$A$ manufacturer produces two models of bikes: Model $X$ and Model $Y$. Model $X$ takes $6$ man-hours to make per unit,while Model $Y$ takes $10$ man-hours per unit. There is a total of $450$ man-hours available per week. Handling and marketing costs are $Rs. 2000$ and $Rs. 1000$ per unit for Models $X$ and $Y$ respectively. The total funds available for these purposes are $Rs. 80,000$ per week. Profits per unit for Models $X$ and $Y$ are $Rs. 1000$ and $Rs. 500$ respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

(A) Let the manufacturer produce $x$ number of Model $X$ bikes and $y$ number of Model $Y$ bikes.
Model $X$ takes $6$ man-hours and Model $Y$ takes $10$ man-hours per unit. Total man-hours available per week is $450$.
$\therefore 6x + 10y \leq 450 \Rightarrow 3x + 5y \leq 225$
Handling and marketing costs are $Rs. 2000$ and $Rs. 1000$ per unit respectively,with total funds of $Rs. 80,000$ per week.
$\therefore 2000x + 1000y \leq 80000 \Rightarrow 2x + y \leq 80$
Also,$x \geq 0, y \geq 0$.
We need to maximize the profit function $Z = 1000x + 500y$ subject to the constraints:
$3x + 5y \leq 225$
$2x + y \leq 80$
$x \geq 0, y \geq 0$
The feasible region is determined by the corner points $(0,0), (40,0), (25,30),$ and $(0,45)$.

Corner Points	Value of $Z = 1000x + 500y$
$(0,0)$	$0$
$(40,0)$	$1000(40) + 500(0) = 40000$
$(25,30)$	$1000(25) + 500(30) = 25000 + 15000 = 40000$
$(0,45)$	$1000(0) + 500(45) = 22500$

The maximum profit is $Rs. 40,000$. This occurs at any point on the line segment joining $(40,0)$ and $(25,30)$. Thus,the manufacturer can produce $40$ units of Model $X$ and $0$ units of Model $Y$,or $25$ units of Model $X$ and $30$ units of Model $Y$ to achieve the maximum profit.