Problem 27 - Algebra and Trigonometry

Question

21-46 . 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 are (0, 0), (5, 0), (1, 2), (0, 3). The solution set is bounded.

Step 1: Graph the Coordinate Axes Inequalities

To begin, we need to understand that the inequalities \(x \geq 0\) and \(y \geq 0\) indicate that the solution set is in the first quadrant. This means any solution must lie above the x-axis and to the right of the y-axis.

Step 2: Graph the Line \(3x + 5y = 15\)

Convert the inequality \(3x + 5y \leq 15\) into equation form \(3x + 5y = 15\) to graph the line. Find the intercepts by setting \(x\) and \(y\) to zero in turn: when \(x=0\), \(y=3\), and when \(y=0\), \(x=5\). Connect these points ((0,3) and (5,0)) to graph the line. The area below this line represents the points satisfying the inequality.

Step 3: Graph the Line \(3x + 2y = 9\)

Similarly, convert \(3x + 2y \leq 9\) into equation form \(3x + 2y = 9\). Find intercepts: when \(x=0\), \(y=4.5\), and when \(y=0\), \(x=3\). Draw the line through these points ((0,4.5) and (3,0)). The region below this line satisfies the inequality.

Step 4: Find the Intersection Points

Calculate the vertices by finding the intersection points of the lines. Solve the system of equations formed by the lines: \(3x + 5y = 15\) and \(3x + 2y = 9\). By subtracting the second from the first, we find \(3y = 6\), resulting in \(y = 2\). Substituting \(y = 2\) into the first equation gives \(x = 1\). So, one intersection point is \((1, 2)\).

Step 5: Determine the Other Vertices

List all intersection points including those calculated from the boundary lines with axes. The vertices are at points \((0, 0)\), \((5, 0)\), \((1, 2)\), and \((0, 3)\).

Step 6: Check if the Solution is Bounded

A set is bounded if it can be enclosed within a finite circle. The solution set, a polygon formed by the vertices \((0,0)\), \((5,0)\), \((1,2)\), and \((0,3)\), creates a bounded area in the first quadrant.

Key Concepts

Graphing Inequalities

When it comes to graphing inequalities, the main aim is to find the region that satisfies all the given conditions. Inequalities are like regular equations, but they involve a range of values rather than a single solution. For example, the inequality \(3x + 5y \leq 15\) includes all the points lying on and below the line \(3x + 5y = 15\).
We start by graphing the "equal to" version of the inequality to form a boundary line.

• If the inequality is 'less than or equal to' (\(\leq\)), you shade below the line.
• If it is 'greater than or equal to' (\(\geq\)), shade above the line.

To graph effectively, find and plot the intercepts or any two points, then draw the line through these points. This gives us a visual representation, allowing us to identify regions where the solutions to the inequalities exist.

Bounded Solution Sets

A bounded solution set represents a region on the graph that is confined within certain limits, meaning it occupies a finite area. In the context of inequalities, this bounded area is typically formed by intersections of lines and/or axes.
The solution set of our given system is bounded if:

• All inequalities form intersecting lines that create a closed shape.
• The region formed by these lines is fully enclosed and does not extend to infinity.

In our case, the points \((0, 0)\), \((5, 0)\), \((1, 2)\), and \((0, 3)\) create a polygon in the first quadrant, showing the bounded nature of the solution set.

Intersection Points

Intersection points are crucial in solving systems of inequalities, as they represent the vertices of the bounded solution set. You find intersection points by solving the system of equations that are formed by transforming each inequality into an equation.
In this exercise:

• We found an intersection by solving \(3x + 5y = 15\) and \(3x + 2y = 9\) simultaneously.
• By eliminating \(x\) through subtraction, we got a simple expression for \(y\).
• This resulted in \(y = 2\) and with substitution, \(x = 1\).

Thus, the intersection at \((1, 2)\) becomes a vertex within the solution area. Each intersection point gives us a possible boundary for the solution set.

Linear Inequalities

Linear inequalities are statements that one side of an equation is less than or greater than the other. They form half-planes on a Cartesian plane:

• A boundary line splits the plane into two regions.
• The solution meets the inequality conditions on one side of this line.

For example, \(3x + 5y \leq 15\) forms a boundary with the line \(3x + 5y = 15\). The half-plane under this line is the region where the inequality holds true.
By solving and graphing multiple inequalities together, we find where the solution regions overlap. This overlapping area, or feasible region, contains all solutions that satisfy every inequality in the system.

Quadrant Analysis

In quadrant analysis, we determine which sections of the Cartesian plane contain solutions for both inequalities. The Cartesian plane has four quadrants, but inequality conditions may restrict solutions to specific ones.
For this problem:

• The conditions \(x \geq 0\) and \(y \geq 0\) confine solutions to the first quadrant.
• This quadrant is where both \(x\) and \(y\) are positive, reflecting real-world constraints in some contexts (like quantities that can't be negative).

By analyzing quadrants, we efficiently narrow down where to look for solution sets, saving time and making calculations simpler.