Two tailors A and B are paid Rs. 150 and Rs. 200 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. Form a linear programming problem to minimise the labour cost to produce at least 60 shirts and 32 pants. Solve the problem graphically.

 Let the tailor A work for x days tailor B work for y days where x ≥ 0, y ≥ 0.
Let z be the total cost.

Table

We are to minimise
z = 150 x + 200 y
subject to the constraints
6 x + 10 y ≥ 60
4 x + 4 y ≥ 32
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.
Now we draw the graph of 6 x + 10 y = 60
For x = 0, 10 y = 60 or y = 6
For y = 0, 6 x = 60 or x = 10
∴ line meets OX in A(10, 0) and OY in L(0, 6).
Again we draw the graph of
4 x + 4 y = 32
For x = 0, 4 y = 32 or y = 8
For y = 0, 4 x = 32 or x = 8
∴ line meets OX in B(8, 0) and OY in M(0, 8).
Since feasible region satisfies all the constraints.

∴ shaded region is the feasible region, which is unbounded , and comer points are A(10, 0), C(5, 3), M(0, 8).
At A(10, 0), z = 150 (10) + 200 (0) = 1500 + 0 = 1500
At C(5, 3), z = 150 (5) + 200 (3) = 750 + 600 = 1350
At M(0, 8), z = 150 (0) + 200 (8) = 0 + 1600 = 1600
∴ least cost = Rs. 1350 at (5, 3).
Since feasible region is unbounded.
∴ we are to check whether this value is minimum.
For this we draw the graph of
150 x + 200 y < 1350    ...(1)
Since (1) has no common point with feasible region.
∴ minimum value = Rs. 1350 at (5, 3)
∴ minimum labour cost is Rs. 1350 when tailor A works for 5 days and tailor B works for 3 days.