Question

Un departamento de pesca y caza del estado proporcionando tres tipos de comida a un lago que

alberga a tres especies de peces. Cada pez de la especie 1 consume cada semana un promedio

de 7 unidad del alimento A, 4 unidad de alimento B, y 1 unidades del alimento C. cada pez de la

especie 2 consume cada semana un promedio de 2 unidades del alimento A, 2 del B y 6 del C.

para un pez de la especie 3, el promedio semanal de consumo es de 2 unidades del alimento A,

1 del B y 6 del C. cada semana se proporciona al lago 25.000 unidades del alimento A 20.000

del B y 55.000 del C. si se supone que los peces se comen todo el alimento. ¿Cuántos peces de

cada especie pueden coexistir en el lago?​

Answer 1

Given : Each fish of species 1 consumes an average each week of 7 units of food A, 4 units of food B, and 1 units of food C.

each fish in the Species 2 consumes an average of 2 units of food A, 2 of B and 6 of C . each week.

for a fish of species 3, the weekly average consumption is 2 units of food A, 1 of B and 6 of C.

Each week 25,000 units of food A 20,000 are provided to the lake

of B and 55,000 of C.

the fish are supposed to eat all the food.

To Find : How many fish of each species can coexist in the lake?

Solution:

Species 1  = x

Species 2  = y

Species 3  = z

7x + 2y + 2z   = 25000    Eq1

4x + 2y  +  z   = 20000   Eq2

x  +   y  +  6z  =  55000     Eq3

Eq1 - Eq3

=> 3x  + z  = 5000    Eq4

 Eq2 - 2 * Eq3

=> 2x - 11z  = -90000   Eq5

11 * Eq4 + Eq5

=> 35x  = -35000

=> x = - 1000  

3x  + z  = 5000

=> z = 8000

x  +   y  +  6z  =  55000  

=> -1000 + y + 48000 = 55000

=> y = 8000

x = - 1000

y = 8000

z = 8000

but species should not be -ve hence looks problem in data

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