Given Z=80 x+120 y , subject to constraints are x+3 y 30 ; 3 x+4 y 60 ; x 0 ; y 0 . P is one of the corner points of the feasible region for the given Linear Programming Problem. Then the coordinate of P is

Step-by-step solution

The constraints are x + 3y 30 , 3x + 4y 60 , x 0 , and y 0 . First, identify the intersection points of the boundary lines with the axes: For x + 3y = 30 : If x = 0 , y = 10 . Point is (0, 10) . If y = 0 , x = 30 . Point is (30, 0) . For 3x + 4y = 60 : If x = 0 , y = 15 . Point is (0, 15) . If y = 0 , x = 20 . Point is (20, 0) . Next, find the intersection of the two lines x + 3y = 30 and 3x + 4y = 60 . Multiply the first equation by 3: 3x + 9y = 90 . Subtract the second equation from this: (3x + 9y) - (3x + 4y) =