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  • Solve the following Linear Programming Problems graphically Z = to 3x + 5y

4.    Solve the following Linear Programming Problems graphically:

         Minimise  Z = 3x + 5y

         Such that  x+3y\geq 3,x+y\geq 2,x,y\geq 0.

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

Answers (1)

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

               

The feasible region is unbounded as shown.

The corner points of the feasible region are A(3,0),B(\frac{3}{2},\frac{1}{2}),C(0,2)

 The value of these points at these corner points are : 

Corner points            Z = 3x + 5y  
        A(3,0)                9  

        B(\frac{3}{2},\frac{1}{2})     

               7 Minimum
       C(0,2)               10  
                          

  The feasible region is unbounded, therefore 7 may or may not be the minimum value of Z .

For this, we draw 3x + 5y< 7 and check whether resulting half plane has a point in common with the feasible region or not.

We can see a feasible region has no common point with.Z = 3x + 5y

Hence, Z has a minimum value of 7 at B(\frac{3}{2},\frac{1}{2})

 

 

 

Posted by

seema garhwal

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