There are three factories on the Moniss River (1, 2, and 3). Each emits two types of pollutants (1 and 2) into the river. If the waste from each factory is processed, the pollution in the river can be reduced. It costs $15 to process a ton of factory 1 waste, and each ton processed reduces the amount of pollutant 1 by 0.10 ton and amount of pollutant 2 by 0.45 ton. It costs $10 to process a ton of factory 2 waste, and each ton processed will reduce the amount of pollutant 1 by 0.20 ton and the amount of pollutant 2 by 0.25 ton. It costs $20 to process a ton of factory 3 waste, and each ton processed will reduce the amount of pollutant 1 by 0.40 ton and the amount of pollutant 2 by 0.30 ton. The state wants to reduce the amount of pollutant 1 in the river by at least 30 tons and the amount of pollutant 2 in the river by at least 40 tons. Formulate and LP that will minimize the cost of reducing pollution by the desired amounts.

Question

Maeve Braun

Mathematics

17 February, 17:04

There are three factories on the Moniss River (1, 2, and 3). Each emits two types of pollutants (1 and 2) into the river. If the waste from each factory is processed, the pollution in the river can be reduced. It costs $15 to process a ton of factory 1 waste, and each ton processed reduces the amount of pollutant 1 by 0.10 ton and amount of pollutant 2 by 0.45 ton. It costs $10 to process a ton of factory 2 waste, and each ton processed will reduce the amount of pollutant 1 by 0.20 ton and the amount of pollutant 2 by 0.25 ton. It costs $20 to process a ton of factory 3 waste, and each ton processed will reduce the amount of pollutant 1 by 0.40 ton and the amount of pollutant 2 by 0.30 ton. The state wants to reduce the amount of pollutant 1 in the river by at least 30 tons and the amount of pollutant 2 in the river by at least 40 tons. Formulate and LP that will minimize the cost of reducing pollution by the desired amounts.

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Answers (1)

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Jonathon Stephens · Answer 1

The required LP is:

Min z = 15 x1 + 10 x2 + 20 x3

0.1 x1 + 0.2 x2 + 0.4 x3 ≥ 30 (Pollutant 1 constraint)

0.45 x1 + 0.25 x2 + 0.3 x3 ≥ 40

(Pollutant 2 constraint)

x1, x2, x3 ≥ 0

Step-by-step explanation:

Let x1, x2 and x3 represent the amount of waste produced by factory 1,2 and 3 respectively.

Min z = 15 x1 + 10 x2 + 20 x3

0.1 x1 + 0.2 x2 + 0.4 x3 ≥ 30 (Pollutant 1 constraint)

0.45 x1 + 0.25 x2 + 0.3 x3 ≥ 40

(Pollutant 2 constraint)

x1, x2, x3 ≥ 0