Chapter 7, Problem 64RE

Solution

To calculate: The solution of the system of inequality and given he coordinates of any corner points.

The common region in the graph is the solution of the inequality and the corner points does not exists for the inequality.

Given Information:

The system of inequality is,

  $\begin{array}{c}y\le x^2+4\\ x^2+y^2\ge 4\end{array}$

Calculation:

Consider the given system of inequality,

  $\begin{array}{c}y\le x^2+4\\ x^2+y^2\ge 4\end{array}$

The boundary point is included if the inequality containing $\le \text{ or }\ge$ symbols and the boundary point is excluded if the symbols $<\text{ or }>$  is used.

In the first inequality, the boundary is $y=x^2+4$ , which is parabola opens to upward. Find the intercept of the line. Substitute 0 for x and then 0 for y in the line one by one.

  $\begin{array}{c}0=x^2+4\\ x^2=-4\\ =\text{not exists}\end{array}$

And,

  $\begin{array}{c}y=0^2+4\\ =4\end{array}$

The intercept of the equation is $\left(0,4\right)$ .

To determine the half-plane to be shaded, use a test point not on the boundary, say  $\left(0,0\right)$ . Then,

  $\begin{array}{l}0\le 0^2+4\\ 0\le 4\end{array}$

The statement is true so the shade the half-plane where  $\left(0,0\right)$  is located.

In the second inequality, the boundary is $x^2+y^2=4$ , which is circle with radius 2. Find the intercept of the line. Substitute 0 for x and then 0 for y in the line one by one.

  $\begin{array}{c}{x}^2+0^2=4\\ x=\pm 2\end{array}$

And,

  $\begin{array}{c}{0}^2+y^2=4\\ y=\pm 2\end{array}$

The intercepts of the equation are $\left(2,0\right),\left(-2,0\right),\text{ and }\left(0,2\right),\left(0,-2\right)$ .

To determine the half-plane to be shaded, use a test point not on the boundary, say  $\left(0,0\right)$ . Then,

  $\begin{array}{c}{0}^2+0^2\ge 4\\ 0\ge 4\end{array}$

The statement is not true so the shade the half-plane where  $\left(0,0\right)$  is not located.

Now, make the graph of the inequality.

  

Thus, the shaded part by both the inequality is the solution of the inequality.

Hence, the corner points does not exist for the inequality.