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 resolve this Linear Programming Problem (LPP), the feasible region must be identified by plotting the constraints. Subsequently, the objective function \( Z = 5x + 10y \) is evaluated at the vertices of this region. 1. Plot the Constraints: - \( x + 2y \leq 120 \) - \( x + y \geq 60 \) - \( x - 2y \geq 0 \) (equivalent to \( x \geq 2y \)) - \( x \geq 0 \) - \( y \geq 0 \) 2. Determine Intersection Points (Vertices): - The vertices are found by solving the systems of equations formed by the intersections of these boundary lines: 1. \( x + 2y = 120 \) 2. \( x + y = 60 \) 3. \( x = 2y \) - Solving pairs of these equations yields the following vertices: - \( (x, y) = (60, 0) \) - \( (x, y) = (80, 40) \) - \( (x, y) = (60, 30) \) 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, occurring at the point \( (60, 30) \). 4. Conclusion: The maximum value of \( Z = 5x + 10y \) is 600, achieved 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}