Question

A privately 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.

Answer 1

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

Answer 2

The lake can support a total of 375 fishes.

Equation

An equation is an expression used to show the relationship between two and more variables and numbers.

Let x represent the amount of food A and y represent the amount of food B, hence:

2x + 5y = 1200   (1)

Also:

4x + 2y = 1200   (2)

x = 225, y = 150

The lake can support a total of 375 fishes.

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