An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distance(in km) between the depots and the petrol pumps is given in the fo

Question

An oil company has two depots A and B with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three petrol pumps, D, E and F whose requirements are 4500L, 3000L and 3500L respectively. The distance(in km) between the depots and the petrol pumps is given in the following table:

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Assuming that the transportation cost of 10 litres of oil is Rs. 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum ? What is the minimum cost ?

Accepted Answer

.: Let from depot A, x litres of oil be transported to pump D,

y litres of oil to pump E and $[7000 - \left( {x + y} \right)]$ litres of oil to pump F.

And from depot B, $(4500 - x)$ litres of oil be transported to pump D,

$(3000 - y)$ litres of oil to pump E and $[3500 - (700 - (x + y))] = (x + y) - 3500)$

litres to pump F. Thus, we have the table.

$figure$

Thus, LPP problem is as below :
Minimize : $Z = \cfrac{{3x}}{{10}} + \cfrac{y}{{10}} + 3950$ ...(1)

Subject to constraints :$4500 - x \ge 0 \Leftrightarrow x4500$ ...(2)

$3000 - y \ge 0 \Leftrightarrow 3000$ ...(3)

$x + y - 3500 \ge 0 \Leftrightarrow x + y \ge 3500$ ...(4)

$7000 - (x + y) \ge 0 \Leftrightarrow x + y \le 7000$ ...(5)

and $x,y \ge 0$ ...(6)

${l_1}:x + y = 3500;{l_2}:x + y = 7000;{l_3}:x = 4500;{l_4}:y = 3000$

Let us graph the inequalities (2) to (6).

The shaded portion represents feasible region which is bounded.

$figure$

Let us evaluate Z at coma- points A (3500, 0),

E(4500,0), F (4500, 2500),G(4000, 3000) and H(500, 3000).

Hence minimum cost = Rs. 4400. From depot A ;

500, 3000 and 3500 litres and depot B; 4000, 0, 0 litres to pumps D, E and F respectively.

Linear Programming — Class 12 Maths Solution