To find the maximum value of \( Z = 3x + 4y \) subject to the constraint \( x + y \leq 6 \) and \( x, y \geq 0 \), we need to determine the feasible region and evaluate the objective function at the vertices of this region. The constraint can be rewritten as \( y \leq 6 - x \).
1. Graphical Representation:
Let's plot the line \( y = 6 - x \) and consider the constraints:
The feasible region is a triangle with vertices at \( (0,0) \), \( (6,0) \), and \( (0,6) \).
2. Evaluate the Objective Function at Each Vertex:
\(Z = 3x + 4y\) | Vertex \( (x,y) \) | Value of \( Z \) |
\((0,0)\) | \(Z = 3(0) + 4(0) = 0\) | |
\((6,0)\) | \(Z = 3(6) + 4(0) = 18\) | |
\((0,6)\) | \(Z = 3(0) + 4(6) = 24\) |
3. Determine the Maximum Value:
The maximum value of \( Z \) is \( 24 \) at the vertex \( (0,6) \).
Therefore, the maximum value of \( Z = 3x + 4y \) under the given constraints is \( \mathbf{24} \).
Assertion (A): The corner points of the bounded feasible region of a L.P.P. are shown below. The maximum value of \( Z = x + 2y \) occurs at infinite points.
Reason (R): The optimal solution of a LPP having bounded feasible region must occur at corner points.