Problem 34 - Precalculus Mathematics for Calculus

Question

Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} x &>0 \\ y &>0 \\ x+y &<10 \\ x^{2}+y^{2} &>9 \end{aligned}\right.$$

Step-by-Step Solution

Answer

Vertices are approximately (4.45, 5.55) and (5.55, 4.45). The solution set is bounded.

Step 1: Identify the Constraints

We begin with the system of inequalities: \( x > 0 \), \( y > 0 \), \( x + y < 10 \), and \( x^2 + y^2 > 9 \). The first two inequalities indicate that the solution lies in the first quadrant of the Cartesian plane. The third inequality, \( x + y < 10 \), represents a region below the line \( x + y = 10 \). The fourth inequality \( x^2 + y^2 > 9 \) represents outside the circle centered at the origin with radius 3.

Step 2: Graph the Inequality Regions

Graph each inequality on the Cartesian plane. Start with the line \( x + y = 10 \); shade the region below this line. Next, plot the circle \( x^2 + y^2 = 9 \); shade the region outside this circle. Be sure to mark the first quadrant, where \( x > 0 \) and \( y > 0 \), as the feasible region.

Step 3: Find Intersection Points of Boundaries

Determine where the line \( x + y = 10 \) intersects the circle \( x^2 + y^2 = 9 \). Substitute \( y = 10 - x \) into \( x^2 + (10-x)^2 = 9 \) and solve for \( x \). This gives \( x^2 + (10-x)^2 = 9 \), which simplifies to a quadratic equation \( 2x^2 - 20x + 91 = 0 \). Solving this equation gives \( x \approx 4.45 \) or \( x \approx 5.55 \). Corresponding \( y \) values will be \( y = 10 - x \), giving points \((4.45, 5.55)\) and \((5.55, 4.45)\) as intersection points.

Step 4: List All Vertices

The vertices of the solution region are determined by the intersection points and the axes. First quadrant constraints \( x > 0 \) and \( y > 0 \) give the axes. The only intersection with the boundaries within the constraints gives the points: \((4.45, 5.55)\) and \((5.55, 4.45)\). Along with the constraint lines, these points help to form the region.

Step 5: Determine if the Solution Set is Bounded

A set is bounded if it can fit within a finite circle. Checking the vertices and the boundary lines, we observe that the region defined is bounded within the first quadrant by \( x = 0 \), \( y = 0 \) and bounded above by the line \( x + y = 10 \) and outside the circle \( x^2 + y^2 = 9 \). Thus, the solution region is bounded.

Key Concepts

Graphing Inequalities

Graphing inequalities involves shading regions on the Cartesian plane to represent solutions that satisfy the inequalities. Let's break down what this means in simple terms.
First, inequalities like \( x > 0 \) and \( y > 0 \) tell us that our solutions are only in the first quadrant, where both \( x \) and \( y \) are positive. This is because the first quadrant of the Cartesian plane is where both coordinate values are positive.
When we have an inequality such as \( x + y < 10 \), it suggests a region below the line \( x + y = 10 \). To graph this, draw the boundary line \( x + y = 10 \) and shade the area below it. Similarly, the inequality \( x^2 + y^2 > 9 \) represents regions outside a circle centered at the origin with a radius of 3. This means we will shade outside the circle's boundary.
In essence, graphing these inequalities involves drawing each condition and clearly marking the areas that satisfy all of them collectively.

Bounded Solution Set

A solution set is considered 'bounded' if it can be enclosed within some finite space, like drawing a circle around it that contains all possible solutions. In our exercise, we are dealing with the intersection of several inequalities.
To determine boundedness, we consider the intersections and the overall shape formed by these inequalities. We are confined to the first quadrant due to the conditions \( x > 0 \) and \( y > 0 \). Additionally, we are limited by the line \( x + y = 10 \) and the circle \( x^2 + y^2 = 9 \).
The solution set cannot extend infinitely in any direction because it is trapped between these boundaries, such as within the right-angled triangle and the section outside the circle, giving us a distinct enclosed region. This bounded set ensures a particular area where all conditions are satisfied without extending into infinity.

Intersection Points

Intersection points are where two or more graphs meet, and in our exercise, finding these points is crucial. They help define the edges of our solution set.
Consider the line \( x + y = 10 \) and the circle \( x^2 + y^2 = 9 \). To find where they intersect, we substitute \( y = 10 - x \) into the circle's equation: \( x^2 + (10-x)^2 = 9 \). Solving this, we find the points are approximately \((4.45, 5.55)\) and \((5.55, 4.45)\).
These intersection points are part of the vertices forming the bounds of our solution area, supporting our conclusion about the region's shape. By analyzing these coordinates, we understand not just where the graphs meet, but also how they divide space within the plane.

Cartesian Plane

The Cartesian plane is a two-dimensional surface, formed by the perpendicular intersection of two number lines, one for the 'x' coordinate and another for the 'y' coordinate. It is a fundamental tool for graphing equations and inequalities.
In our exercise, all actions take place on this plane. Each point within it has an \( (x, y) \) format, providing exact locations for graphing purposes. The plane is divided into four quadrants. Inequalities constrain us to the first quadrant, where both \( x \) and \( y \) are positive.
Understanding this layout allows us to apply algebraic rules visually, with each line, curve, and region of shading meaningfully acting under specified constraints. This visual representation aids in interpreting complex systems of inequalities and visually confirms the bounded nature of solutions.