Question

A diet of two foods F₁ and F₂ contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per 25 gms) is given in the following table:

Nutrients	Food	F1	F2
Thiamine		0.25	0.10
Phosphorous		0.75	1.50
Iron		1.60	0.80

The minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron. The cost of F₁ is 20 paise per 25 gms while the cost of F₂ is 15 paise per 25 gms. Find the minimum cost of diet.

Answer 1

Let 25x grams of food F₁ and 25y grams of food F₂ be used to fulfil the minimum requirement of thiamine, phosphorus and iron.
As, we are given

Nutrients	Food	F1	F2
Thiamine		0.25	0.10
Phosphorous		0.75	1.50
Iron		1.60	0.80

And the minimum requirement of the nutrients in the diet are 1.00 mg of thiamine, 7.50 mg of phosphorous and 10.00 mg of iron.

$$\begin{gathered} \mathrm{Therefore}, \\ 0.25x+0.10y\ge 1 \\ 0.75x+1.50y\ge 7.5 \\ 1.6x+0.8y\ge 10 \\ \mathrm{Since},\mathrm{the}\mathrm{quantity}\mathrm{cannot}\mathrm{be}\mathrm{negative} \\ \therefore x,y\ge 0 \end{gathered}$$

The cost of F₁ is 20 paise per 25 gms while the cost of F₂ is 15 paise per 25 gms.Therefore, the cost of 25x grams of food F₁ and 25y grams of food F₂ is
Rs (0.20x + 0.15y).
Hence,
Minimize Z = $0.20x+0.15y$
subject to

$$\begin{gathered} 0.25x+0.10y\ge 1 \\ 0.75x+1.50y\ge 7.5 \\ 1.6x+0.8y\ge 10 \\ x,y\ge 0 \end{gathered}$$

First, we will convert the given inequations into equations, we obtain the following equations:
0.25x + 0.10y = 1, 0.75x + 1.50y = 7.5, 1.6x + 0.8y = 10, x = 0 and y = 0.

The line 0.25x + 0.10y = 1 meets the coordinate axis at $A\left(4,0\right)$ and B(0, 10). Join these points to obtain the line 0.25x + 0.10y = 1.
Clearly, (0, 0) does not satisfies the inequation 0.25x + 0.10y ≥ 1. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line 0.75x + 1.50y = 7.5. meets the coordinate axis at C(10, 0) and D(0, 5). Join these points to obtain the line 0.75x + 1.50y = 7.5.
Clearly, (0, 0) does not satisfies the inequation 0.75x + 1.50y ≥ 0.75. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

The line 1.6x + 0.8y = 10 meets the coordinate axis at $E\left(\frac{25}{4},0\right)$ and $F\left(0,\frac{25}{2}\right)$. Join these points to obtain the line 1.6x + 0.8y = 10.
Clearly, (0, 0) does not satisfies the inequation 1.6x + 0.8y ≥ 10. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.
These lines are drawn using a suitable scale.

The corner points of the feasible region are F(0, 12.5), G(5, 2.5), C(10, 0)

The value of the objective function at these points are given by the following table

Points	Value of Z
F	0.20(0)+0.15(12.5) = 1.875
G	0.20(5)+0.15(2.5) = 1.375
C	0.20(10) + 0.15(0) = 200

Thus, the minimum cost is at G which is Rs 1.375