For the linear programming problem, x + 2y le | vedclass

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

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

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(D) Step $1$: Identify the corner points of the feasible region defined by $x + 2y \leq 10$, $3x + y \leq 12$, $x \geq 0$, $y \geq 0$.Step $2$: The intersection of $x + 2y = 10$ and $3x + y = 12$ is found by solving the system: $y = 12 - 3x$. Substituting into the first equation: $x + 2(12 - 3x) = 10 \implies x + 24 - 6x = 10 \implies -5x = -14 \implies x = \frac{14}{5}$. Then $y = 12 - 3(\frac{14}{5}) = \frac{60 - 42}{5} = \frac{18}{5}$. The intersection point is $(\frac{14}{5}, \frac{18}{5})$.Step $3$: Evaluate $z = 5x + 10y$ at corner points: $(0, 0) \implies z = 0$; $(4, 0) \implies z = 20$; $(0, 5) \implies z = 50$; $(\frac{14}{5}, \frac{18}{5}) \implies z = 5(\frac{14}{5}) + 10(\frac{18}{5}) = 14 + 36 = 50$.Step $4$: Since $z$ is maximum at $(0, 5)$ and $(\frac{14}{5}, \frac{18}{5})$ with value $50$, the maximum 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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