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(D) $1$. Identify the corner points of the feasible region defined by $x + 2y \leq 10$, $3x + y \leq 12$, $x, y \geq 0$.$2$. The intersection of $x +

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$(0, 5)$ and $(\frac{14}{5}, \frac{18}{5})$

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(D) $1$. Identify the corner points of the feasible region defined by $x + 2y \leq 10$, $3x + y \leq 12$, $x, y \geq 0$.$2$. The intersection of $x + 2y = 10$ and $3x + y = 12$ is found by solving the system: $y = 12 - 3x$, so $x + 2(12 - 3x) = 10 \implies x + 24 - 6x = 10 \implies -5x = -14 \implies x = \frac{14}{5}$. Then $y = 12 - 3(\frac{14}{5}) = \frac{60 - 42}{5} = \frac{18}{5}$.$3$. The corner points are $(0, 0)$, $(4, 0)$, $(\frac{14}{5}, \frac{18}{5})$, and $(0, 5)$.$4$. Evaluate $z = 5x + 10y$ at each point:- At $(0, 0)$, $z = 0$.- At $(4, 0)$, $z = 5(4) + 10(0) = 20$.- At $(\frac{14}{5}, \frac{18}{5})$, $z = 5(\frac{14}{5}) + 10(\frac{18}{5}) = 14 + 36 = 50$.- At $(0, 5)$, $z = 5(0) + 10(5) = 50$.$5$. Since $z$ is maximum $(50)$ at both $(0, 5)$ and $(\frac{14}{5}, \frac{18}{5})$, the maximum value occurs at every point on the line segment joining these two points.

For the linear programming problem, $x + 2y \leq 10$, $3x + y \leq 12$, $x, y \geq 0$, the maximum value of $z = 5x + 10y$ occurs at every point on the line segment joining the points..

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