Answer to Question #284385 in Management for Abreham

Question

There are two types of fertilizers 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 Birr 6/kg and F2 costs Birr 5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost? (solve through simplex method)

Answer 1

Answer

There are two types of fertilizers F1 and F2

​

. F1consists 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 14kg of nitrogen and 14kg of phosphoric acid for her crop. If F 1cost Rs.6/kg and F2 costs Rs.5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?

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Solution

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Let the farmer buy xkg of fertilizer F 

1

​

 and y kg of fertilizer F 

2

​

 . Therefore, x≥0 and y≥0

The given information can be compiled in a table as follows.

Nitrogen (%) Phosphoric Acid (%) Cost (rs/kg)

F 1(x) 10 6 6

F2(y) 5 10 5

Requirement 14 14

F1consists of 10% nitrogen and F2 consists of 5% nitrogen. However, the farmer requires at least 14kg of nitrogen.

∴ 10% of x + 5% of y≥14

x/10+ Y/20≥14

2x+y≥280

F1nconsists of 6% phosphoric acid and F2 consists of 10% phosphoric acid. However, the farmer requires at least 14kg of phosphoric acid

∴ 6% of x + 10% of y≥14

100

6x​/100+10y/100≥14

3x+56y≥700

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

The mathematical formulation of the given problem is

Minimise Z=6x+5y.....(1)

subject to the constraints

2x+y≥280......(2)

3x+5y≥700.....(3)

x,y≥0......(4)

The feasible region determined by the system of constraints is as shown

It can be seen that the feasible region is unbounded.

The corner points are A(3700),B(100,80) and C(0,280)

As the feasible region is unbounded, therefore, 1000 may or may ot be the minimum value ofZ

For this, we draw a graph of the inequality 6x+5y<1000 and check whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with

6x+5y<1000

Therefore, 100kg of fertilizer F1 and 80 kg of fertilizer F2 should be used to minimize the cost. The minimum cost is Rs.1000.