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

The maximum value of $2x + y$ subject to $3x + 5y \leq 26$ and $5x + 3y \leq 30, x \geq 0, y \geq 0$ is

The cost function $Z$ is given by $Z = 4x + 6y$. It is to be minimized. The feasible region for this function $Z$ is the shaded region represented in the following figure. Then the minimum value of $Z$ is and occurs at the point:

The graphical solution set of the system of in-equations $x+y \leq 70, x+2y \leq 100, 2x+y \leq 120, x \geq 0, y \geq 0$ is given by:

The maximum value of $z=50x+15y$ subject to the constraints $x+y \leq 60$,$5x+y \leq 100$,$x \geq 0$,$y \geq 0$ is at the point:

$A$ cottage industry manufactures pedestal lamps and wooden shades,each requiring the use of a grinding/cutting machine and a sprayer. It takes $2 \text{ hours}$ on the grinding/cutting machine and $3 \text{ hours}$ on the sprayer to manufacture a pedestal lamp. It takes $1 \text{ hour}$ on the grinding/cutting machine and $2 \text{ hours}$ on the sprayer to manufacture a shade. On any day,the sprayer is available for at most $20 \text{ hours}$ and the grinding/cutting machine for at most $12 \text{ hours}$. The profit from the sale of a lamp is $Rs. 5$ and that from a shade is $Rs. 3$. Assuming that the manufacturer can sell all the lamps and shades that he produces,how should he schedule his daily production in order to maximise his profit?