The minimum value of the objective function z | vedclass

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Question

(D) To find the minimum value of the objective function $z = 4x + 6y$,we first identify the feasible region defined by the constraints $x + 2y \geq 80

Vedclass · Accepted Answer

$254$

Vedclass · Answer

(D) To find the minimum value of the objective function $z = 4x + 6y$,we first identify the feasible region defined by the constraints $x + 2y \geq 80$,$3x + y \geq 75$,and $x, y \geq 0$.$1$. Find the vertices of the feasible region:- The line $x + 2y = 80$ intersects the $x$-axis at $(80, 0)$ and the $y$-axis at $(0, 40)$.- The line $3x + y = 75$ intersects the $x$-axis at $(25, 0)$ and the $y$-axis at $(0, 75)$.- The intersection point $B$ of $x + 2y = 80$ and $3x + y = 75$ is found by solving the system:$x = 80 - 2y$$3(80 - 2y) + y = 75 \implies 240 - 6y + y = 75 \implies 5y = 165 \implies y = 33$.$x = 80 - 2(33) = 80 - 66 = 14$.So,$B = (14, 33)$.$2$. The vertices of the unbounded feasible region are $A(80, 0)$,$B(14, 33)$,and $C(0, 75)$.$3$. Evaluate $z = 4x + 6y$ at these vertices:- At $A(80, 0)$: $z = 4(80) + 6(0) = 320$.- At $B(14, 33)$: $z = 4(14) + 6(33) = 56 + 198 = 254$.- At $C(0, 75)$: $z = 4(0) + 6(75) = 450$.The minimum value is $254$.

The minimum value of the objective function $z = 4x + 6y$ subject to the constraints $x + 2y \geq 80$,$3x + y \geq 75$,and $x, y \geq 0$ is:

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