Let $x$ bags of brand P and y bags of brand Q are needed to minimize nitrogen contents.
$\therefore$ the objective function is to minimize.
We need at least 240kg of phosphoric acid
$\therefore x+2y\geq 240$
We require at least of 270kg of potash
$\therefore 3x+1.5y\geq 270$
We require at most of 310kg of chlorine
$\therefore 1.5x+2y\leq 310$
Hence the objective function is $Z=3x+3.5y$
$x+2y\geq 240,3x+1.5y\geq 270,1.5x+2y\leq 310,x,y\geq 0$
Now let us draw the graph for the above equations,
(i) the line $x+2y=240$ passes through the point $A(240,0)$ and $(0,120)$
Put $x=0,y=0$ in $x+2y\geq 240$,we get $0\geq 240$ which is not true.
This implies the region $x+2y\geq 240$ lies on and above AB
(ii) Consider the line $3x+1.5y \geq 270$
Clearly this passes through $C(90,0),D(0,180)$
Put $x=0,y=0$,we get
$0\geq 270$ which is not true.
This implies the region $3x+1.5y\geq 270$ lies on CD and above it.
$\Rightarrow 0\leq 310$ which is not true.
$\Rightarrow$ region $1.5x+2y \leq 310$ lies on and below EF
(iv) $x\geq 0$ lies on and to the right of $y$-axis
(v) $y\geq 0$ lies on and above the $x$-axis.
The shaded area $P$ and $R$ represents the feasible region.
Now $P$ is the point of intersection of the lines,
Let us solve the above equation to find the co-ordinate of P
Multiply equ(2) by 2
Hence the coordinate of P are (20,140)
$Q$ is the point of intersection of the line
Multiply equ(3) by 3
Hence the coordinates of $Q$ is $(40,100)$
$R$ is the point of intersection of lines
Hence the coordinates of $R$ is $(140,50)$
Hence the corner points of the feasible region is $P(20,140),Q(40,100),R(140,50)$
Now let us calculate the minimum value of $Z$ as follows :
At the points (x,y) the value of the objective function subject to $Z=3x+3.5y$
At point $P(20,140)$ the value of the objective function $Z=3\times 20+3.5\times 140=550$
At point $Q(40,100)$ the value of the objective function $Z=3\times 40+3.5\times 100=470$
At point $R(140,50)$ the value of the objective function $Z=3\times 140+3.5\times 50=595$
Hence $Z$ is maximum at $R(140,50)$
To maximize the amount of nitrogen,140 bags of brand P and 50 bags of brand Q are required.
Maximum amount of nitrogen required=595kgs.