Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively.

Solution:

Let godown A supply $x$ and $y$ quintals of grain to the shops $D$ and $E$ respectively. Then, $(100-x-y)$ will be supplied to shop $F$.

The requirement at shop $D$ is 60 quintals since $x$ quintals are transported from godown $A$. Therefore, the remaining $(60-x)$ quintals will be transported from godown $B$.

Similarly, $(50-y)$ quintals and $40-(100-x-y)=(x+y-60)$ quintals will be transported from godown B to shop E and F respectively.

The given problem can be represented diagrammatically as follows.

$x \geq 0, y \geq 0$, and $100-x-y \geq 0$

$\Rightarrow x \geq 0, y \geq 0$, and $x+y \leq 100$

$60-x \geq 0,50-y \geq 0$, and $x+y-60 \geq 0$

$\Rightarrow x \leq 60, y \leq 50$, and $x+y \geq 60$

Total transportation cost z is given by,

$z=6 x+3 y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)$

$=6 x+3 y+250-2.5 x-2.5 y+240-4 x+100-2 y+3 x+3 y-180$

$=2.5 x+1.5 y+410$

The given problem can be formulated as

Minimize $z=2.5 x+1.5 y+410$         (1)

subject to the constraints,

$x+y \leq 100$                                   (2)

$x \leq 60$                                         (3)

$y \leq 50$                                         (4)

$x+y \geq 60$                                    (5)

$x, y \geq 0$                                      (6)

The feasible region determined by the system of constraints is as follows.

The corner points are A (60, 0), B (60, 40), C (50, 50), and D (10, 50).

The values of z at these corner points are as follows.

The minimum value of z is 510 at (10, 50).

Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.

The minimum cost is Rs 510.