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

Two tailors A and B earn ₹150 and ₹200 per day respectively. A can

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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