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Question

A house wife wishes to mix together two kinds of food, I and II, in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg of food is given below:
Vitamin AVitamin BVitamin C
Food I123
Food II221
One Kg of food I costs E6 and one Kg of food II costs E10. Formulate the above problem as a linear programming problem and find the least cost of the mixture which will product the diet.

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Solution

Given
Vitamin A Vitamin B Vitamin C Cost Rs
Food i 1 2 3 16
Food ii 2 2 1 10
Requirement Atleast 10 Atleast 12 Atleast 8
Solution:We need to maximise z=6x+10y for the following contraints
i)x+2y≥10
ii)2x+2y≥12⇒ x+y≥6
iii)3x+y≥8
iv)x≥0,y≥0
Finding the feasible region
Since x and y are≥,0,,the fasible region is in first quadrant

i)On solving equation x+y=6 and x+2y=10
We get
x=6−y
⇒6−y+2y=10
y=4
x=6−y
⇒=6−4
x=2
ii)On standing equation $x+2y=10and3x+y =8$ we get
x+2y=10
3x+y=8––––––––––––
x+2y=10
6x+2y=16––––––––––––––
−5x=−6
x=65
x+2y=10
65+2y=10
y=225
Point H(65,225)

iii)On solving equation x+y=6 and 3x+y=8 we get

x+y=6
(−)3x+y=8––––––––––––––––
−2x=−2
x=1

x+y=6
1+y=6
y=5
PointI(1,5)

i)x+2y≥10
x 0 10
y 5 0
Point A(0,5) B(10,0)

ii)x+y≥ 6
x 0 6
y 6 0
Point C(0,6) D(6,0)

iii)3x+y≥8
x 08/3
y80
Point E(0,8) F(8/3,0)
Corner point cost
z=6x+10y
(0,8) 80
(135)56→ MInimum
(2,4) 66
(0,8) 80

The least mixture cost is 56 for point,
x=1and y=5
Least cost=56

882128_847842_ans_77baa12d3c684b34bf59cdd1da3dfd4a.JPG

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