The inequalities are 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0 (i) The line $l_1$ : 3x + 2y = 150 passes through the points (50, 0) and (0, 75). AB represents the line. Consider the inequality 3x + 2y ≤ 150. Putting x = 0, y = 0 in 3x + 2y ≤ 150 ⇒ 0 ≤ 150 which is true, shows that origin lies in this region. The region lying below the line AB and the points lying on AB represents the inequality 3x + 2y ≤ 150. (ii) The line $l_2$ : x + 4y = 80 passes through the points (80, 0), (0, 20). This is represented by CD. Consider the inequality x + 4y ≤ 80 putting x = 0, y = 0, we get 0 ≤ 80, which is true. ⇒ Region lying below the line CD and the points on the line CD represents the inequality x + 4y ≤ 80 (iii) x ≤ 15 is the region lying on the left to $l_3$ : x = 15 represented by EF and the points lying on EF. (iv) x ≥ 0 is the region lying on the right side of Y-axis and all the points on Y-axis. (v) y ≥ 0 is the region lying above the X-axis and all the points on X-axis. Thus, the shaded region in the figure is the solution of the given inequalities.