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(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The problem can be formulated as follows:Minimize $z = 3x + 3.5y$ (Object

Vedclass · Accepted Answer

$470$

Vedclass · Answer

(A) Let the fruit grower use $x$ bags of brand $P$ and $y$ bags of brand $Q$.The problem can be formulated as follows:Minimize $z = 3x + 3.5y$ (Objective function for nitrogen)Subject to the constraints:$x + 2y \geq 240$ (Phosphoric acid)$3x + 1.5y \geq 270 \implies 2x + y \geq 180$ (Potash)$1.5x + 2y \leq 310$ (Chlorine)$x, y \geq 0$The feasible region is bounded by the lines $x + 2y = 240$,$2x + y = 180$,and $1.5x + 2y = 310$.The corner points of the feasible region are found by solving the intersection of these lines:$1$. Intersection of $x + 2y = 240$ and $1.5x + 2y = 310$: Subtracting gives $0.5x = 70 \implies x = 140$. Then $140 + 2y = 240 \implies 2y = 100 \implies y = 50$. Point $A(140, 50)$.$2$. Intersection of $2x + y = 180$ and $1.5x + 2y = 310$: Multiply first by $2$: $4x + 2y = 360$. Subtracting gives $2.5x = 50 \implies x = 20$. Then $2(20) + y = 180 \implies y = 140$. Point $B(20, 140)$.$3$. Intersection of $x + 2y = 240$ and $2x + y = 180$: $y = 180 - 2x$. Substitute: $x + 2(180 - 2x) = 240 \implies x + 360 - 4x = 240 \implies -3x = -120 \implies x = 40$. Then $y = 180 - 2(40) = 100$. Point $C(40, 100)$.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 minimum value of $z$ is $470$ at $(40, 100)$.Thus,$40$ bags of brand $P$ and $100$ bags of brand $Q$ should be used,and the minimum nitrogen is $470\,kg$.

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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	$1.5$	$2$