Solve the following linear programming problem graphically:
Minimise    Z = 3x + 5y subject to the constraints:x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0

We are to minimise,  Z = 3x + 5y subject to the constraints x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + 3y = 3
For x = 0, 3 y = 3 or y = 1
For y = 0, x = 3
∴ line meets OX in A(3, 0) and OY in L(0, 1).
Again we draw the graph of x + y = 2
For x = 0, y = 2
For y = 0, x = 2
∴ line meets OX in B(2, 0) and OY in M(0, 2).
Since feasible region is the region which satisfies all the constraints.
∴ shaded region is the feasible region which is unbounded and has comer points
   
                  

Since feasible region is unbounded.
∴  we are to check whether this value is minimum.
For this we draw the graph of
3x + 5y < 7    ...(1)
Since (1) has n.o common point with feasible region.