Answer:
Let
number of packets of food P = x
and
number of packets of food Q = y
The
contents of each packet of food P and Q is given as
The
above L.P.P. is given as
Maximize,
Z = 6x + 3y, subject to the constraints 12x + 3y 240, 4x + 20y 460
6x
+ 4y 300, x, y
i.e.,
4x + y
3x
+ 2y
L1
: 4x + y = 80 L2 : x + 5y = 115
L3
: 3x + 2y = 150
Here
Z is manimum at point G(40, 15)
Hence
amount of vitamin A wil be maximum if 40 packets of food P and 15 packets of
food Q are to be used.
Food
Calcium
Iron
Cholesterol
Vitamin
A
P
12
4
6
6
Q
3
20
4
3
Minimum
requirement
240
460
At
most 300
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