You May Use Either Level Curves Or The Corner Point M

You May Employ Either The Level Curves Or The Corner Point Method The

You may employ either the level curves or the corner-point method to solve the given linear programming problem. The problem involves maximizing the objective function 4X1 + 5X2 subject to the constraints: X1 + 2X2 ≤ 10, 6X1 + 6X2 ≤ 36, X1 ≤ 4, and X1, X2 ≥ 0. The solution can be visualized by sketching the feasible region using Excel's drawing tools or hand sketching, then scanning or copying the graph into a PDF or image file. Additionally, shading partial cells in Excel can be done by copying the sketch into Windows Paint, applying shading with brush tools, and then pasting the image back into Excel.

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The problem presented requires finding the optimal values of the decision variables X1 and X2 to maximize the objective function 4X1 + 5X2, given a set of constraints. The solution involves graphical analysis, which is an effective method for visualizing feasible regions and identifying optimal solutions for linear programming problems with two variables.

Understanding the Problem and Constraints

The problem constraints define a feasible region in the X1-X2 plane:

1. X1 + 2X2 ≤ 10 (Linear boundary line)

2. 6X1 + 6X2 ≤ 36, which simplifies to X1 + X2 ≤ 6 (Linear boundary line)

3. X1 ≤ 4

4. X1, X2 ≥ 0 (non-negativity constraints)

These constraints, when graphed, create a polygon—the feasible region—that contains all the points where the conditions hold true simultaneously.

Methodology

The problem can be solved either through the level curve method or the corner point method. The level curve method involves drawing lines representing points of equal objective function value and moving these lines outward until they touch the last point in the feasible region. The corner point method involves identifying the vertices (corner points) of the feasible region and evaluating the objective function at each vertex to find the maximum.

Graphing the Constraints

To visualize the feasible region, each constraint line is plotted:

- For X1 + 2X2 = 10, when X1=0, X2=5; when X2=0, X1=10

- For X1 + X2 = 6, when X1=0, X2=6; when X2=0, X1=6

- For X1 = 4, a vertical line at X1=4

- The axes X1=0 and X2=0 are the boundaries of the feasible region in the first quadrant.

The feasible region is found at the intersection of all inequalities, forming a convex polygon. Key vertices are the points of intersection between boundary lines, which can be calculated algebraically or identified graphically.

Finding the Corner Points

The critical corners are the intersections of:

- The lines X1=0 and X1=4 with other constraints

- The intersections of X1 + 2X2 = 10 and X1 + X2 = 6

Calculations yield the following points:

- Intersection of X1=0 and X1 + 2X2 = 10: (0, 5)

- Intersection of X1=0 and X1 + X2 = 6: (0, 6), but since X2=6 violates the first constraint at (0,6) because X1 + 2X2 = 12 > 10, so (0,6) is outside feasible. The feasible point is (0,5).

- Intersection of X1=4 and X1 + 2X2 = 10:

At X1=4, 4 + 2X2=10 → 2X2=6 → X2=3, thus point (4,3).

- Intersection of X1=4 and X1 + X2 = 6:

At X1=4, 4 + X2=6 → X2=2, thus point (4,2).

- Intersection of X1 + 2X2=10 and X1 + X2=6:

From these equations:

1. X1 + 2X2 = 10

2. X1 + X2 = 6 → X1=6 - X2

Substitute into the first:

(6 - X2)+ 2X2=10 →6 + X2=10 → X2=4

Then X1=6 - 4=2

So, point (2,4).

The feasible vertices are: (0,5), (4,3), (2,4), and the intercepts at axes: (0,0) and possibly (4,0), depending on whether it satisfies all constraints.

Evaluating the Objective Function

To identify the maximum, evaluate the objective function at each vertex:

- At (0,5): 4(0)+5(5)=25

- At (4,3): 4(4)+5(3)=16+15=31

- At (2,4): 4(2)+5(4)=8+20=28

- At (0,0): 0+0=0 (not likely optimal)

- At (4,0): check constraints: 4+0=4 ≤10, 6+0=6 ≤36, and X1=4, within bounds, so evaluate: 4(4)+5(0)=16. Not as high as 31.

The highest value is at point (4,3), with the objective function value of 31. This suggests the optimal solution is X1=4, X2=3.

Conclusion

By graphically analyzing the feasible region and evaluating the objective function at the vertices, the optimal solution to the LP problem is to set X1=4 and X2=3, achieving a maximum value of 31 for the objective function. This process exemplifies the effectiveness of the corner point method in linear programming, especially in problems with two variables where graphical analysis is feasible.

Practical Considerations and Implementation

Excel or manual hand drawing can be used to sketch the constraints accurately. Shading partial cells in Excel involves copying the graph into Paint to apply shading effectively, then pasting the image back into Excel. This visual approach aids in understanding the feasible region and facilitates communication of results.

In practice, decision-makers can use this graphical approach for quick analysis or verify by setting up LP models in solver tools like Excel Solver or specialized optimization software such as LINDO or Gurobi, which can handle larger or more complex problems efficiently.

Limitations and Extensions

While the graphical method works well with two variables, it becomes impractical for higher-dimensional problems. In such cases, simplex or interior-point algorithms are used. However, understanding the geometric approach provides foundational insights into LP solutions and constraint interactions.

References

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