Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Linear Programming
0 votes

A manufacture of electronic circuits has a stock of 200 resistors,120 transistors and 150 capacitors and is required to produce two types of circuits A and B .Type A requires 20 resistors,10 transistors and 10 capacitors .Type B requires 10 resistors,20 transistors and 30 capacitors.If the profit on type A circuit is Rs 50 and that an type B circuit is Rs 60,formulate this problem as a LPP so that the manufacturer can maximize his profit.

$\begin{array}{1 1} (A)\;Z=50x+60y \\(B)\;Z=200x+60y \\ (C)\;Z=20x+30y \\(D)\;Z=20x+60y \end{array} $

Can you answer this question?

1 Answer

0 votes
  • Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
  • If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
Step 1:
Let $X$ be the resistors,$Y$ be the transistors and $Z$ be the capacitors produced by the electronic circuits of two types $A$ and $B$
The data in the given problem can be written as follows :
For Type A: $x=20,y=10,z=10$.The profit on type $A$ circuit is Rs.50
For Type B: $x=10,y=20,z=30$.The profit on type $B$ circuit is Rs.60
The maximum requirement is $x=200,y=120,z=150$
Step 2:
These units are bought to fulfill the maximum requirement of $x,y,z$ and to maximize the profit.
The mathematic formulation of the above problem is as follows :
Maximize $Z=50x+60y$ subject to $2x+y\leq 20,x+2y\leq 12,x+3y\leq 15,x\geq 0,y\geq 0$
answered Aug 21, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App