Area Of ​​Inequality System Solution

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Let's find out how to calculate the area of ​​the region that satisfies a system of inequalities. This is a cool topic in math that combines algebra and geometry! We'll break it down step by step so it's easy to follow.

Understanding the Inequalities

First, let's understand each inequality:

  1. x ≥ 0: This means we're only looking at the right side of the y-axis, including the y-axis itself. Basically, x has to be zero or a positive number.
  2. 0 ≤ y ≤ 5: This tells us that y is between 0 and 5, including both 0 and 5. So, we're looking at a region bounded by the x-axis (y = 0) and the horizontal line y = 5.
  3. 5x + 2y ≤ 20: This is a linear inequality. To understand it better, let's rewrite it as an equation: 5x + 2y = 20. This is a straight line. The inequality means we're looking at the region below or on this line.

Graphing the Inequalities

Next, let's put on our graphing hats and draw these inequalities on a coordinate plane.

  1. Draw the line 5x + 2y = 20:
    • To draw this line, find two points. Let's find the x and y intercepts.
    • When x = 0, 2y = 20, so y = 10. Point (0, 10).
    • When y = 0, 5x = 20, so x = 4. Point (4, 0).
    • Draw the line through (0, 10) and (4, 0).
  2. Shade the region 5x + 2y ≤ 20:
    • Since it's "less than or equal to," we shade the region below the line. A simple test is to pick the origin (0, 0). Does 5(0) + 2(0) ≤ 20? Yes, 0 ≤ 20, so we shade the side that includes the origin.
  3. Consider x ≥ 0:
    • This means we only consider the region to the right of the y-axis.
  4. Consider 0 ≤ y ≤ 5:
    • This restricts our region between the lines y = 0 (the x-axis) and y = 5. Draw a horizontal line at y = 5.

Finding the Feasible Region

The feasible region is where all the shaded areas overlap. It's the area that satisfies all three inequalities at the same time. In this case, it's a quadrilateral (a four-sided shape) bounded by the x-axis, the line y = 5, the y-axis, and the line 5x + 2y = 20.

Calculating the Area

Now for the fun part: finding the area of this feasible region!

Identifying the Vertices

First, we need to find the vertices (corners) of our quadrilateral. These are the points where the lines intersect.

  1. (0, 0): Intersection of x = 0 and y = 0.
  2. (4, 0): Intersection of 5x + 2y = 20 and y = 0.
  3. (0, 5): Intersection of x = 0 and y = 5.
  4. To find the last vertex, we need to find where 5x + 2y = 20 intersects y = 5.
    • Substitute y = 5 into the equation: 5x + 2(5) = 20
    • 5x + 10 = 20
    • 5x = 10
    • x = 2
    • So, the point is (2, 5).

Breaking Down the Shape

Our quadrilateral isn't a standard shape like a rectangle or a triangle, so we can divide it into simpler shapes. The easiest way is to see it as a trapezoid. The vertices are (0, 0), (4, 0), (2, 5), and (0, 5).

Calculating the Area of the Trapezoid

The area of a trapezoid is given by the formula:

Area = (1/2) * (base1 + base2) * height

In our case:

  • Base1 = distance from (0, 0) to (4, 0) = 4
  • Base2 = distance from (0, 5) to (2, 5) = 2
  • Height = distance from the x-axis to the line y = 5 = 5

Plugging these values into the formula:

Area = (1/2) * (4 + 2) * 5

Area = (1/2) * 6 * 5

Area = 3 * 5

Area = 15

So, the area of the region that satisfies the system of inequalities is 15 square units.

Alternative Method: Subtracting a Triangle

Another way to think about this problem is to consider the rectangle formed by the points (0, 0), (4, 0), (4, 5), and (0, 5). This rectangle has an area of 4 * 5 = 20 square units. However, our feasible region is smaller because it's cut off by the line 5x + 2y = 20.

The area we need to subtract is the area of the triangle above our feasible region, bounded by the points (2, 5), (0, 5), and the intersection of x = 4 and y = 5. Let's call the intersection point of x=4 and 5x+2y=20 as point A. We know that our last point is where y=5 intersects with 5x+2y=20 which is (2,5). The triangle to be removed is with points (2,5), (4,5), and (4,0) to (4,5) and (0,10) to (4,0). This is incorrect. We should calculate the triangle ABOVE the trapezoid.

We actually need to subtract a small triangle from a rectangle with vertices (0,0), (4,0), (0,5) and (4,5). The line 5x + 2y = 20 intersects the rectangle at (4,0) and (2,5). The vertices of the small triangle that is cut off are (0,5), (2,5), and the y-intercept of the line 5x + 2y = 20, which is (0,10). The base of the triangle is the distance between (0,5) and (2,5), which is 2. The height of the triangle is the distance between (0,5) and (0,10), which is 5. Therefore, the area of the triangle is (1/2) * 2 * 5 = 5.

The area of the rectangle is 4 * 5 = 20. Subtracting the area of the triangle, we get 20 - 5 = 15 square units. This confirms our previous result.

Conclusion

The area of the region that satisfies the system of inequalities is 15 square units. We found this by graphing the inequalities, identifying the feasible region, and then calculating the area of that region. You can approach these types of problems by dividing the feasible region into simpler shapes or using alternative methods like subtracting areas.

Keep practicing, and you'll become a master at solving these problems! Remember, math is all about understanding the concepts and applying them in creative ways. Guys, you nailed it!