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

The shaded part of the given figure indicates the feasible region. Then the constraints are

$A$ 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.

The solution of the linear programming problem,maximize $Z = 3x_{1} + 5x_{2}$ subject to $3x_{1} + 2x_{2} \leq 18$,$x_{1} \leq 4$,$x_{2} \leq 6$,$x_{1} \geq 0$,$x_{2} \geq 0$ is:

The feasible region of an $LPP$ is shown in the figure. If $z = 3x + 9y$,then the minimum value of $z$ occurs at

The maximum value of $z=9x+11y$ subject to $3x+2y \leq 12$,$2x+3y \leq 12$,$x \geq 0$,$y \geq 0$ is . . . . . . .