Answer:
Solution
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Let the tailor P work for x days and the tailor Q work for y days respectively.
Z=150x+200y
Here the problem can be formulated as an L.P.P. as follows
Minimize Z=150x+200y
Subject to the constraints
6x+10y≥60
⇒3x+5y≥30
4x+4y≥32
x+y≥8
and x≥0,y≥0
Converting them into equations we obtain the following equations
3x+5y=30,
⇒y=
5
30−3x
x 0 10 5
y 6 0 3
⇒x+y=8
y=8−x
x 0 8 5
y 8 0 3
The lines are shown on the graph paper and the feasible region (Unbounded convex) is shown shaded.
The corner points are
A(10,0),B(5,3) and C(0,8)
At the corner point the value of Z=150x+200y
At A(10,0)Z=1500
At B(5, 3), Z=150×5+200×3=750+600=1350
At C (0, 8), Z=1600
As the feasible region is unbounded, we draw the graph of the half-plane
150x+200y<1350
3x+4y<27
There is no point common with the feasible region, therefore, Z has minimum value.
Minimum value of Z is Rs. 1350 and it occurs at the point B(5,3).
Hence, the labour cost is Rs. 1350 when P works for 5 days and Q works for 3 days.