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Minimise and minise Z=3x-4y subject to x-2y le 0, -3x+y le 4, x-y le 6 and x,y le 0 |
Answer» From the shown graph, for the feasible region, we see that it is unbounded and coordinates of corner points are (0,0),(12,6) and (0,4) For given, unbounded reigon the MINIMUM value of Z may or may be -16 So, for deciding this, we graph the inequality 3x-4y larr16 and check whether the resulting open half plane has common points with feasible region or not. Thus, from the figure it shows it has common points with feasible region, so it does not have any minimum value. Also, SIMILARLY for maximum value, we graph the in equality `3x-4y gt 12` nad see that resulting open half plane has no common points with the feasibleregion and HENCE maximum value for Z=3x-4y. |
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