Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.1k views
in Linear Programming by (27.7k points)
closed by

A gardener has a supply of fertilizers of the type 1 which consist of 10% nitrogen and 6% phosphoric acid, and of the type II which consist of 5% nitrogen and 10% phosphoric acid. After testing the soil condition, he finds that he needs at least 14kg of nitrogen and 14 kg of phosphoric acid for his crop. If the type - I fertilizer costs 60 paise per kg and the type - II fertilizer costs 40 paise per kg, determine how many kilograms of each type of fertilizer should be used so that the nutrient requirement are met at a minimum cost. What is the minimum cost?

1 Answer

+1 vote
by (25.7k points)
selected by
 
Best answer

Let x and y be number of kilograms of fertilizer I and II

∴According to the question,

0.10x + 0.05y ≥ 14, 0.06x + 0.10y ≥ 14, x ≥ 0, y ≥ 0

Minimize Z = 0.60x + 0.40y

The feasible region determined by 0.10x + 0.05y ≥ 14, 0.06x + 0.10y ≥ 14, x ≥ 0, y ≥ 0 is given by

The feasible region is unbounded. The corner points of feasible region are A(0,280) , B(100,80) , C(700/3,0).

The value of Z at corner points are

Corner Point Z = 0.60x + 0.40y
A(0. 280) 112
B(100, 80) 92 Maximum
C(700/3, 0) 140

The minimum value of Z is 92 at point (100,80). 

Hence, the gardener should by 100 kilograms o fertilizer I and 80 kg of fertilizer II to minimize the cost which is Rs.92.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...