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9.    Solve the following Linear Programming Problems graphically:

           Maximise Z = -x+2y

           Subject to the constraints:x\geq 3,x+y\geq 5,x+2y\geq 6,y\geq 0.

           Show that the minimum of Z occurs at more than two points.

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The region determined by constraints x\geq 3,x+y\geq 5,x+2y\geq 6,y\geq 0.is as follows,

               

The corner points of the feasible region are A(6,0),B(4,1),C(3,2)

 The value of these points at these corner points are : 

Corner points            Z = -x+2y  
        A(6,0)               - 6 minimum

        B(4,1)

              -2  
       C(3,2)                 1 maximum 
                          

The feasible region is unbounded, therefore 1 may or may not be the maximum value of Z.

For this, we draw -x+2y> 1 and check whether resulting half plane has a point in common with a feasible region or not.

We can see the resulting feasible region has a common point with a feasible region.

Hence , Z =1 is  not maximum value , Z has no maximum value. 

 

 

 

Posted by

seema garhwal

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