Question

A house wife wishes to mix two types of food F1 and F2 in such a way that the
vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food F1 costs
E60/Kg and Food F2 costs E80/kg. Food F1 contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while
Food F2 contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Formulate this problem as a linear
programming problem to minimize the cost of the mixtures,

Answer 1

problem 21457-7776655

Answer 2

Given:

Amount of Vitamin A in the mixture = 8 units

Amount of Vitamin B in the mixture = 11 units

Cost of food F1 = E60/kg

Cost of food F2 = E80/kg

Amount of Vitamin A in food F1 = 3 units/kg

Amount of Vitamin B in food F1 = 5 units/kg

Amount of Vitamin A in food F2 = 4 units/kg

Amount of Vitamin B in food F2 = 2 units/kg

To find:

The minimum cost of the mixtures.

Solution:

The decision variables are the weights of F1 and F2 in the mixture. Let $x$ kg be the weight of food F1 and $y$ kg be the weight of food F2.

The minimum requirement for the vitamins is the constraints. At least 8 units of vitamin A is required. Hence,

$3x+4y\geq 8$

Similarly, at least 11 units of vitamin B is required. Hence,

$5x+2y\geq 11$

For 1kg of food F1, the cost is E60 and for 1kg of food F2, the cost is E80. Thus, the total cost of $x$ kg of food F1 and $y$ kg of food F2 is given by the objective function.

$C=60x+80y$

$x\geq 0,y\geq 0$ are the non-negativity conditions.

Mathematically formulating the linear programming problem,

Minimize, $C=60x+80y$

Subject to: $3x+4y\geq 8$

$5x+2y\geq 11$

$x\geq 0,y\geq 0$

We find the optimal solution using the graphical method to identify a feasible region and its corner points. The graph is shown below. The corner points obtained are A(0,5.5), B(2,0.5), C(2.667,0)

Testing these corner points on $C=60x+80y$ to check which corner point gives minimum cost for the mixture.

For A(0,5.5)

$C=60(0)+80(\frac{11}{2})=0+40*11=440$

For B(2,0.5)

$C=60(2)+80(\frac{1}{2})=120+40=160$

For C(2.667,0)

$C=60(\frac{8}{3})+80(0)=20*8+0=160$

Though points B and C provide the same cost, point C is not considered because at point C, $x=\frac{8}{3},y=0$ that means the weight of food F2 should be 0 which is not possible. Hence, we consider point B where $x=2,y=\frac{1}{2}$ which means the weight of food F1 should be 2kg and the weight of food F2 should be 0.5kg in order to minimize the cost of mixtures.

The weight of food F1 should be 2kg and the weight of food F2 should be 0.5kg in order to minimize the cost of mixtures to E160.