Problem based on graphical solution of a given LPP when feasible region is bounded.

1.    Solve the following linear programming graphically;

Maximize and minimize z = 60x+ 15y subject to constraints

     X+y≤50 ; 3x+y≤90; x,y≥ 0

Script 1.    First of all let us draw the lines x+y=50 and 3x+y = 90 using suitable points on the graphs. 2.   The lines intersect at the point 20,30 3.   Now shade the region of intersection of the two lines. 4.   The shaded region gives the feasible region determined by the constraints 5.   Hence OABC is the bounded region 6.   Therefore we use corner point method to calculate the maximum and minimum values 7.   Vertices of feasible region are 8.   Now let us find the maximum and minimum values 9.   For the Corner point O (0,0) 10.                For the corner point Corner point  A(30,0) 11. For the corner point B (20,50) 12.                C(0,50) 13.                Hence the minimum value of Z is ) at (0,0) and 14.                Maximum value of Z is at (30,0)

Solution                                       7.O(0,0), A(30,0), B(20,30), C(0,50)       60(00+15(0) = =0 is minimum   60(30) + 15(0)=1800 is maximum 60(20)+15(50) = 1650   60(0)+15(50)=750

Posted Date: 7/23/2012 4:10:32 AM | Location : United States

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