Problem 95 - Precalculus

Question

In calculus, the first steps when solving the problem of finding the area enclosed by a set of curves are similar to those for finding the feasible region in a linear programming problem. In Exercises \(95-98,\) graph the system of inequalities and identify the vertices, that is, the points of intersection of the given curves. $$\begin{aligned} &y \leq x+2\\\ &y \geq x^{2} \end{aligned}$$

Step-by-Step Solution

Answer

The vertices of the feasible region are (2, 4) and (-1, 1).

Step 1: Graph the first inequality

Start by graphing the line for the inequality \( y \leq x + 2 \). The line itself, \( y = x + 2 \), should be drawn as a solid line since the inequality is "less than or equal to." This means the line is included in the solution set. Shade the region below this line as it represents all points \((x, y)\) where \( y \) is less than or equal to \( x + 2 \).

Step 2: Graph the second inequality

Next, graph the parabola for the inequality \( y \geq x^2 \). Draw the parabola \( y = x^2 \) with a solid curve, indicating that points on the parabola are included in the solution set. Shade the region above this parabola, representing all points \((x, y)\) where \( y \) is greater than or equal to \( x^2 \).

Step 3: Identify the feasible region

The feasible region is where the shaded areas from Step 1 and Step 2 overlap. This overlapping region represents all points that satisfy both inequalities: \( y \leq x + 2 \) and \( y \geq x^2 \). Review the graph to ensure that the correct area is shaded.

Step 4: Find the vertices of the feasible region

The vertices of the feasible region are the points of intersection between the boundaries of the inequalities. Solve \( y = x+2 \) and \( y = x^2 \) simultaneously to find these points. Set \( x+2 = x^2 \), which simplifies to \( x^2 - x - 2 = 0 \). Factor this as \((x - 2)(x + 1) = 0\), giving potential solutions \( x = 2 \) and \( x = -1 \). Substitute these values back into either equation to find the corresponding \( y \)-coordinates: for \( x = 2 \), \( y = 4 \), and for \( x = -1 \), \( y = 1 \). Therefore, the vertices are \((2, 4)\) and \((-1, 1)\).

Step 5: Confirm the vertices satisfy both inequalities

Finally, confirm that these intersection points are indeed vertices of the feasible region by ensuring they satisfy both inequalities. Substituting \( (2, 4) \) and \( (-1, 1) \) into both \( y \leq x + 2 \) and \( y \geq x^2 \) confirms that both points lie within the feasible region.

Key Concepts

Feasible Region

A feasible region is an important concept when working with systems of inequalities in algebra and linear programming. It represents the set of all possible points that satisfy all the inequalities in the system simultaneously. For this particular exercise, we are dealing with two inequalities:

• \( y \leq x + 2 \)
• \( y \geq x^2 \)

To find the feasible region, begin by graphing each inequality on the same coordinate plane. The area where the shading from each inequality overlaps is where all the conditions are met, making it the feasible region.
This region is quite important because it helps visualize where the solutions to the inequalities lie.
In contexts like optimization, solutions within this region can help identify maximum or minimum values of a given objective function. That's why knowing how to accurately graph and interpret feasible regions is critical for solving many mathematical problems.

Vertices

Vertices of a feasible region mark the key intersection points where the boundary lines or curves of the inequalities meet. Identifying these points is crucial because they often represent potential solutions for optimization problems, such as finding maximum or minimum values.
In our exercise, the boundaries are given by the line and the parabola:

• \( y = x + 2 \)
• \( y = x^2 \)

To find the vertices, we set the equations of the boundaries equal: \[ x + 2 = x^2 \] Rearranging gives us a quadratic equation: \[ x^2 - x - 2 = 0 \]
By factoring, we find:
\[(x - 2)(x + 1) = 0\]
This gives the solutions \( x = 2 \) and \( x = -1 \).
Plugging these back into the equations tells us the corresponding \( y \)-values, hence, our vertices are \((2, 4)\) and \((-1, 1)\).
These vertices are checked against the original inequalities to confirm they lie within the feasible region.

Graphing Inequalities

Graphing inequalities is a straightforward way of visualizing the solution set to a system of inequalities. The graph shows which areas of the coordinate plane satisfy the inequalities, making it easier to identify the feasible region.
Let’s outline the steps to graph each inequality:

• For the inequality \( y \leq x + 2 \): Start by drawing the line \( y = x + 2 \). Use a solid line, as it includes equality. Shade below the line to represent \( y \leq x + 2 \).
• For the inequality \( y \geq x^2 \): Plot the parabola \( y = x^2 \). Again, use a solid curve, since the parabola itself is part of the solution. Shade above the curve to represent \( y \geq x^2 \).

When both inequalities are graphed on the same axes, look for the overlapped shaded area. This is your feasible region. Graphing accurately requires careful attention to detail when drawing lines and curves, as well as shading the correct areas.
By doing so, you’ll clearly see where the solutions to the system of inequalities exist.