Answer:
Let
x kg of fertilizer I and y kg of fertilizer II and used. The contents of
fertilizer and given below :
Therefore,
the above L.P.P. is given as
Minimum
cost, C = 6x + 5y, subject to the constraints,
i.e.
2x + y
L1
: 2x + y = 280 L2 : 3x + 5y = 700
Here cost is minimum
at E (100, 80) and is Rs.100
Since
the region is unbounded, thereforeRs.1000 may or not may not be the minimum
value of C. For this draw graph of inequality.
L
: 6x + 5y = 1000
Clearly
open half plane has no common point with the feasible region so minimum value
of C = Rs.1000.
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