Answer to Question #173342 in Linear Algebra for hazel

Question

Three water purification facilities can handle at most 10 million gallons in a certain time period. Plant 1 leaves 20% of certain impurities, and costs P20,000 per million gallons. Plant 2 leaves 15% of these impurities and costs P30,000 per million gallons. Plant 3 leaves 10% impurities and costs P40,000 per million gallons. The desired level of impurities in the water from all three plants is at most 15%. If Plant 1 and Plant 3 combined must handle at least 6 million gallons, find the number of gallons each plant should handle so as to achieve the desired level of purity at minimum cost.

Answer 1

Solution:

Let Plant 1 should handle x million gallons, Plant 2 should handle y million gallons, Plant 3 should handle z million gallons in a certain time period.

Considering that a higher level of water purification costs more we can conclude that the minimum cost will be achieved with a maximum acceptable level of impurities of 15%. A level

of impurities in the water from all three plants:

$x\cdot0.2+y\cdot0.15+z\cdot0.1=(x+y+z)\cdot0.15$

$x=z$

If

Plant 1 and Plant 3 combined must handle at least 6 million gallons and $x=z$

Then

Plant 1 and Plant 3 should handle the same amount of water (each at least 3 million gallons).

Answer:

Plant 1 should handle at least 3 million gallons;

Plant 3 should handle at least 3 million gallons (the same amount as Plant 1)

Plant 2 should handle the rest of water (at most 10 million gallons minus water that should be handled by Plant 1 and Plant 2)