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?

• A. $\$2500$
• B. $\$2800$
• C. $\$3100$
• D. $\$3200$

Answer 1

The correct option is D $\$3200$

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, an 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 $P=500x+300y$

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

substituting $(2,5)⟹P=500x+300y=500\left(2\right)+300\left(5\right)=2500$

substituting $(5,2)⟹P=500x+300y=500\left(5\right)+300\left(2\right)=3100$

$\$3200$ is the maximum profit.