Q. Every gram of wheat provides 0.1 g of proteins and 0.25 g of carbohydrates. The corresponding values of rice are 0.05 g and 0.5 g respectively. Wheat costs ₹ 4 per kg and rice ₹ 6. The minimum daily requirements of proteins and carbohydrates for an average child are 50 g and 200 g respectively. Then in what quantities should wheat and rice be mixed in the daily diet to provide minimum daily requirement of proteins and carbohydrates at minimum cost
Linear Programming
Solution:
Suppose $x$ grams of wheat and $y$ grams of rice are mixed in the daily diet.
Since every grams of wheat provides $0.1 g$ of proteins and every gram of rice gives $0.05 g$ of proteins. Therefore, $x$ gms of wheat and $y$ grams of rice will provide $0.1x + 0.05y$ g of proteins. But the minimum daily requirement of proteins is of $50 g$.
$\therefore \quad0.1x + 0.05y \ge 50\quad\Rightarrow \quad \frac{x}{10}+ \frac{y}{20} \ge 50$
Similarly, $x$ grams of wheat and $y$ grams of rice will provide $0.25x + 0.5y$ g of carbohydrates and the minimum daily requirement of carbohydrates is of $200 g$.
$\therefore \quad0.25x + 0.5y \ge 200\quad\Rightarrow \quad \frac{x}{4}+ \frac{y}{2} \ge 200$
Since, the quantities of wheat and rice cannot be negative.
Therefore,$\quad $x$ \ge 0,\quad y \ge 0$
It is given that wheat costs ₹ $4$ per kg and rice ₹ $6$ per kg. So, $x$ grams of wheat and $y$ grams of rice will cost
$\quad ₹ \frac{4x}{1000} + \frac{6y}{1000}$
Subject to the constraints
$\frac{x}{10} + \frac{y }{20} \ge 50, \frac{x}{4} + \frac{y}{2} \ge 200,\quad$ and $x \ge 0, y \ge 0$
The solution set of the linear constraints is shaded in figure. The vertices of the shaded region are
$A_2 (800, \,0), \,P \,(400,\, 200)$ and $B_1(0,\, 1000)$.
The values of the objective function at these points are given in the following table.
Point $( x_1,\, x_2)$ Valueof objectivefunction
$ Z = \frac{4x}{1000} + \frac{6y}{1000}$ $A_2 (800,0)$ $ Z = \frac{4}{1000} \times800+ \frac{8}{1000}\times0 = 3.2$ $ P(400,\,200)$ $Z = \frac{4}{1000} \times400+ \frac{6}{1000}\times200 = 2.8$ $B_1(0,\,1000)$ $Z = \frac{4}{1000} \times0+ \frac{6}{1000}\times1000 =6$
Clearly, $Z$ is minimum for $x = 400$ and $y = 200$. The minimum diet cost is $2.8.$
| Point $( x_1,\, x_2)$ | Valueof objectivefunction $ Z = \frac{4x}{1000} + \frac{6y}{1000}$ |
|---|---|
| $A_2 (800,0)$ | $ Z = \frac{4}{1000} \times800+ \frac{8}{1000}\times0 = 3.2$ |
| $ P(400,\,200)$ | $Z = \frac{4}{1000} \times400+ \frac{6}{1000}\times200 = 2.8$ |
| $B_1(0,\,1000)$ | $Z = \frac{4}{1000} \times0+ \frac{6}{1000}\times1000 =6$ |