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(A) Let the diet contain $x$ and $y$ packets of foods $P$ and $Q$ respectively. The objective is to maximize $Z = 6x + 3y$. The constraints are: $12x

Vedclass · Accepted Answer

$40$ packets of $P$ and $15$ packets of $Q$; Maximum vitamin $A = 285$ units

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(A) Let the diet contain $x$ and $y$ packets of foods $P$ and $Q$ respectively. The objective is to maximize $Z = 6x + 3y$. The constraints are: $12x + 3y \geq 240 \implies 4x + y \geq 80$ $4x + 20y \geq 460 \implies x + 5y \geq 115$ $6x + 4y \leq 300 \implies 3x + 2y \leq 150$ $x, y \geq 0$. Solving the intersection points of the lines: $4x + y = 80$ and $x + 5y = 115$ gives $A(15, 20)$. $4x + y = 80$ and $3x + 2y = 150$ gives $B(2, 72)$. $x + 5y = 115$ and $3x + 2y = 150$ gives $C(40, 15)$. Evaluating $Z = 6x + 3y$ at corner points: Corner Point$Z = 6x + 3y$$A(15, 20)$$6(15) + 3(20) = 150$$B(2, 72)$$6(2) + 3(72) = 228$$C(40, 15)$$6(40) + 3(15) = 285$ The maximum value is $285$ at $(40, 15)$. Thus,$40$ packets of $P$ and $15$ packets of $Q$ are required.

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