The maximum value of $z = 5x + 3y$ subject to constraints $3x + 5y \leq 15, x \geq 0, y \geq 0$ is :

The feasible region for the constraints $x-2 \leqslant y$,$x \geqslant y-1$,$x \geqslant 2$,$y \leqslant 4$,$x, y \geqslant 0$ is represented by:

The objective function $Z = 4 x_1 + 5 x_2$,subject to $2 x_1 + x_2 \geq 7$,$2 x_1 + 3 x_2 \leq 15$,$x_2 \leq 3$,$x_1, x_2 \geq 0$ has minimum value at the point

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

$A$ 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 . . . . . . .

$A$ manufacturer has three machines $I, II$ and $III$ installed in his factory. Machines $I$ and $II$ are capable of being operated for at most $12 \, hours$ whereas machine $III$ must be operated for at least $5 \, hours$ a day. She produces only two items $M$ and $N$ each requiring the use of all the three machines. The number of hours required for producing $1$ unit of each of $M$ and $N$ on the three machines are given in the following table:

Items	Machine $I$	Machine $II$	Machine $III$
$M$	$1$	$2$	$1$
$N$	$2$	$1$	$1.25$

She makes a profit of $Rs. \, 600$ and $Rs. \, 400$ on items $M$ and $N$ respectively. How many of each item should she produce so as to maximise her profit assuming that she can sell all the items that she produced? What will be the maximum profit?