A linear programming problem (LPP) is as follows: Min z=30x-18y, subject to…

Solution:

We minimize z = 30x - 18y subject to 3x + 4y ≤ 60, 5x - 3y ≥ 20, x, y ≥ 0. Check corner (boundary) points of the feasible region and evaluate the objective.

• Find relevant intersections and axis intercepts:

  • x-axis intercepts from constraints: 3x + 4(0) = 60 → (20, 0). From 5x - 3(0) = 20 → (4, 0).

  • Intersection of the two constraint lines: solve 3x + 4y = 60 and 5x - 3y = 20.

  Solve: multiply first by 3 → 9x + 12y = 180. Multiply second by 4 → 20x - 12y = 80. Add: 29x = 260 → x = 260/29. Then 4y = 60 - 3x = 60 - 780/29 = 960/29 → y = 240/29.

  So the intersection is (260/29, 240/29) ≈ (8.9655, 8.2759).

• Evaluate z = 30x - 18y at candidate vertices:

  • At (4, 0): z = 30*4 - 18*0 = 120.

  • At (20, 0): z = 30*20 - 18*0 = 600.

  • At (260/29, 240/29): z = (30*(260/29) - 18*(240/29)) = (7800 - 4320)/29 = 3480/29 = 120.

• Conclusion:

  • The minimum objective value is z = 120.

  • The value z = 120 is achieved at (4, 0) and at the intersection (260/29, 240/29). Because the line 5x - 3y = 20 is binding and the objective is constant along the direction between these points, every point on the segment of 5x - 3y = 20 between (4, 0) and (260/29, 240/29) is also optimal.

  • Among the listed points, (4, 0) and (7, 5) lie on 5x - 3y = 20 and both give z = 120, so those listed points are optimal.