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

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

Boundary lines:

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

All inequalities are of the \(\ge\) type, so the feasible region lies on the side of each line away from the origin, with \(x \ge 0\), \(y \ge 2\). It is unbounded. [Figure: rough sketch of the three lines with the unbounded feasible region shaded above them, corner points P(0, 7), Q(1.5, 4), R(4.5, 2).]

Corner points:

• P: \(x = 0\) and \(2x + y = 7\) give \((0, 7)\).
• Q: \(2x + y = 7\) and \(2x + 3y = 15\); subtracting, \(2y = 8\), \(y = 4\), \(x = 1.5\): \((1.5, 4)\).
• R: \(2x + 3y = 15\) and \(y = 2\) give \(x = 4.5\): \((4.5, 2)\).

The smallest value is 52 at Q. Since the region is unbounded, check: \(8x + 10y = (2x + y) + 3(2x + 3y) \ge 7 + 45 = 52\) for every feasible point, so 52 is indeed the minimum.

Hence \(z_{\min} = 52\) at \(x = 1.5,\ y = 4\).