(D) 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 $A$,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.
Total transportation cost $z$ is given by:
$z = 6x + 3y + 2.5(100-x-y) + 4(60-x) + 2(50-y) + 3(x+y-60)$
$z = 6x + 3y + 250 - 2.5x - 2.5y + 240 - 4x + 100 - 2y + 3x + 3y - 180$
$z = 2.5x + 1.5y + 410$
The problem is to minimize $z = 2.5x + 1.5y + 410$ subject to:
$x+y \leq 100, x \leq 60, y \leq 50, x+y \geq 60, x, y \geq 0$.
The corner points of the feasible region are $A(60, 0), B(60, 40), C(50, 50),$ and $D(10, 50)$.
| Corner point | $z = 2.5x + 1.5y + 410$ |
|---|
| $A(60, 0)$ | $560$ |
| $B(60, 40)$ | $620$ |
| $C(50, 50)$ | $610$ |
| $D(10, 50)$ | $510$ (Minimum) |
The minimum value of $z$ is $510$ at $(10, 50)$.
Thus,the amount of grain transported from $A$ to $D, E, F$ is $10, 50, 40$ quintals respectively,and from $B$ to $D, E, F$ is $50, 0, 0$ quintals respectively.