Problem 23 - Algebra and Trigonometry

Question

19–40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} x & \geq 0 \\ y & \geq 0 \\ 3 x+5 y & \leq 15 \\\ 3 x+2 y & \leq 9 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices: (0,3), (5,0), (3,0), (\(\frac{5}{3},\frac{19}{6}\)); Bounded.

Step 1: Graph the Line for Each Inequality

Take each inequality and first convert it to an equation. For \(3x + 5y \leq 15\), draw the line \(3x + 5y = 15\). For \(3x + 2y \leq 9\), draw the line \(3x + 2y = 9\). These will form the boundaries of the region. Use a solid line for these boundaries, as the inequalities are \(\leq\). Since both \(x\) and \(y\) are restricted to be non-negative (\(x \geq 0\), \(y \geq 0\)), draw the axes \(x = 0\) and \(y = 0\).

Step 2: Determine the Feasible Region

The feasible region is the common area that satisfies all inequalities, including \( x \geq 0 \) and \( y \geq 0 \). It's the area that is on or below both the lines \(3x + 5y = 15\) and \(3x + 2y = 9\), and in the first quadrant where both \(x\) and \(y\) are non-negative.

Step 3: Find Intersection Points (Vertices)

Find the intersection of the lines to determine the potential vertices of the bounded region. Solve the system of equations \(3x + 5y = 15 \3x + 2y = 9\) by multiplying the second equation by 5 and the first by 2, and subtract:\[15x + 10y = 45\]\[6x + 10y = 30\]Subtract to get \(9x = 15\), so \(x = \frac{5}{3}\). Substitute \(x\) back into \(3x + 2y = 9\) to find \(y = \frac{12}{3} - x\), resulting in \(y = \frac{19}{6}\). Intersection at \((\frac{5}{3}, \frac{19}{6})\).

Step 4: Find Additional Intersection Points with Axes

Determine where the lines intersect the axes. For \(3x + 5y = 15\), intersect with \(x = 0\) gives \(y = 3\); intersect with \(y = 0\) gives \(x = 5\). For \(3x + 2y = 9\), intersect with \(x = 0\) gives \(y = \frac{9}{2}\); intersect with \(y = 0\) gives \(x = 3\). Collected vertices are \((0, 3)\), \((5, 0)\), and \((3, 0)\). Alongside point \((\frac{5}{3}, \frac{19}{6})\), these form the boundary.

Step 5: Determine if the Solution Set is Bounded

The feasible region is bounded if it forms a closed area. Since we found all vertices and the region formed in the first quadrant is a polygon with these vertices, the solution set is bounded.

Key Concepts

Solution Set

When graphing systems of inequalities, the solution set refers to all the points that satisfy each inequality in the system. For our exercise, this involves graphing each inequality:

• For \(3x + 5y \leq 15\), graph the boundary line \(3x + 5y = 15\) and shade below it.
• Similarly, for \(3x + 2y \leq 9\), graph the boundary line \(3x + 2y = 9\) and shade below it.
• Include the restrictions \(x \geq 0\) and \(y \geq 0\) to confine the solution to the first quadrant.

Together, these shaded areas represent where all conditions are true. This overlapping shaded region on the graph is your solution set. It's crucial for determining the area where conditions overlap, as this identifies potential solutions to the system.

Feasible Region

The feasible region is the area on the graph where all the inequalities intersect and the conditions are satisfied. In the context of the exercise:

• The feasible region is where the shaded parts of all inequalities overlap, specifically focused in the first quadrant due to \(x \geq 0\) and \(y \geq 0\).
• For our set, it's the part of the graph that lies at or below the lines \(3x + 5y = 15\) and \(3x + 2y = 9\).

Finding the feasible region is critical because it visually represents all possible solutions to the inequality system.

Vertices

Vertices are the corner points that define the feasible region. These points occur at intersections of the boundary lines. For our system, to find the vertices:

• Solve \(3x + 5y = 15\) and \(3x + 2y = 9\) simultaneously to find intersecting points like \(x = \frac{5}{3}, y = \frac{19}{6}\).
• Check where each line intersects the axes. The line \(3x + 5y = 15\) intersects at (0,3) and (5,0), while \(3x + 2y = 9\) intersects at (0, \(\frac{9}{2}\)) and (3,0).

By plotting these points, you can clearly see the shape of the feasible region. Vertices are essential as they outline the polygon formed by the feasible region.

Bounded Region

A bounded region is one that forms a closed and finite area. For our system of inequalities, determining if the solution set is bounded involves:

• Identifying if the feasible region forms a closed shape with the identified vertices.
• Since our solution area forms a polygon in the positive quadrant using the vertices \((0, 3)\), \((5, 0)\), \((3, 0)\), and \((\frac{5}{3}, \frac{19}{6})\), it is a closed region.

A bounded region signifies that within these lines and constraints, there is a maximum area where solutions can exist, confirming the system's finite nature in the given space. It plays a fundamental role in optimization and linear programming.