Consider the following linear optimization problem: Maximize Z = 6x+5y Subject…

Solution and final answer: Evaluate the objective at all feasible corner points.

• Constraints: 2x - 3y ≤ 5, x + 3y ≤ 11, 4x + y ≤ 15, with x ≥ 0, y ≥ 0.

• Find corner points (intersections and axis intercepts):

  • (0, 0) — origin.

  • (0, 11/3) from x = 0 and x + 3y = 11.

  • (2.5, 0) from y = 0 and 2x - 3y = 5 (gives x = 5/2).

  • Intersection of x + 3y = 11 and 4x + y = 15: solve y = 15 - 4x, substitute into x + 3y = 11 -> x + 3(15 - 4x) = 11 -> -11x = -34 -> x = 34/11, y = 29/11.

  • Intersection of 2x - 3y = 5 and 4x + y = 15: solve y = 15 - 4x, substitute -> 2x - 3(15 - 4x) = 5 -> 14x = 50 -> x = 25/7, y = 5/7.

  • The intersection of 2x - 3y = 5 and x + 3y = 11 gives x = 16/3, y = 17/9 but that point violates 4x + y ≤ 15, so it is not feasible.

• Evaluate Z = 6x + 5y at each feasible vertex:

  • (0, 0) -> Z = 0

  • (0, 11/3) -> Z = 5*(11/3) = 55/3 ≈ 18.333

  • (2.5, 0) -> Z = 6*(2.5) = 15

  • (25/7, 5/7) -> Z = 6*(25/7) + 5*(5/7) = 175/7 = 25

  • (34/11, 29/11) -> Z = 6*(34/11) + 5*(29/11) = 349/11 ≈ 31.727

Conclusion: The maximum value is 349/11 ≈ 31.727, attained at x = 34/11, y = 29/11.