y\)

Substitute into \(X + 3Y = 62\):

\((43 - Y) + 3Y = 62\)

\(43 + 2Y = 62\)

\(2Y = 19 \Rightarrow Y = 9.5\)

Then, \(X = 43 - 9.5 = 33.5\)

- Intersection point: (X, Y) = (33.5, 9.5)

Because the inequalities are "greater than or equal to," the feasible region encompasses points on or above these lines. The critical feasible region corner in this case is at (33.5, 9.5). Given the problem's constraints and the structure of the feasible region, the optimal solution typically occurs at one of the vertices of the feasible region.

Evaluating the Objective Function at Vertices

The potential optimal points are at the intersection of the constraints:

- At the intersection point (33.5, 9.5):

\(Z = 8(33.5) + 12(9.5) = 268 + 114 = 382\)

- Check other key points:

- At \(X = 0\):

From \(X + 3Y \geq 62\):

\(0 + 3Y \geq 62 \Rightarrow Y \geq 20.67\)

From \(X + Y \geq 43\):

\(0 + Y \geq 43 \Rightarrow Y \geq 43\)

The higher is \(Y \geq 43\), so feasible point at (0, 43).

Objective: \(Z = 80 + 1243 = 516\)

- At \(Y = 0\):

From \(X + 3*0 \geq 62 \Rightarrow X \geq 62\).

From \(X + Y \geq 43\):

\(X \geq 43 \Rightarrow X \geq 62\) (since 62 > 43).

Objective: \(Z = 862 + 120 = 496\)

Among these points, the minimal objective is at (33.5, 9.5), with a value of 382. The other points yield higher costs, so the optimal solution for the minimization problem is approximately at (33.5, 9.5).

Conclusion

Therefore, the optimal solution is approximately:

\[

X \approx 33.5, \quad Y \approx 9.5,

\]

with the minimum objective function value:

\[

Z_{min} \approx 382.

\]

This point satisfies all constraints and yields the lowest cost according to the graphical method.

Additional Note:

In a more formal setting, validating the constraints’ feasibility at this point ensures no violations occur. Also, recognizing redundancy in constraints simplifies the process, emphasizing the key boundaries shaping the feasible region.

References

- Winston, W. L. (2004). Operations Research: Applications and Algorithms. Duxbury Press.

- Hillier, F. S., & Lieberman, G. J. (2001). Introduction to Operations Research. McGraw-Hill.

- Taha, H. A. (2017). Operations Research: An Introduction. Pearson.

- Bazaraa, M. S., Jarvis, J. J., & Sherali, H. D. (2010). Linear Programming and Network Flows. Wiley.

- Ravindran, A., Ragsdell, K., & Rekik, Y. (2002). Operations Research: Principles and Practice. Wiley.

- Hillier, F., & Lieberman, G. (2010). Introduction to Operations Research. McGraw-Hill Education.

- Murty, K. G. (2002). Operations Research: Deterministic Optimization Models. Prentice Hall.

- Luce, R. D., & Raiffa, H. (1957). Games and Decisions: Introduction and Critical Survey. Wiley.

- Gass, S. I. (2003). Linear Programming: Methods and Applications. Dover Publications.

- Papadopoulos, A. V., & Mara, A. (Eds.). (2014). Linear Programming and Extensions. Springer.