Shade the feasible region for the inequations 6 x+4 y 120,3 x+10 y 180, x, y 0 in a rough figure.

Step-by-step solution

The given inequations are 6x + 4y 120 and 3x + 10y 180 with x, y 0 . First, simplify the inequalities: 6x + 4y 120 3x + 2y 60 3x + 10y 180 Find the intercepts for the lines: For 3x + 2y = 60 : If x = 0 , y = 30 . Point is (0, 30) . If y = 0 , x = 20 . Point is (20, 0) . For 3x + 10y = 180 : If x = 0 , y = 18 . Point is (0, 18) . If y = 0 , x = 60 . Point is (60, 0) . Find the intersection point of the two lines: 3x + 2y = 60 3x + 10y = 180 Subtracting the first from the second: 8y = 120 y = 15 . Substituting y = 15