(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

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

Class 12 Mathematics Linear Programming Word problem of Linear programming

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$A$ dietician has to develop a special diet using two foods $X$ and $Y$. Each packet (containing $30 \ g$) of food $X$ contains $12$ units of calcium,$4$ units of iron,$6$ units of cholesterol and $6$ units of vitamin $A$. Each packet of the same quantity of food $Y$ contains $3$ units of calcium,$20$ units of iron,$4$ units of cholesterol and $3$ units of vitamin $A$. The diet requires at least $240$ units of calcium,at least $460$ units of iron and at most $300$ units of cholesterol. The corner points of the feasible region are:

KCET 2022 All KCET

A
$(2,72), (40,15), (15,20)$
B
$(2,72), (15,20), (0,23)$
C
$(0,23), (40,15), (2,72)$
D
$(2,72), (40,15), (115,0)$

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