We are to minimize
z = 2x + 3y
subject to the constraints
x + 2y ≤ 10
x + 2y ≥ 1
x ≥ 0, 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 + 2y = 10
For x = 0, 2y =10 or y = 5
For y = 0, x = 10
∴  line meets OX in A (10, 0) and OY in L (0, 5).
Again we draw the graph of x + 2y = 1
For x = 0, 2 y = 1 or y = 0.5
For y = 0, x = 1
∴ line meets OX is B (1, 0) and OY in (0, 0.5)

Since feasible region satisfies all the constraints
∴ in this case, BALM is the feasible region and corner points are
B(1, 0), A(10, 0), L(0, 5), M(0, 0.5).
At B(1, 0), z = 2(1) + 3(0) = 2 + 0 = 2
At A(10, 0), z = 2 (10) + 3 (0) = 20 + 0 = 20
At L(0, 5), z = 2(0) + 3(5) = 0 +15 = 15
At M(0, 0.5), z = 2 (0) + 3 (0.5) = 0 + 1.5 = 1.5
∴ minimum value = 1.5 at (1, 0).