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(A) Let $x$ and $y$ liters of oil be supplied from depot $A$ to petrol pumps $D$ and $E$ respectively.Then,$(7000 - x - y) \, L$ will be supplied from

Vedclass · Accepted Answer

$4400$

Vedclass · Answer

(A) Let $x$ and $y$ liters of oil be supplied from depot $A$ to petrol pumps $D$ and $E$ respectively.Then,$(7000 - x - y) \, L$ will be supplied from depot $A$ to petrol pump $F$.The requirements at petrol pumps $D, E, F$ are $4500 \, L, 3000 \, L, 3500 \, L$ respectively.Thus,the amounts supplied from depot $B$ to $D, E, F$ are:$D: 4500 - x \, L$$E: 3000 - y \, L$$F: 3500 - (7000 - x - y) = x + y - 3500 \, L$Constraints:$x \geq 0, y \geq 0, x + y \leq 7000$$x \leq 4500, y \leq 3000, x + y \geq 3500$Transportation cost for $10 \, L$ is $Rs. \, 1$ per $km$,so cost for $1 \, L$ is $Rs. \, 0.1$ per $km$.Total cost $Z = 0.1 [7x + 6y + 3(7000 - x - y) + 3(4500 - x) + 4(3000 - y) + 2(x + y - 3500)]$$Z = 0.1 [7x + 6y + 21000 - 3x - 3y + 13500 - 3x + 12000 - 4y + 2x + 2y - 7000]$$Z = 0.1 [3x + y + 39500] = 0.3x + 0.1y + 3950$Evaluating $Z$ at corner points of the feasible region:$A(3500, 0): Z = 0.3(3500) + 3950 = 1050 + 3950 = 5000$$B(4500, 0): Z = 0.3(4500) + 3950 = 1350 + 3950 = 5300$$C(4500, 2500): Z = 0.3(4500) + 0.1(2500) + 3950 = 1350 + 250 + 3950 = 5550$$D(4000, 3000): Z = 0.3(4000) + 0.1(3000) + 3950 = 1200 + 300 + 3950 = 5450$$E(500, 3000): Z = 0.3(500) + 0.1(3000) + 3950 = 150 + 300 + 3950 = 4400$The minimum cost is $Rs. \, 4400$ at $(500, 3000)$.

From/To	$A$	$B$
$D$	$7$	$3$
$E$	$6$	$4$
$F$	$3$	$2$

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