# Misc 7 - Chapter 12 Class 12 Linear Programming

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 7 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 distances (in km) between the depots and the petrol pumps is given in the following table: Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how should the delivery be scheduled in order that the transportation cost is minimum? What is the minimum cost? Let depot A transport x litres to petrol pump D depot A transport y litres to petrol pump E Transportation Cost of 10 km = Rs 1 Transportation Cost of 1 km = Rs 110 = Rs 0.1 Hence, Since number of litres transported to each depot must be greater than or equal to zero -: ∴ x ≥ 0, y ≥ 0 7000 − (x + y) ≥ 0 ⇒ x + y ≤ 7000 7000 − (x + y) ≥ 0 ⇒ x + y ≤ 7000 4500 − x ≥ 0 ⇒ x ≤ 4500 3000 − y ≥ 0 ⇒ y ≤ 3000 x + y − 3500 ≥ 0 ⇒ x + y ≥ 3500 As, we need to Minimize the cost of transportation, Hence the function used here is minimize Z. Total Transportation Cost : Z = 0.7x + 0.6y + 0.3 (7000 − (x + y)) + 0.3 (4500 − x) + 0.4(3000 – y) + 0.2 (x + y – 3500) Z = 0.3x + 0.1y + 3950 Hence, Minimize Z = 0.3x 0.1y + 3950 Combing all constraints : Minimize Z = 0.3x + 0.1y + 3950 Subject to constraints : x + y ≤ 7000 y ≤ 3500 x ≤ 4500 x + y ≥ 3500 x, y ≥ 0 Hence, transportation cost will be Minimum if : Minimum Cost = Rs 4400

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.