A farmer wants to plant soy beans and corn. He wants to plant at least 2 acres of soy beans. He has only $1200 to spend and each acre of soy beans costs $200 to plant and each acre of corn costs $100 to plant. Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of soy beans and 2 hours to plant an acre of corn. The farmer profits $500 per acre of soy beans and $300 per acre of corn. If the farmer wants to maximize his profit, how many acres of each should he plant?

Question

03-05-2021
Mathematics

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A farmer wants to plant soy beans and corn. He wants to plant at least 2 acres of soy beans. He has only $1200 to spend and each acre of soy beans costs $200 to plant and each acre of corn costs $100 to plant. Moreover, the farmer has to get the planting done in 12 hours and it takes an hour to plant an acre of soy beans and 2 hours to plant an acre of corn. The farmer profits $500 per acre of soy beans and $300 per acre of corn. If the farmer wants to maximize his profit, how many acres of each should he plant?

Sagot :

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Answer:

4 acres of each

Step-by-step explanation:

Let s and c represent the number of acres of soybeans and corn, respectively. The problem statement gives rise to several constraints. There is also an objective function to maximize. The linear programming problem can be described by ...

maximize 500s +300c subject to ...

s ≥ 2

200s +100s ≤ 1200 . . . . cost to plant

s + 2c ≤ 12 . . . . . . . . . . . . hours to plant

__

The attached graph shows the objective function is maximized at (s, c) = (4, 4).

The farmer should plant 4 acres each of soybeans and corn.

__

For the solution here, the constraint that s ≥ 2 is irrelevant. If the graph included it, the vertices of the feasible solution space would be (2,5), (4, 4), (6, 0) and (2, 0). The solution that maximizes profit is (4, 4).

Sagot :

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