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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,how many of circuits of type A and type B,should be produced by the manufacturer so as to maximize his profit?Determine the maximum profit.

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  • 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:
The given problem can be written as follows :
For type of A circuit $\Rightarrow$ 20 resistors,10 trasistors,10 capacitors,Maximum stock is 200.
For type of B circuit $\Rightarrow$ 10 resistors,20 transistors,30 capacitors,Maximum stock is 120.
Profit on type A is Rs.50 and profit on type B is Rs.60.
The objective function which has to be maximized is $Z=50x+60y$
Mathematically the above problem can be formulated as follows :
Let $x$ be the products obtained from type A and $y$ be the products obtained from type B.
The maximum stock of resistors the factory has 200
$\therefore 20x+10y\leq 200\Rightarrow 2x+y\leq20$
The maximum stock of transistors the factory has 120
$\therefore 10x+20y\leq 120\Rightarrow x+3y\leq 15$
Step 2:
Now let us draw the graph for the lines,$AB :2x+y=20;CD :x+2y=12$ and $EF :x+3y=15$
Consider the line AB :$2x+y=20$
Put $x=0,y=0$.Clearly $0\leq 20$ which is true.
Hence $2x+y\leq 20$ lies below the line AB.
Consider the line CD :$x+2y=12$
Put $x=0,y=0$.Clearly $0\leq 12$ which is true.
Hence $x+2y\leq 12$ lies below the line CD.
Consider the line EF :$x+3y=15$
Put $x=0,y=0$.Clearly $0\leq 15$ which is true.
Hence $x+3y\leq 15$ lies below the line EF.
Clearly the feasible region includes the origin also.
Step 3:
The coordinates of the point $Q$ can be obtained by solving the equation $2x+y=20$ and $x+2y=12$
$2x+y=20$------(1)
$x+2y=12$------(2)
Multiply equ(2) by 2 we get,
$2x+\;\;y=20$
$2x+4y=24$
_________________
$-3y=-4$
$y=1.3$
Hence $x=9.3$
Hence $Q(9.3,1.3)$
Step 4:
The coordinates of the point $P$ can be obtained by solving the equation $x+2y=12$ and $x+3y=15$
$x+2y=12$------(3)
$x+3y=15$------(4)
_________________
$-y=-3$
$y=3$
$x=6$
Hence the coordinates of $P$ is $(6,3)$
Step 5:
The feasible region OEPQB is the shaded portion shown the graph.
The corner points are $O(0,0),E(0,5),P(6,3),Q(9.3,1.3),B(10,0)$
Step 6:
Let us find the values of the objective function $Z=50x+60y$
At the points $(x,y)$ the value of the objective function subjected to $Z=50x+60y$
At $O(0,0)$ the value of the objective function $Z=50\times 0+60\times 0=0$
At $E(0,5)$ the value of the objective function $Z=50\times 0+60\times 5=300$
At $P(6,3)$ the value of the objective function $Z=50\times 6+60\times 3=480$
At $Q(9.3,1.3)$ the value of the objective function $Z=50\times 9.3+60\times1.3=443$
At $B(10,0)$ the value of the objective function $Z=50\times 10+60\times 0=500$
The maximum value should be 480 at $P(6,3)$
Hence the maximum profit is Rs480
Type A should be produce 6 circuits and type B should produce 3 circuits.
answered Aug 30, 2013 by sreemathi.v
 

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