Question 32 - CBSE Class 12 Sample Paper for 2020 Boards

Last updated at Oct. 24, 2019 by Teachoo

Two tailors A and B earn ₹150 and ₹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 L.P.P to minimize the labour cost to produce (stitch) at least 60 shirts and 32 pants and solve it graphically

Question 32 Two tailors A and B earn ₹150 and ₹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 L.P.P to minimize the labour cost to produce (stitch) at least 60 shirts and 32 pants and solve it graphically
Let Tailor A work for x days
Tailor B work for y days
Given that
Two tailors A and B earn ₹150 and ₹200 per day respectively.
And, we need to minimize labour cost
So, our equation will be
∴ Z = 150x + 200y
Now, according to question
A can stitch 6 shirts and 4 pants per day, while B can stitch 10 shirts and 4 pants per day. We need to minimize the labour cost to produce at least 60 shirts and 32 pants and solve it graphically
Shirts
Min Shirts = 60
6x + 10y ≥ 60
3x + 5y ≥ 30
Pants
Min Pants = 32
4x + 4y ≥ 32
x + y ≥ 8
Also,
x ≥ 0, y ≥ 0
Combining all constraints :
Min Z = 150x + 200y
Subject to constraints,
3x + 5y ≥ 30
x + y ≥ 8
& x ≥ 0 , y ≥ 0
3x + 5y ≥ 30
x + y ≥ 8
As the region that is feasible is unbounded.
Hence 1350 may or may not be the minimum value of Z.
To check this, we graph inequality.
150x + 200y ≤ 1350
i.e. 3x + 4y ≤ 27
As there is no common point between the feasible region & the inequality.
Hence, 1350 is the minimum value of Z.
Hence, cost will be minimum if
Number of days tailor A works for = 5
Number of days tailor B works for = 3
Minimum Cost = Rs 1350

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.