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Refers to question 12.i.eZ represents cost , if Z=400x+200y, x+2y ≥ 30, x-y=0 and 5x+2y ≥ 30,What will be the minimum cost?

Answer»


Solution :Referring to solution 12, we have minimise Z=400x+200y, subject to `x+2y ge 30`
On solving y ≥ x and 5x+2y=30, we GET
`y(30)/(7),x=(30)/(7)`

On solving x-y=0 and 2x+y=15, we get x=5, y=5
So, from the shaded FEASIBLE region it is clear that CORRDINATES of corner points are (0,15), (5,5) and `((30)/(7),(30)/(7))`

HENCE the minimum cost is 2571.43


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