There are two types of fertilisers F_{1} and | vedclass

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(A) Let the farmer buy $x\,kg$ of fertilizer $F_{1}$ and $y\,kg$ of fertilizer $F_{2}$.Constraints:$1$. Nitrogen requirement: $0.10x + 0.05y \geq 14 \

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$1000$

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(A) Let the farmer buy $x\,kg$ of fertilizer $F_{1}$ and $y\,kg$ of fertilizer $F_{2}$.Constraints:$1$. Nitrogen requirement: $0.10x + 0.05y \geq 14 \implies 10x + 5y \geq 1400 \implies 2x + y \geq 280$.$2$. Phosphoric acid requirement: $0.06x + 0.10y \geq 14 \implies 6x + 10y \geq 1400 \implies 3x + 5y \geq 700$.$3$. Non-negativity: $x \geq 0, y \geq 0$.Objective function to minimize: $Z = 6x + 5y$.Finding corner points of the feasible region:- Intersection of $2x + y = 280$ and $3x + 5y = 700$:Multiply first by $5$: $10x + 5y = 1400$.Subtract second: $(10x - 3x) = 1400 - 700 \implies 7x = 700 \implies x = 100$.Substitute $x=100$ in $2x + y = 280 \implies 200 + y = 280 \implies y = 80$.Corner point $B(100, 80)$.- Intersection with axes:For $2x + y = 280$,intercepts are $(140, 0)$ and $(0, 280)$.For $3x + 5y = 700$,intercepts are $(700/3, 0)$ and $(0, 140)$.The feasible region is unbounded with corner points $A(700/3, 0)$,$B(100, 80)$,and $C(0, 280)$.Evaluating $Z$ at corner points:- $Z(A) = 6(700/3) + 5(0) = 1400$.- $Z(B) = 6(100) + 5(80) = 600 + 400 = 1000$.- $Z(C) = 6(0) + 5(280) = 1400$.Since the region is unbounded,we check $6x + 5y < 1000$. The line $6x + 5y = 1000$ does not intersect the interior of the feasible region. Thus,the minimum cost is $Rs\,1000$ at $x=100, y=80$.

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