Solution of the following $LP$ problem
Minimize $z = -3x + 2y$
subject to $0 \leq x \leq 4, 1 \leq y \leq 6, x + y \leq 5$ is $.....$

In the following figure, the shaded region represents the system of constraints:

The difference between the maximum and minimum values of the objective function $Z = 3x + 5y$,subject to the constraints $x + 3y \leqslant 60$,$x + y \geqslant 10$,$x - y \leqslant 0$,$x \geqslant 0$,$y \geqslant 0$ is

The graph with the correct feasible region of the $L.P.P.$ for the constraints $2x + y \leqslant 10$,$y \leqslant x$,$y \leqslant 2$,$x, y \geqslant 0$ is $\ldots$

$A$ 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.

$A$ company manufactures two types of sweaters: type $A$ and type $B.$ It costs $Rs. 360$ to make a type $A$ sweater and $Rs. 120$ to make a type $B$ sweater. The company can make at most $300$ sweaters and spend at most $Rs. 72000$ a day. The number of sweaters of type $B$ cannot exceed the number of sweaters of type $A$ by more than $100.$ The company makes a profit of $Rs. 200$ for each sweater of type $A$ and $Rs. 120$ for every sweater of type $B.$ Formulate this problem as a $LPP$ to maximize the profit to the company.