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(A) Let $x$ and $y$ be the number of packets of food $X$ and $Y$ respectively. The constraints are given by:$12x + 3y \geq 240 \Rightarrow 4x + y \geq

Vedclass · Accepted Answer

$(2,72), (40,15), (15,20)$

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(A) Let $x$ and $y$ be the number of packets of food $X$ and $Y$ respectively. The constraints are given by:$12x + 3y \geq 240 \Rightarrow 4x + y \geq 80$$4x + 20y \geq 460 \Rightarrow x + 5y \geq 115$$6x + 4y \leq 300 \Rightarrow 3x + 2y \leq 150$$x \geq 0, y \geq 0$To find the corner points,we find the intersection of these lines:$1$. Intersection of $4x + y = 80$ and $x + 5y = 115$: Solving these,we get $x = 15, y = 20$.$2$. Intersection of $4x + y = 80$ and $3x + 2y = 150$: Solving these,we get $x = 2, y = 72$.$3$. Intersection of $x + 5y = 115$ and $3x + 2y = 150$: Solving these,we get $x = 40, y = 15$.Thus,the corner points of the feasible region are $(2,72), (40,15), (15,20)$.

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