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,

$2x+\;\;y=20$

$2x+4y=24$

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$-3y=-4$

$y=1.3$

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)

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$-y=-3$

$y=3$

$x=6$

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.