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 maximum value of $Z = 4x + 5y$, subject to the constraints $3x + y \leq 15$, $3x + 4y \leq 24$, $x \geq 0$, $y \geq 0$ is

The maximum value of the objective function $z=4x+5y$ subject to the constraints $2x+3y \leq 12$,$2x+y \leq 8$ and $x \geq 0, y \geq 0$ 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.

The minimum value of $Z = 3x + y$, subject to the constraints $2x + 3y \leq 6$, $x + y \geq 1$, $x \geq 0$, and $y \geq 0$ is:

$A$ diet is to contain at least $80$ units of vitamin $A$ and $100$ units of minerals. Two foods $F_{1}$ and $F_{2}$ are available. Food $F_{1}$ costs $Rs. 4$ per unit and food $F_{2}$ costs $Rs. 6$ per unit. One unit of food $F_{1}$ contains $3$ units of vitamin $A$ and $4$ units of minerals. One unit of food $F_{2}$ contains $6$ units of vitamin $A$ and $3$ units of minerals. Formulate this as a linear programming problem. Find the minimum cost for a diet that consists of a mixture of these two foods and also meets the minimal nutritional requirements.