The solution set of the constraints 2x + 3y l | vedclass

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(C) The constraints are given by:$1$) $2x + 3y \leq 6$$2$) $x + 4y \leq 4$$3$) $x \geq 0, y \geq 0$To find the corner points of the feasible region,we

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$(\frac{12}{5}, \frac{2}{5})$

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(C) The constraints are given by:$1$) $2x + 3y \leq 6$$2$) $x + 4y \leq 4$$3$) $x \geq 0, y \geq 0$To find the corner points of the feasible region,we identify the intersection points of the boundary lines:- The line $2x + 3y = 6$ intersects the axes at $(3, 0)$ and $(0, 2)$.- The line $x + 4y = 4$ intersects the axes at $(4, 0)$ and $(0, 1)$.Solving the system of equations $2x + 3y = 6$ and $x + 4y = 4$:From the second equation,$x = 4 - 4y$.Substitute into the first: $2(4 - 4y) + 3y = 6 \implies 8 - 8y + 3y = 6 \implies -5y = -2 \implies y = \frac{2}{5}$.Then $x = 4 - 4(\frac{2}{5}) = 4 - \frac{8}{5} = \frac{12}{5}$.The corner points of the feasible region are $(0, 0)$,$(3, 0)$,$(0, 1)$,and $(\frac{12}{5}, \frac{2}{5})$.Comparing these with the given options,the point $(\frac{12}{5}, \frac{2}{5})$ is a corner point.

The solution set of the constraints $2x + 3y \leq 6$,$x + 4y \leq 4$,$x \geq 0$,and $y \geq 0$ includes the point $\ldots$ as a corner point.

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