Question

(Allocation problem) A cooperative society of farmers has $50$ hectare of land to grow two crops $X$ and $Y$. The profit from crops $X$ and $Y$ per hectare are estimated at $Rs.10,500$ and $Rs.9,000$ respectively. To control weeds, a liquid herbicide has to be used for crops $X$ and $Y$ at rates of $20$ litres and $10$ litres per hectare. Further, no more than $800$ litres of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximise the total profit of the society?

Answer 1

Let $x$ hectare of land be allocated to crop $X$ and $y$ hectare to crop $Y$. Obviously, $x\ge 0,y\ge 0$.

Profit per hectare on crop $X=Rs.10500$
Profit per hectare on crop $Y=Rs.9000$

Therefore, total profit $=Rs(10500x+9000y)$
The mathematical formulation of the problem is as follows:

Maximise $Z=10500x+9000y$

subject to the constraints:
$x+y\le 50$ (constraint related to land) $...\left(1\right)$
$20x+10y\le 800$ (constraint related to use of herbicide)
i.e. $2x+y\le 80...\left(2\right)$
$x\ge 0,y\ge 0$ (non negative constraint) $...\left(3\right)$

Let us draw the graph of the system of inequalities $\left(1\right)$ to $\left(3\right)$. The feasible region $OABC$ is shown (shaded) in Fig. Observe that the feasible region is bounded.

The coordinates of the corner points $O,A,B$ and $C$ are $(0,0),(40,0),(30,20)$ and $(0,50)$ respectively. Let us evaluate the objective function $Z=10500x+9000y$ at these vertices to find which one gives the maximum profit.

Corner Point	$Z=10500x+9000y$
$O(0,0)$	$0$
$A(40,0)$	$42000$
$B(30,20)$	$495000\leftarrow Maximum$
$C(0,50)$	$450000$

Hence, the society will get the maximum profit of $Rs.4,95,000$ by allocating $30$ hectares for crop $X$ and $20$ hectares for crop $Y$.