A diet for a sick person must contain at least 4000 units of vitamins , 50 units of minerals , & 1400 of calories . Two foods A & B , are available at a cost of Rs, 4 & Rs , 3 per unit respec. If one unit of A containa 200 units of vitamin , 1 unit of mineral & 40 calories & one unit of food B contains 100 units of vitamin , 2 units of minerals & 40 calories , find what combination of foods should be used to have the least cost ?

Here is the answer to your question.

The given data may be put in the following tabular form:

Food | Vitamins | Minerals | Calories | Cost per unit |

A | 200 | 1 | 40 | Rs 4 |

B | 100 | 2 | 40 | Rs 3 |

Minimum requirement | 4000 | 50 | 1400 |

Suppose the diet contains* x *units of food A and *y* units of food B. Since, one unit of food A costs Rs 4 and one unit of food B costs Rs 3. Therefore, total cost of* x* units of food A and* y* units of food B is Rs (4*x* + 3*y*).

Let *Z* denotes the total cost.

*Z* = 4*x* + 3*y*

Since each unit of food A contains 200 units of vitamins. Therefore, *x* units of food A contain 200 *x* units of vitamins. A unit of food B contains 100 units of vitamins. Therefore, *y* units of food B contain 100 *y* units of vitamins.

Thus, *x* units of food A and *y* units of food B contain 200 *x* + 100 *y* units of vitamins. But, the minimum requirement of vitamins is 4000 units.

∴ 200 *x* + 100 *y* ≥ 4000

Similarly, the total amount of minerals supplied by *x* units of food A and *y* units of food B is *x* + 2*y *and the minimum requirement is of 50 units.

∴ *x* + 2*y* ≥50

Finally, the total calories in *x* units of food A and *y* units of food B is 40 *x* + 40 *y* and the minimum calories required is 1400 units.

∴ 40 *x* + 40 *y* ≥ 1400

Clearly, *x* ≥ 0 and *y* ≥ 0

Since, we have to minimize the total cost Z. Thus, the mathematical form of the given LPP is as follows:

Minimize Z = 4*x* + 3*y*

Subject to

200*x* + 100*y* ≥ 4000

*x* + 2*y* ≥ 50

40*x* + 40*y* ≥ 1400

*x*, *y *≥ 0

The inequalities can now be represented on graph as:

The solution set of the linear constraints is shaded in the figure. The vertices of the shaded region are A (0, 40), P_{1} (5, 30), P_{2} (20, 15) and B (50, 0).

The value of the objective function at these points are given in the following table.

Point (x, y) | Value of objective function Z = 4x + 3y |

A (0, 40) P P B (50, 0) | Z = 4 × 0 + 3 × 40 = 0 + 120 = 120 Z = 4 × 5 + 3 × 30 = 20 + 90 = 110 Z = 4 × 20 + 3 × 15 = 80 + 45 = 125 Z = 4 × 50 + 3 × 0 = 200 + 0 = 200 |

Clearly, Z is minimum for *x* = 5 and *y* = 30 and the minimum value of Z is 110.

Hence, the diet cost is minimum when 5 units of food A and 30 units of food B are taken.

The minimum diet cost is Rs 110.