CBSE Class 12 · 2005 · Linear ProgrammingLong Answer

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Two tailors A and B earn Rs. 150 and Rs. 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 linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 pants.

Answer: Minimize Z = 150x + 200y subject to 3x + 5y ≥ 30, x + y ≥ 8, x, y ≥ 0 (min cost Rs 1350 at A = 5 days, B = 3 days)

Solution

Let tailor A work for

x

days and tailor B for

y

days.
Labour cost to minimise:

Z = 150 x + 200 y

.
Shirts:

6 x + 10 y \geq 60 ⟹ 3 x + 5 y \geq 30

Pants:

4 x + 4 y \geq 32 ⟹ x + y \geq 8

x \geq 0, y \geq 0

This is the required LPP. Solving it, the corner points of the unbounded feasible region are

(10, 0)

(5, 3)

(intersection of

3 x + 5 y = 30

and

x + y = 8

) and

(0, 8)

.
Evaluate

Z

(10, 0)

Z = 1500

(5, 3)

Z = 750 + 600 = 1350

← minimum

(0, 8)

Z = 1600

∴ Z_{m i n} = Rs 1350 at (5, 3)

Tailor A should work 5 days and tailor B 3 days.

CBSE Class 12 Board · 2005 · Subjective

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CBSE Class 12 · 2005 · Linear ProgrammingLong Answer

Solution

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