Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{array}{r}2 x+3 y \geq 6 \\ 0 \leq x \leq 5 \\ 0 \leq y \leq 4\end{array}\right.\)

Step-by-Step Solution

Answer

The feasible region is bounded by \(x = 0, x = 5, y = 0, y = 4\), and \(2x + 3y \geq 6\).

Step 1: Graph the Inequality Region

To graph the inequality \(2x + 3y \geq 6\), first express it as an equation: \(2x + 3y = 6\). Find the intercepts by setting \(x = 0\) to get \(y = 2\), and \(y = 0\) to get \(x = 3\). Plot these points and draw the line. The inequality states \(2x + 3y\) should be greater than or equal to 6, so shade above the line.

Step 2: Define the Constraints for x

The inequality \(0 \leq x \leq 5\) tells us that \(x\) is bounded between 0 and 5. This means the solution region is confined within these vertical lines on the graph.

Step 3: Define the Constraints for y

Similarly, the inequality \(0 \leq y \leq 4\) restricts \(y\) to be between 0 and 4. This confines the solution region within these horizontal bounds on the graph.

Step 4: Identify the Feasible Region

The feasible region is where all the conditions are satisfied, including the inequality and the restrictions on the values of \(x\) and \(y\). It is confined by the lines \(x = 0\), \(x = 5\), \(y = 0\), \(y = 4\), and the shaded area for \(2x + 3y \geq 6\).

Step 5: Determine the Vertices of the Feasible Region

Find the intersection points of the boundary lines to determine the vertices of the feasible region. Check intersections with the inequality line and the constraint boundaries: \((x, y) = (0, 2)\), \((3, 0)\), and within the bounds, \((5, 0)\) and \((0, 4)\). However, check which satisfy \(2x + 3y \geq 6\).

Key Concepts

Graphing Inequalities

Graphing inequalities involves visualizing areas that satisfy a mathematical condition. Let's take the inequality \(2x + 3y \geq 6\). This inequality is graphically represented by first plotting the corresponding equation \(2x + 3y = 6\). This equation forms a straight line which acts as a boundary.To graph it:

• Determine the x-intercept by setting \(y = 0\). Solve for \(x\), giving \((3, 0)\).
• Find the y-intercept by setting \(x = 0\). Solve for \(y\), giving \((0, 2)\).
• Draw a line through these intercepts.

Since the inequality is "greater than or equal to," you shade above the line where the region satisfies the inequality condition. The shading indicates points where \(2x + 3y \geq 6\). This helps visualize all potential solutions.

Feasible Region

The feasible region represents all the points that satisfy every part of a system of inequalities. In this exercise, the feasible region combines linear inequality \(2x + 3y \geq 6\) with constraints on \(x\) and \(y\).The shaded portion created from graphing the inequality line is further restricted by:

• Vertical lines \(x = 0\) and \(x = 5\), making sure \(x\) values stay within these limits.
• Horizontal lines \(y = 0\) and \(y = 4\), ensuring \(y\) values fall within this range.

The feasible region is where these boundaries intersect, creating a polygonal region on the graph. It is crucial because it contains all possible solutions fitting both the main inequality and any additional constraints.

Constraints

Constraints are conditions or limits placed on variables within inequalities, crucial for shaping the solution set. In our example, constraints are:*Distribution of X*: \(0 \leq x \leq 5\) sets \(x\) to only values between 0 and 5. These are vertical boundaries on the graph. They're essential because they limit where the feasible region can extend.*Distribution of Y*: \(0 \leq y \leq 4\) limits \(y\) values to between 0 and 4. They form horizontal lines on the graph.These constraints are non-negotiable lines that contain the regions, alongside the inequality. They ensure the bounding of the graph is realistic and practical, stopping the feasible region from extending indefinitely.

Intersection Points

Intersection points occur where the lines determined by the inequalities and constraints cross each other. These points are crucial as they define the boundaries of the feasible region.For our problem, potential intersection points might be found at:

• The intersection of \(x = 0\) with \(y = 4\), resulting in \((0, 4)\).
• Where \(x = 5\) intersects \(y = 0\), leading to \((5, 0)\).
• Intersection with the inequality line \(2x + 3y = 6\) at points like \((0, 2)\) and \((3, 0)\).

Not all intersection points may belong to the feasible region. Each must be checked against all conditions to ensure they meet \(2x + 3y \geq 6\) and other constraints. Valid vertices of the feasible region give us corner points which outline the solution area on the graph.