Q. 14.5( 12 Votes )

# 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.

Answer :

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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Using the method of integration, find the area of the region bounded by the following lines:

5x - 2y - 10 = 0

x + y - 9 = 0

2x - 5y - 4 = 0

Mathematics - Board PapersRefer to Exercise 27. (Maximum value of Z + Minimum value of Z) is equal to

Mathematics - ExemplarMaximise and Minimise Z = 3x – 4y

subject to x – 2y ≤ 0

– 3x + y ≤ 4

x – y ≤ 6

x, y ≥ 0

Mathematics - Exemplar