91Ó°ÊÓ

To sketch the inequalities, we first need to rewrite them as equalities and plot the lines that describe the boundaries. We must also determine which side of the line corresponds to the inequality being less than or equal to or greater than or equal to. 1) \( -3x + 2y = 5 \) Determine the intersection points with the x-axis and the y-axis: x=0: \( y = \frac{5}{2} \) y=0: \( x = -\frac{5}{3} \) As the inequality is less than or equal to, the region is below the line. 2) \( 3x - 2y = 6 \) Determine the intersection points with the x-axis and the y-axis: x=0: \( y = -3 \) y=0: \( x = 2 \) As the inequality is less than or equal to, the region is below the line. 3) \( x = 2y \) Determine the intersection points with the x-axis and the y-axis: x=0: \( y = 0 \) y=0: \( x = 0 \) As the inequality is less than or equal to, the region is on the right of the line. 4) \( x = 0 \) and \( y = 0 \) This inequality represents the constraints that x and y must be greater than or equal to 0. Therefore, we are only considering the first quadrant of the coordinate plane.

Now that we have plotted the lines, we need to identify the region where all inequalities overlap. This is the region that satisfies all given inequalities simultaneously. To do this, consider the signs of the inequalities and the regions they define. Label the feasible region.

Observe the feasible region and its relation to the coordinate plane. If the feasible region is confined within a closed shape, it is bounded. If it extends indefinitely along one or more directions, it is unbounded. In our case, the feasible region is bounded since it is enclosed by the lines.

Identify the points at which the lines intersect and correspond to the bounded region. In our case, there are four intersections that determine the feasible region. To find the coordinates of these corner points, solve the systems of linear equations formed by the pairs of lines: 1) Intersection of the lines \( -3x+2y=5 \) and \( x=2y \): \( \begin{cases} -3 x + 2 y = 5 \\ x = 2y \end{cases} \) Solution: (2,1) 2) Intersection of the lines \( -3x+2y=5 \) and \( 3x - 2y = 6 \): \( \begin{cases} -3 x + 2 y = 5 \\ 3 x - 2 y = 6 \end{cases} \) Solution: (1, 2) 3) Intersection of the lines \( 3x - 2y = 6 \) and \( x = 0 \): \( \begin{cases} 3 x - 2 y = 6 \\ x = 0 \end{cases} \) Solution: (0, -3) 4) Intersection of the lines \( x = 2y \) and \( y = 0 \): \( \begin{cases} x = 2y \\ y = 0 \end{cases} \) Solution: (0, 0) The corner points of the feasible region are (2,1), (1, 2), (0, -3), and (0, 0).