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?
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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