If $Z=10x+25y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x+y \leq 5, x \geq 0, y \geq 0$,then $Z$ is maximum at the point:

$A$ 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 objective function $z=x_1+x_2$,subject to $x_1+x_2 \leq 10, -2x_1+3x_2 \leq 15, x_1 \leq 6, x_1, x_2 \geq 0$,has a maximum value at:

The correct constraints for the given feasible region are:

The difference between the maximum value and minimum value of the objective function $z = 3x + 5y$ of a linear programming problem subject to constraints $5x + 10y \leq 50$, $x + y \geq 1$, $y \leq 4$ and $x \geq 0, y \geq 0$ is $3\lambda$. Then the value of $\lambda$ is

$A$ man rides his motorcycle at the speed of $50 \, km/h$. He has to spend $Rs. \, 2$ per $km$ on petrol. If he rides it at a faster speed of $80 \, km/h$,the petrol cost increases to $Rs. \, 3$ per $km$. He has at most $Rs. \, 120$ to spend on petrol and one hour's time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.