Problem 25 - Algebra 2

Question

GEOMETRY Find the area of the region defined by the system of inequalities \(x \geq-3, y+x \leq 8,\) and \(y-x \geq-2\)

Step-by-Step Solution

Answer

The area of the region is 52 square units.

Step 1: Identify the Inequalities

We need to work with the inequalities: \( x \geq -3 \), \( y + x \leq 8 \), and \( y - x \geq -2 \). These define the boundaries of the region we want to find the area of.

Step 2: Graph the Inequalities

Begin by graphing each linear inequality on a coordinate plane. \( x \geq -3 \) is a vertical line at \( x = -3 \). \( y + x \leq 8 \) can be rewritten as \( y \leq -x + 8 \), which is a line with slope -1 and y-intercept 8. \( y - x \geq -2 \) can be rewritten as \( y \geq x -2 \), which is a line with slope 1 and y-intercept -2.

Step 3: Determine the Feasible Region

The feasible region is where the solutions to these inequalities overlap. You should shade the areas where the conditions of each inequality are true. The feasible region will be the intersection of these shaded areas.

Step 4: Find Intersection Points

Calculate where the lines intersect to find the vertices of the feasible region. 1. \( x = -3 \) and \( y=-x+8 \): Substitute \( x = -3 \) into \( y = -x + 8 \), \( y = -(-3) + 8 = 3 + 8 = 11 \). So, point is \((-3,11)\).2. \( x = -3 \) and \( y=x-2 \): Substitute \( x = -3 \) into \( y = x - 2 \), \( y = -3 - 2 = -5 \). So, point is \((-3,-5)\).3. \( y = -x+8 \) and \( y = x-2 \): Set equal: \(-x + 8 = x - 2\), Solving gives \( 2x = 10 \), \( x = 5 \). Substitute into \( y = x - 2 \): \( y = 5 - 2 = 3 \). So, point is \((5,3)\).

Step 5: Determine the Area of the Region

The vertices found are \((-3, 11)\), \((-3, -5)\), and \((5, 3)\). These points form a triangle. Use the formula for the area of a triangle with vertices (\(x_1, y_1\)), (\(x_2, y_2\)), (\(x_3, y_3\)): \[\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]Plugging in the points:\[x_1 = -3, y_1 = 11; x_2 = -3, y_2 = -5; x_3 = 5, y_3 = 3\]\[\text{Area} = \frac{1}{2} \left| -3(-5 - 3) + (-3)(3 - 11) + 5(11 - (-5)) \right| = \frac{1}{2} \left| 24 - 24 + 80 \right| = 52\]

Step 6: Finalize the Solution

The feasible region is a triangle with vertices \((-3, 11)\), \((-3, -5)\), and \((5, 3)\). The area of this region, calculated using the triangle area formula, is 52 square units.

Key Concepts

Area of a Triangle

Understanding how to calculate the area of a triangle is very useful in solving many geometry problems. A triangle is a three-sided polygon, and its area can be calculated using the coordinates of its vertices. When you have vertices at \(x_1, y_1\), \(x_2, y_2\), and \(x_3, y_3\), you can apply the coordinate geometry formula to find its area:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \]
This formula helps you determine the exact size of the area enclosed by these points without needing to draw the triangle or measure side lengths directly. Always double-check your substitution of points to avoid mistakes in your calculations. Finding the area is particularly important when analyzing feasible regions in optimization problems, especially those involving linear programming.

System of Inequalities

A system of inequalities consists of two or more inequalities that are solved together. In these problems, you're not just finding when a single condition is met, but when all conditions are met simultaneously. This means you're looking for the 'feasible region' where all the inequalities overlap.
When working with systems of inequalities, remember:

• Each inequality divides the plane into two regions: one where the inequality holds and one where it doesn't.
• Graph each inequality carefully to ensure the shaded regions where they overlap are accurate.
• The solution to the system is the set of points that satisfy all inequalities -- this overlap is the 'feasible region'.

Systems of inequalities frequently appear in optimization scenarios where you need to maximize or minimize a value within constraints.

Graphing Linear Inequalities

To solve a problem involving linear inequalities, it's crucial to understand how to graph them on a coordinate plane. Each inequality represents a boundary line that divides the plane, and the true solution lies within or along this boundary.
Follow these steps to effectively graph linear inequalities:

• Convert each inequality to its corresponding linear equation by replacing the inequality sign with an equal sign.
• Graph the equation as a line. For \(y = mx + c\), plot the line using the slope \(m\) and the y-intercept \(c\).
• Determine which side of the line represents the solution to the inequality. Test a point that is not on the line (commonly the origin, if it isn't on the line) to see if it satisfies the inequality.
• Shade the correct side of the line to show all potential solutions for the inequality.
• Repeat this for all inequalities to identify the feasible region where all conditions are satisfied simultaneously.

By mastering this technique, you'll be well-equipped to handle problems that involve finding feasible regions and calculating their areas.