A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F₁ and F₂ are available. Food F₁ costs Rs 4 per unit food and F₂ costs Rs 6 per unit. One unit of food F₁ contains 3 units of vitamin A and 4 units of minerals. One unit of food F₂ contains 6 units of vitaminAand 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Answers
Answer:
24 units of F₁ & 4/3 units of F₂ to be bought
min cost = Rs 104
in case of whole integer
25 units of F₁ & 1 units of F₂ to be bought
cost = 25 × 4 + 6 = Rs 106
Step-by-step explanation:
A = VitaminA M = Mineral
F₁ = 3A + 4M
F₂ = 6A + 3M
F₁ Costs Rs 4 per unit
F₂ Costs Rs 6 per unit
Let say X unit of F₁ & Y units of F₂
total vitamins = 3X + 6 Y = 80 Eq 1
4X + 3Y = 100
8X + 6Y = 200 Eq2 ( multiplying by 2)
3X + 6Y = 80 (Eq2 - Eq 1)
5X = 120
X = 24
3×24 + 6Y = 80 ( putting value of X in eq 1)
72 + 6Y = 80
Y = 8/6
Y = 4/3
24 units of F₁ & 4/3 units of F₂ to be bought
Cost = (4 × 24) + (6 × 4/3)
= 96 + 8 = Rs 104
Now if we say that units can not be bought in Fraction
then F₂ should be bought either 1 Or 2 Unit
Case 1
F₂ bought 1 unit
6 A + 3M ( present Mineral and vitamin )
remaining vitaminA = 80-6 = 74 so min F₁= 74/3 = 24.66 = 25 ( next integer)
Remaining Mineral = 100-3 = 97 so min F₁ = 97/4= 24.25 = 25 (next integer)
So F₂ bought 1 unit & F₁ bought = 25 units
cost = (25 × 4) + (1 × 6) = Rs 106
Case 2
F₂ bought 2 unit
12 A + 6M ( present Mineral and vitamin )
remaining vitaminA = 80-12 = 68 so min F₁= 68/3 = 22.66 = 23 ( next integer)
Remaining Mineral = 100-6 = 94 so min F₁ = 94/4= 23.5 = 24 (next integer)
to meet mineral requirements we need to buy 24F₁ at least
So F₂ bought 2 unit & F₁ bought = 24 units
cost = (24 × 4) + (2 × 6) = Rs 108
106 < 108
So case 1 has less amount
So F₂ bought 1 unit & F₁ bought = 25 units
cost = (25 × 4) + (1 × 6) = Rs 106
This is the detailed solution for your question.
