Linear Programming Question 12
Question: Every gram of wheat provides 0.1 g of proteins and 0.25g of carbohydrates. The corresponding values of rice are 0.05 g and 0.5 g respectively. Wheat costs Rs. 4 per kg and rice Rs. 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
Options:
A) 400, 200
B) 300, 400
C) 200, 400
D) 400, 300
Show Answer
Answer:
Correct Answer: A
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 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 200 g.
$ \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) $ | Value of objective function $ Z=\frac{4x}{1000}+\frac{6y}{1000} $ | |
| $ A_2(800,0) $ $ P(400,200) $ $ B_1(0,1000) $ | $ Z=\frac{4}{1000}\times 800+\frac{8}{1000}\times 0=3.2 $ $ Z=\frac{4}{1000}\times 400+\frac{6}{1000}\times 200=2.8 $ $ 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 2.8. |
BETA
