For an objective function, Min Z = 10x + 20y; if the corner points are (0,18), (2,6), (4,2), (12,0); the optimal value of Z is
Solution:
Let given points be A(0,18), B(2,6), C(4,2), D(12,0)
To minimise Z = 10x + 20y
At A(0,18), Z = 10(0)+20(18) = 360
At B(2, 6), Z = 10(2)+20(6) = 20 + 120 = 140
At C(4, 2), Z = 10(4)+20(2) = 40 + 40 = 80
At D(12,0), Z = 10(12)+20(0) = 120
Clearly, the minimum value is 80 which occurs at C(4, 2).
Comments
Leave a comment