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Question

A farmer has 10 acres of land to plant wheat and rye. He has to plant atleast 7 acres. Each acre of wheat costs $200 and each acre of rye costs $100 to plant. He has only $1200 to spend. Moreover, the farmer has to get the planting done in 12 hours and it takes 1 hour to plant an acre of wheat and 2 hours to plant an acre of rye. An acre of wheat yields a profit of $500 and an acre of rye yields a profit of $300.

(Take x and y as the acres of wheat and rye planted respectively). What is the maximum profit that the farmer can make?

(Take x and y as the acres of wheat and rye planted respectively). What is the maximum profit that the farmer can make?

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Solution

The correct option is **D** $3200

let x be the acres of wheat planted and

let x be the acres of wheat planted and

y be the acres of rye planted

Given that there are a total of 10 acres of land to plant.

Atleast 7 acres is to be planted i.e., x+y≥7

Given that the cost to plant one acre of wheat is $200

Therefore, the cost for x acres of wheat is 200x

Given that the cost to plant one acre of rye is $100

Therefore, the cost for y acres of rye is 100y

Given that, an amount for planting wheat and rye is $1200

Therefore the total cost to plant wheat and rye is 200x+100y≤1200⟹2x+y≤12

Given that, the time taken to plant one acre of wheat is 1 hr

Therefore, the time taken to plant x acres of wheat is x hrs

Given that, the time taken to plant one acre of rye is 2 hrs

Therefore, the time taken to plant y acres of rye is 2y hrs

Given that, the total time for planting is 12 hrs

Therefore, the total time to plant wheat and rye is x+2y≤12

Given that, one acre of wheat yields a profit of $500

Therefore, the profit from x acres of wheat is 500x

Given that, one acre of rye yields a profit of $300

Therefore, the profit from y acres of wheat is 300y

therefore the total profit from the wheat and rye is P=500x+300y

In the above figure, the blue shaded region is the feasible region with three corner points.(4,4),(2,5),(5,2)

Now substituting the corner points the profit equation,

substituting (4,4)⟹P=500x+300y=500(4)+300(4)=3200

substituting (2,5)⟹P=500x+300y=500(2)+300(5)=2500

substituting (5,2)⟹P=500x+300y=500(5)+300(2)=3100

$3200 is the maximum profit.

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