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**Hint:**We solve this by linear programming that is by drawing a graph of inequalities. Let ‘\[x\]’ grams of wheat are mixed and ‘\[y\]’ grams of rice are mixed. Using this we will get two inequities. We find the values by equating ‘\[x\]’ to zero, we get ‘\[y\]’ and vice versa. Plotting a graph we can find the minimum cost.

**Complete step-by-step solution:**

Let \[x\] gram of wheat and y grams of rice are mixed in a daily diet.

Given, every gram of wheat provides 0.1g of proteins and every gram of rice gives 0.05g of proteins. Therefore, \[x\] grams of wheat and \[y\] grams of rice will provide \[0.1x + 0.05y\] grams of proteins. But the minimum requirement of proteins is 50g.

That is, \[0.1x + 0.05y \geqslant 50\].

To get the simplified form multiply and divide by 100 on each term,

\[ \Rightarrow 0.1x \times \dfrac{{100}}{{100}} + 0.05y \times \dfrac{{100}}{{100}} \geqslant 50 \times \dfrac{{100}}{{100}}\]

\[ \Rightarrow \dfrac{{10x}}{{100}} + \dfrac{{5y}}{{100}} \geqslant 50\]

Cancelling, we get,

\[ \Rightarrow \dfrac{x}{{10}} + \dfrac{y}{{20}} \geqslant 50\] ----- (1)

Similarly \[x\]grams of wheat and \[y\]grams of rice will provide\[0.25x + 0.5y\]grams of carbohydrates and the minimum daily requirement is 200 grams.

That is, \[0.25x + 0.5y \geqslant 200\].

To get the simplified form multiply and divide by 100 on each term,

\[ \Rightarrow 0.25x \times \dfrac{{100}}{{100}} + 0.5y \times \dfrac{{100}}{{100}} \geqslant 200 \times \dfrac{{100}}{{100}}\]

\[ \Rightarrow \dfrac{{25x}}{{100}} + \dfrac{{50y}}{{100}} \geqslant 200\]

Cancelling, we get:

\[ \Rightarrow \dfrac{x}{4} + \dfrac{y}{2} \geqslant 200\] ----- (2)

Also quantities of wheat and rice will never be negative.

\[x \geqslant 0\]and \[y \geqslant 0\]

It is given that wheat costs Rs.4 per kg and rice Rs.6 per kg. So we have,

\[\dfrac{{4x}}{{1000}} + \dfrac{{6y}}{{1000}} = z\]minimum.

Subject to the constraints

\[\dfrac{x}{{10}} + \dfrac{y}{{20}} \geqslant 50\], \[\dfrac{x}{4} + \dfrac{y}{2} \geqslant 200\], \[x \geqslant 0\], \[y \geqslant 0\].

Now to find the points ,

Put \[x = 0\] in equation (1) we have,

\[ \Rightarrow \dfrac{0}{{10}} + \dfrac{y}{{20}} = 50\]

\[ \Rightarrow \dfrac{y}{{20}} = 50\]

\[ \Rightarrow y = 1000\]. Thus we have (0, 1000).

Similarly put \[y = 0\] in equation (1)

\[ \Rightarrow \dfrac{x}{{10}} = 50\]

\[ \Rightarrow x = 500\]. Thus we have (500, 0).

Similarly put \[x = 0\] in equation (2), we have:

\[ \Rightarrow \dfrac{y}{2} = 200\]

\[ \Rightarrow y = 400\]. Thus we have (0, 400)

Similarly put \[y = 0\] in equation (2)

\[ \Rightarrow \dfrac{x}{4} = 200\]

\[ \Rightarrow x = 800\]. Thus we have (800, 0).

Thus we have the points, \[(0,1000)\], \[(500,0)\], \[(0,400)\]and \[(800,0)\]. Plotting a graph for this.

Scale: X-axis = 1 unit =100 units, Y-axis = 1 unit =100 units.

The values of the objective function at these points are given in the following table:

Corner points | Value of \[z = \dfrac{{4x}}{{1000}} + \dfrac{{6y}}{{1000}}\] |

\[(800,0)\] | \[z = \dfrac{{4 \times 800}}{{1000}} + \dfrac{{6 \times 0}}{{1000}} = \dfrac{{32}}{{10}} = 3.2\] |

\[(400,200)\] | \[z = \dfrac{{4 \times 400}}{{1000}} + \dfrac{{6 \times 200}}{{1000}} = 1.6 + 1.2 = 2.8\] |

\[(0,1000)\] | \[z = \dfrac{{4 \times 0}}{{1000}} + \dfrac{{6 \times 1000}}{{1000}} = 6\] |

Clearly, Z is minimum for \[x = 400\]grams of wheat and \[y = 200\] grams of rice. The minimum cost is Rs.\[2.8\]

**Note:**All we did in above is converting the given word problem into inequality problems. As we know that we need a point to join a line, hence we find the inequality value at ‘x’ is zero and then at ‘y’ is zero. Also if they mention minimum we take greater than or equal to(\[ \geqslant \]). If they mention maximum we take less than or equal to (\[ \leqslant \]).

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