Question

In the Linear Programming Problem (LPP), find the point/points giving the maximum value for \( Z = 5x + 10y\) subject to the constraints:
\[x + 2y \leq 120 \\ x + y \geq 60 \\ x - 2y \geq 0 \\ x \geq 0, y \geq 0\]

Hint:

Quick Tip: When solving LPP problems, always plot the constraints to find the feasible region. Evaluate the objective function at each vertex of the feasible region to find the maximum or minimum value.

Answer 1

To solve this Linear Programming Problem (LPP), we need to find the feasible region by plotting the constraints and then evaluate \( Z = 5x + 10y \) at the vertices of the feasible region.
1. Plot the Constraints:
- \( x + 2y \leq 120 \)
- \( x + y \geq 60 \)
- \( x - 2y \geq 0 \) (or \( x \geq 2y \))
- \( x \geq 0 \)
- \( y \geq 0 \)
2. Find the Intersection Points (Vertices):
- To find the vertices, solve the system of equations formed by the intersections of these lines:
1. \( x + 2y = 120 \)
2. \( x + y = 60 \)
3. \( x = 2y \)
Solving these pairs of equations gives us the following points:
- \( (x, y) = (60, 0) \)
- \( (x, y) = (80, 40) \)
- \( (x, y) = (60, 30) \)
3. Evaluate \( Z = 5x + 10y \) at the Vertices:
- At \( (60, 0) \): \( Z = 5(60) + 10(0) = 300 \)
- At \( (80, 40) \): \( Z = 5(80) + 10(40) = 400 \)
- At \( (60, 30) \): \( Z = 5(60) + 10(30) = 600 \)
The maximum value of \( Z \) is 600 at the point \( (60, 30) \). 4. Conclusion: The maximum value of \( Z = 5x + 10y \) is 600, and it occurs at the point \( (60, 30) \).
% Graph for Linear Programming Problem \begin{figure}[ht] \centering \includegraphics[width=0.7\textwidth]{Feasible Region and Optimal Solution.png} \end{figure}