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 14kg of phosphoric acid for her crop. If F1 costs Rs6/kg and F2 costs Rs5/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?

Answer 1

Let the farmer buy x kg of fertilizer F1 and y kg of fertilizer F2.

Therefore, x≥0 and y≥0

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

| Nitrogen(%)| Phosphoric Acid(%)| Cost(Rs/kg)
---|---|---|---
F1(x)| 10| 6| 6
F2(y)| 5| 10| 5
Requirement(kg)| 14| 14|

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

∴10% of x+5% of y≥14
$\frac{x}{10}+\frac{y}{20}$≥14
2x+y≥280

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

∴ 6%of x+10%of y≥14
$\frac{6x}{100}+\frac{10y}{100}$≥14
3x+56y≥700

The total cost of fertilizers, Z=6x+5y

The mathematical formulation of the given problem is
Minimize 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 follows.

It can be seen that the feasible region is unbounded.

The corner points are A($\frac{700}{3}$,0), B(100,80), and C(0,280)

The values of Z at these points are as follows:

Corner point| z=6x+5y|
---|---|---
A($\frac{700}{3}$,0)| 1400|
B(100,80)| 1000| $\rightarrow$Minimum
C(0,280)| 1400|

As the feasible region is unbounded, therefore, 1000 may or may not be the minimum value of Z.

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 80kg of fertilizer F2 should be used to minimize the cost. The minimum cost is 1000.