Chapter 4.9, Problem 61PPSE

Solution

To Find: The side of the square with vertical and horizontal sides inscribed in the region.

The square side length is $2x=\sqrt{5}-1$ .

Given:

  $\left\{\begin{array}{l}y\le -x^2+1\hfill \\ y\ge x^2-1\hfill \end{array}\right.$

Explanation:

White Region in the graph below contains the points $\left(x,y\right)$

Such that the $x,y$ values satisfies the Inequalities:

  $\begin{array}{l}y\le -x^2+1\\ y\ge x^2-1\end{array}$

  

In the symmetry of the White Region:

In this case, the square will be symmetrical about both the $x-$ axis and $y-$ axis.

The Image of $\left(a,b\right)$ in $x-$ axis is $\left(a,-b\right)$ . Likewise, the Image of $\left(a,b\right)$ in $y-$ axis is $\left(-a,b\right)$

Consider in $I$ Quadrant, the vertices of the vertex are $\left(x,y\right)$

So, $II$ quadrant vertex is the image of $I$ Quadrant vertex about $y-$ axis.

Hence, the $I$ Quadrant vertex is $\left(-x,y\right)\to \left(1\right)$

  $III$ quadrant vertex is the Image of the $II$ quadrant vertex about $x-$ axis.

Hence, the $I$ Quadrant vertex is $\left(-x,-y\right)\to \left(2\right)$

  $IV$ quadrant vertex is the Image of $I$ Quadrant vertex about $x-$ -axis.

Hence, $I$ Quadrant vertex is $\left(x,-y\right)\to \left(3\right)$

Applying equation $\left(1\right)$ , the distance between $I$ and $II$ Quadrants vertices is $2x$ , which is square's side length.

Applying equation $\left(3\right)$ , The distance between $I$ and $IV$ Quadrants vertices is $2y$ , Which is also square's side length.

The equation of the side length of the square:

  $\begin{array}{c}2x=2y\\ x=y\end{array}$

So, in $II$ quadrant the vertex of the square is $\left(x,x\right)$ (Since $x=y$ )

The parabola lies on the vertex of the square, but the vertex is inscribed in the region:

  $x=-x^2+1$

Combine alike terms:

  $x+x^2=1$

Add ${\left(\frac{1}{2}\right)}^2$ on both sides:

  $\begin{array}{c}{x}^2+\left(2\right)\left(\frac{1}{2}\right)\left(x\right)+{\left(\frac{1}{2}\right)}^2=1+{\left(\frac{1}{2}\right)}^2\\ {\left(x+\frac{1}{4}\right)}^2=\frac{5}{4}\end{array}$

Take square roots on both sides,

  $x=\frac{-1\pm \sqrt{5}}{2}$

Though $x$ is in $I$ Quadrant which is positive, $x>0$ , So:

  $\begin{array}{c}x=\frac{-1+\sqrt{5}}{2}\\ 2x=-1+\sqrt{5}\\ 2x=\sqrt{5}-1\end{array}$

So, the square side length is $2x=\sqrt{5}-1$ .