Advertisement Remove all ads

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. - Mathematics

Advertisement Remove all ads
Advertisement Remove all ads
Advertisement Remove all ads
Sum

There are two types of fertilizers Fand F2. Fconsists of 10% nitrogen and 6% phosphoric acid and ​Fconsists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds the she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If Fcosts ₹6/kg and Fcosts ₹5/kg, determine how much of each type of fertilizer should be used so that the nutrient requirements are met at minimum cost. What is the minimum cost? 

Advertisement Remove all ads

Solution

Suppose x kg of fertilizer Fand and y kg of fertilizer F2 is used to meet the nutrient requirements.

Fconsists of 10% nitrogen and Fconsists of 5% nitrogen. But, the farmer needs atleast 14 kg of nitorgen for the crops.

∴ 10% of x kg + 5% of y kg ≥ 14 kg

\[\Rightarrow \frac{x}{10} + \frac{y}{20} \geq 14\]
\[ \Rightarrow 2x + y \geq 280\] 

Similarly, Fconsists of 6% phosphoric acid and Fconsists of 10% phosphoric acid. But, the farmer needs atleast 14 kg of phosphoric acid for the crops.

∴ 6% of x kg + 10% of y kg ≥ 14 kg

\[\Rightarrow \frac{6x}{100} + \frac{10y}{100} \geq 14\]
\[ \Rightarrow 3x + 5y \geq 700\]
The cost of fertilizer Fis ₹6/kg and fertilizer Fis ₹5/kg, therefore, total cost of x kg of fertilizer Fand and y kg of fertilizer Fis ₹(6x + 5y).

Thus, the given linear programming problem is

Minimise Z = 6+ 5y

subject to the constraints

2x + ≥ 280

3x + 5≥ 700

x, y ≥ 0

The feasible region determined by the given constraints can be diagrammatically represented as,

The coordinates of the corner points of the feasible region are \[A\left( \frac{700}{3}, 0 \right), B ( 100,80 )   \text{ and C }( 0,280 )\] .

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

Corner Point Z = 6+ 5y
 
\[\left( \frac{700}{3}, 0 \right)\]
 
\[6 \times \frac{700}{3} + 5 \times 0 = 1400\]
(100, 80) 6 × 100 + 5 × 80 = 1000 → Minimum
(0, 280) 6 × 0 + 5 × 280 = 1400
The smallest value of Z is 1000 which is obtained at x = 100, y = 80.

It can be seen that the open half-plane represented by 6x + 5y < 1000 has no common points with the feasible region.

So, the minimum value of Z is 1000.

Hence, 100 kg of fertilizer Fand and 80 kg of fertilizer F2 should be used so that the nutrient requirements are met at minimum cost. The minimum cost is ₹1,000.
Concept: Graphical Method of Solving Linear Programming Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.4 | Q 47 | Page 56

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×