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Question
Two tailors, A and B earn Rs 15 and Rs 20 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to produce (at least) 60 shirts and 32 pants at a minimum labour cost?
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Solution
Let tailor A work for x days and tailor B work for y days.
In one day, A can stitch 6 shirts and 4 pants whereas B can stitch 10 shirts and 4 pants
Thus, in x days A can stitch 6x shirts and 4y pants whereas B can stich 10y shirts and 4y pants.
pants.
It is given that the minimum requirement of the shirts and pants are respectively 60 and 32.
Thus,
\[6x + 10y \geq 60\]
\[4x + 4y \geq 32\]
Further it is given that A and B earn Rs 15 and Rs 20 per day respectively.
Thus, A earn Rs 15x and B earn Rs 20y .
Let Z denotes the total cost
\[\therefore Z = 15x + 20y\]
Days cannot be negative.
∴ \[x, y \geq 0\]
Min Z = \[15x + 20y\] subject to
\[6x + 10y \geq 60\]
\[4x + 4y \geq 32\]
6x + 10y = 60, 4x + 4y = 32, x = 0 and y = 0
Region represented by 6x + 10y ≥ 60:
The line 6x + 10y = 60 meets the coordinate axes at A1(10, 0) and B1(0, 6) respectively. By joining these points we obtain the line6x + 10y = 60. Clearly (0,0) does not satisfies the 6x + 10y = 60. So,the region which does not contains the origin represents the solution set of the inequation 6x + 10y ≥ 60.
Region represented by 4x + 4y ≥ 32:
The line 4x + 4y =32 meets the coordinate axes at C1(8, 0) and D1(0, 8) respectively. By joining these points we obtain the line 4x + 4y = 32.Clearly (0,0) does not satisfies the inequation 4x + 4y ≥ 32. So,the region which does not contains the origin represents the solution set of the inequation 4x + 4y ≥ 32.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 6x + 10y ≥ 60,4x + 4y ≥ 32, x ≥ 0, and y ≥ 0 are as follows.
Thus, the mathematical formulation of the given linear programming problem is
The corner points are D1(0, 8), E1(5, 3) and A1(10, 0).The values of Z at these corner points are as follows
| Corner point | Z = 15x + 20y |
| D1 | 160 |
| E1 | 135 |
| A1 | 150 |
The minimum value of Z is 135 which is attained at E1(5, 3).
Thus, for minimum labour cost, A should work for 5 days and B should work for 3 days.
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