Step 1:

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.

$Z=3x+3.5y$

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 $Z=3x+3.5y$

$x+2y\geq 240,3x+1.5y\geq 270,1.5x+2y\leq 310,x,y\geq 0$

Step 2:

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.

(iii)$1.5x+2y\leq 310$

Put $x=0,y=0$

$\Rightarrow 0\leq 310$ 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.

Step 3:

The shaded area $P$ and $R$ represents the feasible region.

Now $P$ is the point of intersection of the lines,

$3x+1.5y=270$----------(1)

$1.5x+2y=310$----------(2)

Let us solve the above equation to find the co-ordinate of P

Multiply equ(2) by 2

$3x+1.5y=270$

$-3x+4y=620$

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$-2.5y=-350$

$y=140$

Therefore $x=20$

Hence the coordinate of P are (20,140)

Step 4:

$Q$ is the point of intersection of the line

$x+2y=240$--------(3)

$3x+1.5y=270$-------(4)

Multiply equ(3) by 3

$3x+6y=720$

$-3x+1.5y=270$

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$4.5y=450$

$y=100$

$\therefore x=40$

Hence the coordinates of $Q$ is $(40,100)$

Step 5:

$R$ is the point of intersection of lines

$\;\;\;\;x+2y=240$

$-1.5x+2y=310$

___________________

$-0.5x=-70$

$x=140$

$\therefore y=50$

Hence the coordinates of $R$ is $(140,50)$

Hence the corner points of the feasible region is $P(20,140),Q(40,100),R(140,50)$

Step 6:

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.