Answer: the maximum number of fish that the lake can support = 375 fish
Step-by-step explanation:
Let b and t represent the bass and trout fish respectively.
Given that;
i. Bass require 2 units of food A and 4 units of food B
ii. trout require 5 units of food A and 2 units of food B
iii. the owner has 1200 units of each food
For food A
2b + 5t = 1200 ....eqn1
For food B
4b + 2t = 1200 ....eqn2
Solving the simultaneous equation
Multiplying eqn1 by 2. We have;
4b + 10t = 2400 .....eqn3
Subtract eqn2 from eqn3
4b-4b +10t-2t = 2400-1200
8t = 1200
t = 1200/8
t = 150
Substituting t= 150 into eqn1
2b + 5(150) = 1200
2b = 1200 - 750 = 450
b = 450/2 = 225
Therefore, the total number of fish is given as
t + b = 150 + 225 = 375
225 bass and 150 trout
