A) 400, 200
B) 300, 400
C) 200, 400
D) 400, 300
Correct Answer: A
Solution :
[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 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\[Z=\frac{4x}{1000}+\frac{6y}{1000}\] |
| \[{{A}_{2}}(800,0)\] \[P(400,200)\] \[{{B}_{1}}(0,1000)\] | \[Z=\frac{4}{1000}\times 800+\frac{8}{1000}\times 0=3.2\] \[Z=\frac{4}{1000}\times 400+\frac{6}{1000}\times 200=2.8\] \[Z=\frac{4}{1000}\times 0+\frac{6}{1000}\times 1000=6\] |
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