The point at which the maximum value of Z = x | vedclass

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(C) Step $1$: Identify the corner points of the feasible region defined by the constraints.Step $2$: The intersection of $x + 2y = 70$ and $2x + y = 9

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$(40, 15)$

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(C) Step $1$: Identify the corner points of the feasible region defined by the constraints.Step $2$: The intersection of $x + 2y = 70$ and $2x + y = 95$ is found by solving the system: $x = 70 - 2y$. Substituting into the second equation: $2(70 - 2y) + y = 95 \implies 140 - 4y + y = 95 \implies 3y = 45 \implies y = 15$. Then $x = 70 - 2(15) = 40$. The intersection point is $(40, 15)$.Step $3$: The corner points are $(0, 0)$, $(47.5, 0)$, $(0, 35)$, and $(40, 15)$.Step $4$: Evaluate $Z = x + y$ at each corner point:At $(0, 0)$, $Z = 0$.At $(47.5, 0)$, $Z = 47.5$.At $(0, 35)$, $Z = 35$.At $(40, 15)$, $Z = 40 + 15 = 55$.Step $5$: The maximum value is $55$ at the point $(40, 15)$.

The point at which the maximum value of $Z = x + y$ subject to the constraints $x + 2y \leq 70$, $2x + y \leq 95$, $x \geq 0$, $y \geq 0$ occurs is:

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