Chapter 4.1, Problem 37E

To determine

To find : extreme values of the function and where they occur.

Answer to Problem 37E

Minimum occur at $x=-\sqrt{2}$ and extreme value is $y=-2$ .

Maximum occur at $x=\sqrt{2}$ and extreme value is $y=2$ .

Maximum $\left(-2,0\right)$ and minimum $\left(2,0\right)$ .

Explanation of Solution

Given information :

  $y=x\sqrt{4-x^2}$

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Product rule of derivative $\frac{d}{dx}\left(uv\right)=u^{\prime }v+u{v}^{\prime }$ .

Differentiate $y=x\sqrt{4-x^2}$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x\sqrt{4-x^2}\right)\\ =x\cdot \frac{d}{dx}\left(\sqrt{4-x^2}\right)+\sqrt{4-x^2}\cdot \frac{d}{dx}\left(x\right)\\ =\frac{x}{2\sqrt{4-x^2}}\cdot \frac{d}{dx}\left(4-x^2\right)+\sqrt{4-x^2}\cdot 1\\ =\frac{x}{2\sqrt{4-x^2}}\cdot \left(-2x\right)+\sqrt{4-x^2}\end{array}$

Simplifying,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{x^2}{\sqrt{4-x^2}}+\sqrt{4-x^2}\\ =\frac{-x^2+\left(4-x^2\right)}{\sqrt{4-x^2}}\\ =\frac{2\left(2-x^2\right)}{\sqrt{4-x^2}}\end{array}$

Take $\frac{dy}{dx}=0$ for critical points,

  $\begin{array}{c}\frac{2\left(2-x^2\right)}{\sqrt{4-x^2}}=0\\ 2-x^2=0\left[\begin{array}{l}4-x^2\ne 0\\ x\ne -2,2\end{array}\right]\\ x=-\sqrt{2},\sqrt{2}\end{array}$

The critical points are $x=-2,-\sqrt{2},\sqrt{2},2$ .

Now plug point $x=-2$ in $y=x\sqrt{4-x^2}$ .

  $\begin{array}{c}y=-2\sqrt{4-{\left(-2\right)}^2}\\ =-2\sqrt{4-4}\\ =0\end{array}$

Now plug point $x=-\sqrt{2}$ in $y=x\sqrt{4-x^2}$ .

  $\begin{array}{c}y=-\sqrt{2}\sqrt{4-{\left(-\sqrt{2}\right)}^2}\\ =-\sqrt{2}\sqrt{4-2}\\ =-\sqrt{2}\sqrt{2}\\ =-2\end{array}$

Now plug point $x=\sqrt{2}$ in $y=x\sqrt{4-x^2}$ .

  $\begin{array}{c}y=\sqrt{2}\sqrt{4-{\left(\sqrt{2}\right)}^2}\\ =\sqrt{2}\sqrt{4-2}\\ =\sqrt{2}\sqrt{2}\\ =2\end{array}$

Now plug point $x=2$ in $y=x\sqrt{4-x^2}$ .

  $\begin{array}{c}y=2\sqrt{4-{\left(2\right)}^2}\\ =2\sqrt{4-4}\\ =0\end{array}$

Examine whether the function is either maxima or minima. Check the surrounding values of critical point.

As $x$ moves away from $-\sqrt{2}$ on either side, the value of $y$ increases.

Thus, minimum occur at $x=-\sqrt{2}$ .

As $x$ moves away from $\sqrt{2}$ on either side, the value of $y$ decreases.

Thus, maximum occur at $x=\sqrt{2}$ .

At end points, maximum occur at $x=-2$ and minimum occur at $x=2$ .