Minimize the objective function $Z = 3x + 2y$ subject to the constraints: $x + y \geq 8$,$x + y \leq 5$,$x \geq 0$,$y \geq 0$.

Corner points of the feasible region for an $\operatorname{LPP}$ are $(0,2), (3,0), (6,0), (6,8)$ and $(0,5)$. Let $F = 4x + 6y$ be the objective function. The minimum value of $F$ occurs at $....$

Maximum value of $Z=100 x+70 y$ subject to $2 x \geq 4, y \leq 3, x+y \leq 8, x, y \geq 0$ is

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 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

$A$ manufacturer makes two types of toys $A$ and $B$. Three machines are needed for this purpose and the time (in $minutes$) required for each toy on the machines is given below:

Types of Toys	Machine-$I$	Machine-$II$	Machine-$III$
$A$	$12$	$18$	$6$
$B$	$6$	$0$	$9$

Each machine is available for a maximum of $6 \, hours$ $(360 \, minutes)$ per day. If the profit on each toy of type $A$ is $Rs. \, 7.50$ and that on each toy of type $B$ is $Rs. \, 5$,find the number of toys of each type that should be manufactured in a day to get maximum profit.