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(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective function is to maximize nitrogen: $Z = 3x + 3.5y$.Constrain

Vedclass · Accepted Answer

$140$ bags of $P$ and $50$ bags of $Q$; Maximum nitrogen = $595 \, kg$

Vedclass · Answer

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective function is to maximize nitrogen: $Z = 3x + 3.5y$.Constraints based on the table:$1$. Phosphoric acid: $x + 2y \geq 240$$2$. Potash: $3x + 1.5y \geq 270 \implies 2x + y \geq 180$$3$. Chlorine: $1.5x + 2y \leq 310$$4$. Non-negativity: $x, y \geq 0$Solving the system of inequalities,the feasible region is bounded by the corner points $A(140, 50)$,$B(20, 140)$,and $C(40, 100)$.Evaluating $Z = 3x + 3.5y$ at corner points:- At $A(140, 50): Z = 3(140) + 3.5(50) = 420 + 175 = 595$- At $B(20, 140): Z = 3(20) + 3.5(140) = 60 + 490 = 550$- At $C(40, 100): Z = 3(40) + 3.5(100) = 120 + 350 = 470$The maximum nitrogen is $595 \, kg$ at $x = 140$ and $y = 50$.

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	Brand $P$ ($kg$ per bag)	Brand $Q$ ($kg$ per bag)
Nitrogen	$3$	$3.5$
Phosphoric acid	$1$	$2$
Potash	$3$	$1.5$
Chlorine	$1.5$	$2$