Mathematics Part-II

Book: Mathematics Part-II

Chapter: 12. Linear Programming

Subject: Maths - Class 12th

Q. No. 10 of Exercise 12.2

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There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

let the farmer buy x kf of fertilizer f1 and y kg of fertilizer f2

x and y ≥ 0

The tabular representation of the data is as follows:

F1 consists of 10% nitrogen and F2 consists of 5% nitrogen. But the farmer requires at least 14 kg of nitrogen.

10% × x + 5% × y ≥ 14

2x+y ≥ 280

food F1 consists of 6% phosphoric acid and food F2 consists of 10% phosphoric acid. But the farmer requires at least 14kg of phosphoric acid

6% × x + 10% of y ≥ 14

3x + 56y ≥ 700

Total cost of fertilizers, Z= 6x+ 5y

The mathematical formulation of the given data is

Minimize, Z = 6x +5y

Subject to constraints

2x+y ≥ 280

3x + 56y ≥ 700

x and y ≥ 0

The feasible region determined by the system of constraint is:

It can be seen from the graph that the feasible region is unbounded

The corner points are

The value of Z at these points are:

As the feasible region is unbounded, 1000 may or may not be the minimum value

For this we will draw graph of inequality

6x+ 5y < 1000

It can be seen there is no common point between feasible region and 6x+ 5y < 1000

100 KG of fertilizer F1 and 80 kg of fertilizerF2 should be used to minimize the cost and the minimum cost is Rs 1000.


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