Q11 of 118 Page 30

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

The data can be represented in a table below



To minimize labour cost means to assume to minimize the earnings i.e,


Min Z = 150x + 200y


With constraints


x, y ≥ 0 at least 1 shirt and 1 pant is required


6x + 10y ≥ 60 require at least 60 shirts


4x + 4y ≥ 32 require at least 32 pants


On solving the above inequalities as equations, we get,


x = 5 and y = 3


other corner points obtained are [0, 6], [10, 0], [0, 8] and [8, 0]



The feasible region is the upper unbounded region A - E - D


Point E(5, 3) may not be minimal value. So, plot 150x + 200y < 1350 to see


If there is a common region with A - E - D.


The green line has no common point, therefore



Thus, stitching 5 shirts and 3 pants minimizes labour cost to Rs 1350/ -


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