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 Linear Programming Problem ($L$.$P$.$P$.) to minimize $z = 30x + 20y$ subject to the constraints $x + y \leqslant 8$,$x + 2y \geqslant 4$,$6x + 4y \geqslant 12$,$x \geqslant 0$,and $y \geqslant 0$ has:

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

The solution set of the constraints $x+2y \leq 2000$,$x+y \leq 1500$,$y \leq 600$ and $x \geq 0$ does not include the point

$A$ farmer mixes two brands $P$ and $Q$ of cattle feed. Brand $P$,costing $Rs. 250$ per bag,contains $3$ units of nutritional element $A$,$2.5$ units of element $B$ and $2$ units of element $C$. Brand $Q$ costing $Rs. 200$ per bag contains $1.5$ units of nutritional element $A$,$11.25$ units of element $B$,and $3$ units of element $C$. The minimum requirements of nutrients $A$,$B$ and $C$ are $18$ units,$45$ units and $24$ units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?

The maximum value of the objective function $z=4x+6y$ subject to the constraints $3x+2y \leq 12$,$x+y \geq 4$,$x, y \geq 0$ is