Find the maximum value of $z = 2x + 6y$ subject to the constraints $-x + y \leq 1$,$2x + y \leq 2$,$x \geq 0$,and $y \geq 0$.

The minimum value of $z = 3x + 5y$, subject to constraints $x \leq 80$, $y \geq 60$, $x + y \leq 200$ and $x, y \geq 0$ occurs at the point:

The production of item $A$ is $x$ and the production of item $B$ is $y$. If the corner points of the bounded feasible region are $(1,0), (2,0), (0,2)$ and $(0,1)$,then the maximum profit $z = 2000x + 5000y$ is $\ldots \ldots$

There are two factories located at place $P$ and place $Q$. From these locations,a certain commodity is to be delivered to each of the three depots situated at $A, B$ and $C$. The weekly requirements of the depots are $5, 5$ and $4$ units respectively,while the production capacities of the factories at $P$ and $Q$ are $8$ and $6$ units respectively. The cost of transportation per unit is given below:

From/To	$A$	$B$	$C$
$P$	$160$	$100$	$150$
$Q$	$100$	$120$	$100$

How many units should be transported from each factory to each depot in order that the transportation cost is minimum? What will be the minimum transportation cost?

$A$ manufacturing company produces two items,$A$ and $B$. Each item must be processed by two machines,$I$ and $II$. Machine $I$ can be operated for a maximum of $10$ hours $40$ minutes ($640$ minutes). It takes $20$ minutes for an item $A$ and $15$ minutes for an item $B$. Machine $II$ can be operated for a maximum of $8$ hours $20$ minutes ($500$ minutes). It takes $5$ minutes for an item $A$ and $8$ minutes for an item $B$. The profit per item of $A$ is ₹ $25$ and per item of $B$ is ₹ $18$. The formulation of an $L.P.P.$ to maximize the profit (where $x$ is the number of items $A$ and $y$ is the number of items $B$) is . . . . . . .

For the Linear Programming Problem ($L$.$P$.$P$.),maximize $z = 4x_1 + 2x_2$ subject to the constraints $3x_1 + 2x_2 \geq 9$,$x_1 - x_2 \leq 3$,$x_1 \geq 0$,$x_2 \geq 0$,the problem has: