Question

How many acres of each (wheat and rye) should the farmer plant in order to get maximum profit?

• A. $(5,5)$
• B. $(4,4)$
• C. $(4,5)$
• D. $(4,3)$

Answer 1

Answer: (B)

$(4,4)$

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., $x+y\ge 7$

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, amount for planting wheat and rye is $\$1200$

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

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 $x+2y\le 12$

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 $P=500x+300y$

Now substituting the options in the profit expression and verifying

Substituting option A $(x,y)=(5,5)$

$x+y\ge 0⟹5+5\ge 7⟹10\ge 7$ True

$2x+y\le 12⟹2\left(5\right)+5\le 12⟹15\le 12$ False

Substituting option B $(x,y)=(4,4)$

$x+y\ge 0⟹4+4\ge 7⟹8\ge 7$ True

$2x+y\le 12⟹2\left(4\right)+4\le 12⟹12\le 12$ True

$x+2y\le 12⟹4+2\left(4\right)\le 12⟹12\le 12$ True

Substituting option C $(x,y)=(4,5)$

$x+y\ge 0⟹4+5\ge 7⟹9\ge 7$ True

$2x+y\le 12⟹2\left(4\right)+5\le 12⟹13\le 12$ False

Substituting option D $(x,y)=(4,3)$

$x+y\ge 0⟹4+3\ge 7⟹7\ge 7$ True

$2x+y\le 12⟹2\left(4\right)+3\le 12⟹11\le 12$ True

$x+2y\le 12⟹4+2\left(3\right)\le 12⟹10\le 12$ True

$(4,4),(4,3)$ satisfies the constraints. Therefore finding the profit

for $(4,4)$, $P=500x+300y=500\left(4\right)+300\left(4\right)=3200$

for $(4,3)$, $P=500x+300y=500\left(4\right)+300\left(3\right)=2900$

Therefore the maximum profit is attained at $(4,4)$