Step 1:
Model of bikes$\quad$X$\;\;\;\;\quad$Y
Hours/unit$\;\;\;\quad\;\;\;$6$\;\;\;\;\quad$10
Cost/unit$\quad\;\;$Rs2000$\quad$Rs1000
Profit/unit$\quad\;\;$Rs1000$\quad$Rs500
It is given model $x$ takes 6-man-hours to make per unit and model $Y$ takes 10 man hours per unit.
No of hours available pr week is 450
$6x+10y\leq 450$
Handling and marketing costs are RS.2000 for model $X$ and Rs 1000 for model $Y$
The total funds available is Rs80,000 per week.
Therefore $2000x+1000y\leq 80,000\Rightarrow 2x+y\leq 80$
Profits per unit of models $X$ and $Y$ are Rs 1000 and Rs.500.
Hence the objective function to be maximized is $Z=1000x+500y$
Step 2:
Now let us draw the graph for the line AB :$6x+10y=450$ and CD :$2x+y=80$
Consider the line AB :$6x+10y=450$
Put $x=0,y=0\Rightarrow 0\leq 450$
Clearly the region $6x+10y=450$ lies below the line $6x+10y=450$
Consider the line CD :$2x+y=80$
Put $x=0,y=0\Rightarrow 0\leq 80$
Clearly the region $2x+y\leq 80$ lies below the line $2x+y=80$
The point of intersection of the two lines AB and CD is (25,30)
Clearly the origin (0,0) is also included in the feasible region.
The shaded portion is the feasible region OAPD
The corner points of the feasible region are $O(0,0),A(0,45),P(25,30),D(40,0)$
Step 3:
Let us find the value of the objective function $Z=1000x+500y$
At the points $(x,y)$ the value of the objective function subjected to $Z=1000x+500y$
At $O(0,0)$ the value of the objective function $Z=0$
At $A(0,45)$ the value of the objective function $Z=1000\times 0+500\times 45=22500$
At $P(25,30)$ the value of the objective function $Z=1000\times 25+500\times 30=40000$
At $D(40,0)$ the value of the objective function $Z=1000\times 40+500\times 0=40000$
Clearly the maximum value is 40000 at P(25,30)
The maximum profit is 40,000
25 bikes of model X and 30 bikes of model Y has to be manufactured.