Problema Solution
Bacteria of species A and species B are kept in a single test tube, where they are fed two nutrients. Each day the test tube is supplied with 28,570 units of the first nutrient and 45,160 units of the second nutrient. Each bacterium of species A requires 6 units of the first nutrient and 7 units of the second, and each bacterium of species B requires 1 unit of the first nutrient and 9 units of the second. What populations of each species can coexist in the test tube so that all the nutrients are consumed each day?
Answer provided by our tutors
Let
a = the population of bacterias A
b = the population of bacterias B
6a + b = 28570
7a + 9b = 45160
........
........
a = 4,510 bacterias
b = 1,510 bacterias
There can be 4,510 bacterias of population A and 1,510 bacterias of population B coexisting in the test tube.