There are two types of fertilisers *'A' *and *'B' . *'A' consists of 12% nitrogen and 5% phosphoric acid whereas *'B'* consists of 4% nitrogen and 5% phosphoric acid. After testing the soil conditions, farmer finds that he needs at least 12 kg of nitrogen and 12 kg of phosphoric acid for his crops. If *'A'* costs ₹10 per kg and *'B' cost ₹*8 per kg, then graphically determine how much of each type of fertiliser should be used so that nutrient requiremnets are met at a minimum cost

#### Solution

The given information can tabulated as follows:

Fertilizer | Nitrogen | Phosphoric Acid | Cost/kg (in ₹) |

A | 12% | 5% | 10 |

B | 4% | 5% | 8 |

Let the requirement of fertilizer A by the farmer be *x* kg and that of B be *y* kg.

It is given that farmer requires atleast 12 kg of nitrogen and 12 kg of phosphoric acid for his crops.

The inequations thus formed based on the given information are as follows:

\[\frac{12}{100}x + \frac{4}{100}y \geq 12\]

\[ \Rightarrow 12x + 4y \geq 1200\]

\[ \Rightarrow 3x + y \geq 300 . . . . . \left( 1 \right)\]

Also,

\[\frac{5x}{100} + \frac{5y}{100} \geq 12\]

\[ \Rightarrow 5x + 5y \geq 1200\]

\[ \Rightarrow x + y \geq 240 . . . . . \left( 2 \right)\]

Total cost of the fertilizer *Z* = *₹ *(10*x* + 8*y*)

Therefore, the mathematical formulation of the given linear programming problem can be stated as:**Minimize** *Z* = 10*x* + 8*y *

Subject to the constraints

3*x* + *y* ≥ 300 .....(1)*x* + *y* ≥ 240 .....(2)*x* ≥ 0, *y* ≥ 0 .....(3)

The feasible region determined by constraints (1) to (3) is graphically represented as:

Here, it is seen that the feasible region is unbounded. The values of *Z* at the corner points of the feasible region are represented in tabular form as:

Corner Point |
Z = 10x + 8y |

A(0, 300) | Z = 10 × 0 + 8 × 300 = 2400 |

B(30, 210) | Z = 10 × 30 + 8 × 210 = 1980 |

C(240, 0) | Z = 10 × 240 + 8 × 0 = 2400 |

The open half plane determined by 10*x* + 8*y* < 1980 has no point in common with the feasible region. So, the minimum value of *Z* is 1980.

The minimum value of *Z* is 1980, which is obtained at *x* = 30 and *y* = 210.

Thus, the minimum requirement of fertilizer of type A will be **30 kg** and that of type B will be **210 kg**.

Also, the total minimum cost of the fertilisers is **₹ 1980**.