Question

Solve the following L.P.P. by graphical method. Maximize : \(z = 4x + 6y\) Subject to : \(3x + 2y \leq 12\), \(x + y \geq 4\), \(x, y \geq 0\)

Hint:

The optimal solution for an L.P.P. always occurs at one of the corner points (vertices) of the feasible region. Once you've graphed the region, just test all the vertices in the objective function to find the maximum or minimum.

Answer 1

Step 1: Convert inequalities to equations and find points.
For \(3x + 2y = 12\):
If \(x=0\), \(2y=12 \implies y=6\). Point is (0, 6).
If \(y=0\), \(3x=12 \implies x=4\). Point is (4, 0).
The region \(3x+2y \leq 12\) is on the origin side of this line.
For \(x + y = 4\):
If \(x=0\), \(y=4\). Point is (0, 4).
If \(y=0\), \(x=4\). Point is (4, 0).
The region \(x+y \geq 4\) is on the non-origin side of this line.
Step 2: Graph the lines and find the feasible region.
The feasible region is the area bounded by the lines that satisfies all constraints, including \(x \geq 0\) and \(y \geq 0\). The corner points of the feasible region are:
A = (0, 4)
B = (4, 0)
C = (0, 6)
The intersection point of \(x+y=4\) and \(3x+2y=12\) is not needed as it lies outside the region defined by the combination of inequalities. The feasible region is the triangle formed by points A, B, and C.
Step 3: Evaluate the objective function at the corner points.
We evaluate \(Z = 4x + 6y\) at each vertex:
At A(0, 4): \(Z = 4(0) + 6(4) = 24\)
At B(4, 0): \(Z = 4(4) + 6(0) = 16\)
At C(0, 6): \(Z = 4(0) + 6(6) = 36\)
Step 4: Determine the maximum value.
The maximum value of Z is 36, which occurs at the point (0, 6). So, the optimal solution is \(x=0\), \(y=6\), with a maximum value of \(Z = 36\).