Step 1:

Let $x$ quintals of grains be transported from godown A to ration shop D and y quintals of grains be transported from godown A to ration shop E.

Therefore the grains transported from godown A to shop F will be $100-(x+y)$ quintals.

Because capacity of godown A is 100 quintals.

Therefore we have $x\geq 0,y\geq 0$ and $100-(x+y)\geq 0$ (i.e)$x+y\leq 100$

As requirements of grains at shop D is 60 quintals,$(60-x)$ quintals of grain will be transported from godown B to D

Step 2:

Similarly $(50-y)$ quintals and $40-[100-(x+y)]$ should be transported from godown B to shops E and F respectively.

Hence $(60-x)\geq 0$

$\Rightarrow x\leq 60$

Similarly $(50-y)\geq 0$

$\Rightarrow y \leq 50$

$40-[100-(x+y)]\geq 0$

$-60+x+y\geq 0$

$x+y\geq 60$

Step 3:

The cost of transportation is

$Z=6x+3y+250(100-x-y)+4(60-x)+2(50-y)+3(40-100-x+y]$

Simplifying this we get,

$\Rightarrow 6x+3y+250-2.5x-2.5y+240-4x+100-2y+120-300+3x+3y$

(i.e) $Z=2.5x+1.5y+410$

Subject to constraints :

$x\leq 60,y\leq 50,x+y \geq 60,x+y \leq 100,y\geq 0$

Step 4:

Let us plot the graph of equations

$x=60,y=50,x+y=60$ and $x+y=100$

The feasible region ABCD shows the shaded area in the figure satisfies the inequalities $x\leq 60,y\leq 50,x+y\geq 60,x+y\leq 100$ and $x,y\geq 0$

Step 5:

Let us calculate the values of $Z$ at the corner points $A(10,50),B(50,50)C(60,40)$ and $D(60,0)$

At the points (x,y) the value of the objective function $Z=2.5x+1.5y+410$

At $A(10,50)$ the value of the objective function $Z=2.5\times 10+1.5\times 50+410=25+75+410=510$

At $B(50,50)$ the value of the objective function $Z=2.5\times 50+1.5\times 50+410=125+75+410=610$

At $C(60,40)$ the value of the objective function $Z=2.5\times 60+1.5\times 40+410=150+60+410=620$

At $D(60,0)$ the value of the objective function $Z=2.5\times 60+1.5\times 0+410=150+0+410=560$

This implies the minimum cost 510 at $A(10,50)$