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?

#### 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 10*y* shirts and 4*y* 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 15*x** *and B earn Rs 20*y* .

Let Z denotes the total cost

\[\therefore Z = 15x + 20y\]

Days cannot be negative.

∴ \[x, y \geq 0\]

Min Z = \[15x + 20y\] subject to

\[6x + 10y \geq 60\]

\[4x + 4y \geq 32\]

6

*x*+ 10

*y*= 60, 4

*x*+ 4

*y*= 32,

*x*= 0 and

*y*= 0

Region represented by 6

*x*+ 10

*y*≥ 60:

The line 6

*x*+ 10

*y*= 60 meets the coordinate axes at

*A*

_{1}(10, 0) and

*B*

_{1}(0, 6) respectively. By joining these points we obtain the line6

*x*+ 10

*y*= 60. Clearly (0,0) does not satisfies the

*6*

*x*+ 10

*y*= 60. So,the region which does not contains the origin represents the solution set of the inequation 6

*x*+ 10

*y*≥ 60.

Region represented by 4

*x*+ 4

*y*≥ 32:

The line 4

*x*+ 4

*y*=32 meets the coordinate axes at

*C*

_{1}(8, 0) and

*D*

_{1}(0, 8) respectively. By joining these points we obtain the line 4

*x*+ 4

*y*= 32.Clearly (0,0) does not satisfies the inequation 4

*x*+ 4

*y*≥ 32. So,the region which does not contains the origin represents the solution set of the inequation 4

*x*+ 4

*y*≥ 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 6

*x*+ 10

*y*≥ 60,4

*x*+ 4

*y*≥ 32,

*x*≥ 0, and

*y*≥ 0 are as follows.

Thus, the mathematical formulation of the given linear programming problem is

*D*

_{1}(0, 8),

*E*

_{1}(5, 3) and

*A*

_{1}(10, 0).

The values of Z at these corner points are as follows

Corner point | Z = 15x + 20y |

D_{1} |
160 |

E_{1} |
135 |

A_{1} |
150 |

The minimum value of Z is 135 which is attained at

*E*

_{1}(5, 3).

Thus, for minimum labour cost,

*A*should work for 5 days and

*B*should work for 3 days.