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(A) Let $x$ hectares of land be allocated to crop $X$ and $y$ hectares to crop $Y$.Obviously,$x \geq 0, y \geq 0.$Profit per hectare on crop $X = 	ex

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$30$ hectares for crop $X$ and $20$ hectares for crop $Y$

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(A) Let $x$ hectares of land be allocated to crop $X$ and $y$ hectares to crop $Y$.Obviously,$x \geq 0, y \geq 0.$Profit per hectare on crop $X = 	ext{Rs } 10,500$Profit per hectare on crop $Y = 	ext{Rs } 9,000$Therefore,total profit $Z = 10,500x + 9,000y$The mathematical formulation of the problem is as follows:Maximise $Z = 10,500x + 9,000y$Subject to the constraints:$x + y \leq 50$ (constraint related to land) $... (1)$$20x + 10y \leq 800$ (constraint related to use of herbicide)i.e.,$2x + y \leq 80$ $... (2)$$x \geq 0, y \geq 0$ (non-negative constraint) $... (3)$The feasible region $OABC$ is bounded by the vertices $O(0,0), A(40,0), B(30,20),$ and $C(0,50)$.Evaluating the objective function $Z = 10,500x + 9,000y$ at these vertices:Corner Point$Z = 10,500x + 9,000y$$O(0,0)$$0$$A(40,0)$$4,20,000$$B(30,20)$$10,500(30) + 9,000(20) = 3,15,000 + 1,80,000 = 4,95,000$ (Maximum)$C(0,50)$$4,50,000$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 $Y$.

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