The difference between the maximum and minimum values of the objective function Z =3 x+5 y , subject to the constraints x+3 y 60, x+ y 10 , x- y 0, x, y 0 is

Step-by-step solution

Finding the range of Z = 3x + 5y subject to constraints The feasible region is bounded by the constraints x + 3y 60 , x + y 10 , y x , with x 0 , y 0 . The corner points occur at intersections: (5,5) from x+y=10 and y=x (15,15) from x+3y=60 and y=x (0,10) from x+y=10 and x=0 (0,20) from x+3y=60 and x=0 Evaluating Z = 3x + 5y at these points: (5,5): Z = 3(5) + 5(5) = 40 (15,15): Z = 3(15) + 5(15) = 120 (0,10): Z = 3(0) + 5(10) = 50 (0,20): Z = 3(0) + 5(20) = 100 The minimum value of Z is 40 at (5,5) The maximum valu