Minimize: \(z=4x+2y\)Subject to: \(3x+y\ge 27\)\(x+y\ge 21\)\(x+2y\ge 30\)\(x\ge 0,\ y\ge 0\)Solve the above L.P.P. by graphical method.

Answer: Minimum \(z=48\) at \(x=3,\ y=18\)

Draw the lines:

• \(3x+y=27\): through \((9,0)\) and \((0,27)\)
• \(x+y=21\): through \((21,0)\) and \((0,21)\)
• \(x+2y=30\): through \((30,0)\) and \((0,15)\)

All constraints are of the \(\ge\) type and the origin does not satisfy any of them, so the feasible region lies on the side away from the origin, in the first quadrant. It is unbounded, with corner points A(0, 27), B, C, D(30, 0). [Rough sketch: the region above the boundary A–B–C–D.]

B: \(3x+y=27\) and \(x+y=21\) give \(2x=6\), \(x=3,\ y=18\).
C: \(x+y=21\) and \(x+2y=30\) give \(y=9,\ x=12\).

The minimum value of \(z\) is 48 at \(x=3,\ y=18\).