Answer

Verified

373.2k+ views

**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 (\[ \leqslant \]).

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Name 10 Living and Non living things class 9 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Write the 6 fundamental rights of India and explain in detail