A hospital dietician wishes to find the cheapest combination of two foods, *A* and *B*, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of *A*contains 0.12 milligram of thiamin and 100 calories, while each unit of *B* contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?

#### Solution

Let the dietician wishes to mix *x *units of food *A* and *y* units of food *B.*

Therefore,

Thiamine(mg) | Calories | |

Food A |
0.12 | 100 |

Food B |
0.1 | 150 |

Minimum requirement | 0.5 | 600 |

According to the question,

The constraints are

\[0 . 12x + 0 . 1y \geq 0 . 5\]

\[100x + 150y \geq 600\]

It is given that each food costs 10 paise per units

Therefore,

Total cost, Z = \[10x + 10y\]

Thus, the mathematical formulation of the given linear programmimg problem is

\[0 . 12x + 0 . 1y \geq 0 . 5\]

\[100x + 150y \geq 600\]

Region represented by 0.12*x** *+0.1*y* ≥ 0.5:

The line 0.12*x** *+ 0.6*y* = 20 meets the coordinate axes at\[A_1 \left( \frac{25}{6}, 0 \right)\] and \[B_1 \left( 0, 5 \right)\]respectively. By joining these points we obtain the line 0.12*x** *+ 0.6*y* = 20.Clearly (0,0) does not satisfies the 0.12*x** *+ 0.6*y* = 20. So,the region which does not contains the origin represents the solution set of the inequation 0.12*x*+0.1*y* ≥ 0.5.

Region represented by 100*x** *+ 150*y* ≥ 600:

The line 100*x** *+ 150*y* = 600 meets the coordinate axes at\[C_1 \left( 6, 0 \right)\] and \[D_1 \left( 0, 4 \right)\] respectively. By joining these points we obtain the line 100*x** *+ 150*y* = 600. Clearly (0,0) does not satisfies the inequation 100*x** *+ 150*y* ≥ 600. So,the region which does not contains the origin represents the solution set of the inequation 100*x** *+ 150*y* ≥ 600.

Region represented by *x* ≥ 0 and* y* ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations *x* ≥ 0, and *y *≥ 0.

The feasible region determined by the system of constraints 0.12*x** *+0.1*y* ≥ 0.5, 100*x** *+ 150*y* ≥ 600, *x* ≥ 0, and *y* ≥ 0 are as follows.

The corner points are *B*_{1}(0, 5),

Corner point | Z= 10x +10y |

B_{1} |
50 |

E_{1} |
46.2 |

C_{1} |
60 |

*A*and 2.75 units of food

*B.*