Two tailors A and B earn ₹150 and ₹200 per day respectively. A can stich 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. Form a L.P.P to minimize the labour cost to produce (stitch) at least 60 shirts and 32 pants and solve it graphically.

Given: Two tailors A and B earn ₹150 and ₹200 per day respectively. A can stich 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day

To Find: Minimum labour cost to produce (stitch) at least 60 shirts and 32 pants.

Let tailor A works for x days and tailor B works for y days.

We need to minimize the labour cost.

The net equation is, Z = 150x + 200y

A can stich 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day.

For shirt,

6x + 10y 60

⇒ 3x + 5y 30

And for pant,

4x + 4y 32

⇒ x + y 8

Plotting the lines on graph we get,

The corner points are (0,8), (5,3) and (10,0). We have to minimize Z.

So, for the point (5, 3) the value of Z is minimum. So,

x = 5 and y = 3.

So, A need to work for 5 days and B need to work for 3 days. And the cost is 1350 Rs.