Q. 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 $20 L$ and $10 L$ per hectare. Further no more than $800 L$ 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
Linear Programming
Solution:
Let $x$ hec of land be allocated to crop $X$ and $y$ hec to crop $Y$. Obviously, $x \geq 0, y \geq 0$.
Profit per hectare on crop $X=₹ 10500$
Profit per hectare on crop $Y=₹ 9000$
Therefore, total profit $=₹(10500 x+9000 y)$
The mathematical formulation of the problem is as follows Maximise
$ Z=10500 x+9000 y$
Subjectl lo lite couristrainls are
$x+y \leq 50$ ( constraint related to land)...(i)
$20 x+10 y \leq 800$ (constraint related to use of herbicide)
i.e., $2 x+y \leq 80$...(ii)
$x \geq 0, y \geq 0$ (non-negative constraint)...(iii)
Let us draw the graph of the system of inequalities (i) to (iii). The feasible region $O A B C$ 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 $Z=10500 x+9000 y$ at these vertices to find which one gives the maximum profit.
Corner point
$Z-10500 x+9000 y$
$O(0,0)$
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 $Y$.
| Corner point | $Z-10500 x+9000 y$ |
|---|---|
| $O(0,0)$ | 0 |
| $A(40,0)$ | 420000 |
| $B(30,20)$ | $495000 \leftarrow$ Maximum |
| $C(0,50)$ | 450000 |