Chapter 4, Problem 1RE

To determine

To find: To find the global extreme values.

Answer to Problem 1RE

The global extreme values are

Global maximum: $\frac{\text{4}\sqrt{\text{6}}}{\text{9}}\text{ at x = }\frac{\text{4}}{\text{3}}$

Global minimum: $\text{-4 at x = -2}$

Explanation of Solution

Given information:

The given function is $\text{y = x}\sqrt{\text{2 - x}}$

The interval for the given function is $\text{-2 }\le \text{ x }\le \text{ 2}$

Calculation:

The given function is,

  $\text{y = x}\sqrt{\text{2 - x}}$

The global extremum is found at the point where the first derivative is zero. These sites are referred to as critical points, and global extrema can be found at the end points.

Differentiating the above equation with respect to $\text{x}$ ,

  $\begin{array}{c}\frac{\text{dy}}{\text{dx}}\text{ = }\frac{\text{d}}{\text{dx}}\text{(x}\sqrt{\text{2 - x}}\text{)}\\ \text{= }\frac{\text{d}}{\text{dx}}\text{(x) }\cdot \sqrt{\text{2 - x }}\text{+ x }\cdot \frac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{2 - x}}\text{)}\\ \text{= 1 }\cdot \sqrt{\text{2 - x}}\cdot \text{ x }\cdot \left(\frac{\text{1}}{\text{2}}{\text{(2 - x)}}^{\text{1/2-1}}\cdot \frac{\text{d}}{\text{dx}}\text{(2 - x)}\right)\\ \text{= }\sqrt{\text{2 - x }}\text{+ x}\left(\frac{\text{1}}{\text{2}\sqrt{\text{2-x}}}\cdot \text{ (0 - 1)}\right)\\ \frac{\text{dy}}{\text{dx}}\text{=}\sqrt{\text{2 - x }}\text{- }\frac{\text{x}}{\text{2}\sqrt{\text{2 - x}}}\end{array}$

To get the critical value,

  $\begin{array}{c}\frac{\text{dy}}{\text{dx}}\text{ = 0}\\ \sqrt{\text{2 - x }}\text{- }\frac{\text{x}}{\text{2}\sqrt{\text{2 - x}}}\text{ = 0}\\ \text{2(2 - x) - x = 0}\\ \text{4 - 2x - x = 0}\\ \text{4 - 3x = 0}\\ \text{3x = 4}\\ \text{x = }\frac{4}{3}\end{array}$

To find the values $\text{y(x)}$ at $\text{x = -2, 2, }\frac{\text{4}}{\text{3}}$

At $\text{x = -2}$ ,

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

At $\text{x = 2}$ ,

  $\begin{array}{c}\text{y}\left(\frac{\text{4}}{\text{3}}\right)\text{ = }\frac{\text{4}}{\text{3}}\sqrt{\text{2 - }\frac{\text{4}}{\text{3}}}\\ \text{= }\frac{\text{4}}{\text{3}}\sqrt{\frac{\text{6 - 4}}{\text{3}}}\\ \text{=}\frac{\text{4}}{\text{3}}\cdot \sqrt{\frac{\text{2}}{\text{3}}}\\ \text{=}\frac{\text{4}\sqrt{\text{6}}}{\text{9}}\\ \text{y}\left(\frac{\text{4}}{\text{3}}\right)\approx \text{1}\text{.09}\end{array}$

At $\text{x = }\frac{\text{4}}{\text{3}}$ ,

  $\begin{array}{c}\text{y(2) = 2}\sqrt{\text{2 - 2}}\\ \text{= 2 }\cdot \text{ 0}\\ \text{y(2) = 0}\end{array}$

Therefore, the global minimum at $\text{x = -2}$ is $-4$ and the global maximum at $\text{x = }\frac{\text{4}}{\text{3}}$ is $\frac{\text{4}\sqrt{\text{6}}}{\text{9}}$ .