The maximum value of z=3x+4y,subject to the c | vedclass

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(C) To find the maximum value of the objective function $z=3x+4y$,we identify the feasible region defined by the constraints:$1$. $x+y \leq 40$$2$. $x

Vedclass · Accepted Answer

$140$

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(C) To find the maximum value of the objective function $z=3x+4y$,we identify the feasible region defined by the constraints:$1$. $x+y \leq 40$$2$. $x+2y \leq 60$$3$. $x, y \geq 0$The corner points of the feasible region are determined by the intersection of these lines and the axes:- Intersection of $x+y=40$ and $x+2y=60$: Subtracting the first from the second gives $y=20$,which implies $x=20$. Point: $(20, 20)$.- Intersection of $x+y=40$ with the $x$-axis $(y=0)$: Point $(40, 0)$.- Intersection of $x+2y=60$ with the $y$-axis $(x=0)$: Point $(0, 30)$.- The origin $(0, 0)$ is also a corner point.Now,evaluate $z=3x+4y$ at each corner point:- At $(0, 0)$: $z = 3(0) + 4(0) = 0$- At $(40, 0)$: $z = 3(40) + 4(0) = 120$- At $(0, 30)$: $z = 3(0) + 4(30) = 120$- At $(20, 20)$: $z = 3(20) + 4(20) = 60 + 80 = 140$The maximum value is $140$ at the point $(20, 20)$.

The maximum value of $z=3x+4y$,subject to the constraints $x+y \leq 40$,$x+2y \leq 60$ and $x, y \geq 0$ is

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