Asked by Selam
A farmer has 1000 acres of land on which he can grow corn, wheat or soya beans. Each acre of corn costs birr 100 for preparation, requires 7 man days of work and yields a profit of birr 30. An acre of wheat costs birr 120 for preparation, requires 10 man days of work and yields a profit of birr 40. Soya beans cost birr 70 to prepare require 8 man days of work and yields a profit of birr 20. If the farmer has birr 100,000 and can count on 80 man days work, formulate the linear programming model.
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Answered by
GPT 3.5
Let:
- x be the number of acres of corn.
- y be the number of acres of wheat.
- z be the number of acres of soya beans.
Objective function: Profit
The profit from corn is 30 birr per acre, so the profit from growing x acres of corn is 30x.
The profit from wheat is 40 birr per acre, so the profit from growing y acres of wheat is 40y.
The profit from soya beans is 20 birr per acre, so the profit from growing z acres of soya beans is 20z.
The total profit is 30x + 40y + 20z.
Constraints:
The costs for preparing corn is 100 birr per acre, so the cost of preparing x acres of corn is 100x.
The costs for preparing wheat is 120 birr per acre, so the cost of preparing y acres of wheat is 120y.
The costs for preparing soya beans is 70 birr per acre, so the cost of preparing z acres of soya beans is 70z.
The total cost cannot exceed the available budget of 100,000 birr, so the constraint is: 100x + 120y + 70z ≤ 100,000.
The labor requirement for corn is 7 man days per acre, so the labor required for x acres of corn is 7x man days.
The labor requirement for wheat is 10 man days per acre, so the labor required for y acres of wheat is 10y man days.
The labor requirement for soya beans is 8 man days per acre, so the labor required for z acres of soya beans is 8z man days.
The total man days of work cannot exceed the available 80 man days, so the constraint is: 7x + 10y + 8z ≤ 80.
The number of acres of corn, wheat, and soya beans cannot be negative, so the constraints are: x ≥ 0, y ≥ 0, z ≥ 0.
Therefore, the Linear Programming model is:
Maximize: 30x + 40y + 20z
Subject to:
100x + 120y + 70z ≤ 100,000
7x + 10y + 8z ≤ 80
x ≥ 0
y ≥ 0
z ≥ 0
- x be the number of acres of corn.
- y be the number of acres of wheat.
- z be the number of acres of soya beans.
Objective function: Profit
The profit from corn is 30 birr per acre, so the profit from growing x acres of corn is 30x.
The profit from wheat is 40 birr per acre, so the profit from growing y acres of wheat is 40y.
The profit from soya beans is 20 birr per acre, so the profit from growing z acres of soya beans is 20z.
The total profit is 30x + 40y + 20z.
Constraints:
The costs for preparing corn is 100 birr per acre, so the cost of preparing x acres of corn is 100x.
The costs for preparing wheat is 120 birr per acre, so the cost of preparing y acres of wheat is 120y.
The costs for preparing soya beans is 70 birr per acre, so the cost of preparing z acres of soya beans is 70z.
The total cost cannot exceed the available budget of 100,000 birr, so the constraint is: 100x + 120y + 70z ≤ 100,000.
The labor requirement for corn is 7 man days per acre, so the labor required for x acres of corn is 7x man days.
The labor requirement for wheat is 10 man days per acre, so the labor required for y acres of wheat is 10y man days.
The labor requirement for soya beans is 8 man days per acre, so the labor required for z acres of soya beans is 8z man days.
The total man days of work cannot exceed the available 80 man days, so the constraint is: 7x + 10y + 8z ≤ 80.
The number of acres of corn, wheat, and soya beans cannot be negative, so the constraints are: x ≥ 0, y ≥ 0, z ≥ 0.
Therefore, the Linear Programming model is:
Maximize: 30x + 40y + 20z
Subject to:
100x + 120y + 70z ≤ 100,000
7x + 10y + 8z ≤ 80
x ≥ 0
y ≥ 0
z ≥ 0
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