Answer:
Let
number of units of food type P = x
and
number of units of food type Q = y
The
contents of food are given below :
Therefore
the above L.P.P. is given as
Minimize,
C = 60x + 80y, subject to the containts
3x
+ 4y 8; 5x + 2y 11, x, y 0
L1 :
3x + 4y = 8 L2 : 5x + 2y = 11
Here
cost C is minimum at and and is Rs.160.
Since
the region is unbounded, therefore, 160 may or may not be the minimum value of
C. For this draw graph of inequality 60x + 80y < 160.
i.e.
3x + 4y < 8
L
: 3x + 4y = 8
Clearly
open half plane has no common points with the feasible region, so minimum value
of C is 160.
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