Minimize : \(z = 6x + 2y\)Subject to : \(x + 2y \ge 3,\ x + 4y \ge 4,\ 3x + y \ge 3,\ x \ge 0,\ y \ge 0\)

Answer: Minimum \(z = 6\), at \((0, 3)\) and \(\left(\dfrac{3}{5}, \dfrac{6}{5}\right)\), and in fact at every point of the segment joining them (infinitely many optimal solutions).

Boundary lines and their intercepts:

[Rough sketch: the feasible region is the unbounded region in the first quadrant lying on or above all three lines, with corner points P(0, 3), Q, R and S(4, 0).]

Q is the intersection of \(3x + y = 3\) and \(x + 2y = 3\): from \(y = 3 - 3x\), \(x + 6 - 6x = 3 \Rightarrow x = \dfrac{3}{5},\ y = \dfrac{6}{5}\).

R is the intersection of \(x + 2y = 3\) and \(x + 4y = 4\): subtracting, \(2y = 1 \Rightarrow y = \dfrac{1}{2},\ x = 2\).

The smallest value is 6, at both P and Q. Since \(z = 6x + 2y = 2(3x + y)\) and the constraint \(3x + y \ge 3\) gives \(z \ge 6\) everywhere in the region, 6 is the minimum even though the region is unbounded. Hence \(z_{\min} = 6\) at \(x = 0, y = 3\) (or \(x = \dfrac{3}{5}, y = \dfrac{6}{5}\), or any point on segment PQ).