Question 32 - CBSE Class 12 Sample Paper for 2020 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at April 16, 2024 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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