Bacteria of species A and species B are kept in a single environment, where they are fed two nutrients. Each day the environment is supplied with 10,560 units of the first nutrient and 31,510 units of the second nutrient. Each bacterium of species A requires 2 units of the first nutrient and 5 units of the second, and each bacterium of species B requires 1 unit of the first nutrient and 6 units of the second. What populations of each species can coexist in the environment so that all the nutrients are consumed each day? Incorrect: Your answer is incorrect. Of species A

Now we require 2 units of the first nutrient for species A and one unit of the first nutrient for species B. The total nutrients required by species A is 2n₁ and that by species B is 1n₂ = n₂. So, the total nutrients required by both species A and B is 2n₁ + n₂. Since this equals the quantity of the first nutrient which is 10,560, then 2n₁ + n₂ = 10,560 (1)

Now we require 5 units of the second nutrient for species A and 6 units of the second nutrient for species B. The total nutrients required by species A is 5n₁ and that by species B is 6n₂. So, the total nutrients required by both species A and B is 5n₁ + 6n₂. Since this equals the quantity of the first nutrient which is 31,510, then 5n₁ + 6n₂ = 31,510 (2).

So, we have two simultaneous equations which we would solve to find the populations of A and B which satisfy both equations.

2n₁ + n₂ = 10,560 (1)

5n₁ + 6n₂ = 31,510 (2)

From (1) n₂ = 10,560 - 2n₁ (3)

Substituting equation (3) into (2), we have

5n₁ + 6(10,560 - 2n₁) = 31,510

expanding the brackets, we have

5n₁ + 63,360 - 12n₁ = 31,510

collecting like terms, we have

5n₁ - 12n₁ = 31,510 - 63,360

simplifying, we have

- 7n₁ = -31,850

dividing both sides by -7, we have

n₁ = -31,850/-7

n₁ = 4,550

Substituting n₁ = 4,550 into (3), we have

n₂ = 10,560 - 2(4,550)

n₂ = 10,560 - 9,100

n₂ = 1,460

So, we require 4,550 of species A and 1,460 of species B that can coexist in the environment so that all the nutrients are consumed each day

Palky Palky · Answer 2 · 2022-01-28T14:18:48+01:00

Number of Bacteria A and Bacteria B = 4550 and 1460 respectively,

Let the number of bacterias of Specie A and Specie B be = x and y respectively.

Given : 2 Nutrient Requirements of both species of Bacteria A and B , Total Nutrients available

To find : Number of Bacteria 1 and Bacteria 2

Nutrient 1 Constraint Equation : 2x + y = 10560

Nutrient 2 Constraint Equation : 5x + 6y = 31510

Putting the value of y from 1st equation in 2nd equation ;

5x + 6 (10560 - 2x) = 31510

5x + 63360 - 12x = 31510

63360 - 31510 = 7x

7x = 31850

x = 31850 / 7 = 4550 { Bacteria A}

Putting the value of x in 1st equation,

2 (4550) + y = 10560

9100 + y = 10560

y = 10560 - 9100 = 1460 {Bacteria B}

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