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(A) Let $x$ and $y$ be the number of packets of food $P$ and $Q$ respectively. The objective is to minimise $Z = 6x + 3y$ (vitamin $A$).Constraints ar

Vedclass · Accepted Answer

$15$ packets of $P$,$20$ packets of $Q$; Minimum vitamin $A = 150$ units

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(A) Let $x$ and $y$ be the number of packets of food $P$ and $Q$ respectively. The objective is to minimise $Z = 6x + 3y$ (vitamin $A$).Constraints are:$12x + 3y \geq 240 \implies 4x + y \geq 80$$4x + 20y \geq 460 \implies x + 5y \geq 115$$6x + 4y \leq 300 \implies 3x + 2y \leq 150$$x, y \geq 0$The feasible region is bounded by the corner points $L(2, 72)$,$M(15, 20)$,and $N(40, 15)$.Corner Point$Z = 6x + 3y$$L(2, 72)$$6(2) + 3(72) = 12 + 216 = 228$$M(15, 20)$$6(15) + 3(20) = 90 + 60 = 150$$N(40, 15)$$6(40) + 3(15) = 240 + 45 = 285$The minimum value of $Z$ is $150$ at point $(15, 20)$. Thus,$15$ packets of food $P$ and $20$ packets of food $Q$ should be used.

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