Let $x$ litres oil is supplied from depot A to petrol pump D and y litres of oil is supplied from depot A to petrol pump E, Now,
Subject to Constraints are :
$$
\begin{aligned}
&x \leq 4500, y \leq 3000 \\
&x+y \leq 7000, x+y \geq 3500, x, y \geq 0
\end{aligned}
$$


$\therefore$ The cost of transportation per km per litre $=₹ \frac{1}{10}=₹ 0-1$.

$\Rightarrow$ The $\min ^{\mathrm{m}}$ cost of transportation
$\mathrm{Z}=0 \cdot 7 \mathrm{x}+0 \cdot 6 \mathrm{y}+0 \cdot 3[7000-(\mathrm{x}+\mathrm{y})]+0 \cdot 3(4500-\mathrm{x})$
$+0-4(3000-y)+0 \cdot 2[(x+y)-3500]$
$=0.3 \mathrm{x}+0.1 \mathrm{y}+3950$.
The feasible area is ABECD. Now the points A,B,E,C,D
are A $(500,3000)$, B $(3500,0)$, E $(4500,0), C(4500,2500)$,
D $(400,3000)$
Now, $\mathrm{Z}=0.3 \mathrm{x}+0-1 \mathrm{y}+3950$
At A $(500,3000)$,
$\mathrm{Z}=0.3 \times 500+0.1 \times 3000+3950=4400 \mathrm{~min}^{\mathrm{m}}$
At B $(3500,0)$,
$\mathrm{Z}=0.3 \times 3500+0+3950=5000$
AtE $(4500,0)$,
$\mathrm{Z}=0.3 \times 4500+0+3950=5300$
AtC (4500, 2500),
$\mathrm{Z}=0 \cdot 3 \times 4500+0 \cdot 1 \times 2500+3950=5550$
At D $(4000,3000)$,
$Z=0.3 \times 4000+0.1 \times 3000+3950=5450$
Minimum transport charges are $₹ 4400$ at $\mathrm{A}(500,3000)$ when $x=500$ and $y=3000$. 500 litres, 3000 litres and 3500 litres of oil should be transported from depot A to petrol pumps D, E and F and 400 litres, 0 litres and 0 litres of oil be transported from depot B to petrol pumps D, E and F with minimum cost of transpotation of $₹ 4400$.