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

Answer: (A)

Suppose $x$ grams of wheat and $y$ grams of rice are mixed in the daily diet.
Since every grams of wheat provides $0.1g$ of proteins and every gram of rice gives $0.05g$ 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 $50g$.
$\therefore 0.1x+0.05y\ge 50\Rightarrow \frac{x}{10}+\frac{y}{20}\ge 50$
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 $200g$.
$\therefore 0.25x+0.5y\ge 200\Rightarrow \frac{x}{4}+\frac{y}{2}\ge 200$
Since, the quantities of wheat and rice cannot be negative.
Therefore,x$\ge 0,y\ge 0$
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
$₹\frac{4x}{1000}+\frac{6y}{1000}$
Subject to the constraints
$\frac{x}{10}+\frac{y}{20}\ge 50,\frac{x}{4}+\frac{y}{2}\ge 200,$ 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 $Z=\frac{4x}{1000}+\frac{6y}{1000}$
$A_2(800,0)$	$Z=\frac{4}{1000}\times 800+\frac{8}{1000}\times 0=3.2$
$P(400,200)$	$Z=\frac{4}{1000}\times 400+\frac{6}{1000}\times 200=2.8$
$B_1(0,1000)$	$Z=\frac{4}{1000}\times 0+\frac{6}{1000}\times 1000=6$

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