The minimum value of z = 3x + 5y, subject to | vedclass

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(C) Step $1$: Identify the feasible region defined by the constraints.Step $2$: The intersection points of the lines $x=80$, $y=60$, $x+y=200$, and th

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$(0, 60)$

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(C) Step $1$: Identify the feasible region defined by the constraints.Step $2$: The intersection points of the lines $x=80$, $y=60$, $x+y=200$, and the axes are the vertices of the feasible region.Step $3$: The vertices are found by solving the system of equations:- Intersection of $y=60$ and $x=0$ gives $(0, 60)$.- Intersection of $y=60$ and $x+y=200$ gives $(140, 60)$, but $x \leq 80$, so we use $(80, 60)$.- Intersection of $x=80$ and $x+y=200$ gives $(80, 120)$.- Intersection of $x=0$ and $x+y=200$ gives $(0, 200)$.Step $4$: Evaluate $z = 3x + 5y$ at each vertex:- At $(0, 60)$: $z = 3(0) + 5(60) = 300$.- At $(80, 60)$: $z = 3(80) + 5(60) = 240 + 300 = 540$.- At $(80, 120)$: $z = 3(80) + 5(120) = 240 + 600 = 840$.- At $(0, 200)$: $z = 3(0) + 5(200) = 1000$.Step $5$: The minimum value is $300$ at point $(0, 60)$.

The minimum value of $z = 3x + 5y$, subject to constraints $x \leq 80$, $y \geq 60$, $x + y \leq 200$ and $x, y \geq 0$ occurs at the point:

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