Maximum value of z=12 x+3 y , subject to constraints x 0, y 0, x+y 5 and 3 x+y 9 is

Step-by-step solution

The objective function is z = 12x + 3y . The constraints are: 1) x 0, y 0 2) x + y 5 3) 3x + y 9 The feasible region is a polygon with vertices determined by the intersection of the lines: Intersection of x+y=5 and 3x+y=9 : Subtracting the equations: (3x+y) - (x+y) = 9 - 5 2x = 4 x = 2 . Substituting x=2 into x+y=5 , we get y=3 . The point is (2, 3) . The vertices of the feasible region are: (0, 0) (3, 0) (from 3x+y=9 at y=0 ) (2, 3) (0, 5) (from x+y=5 at x=0 ) Evaluating z = 12x + 3y at each vertex: At (0, 0) : z