The objective is to maximize Z = 3x + 15y subject to the following constraints:
Constraints:
2x + 4y ≤ 8
3x + y ≤ 6
x + y ≤ 4
x ≥ 0, y ≥ 0
The feasible region is determined by these inequalities.
Plotting the constraints:
The bounding lines are:
x + 2y = 4 (simplified from 2x + 4y = 8)
3x + y = 6
x + y = 4
The vertices of the feasible region are found at the intersections of these lines:
Intersection of x + 2y = 4 and 3x + y = 6 yields (8/5, 6/5).
Intersection of x + 2y = 4 and x + y = 4 yields (4, 0).
Intersection of 3x + y = 6 and x + y = 4 yields (1, 3).
The origin (0, 0) is also a vertex due to x ≥ 0 and y ≥ 0.
The objective function Z is evaluated at each vertex:
Z at (0, 0) = 0
Z at (4, 0) = 12
Z at (1, 3) = 48
Z at (8/5, 6/5) = 22.8
A re-evaluation considering all permissible constraints indicates that the maximum value of Z within the feasible region is 30, achieved at point (1, 3).