Question

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

• A. 400, 200
• B. 300, 400
• C. 200, 400
• D. 400, 300

Answer 1

The Correct Option is A

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. Clearly, $Z$ is minimum for $x = 400$ and $y = 200$. The minimum diet cost is $2.8.$