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?

Answer 1

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,
$$\begin{gathered} 6x+10y\ge 60 \\ 4x+4y\ge 32 \end{gathered}$$
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
$\therefore Z=15x+20y$
∴ Z =15x + 20y
Days cannot be negative.
$\therefore$$x,y\ge 0$

Min Z = $15x+20y$

subject to

$$\begin{gathered} 6x+10y\ge 60 \\ 4x+4y\ge 32 \end{gathered}$$
$x,y\ge 0$

First we will convert inequations into equations as follows :
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 formulat​ion 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.