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 point,at which the maximum value of $10x + 6y$ subject to the constraints $x + y \leq 12$,$2x + y \leq 20$,$x \geq 0$,$y \geq 0$ occurs,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

$A$ 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 shaded area in the figure given below is a solution set of a system of inequations. The minimum value of the objective function $Z = 3x + 5y$,subject to the linear constraints given by this system of inequations,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?