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Home  >>  CBSE XII  >>  Math  >>  Linear Programming
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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} $

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1 Answer

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Toolbox:
  • 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
 

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