Aprivately owned lake contains two types of game fish, bass and trout. the owner provides two types of food, a and b, for these fish. bass require 2 units of food a and 4 units of food b, and trout require 5 units of food a and 2 units of food b. if the owner has 1200 units of each food, find the maximum number of fish that the lake can support.

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

x = -2

y = 848

step-by-step explanation:

plug in the numbers