The maximum of Z is where, Z=4 x+2 y subject to constraints 4 x+2 y 46, x+3 y 24 and x, y 0 is

Step-by-step solution

The objective function is Z = 4x + 2y . The constraints are: 1) 4x + 2y 46 2x + y 23 2) x + 3y 24 3) x, y 0 We analyze the feasible region defined by these inequalities. The line 2x + y = 23 passes through (11.5, 0) and (0, 23) . The line x + 3y = 24 passes through (24, 0) and (0, 8) . To find the intersection point of 2x + y = 23 and x + 3y = 24 : From the first equation, y = 23 - 2x . Substituting into the second: x + 3(23 - 2x) = 24 x + 69 - 6x = 24 -5x = -45 x = 9 y = 23 - 2(9) = 23 - 18 = 5 . The intersection