Question

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 as ₹ 10500 and ₹ 9000 , respectively. To control weeds, a liquid herbicide has to be used for crops $X$ and $Y$ at rates of $20L$ and $10L$ per hectare. Further no more than $800L$ of herbicide should be used in order to protect fish and wildlife using a pond which collects drainage from his land. To maximise the total profit of the society, the land should be allocated to each crop is

• A. 20 hec for crop $X$ and 30 hec for $\mathrm{crop}Y$
• B. 20 hec each for both crop $X$ and crop $Y$
• C. 30 hec for crop $X$ and 20 hec for crop $Y$
• D. 30 hec each for both crop $X$ and crop $Y$

Answer 1

Answer: (C)

Let $x$ hec of land be allocated to crop $X$ and $y$ hec to crop $Y$. Obviously, $x\ge 0,y\ge 0$.
Profit per hectare on crop $X=₹10500$
Profit per hectare on crop $Y=₹9000$
Therefore, total profit $=₹(10500x+9000y)$
The mathematical formulation of the problem is as follows Maximise
$Z=10500x+9000y$
Subjectl lo lite couristrainls are
$x+y\le 50$ ( constraint related to land)...(i)
$20x+10y\le 800$ (constraint related to use of herbicide)
i.e., $2x+y\le 80$...(ii)
$x\ge 0,y\ge 0$ (non-negative constraint)...(iii)
Let us draw the graph of the system of inequalities (i) to (iii). The feasible region $OABC$ is shown (shaded) in the figure. 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
$A(40,0)$	420000
$B(30,20)$	$495000\leftarrow$ Maximum
$C(0,50)$	450000

Hence, the society will get the maximum profit of ₹ 495000 by allocating 30 heo for crop $X$ and 20 hoc for crop $Y$.