Question

A cooperrative society of farmers has $50$ hectare of land to grow two crops $X$ and $Y$. The profit from crop $X$ and $Y$ per hectare are estimated as 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 wild life 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 of land be allocated to crop $Y$
$\Rightarrow x\ge 0,y\ge 0$
Profit per hectare on crop $X=$ Rs $10,500$
Profit per hectare on crop $Y=$ Rs $9,000$
Therefore, total profit $=$ Rs $10500x+9000y$

The mathematical formulation of the problem is as follows:
Maximise $Z=10500x+9000y$
Subject to constraints:
$x+y\le 50\cdots \left(1\right)20x+10y\le 800\Rightarrow 2x+y\le 80\cdots \left(2\right)x,y\ge 0\cdots \left(3\right)$

Now we can plot the graph for the above constraints.

Our feasible region is bounded with corner points $O(0,0),A(40,0),B(30,20)\&C(0,50).$

Corner Points	$Z=10500x+9000y$
$O(0,0)$	$0$
$A(40,0)$	$420000$
$B(30,20)$	$495000$$\leftarrow$Maximum
$C(0,50)$	$450000$

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