Solve the following L.P.P. by graphical method.Minimize : \(z=8x+10y\)Subject to : \(2x+y\ge 7,\ 2x+3y\ge 15,\ y\ge 2,\ x\ge 0,\ y\ge 0\)

Answer: Minimum \(z=52\) at \(x=1.5,\ y=4\)

Draw the boundary lines:

• \(2x+y=7\): passes through (3.5, 0) and (0, 7).
• \(2x+3y=15\): passes through (7.5, 0) and (0, 5).
• \(y=2\): line parallel to the X-axis.

All constraints are of the '\(\ge\)' type, so the feasible region lies on the side of each line away from the origin, in the first quadrant. It is unbounded, with corner points:

• A (0, 7): \(2x+y=7\) meets the Y-axis.
• B: \(2x+y=7\) and \(2x+3y=15\) give \(2y=8\), \(y=4\), \(x=1.5\); B (1.5, 4).
• C: \(2x+3y=15\) and \(y=2\) give \(x=4.5\); C (4.5, 2).

[Rough sketch: the shaded feasible region is above the line through A and B, above the line through B and C, and above \(y=2\) to the right of C, extending without bound upwards and to the right.]

The least value is 52 at B. Since both coefficients of z are positive, z only increases as we move further into the unbounded region, so the minimum exists. Minimum \(z=52\) at \(x=1.5,\ y=4\).