A farmer mixes two brands P and Q of cattle f | vedclass

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(A) Let the farmer mix $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective is to minimize the cost $Z = 250x + 200y$.Subject to the constrai

Vedclass · Accepted Answer

$3$ bags of $P$ and $6$ bags of $Q$,Minimum cost $= Rs. 1950$

Vedclass · Answer

(A) Let the farmer mix $x$ bags of brand $P$ and $y$ bags of brand $Q$.The objective is to minimize the cost $Z = 250x + 200y$.Subject to the constraints:$3x + 1.5y \geq 18$ (Nutrient $A$)$2.5x + 11.25y \geq 45$ (Nutrient $B$)$2x + 3y \geq 24$ (Nutrient $C$)$x, y \geq 0$Solving the intersection points of the lines:$1$. $3x + 1.5y = 18$ and $2x + 3y = 24$ intersect at $(3, 6)$.$2$. $2.5x + 11.25y = 45$ and $2x + 3y = 24$ intersect at $(9, 2)$.$3$. The boundary lines intersect the axes at $(18, 0)$ and $(0, 12)$.The corner points of the feasible region are $(18, 0), (9, 2), (3, 6), (0, 12)$.Evaluating $Z = 250x + 200y$ at these points:- At $(18, 0): Z = 250(18) + 200(0) = 4500$- At $(9, 2): Z = 250(9) + 200(2) = 2250 + 400 = 2650$- At $(3, 6): Z = 250(3) + 200(6) = 750 + 1200 = 1950$- At $(0, 12): Z = 250(0) + 200(12) = 2400$Since the feasible region is unbounded,we check if $250x + 200y < 1950$ has any common points with the feasible region. The line $250x + 200y = 1950$ (or $5x + 4y = 39$) does not intersect the feasible region except at $(3, 6)$. Thus,the minimum cost is $Rs. 1950$ at $x=3, y=6$.

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