Solve the following L.P.P. by graphical method :Maximize : \(z = 7x + 11y\)Subject to : \(3x + 4y \le 24\), \(5x + 3y \le 30\), \(x \ge 0\), \(y \ge 0\)

Answer: \(z_{\max} = 66\) at \(x = 0\), \(y = 6\)

Draw the lines:

• \(3x + 4y = 24\): passes through \((8, 0)\) and \((0, 6)\).
• \(5x + 3y = 30\): passes through \((6, 0)\) and \((0, 10)\).

Both inequalities are satisfied by the origin, so the feasible region lies on the origin side of both lines, in the first quadrant.

Point of intersection: \(9x + 12y = 72\) and \(20x + 12y = 120\) give \(11x = 48\), \(x = \dfrac{48}{11}\), \(y = \dfrac{30}{11}\).

[Rough sketch: the feasible region is the quadrilateral OABC with O(0, 0), A(6, 0), B(48/11, 30/11), C(0, 6).]

The maximum value of \(z\) is 66, at C(0, 6), i.e. when \(x = 0\) and \(y = 6\).