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

Transcript

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