Chapter 1.2, Problem 51PS

Solution

To calculate the Kangaroo’s maximum height and jump length.

Kangaroo’s maximum height is $6\text{ feet}$ and jump length is $20\sqrt{2}\text{ feet}\text{.}$

Given:

The jump of the Kangaroo is modelled by the following function:

  $y=-0.03{\left(x-14\right)}^2+6$

where

  $\begin{array}{l}x=\text{ horizontal distance (in feet)}\\ y=\text{ corresponding height (in feet)}\end{array}$

Graph:

  

Calculation :

To determine the jump length, the roots of the quadratic function is calculated and the length is simply the difference between those two root values. At these points, the function will have a value of zero, i.e.

  $y=0$

Now, solving the roots as:

  $\begin{array}{l}y=-0.03{\left(x-14\right)}^2+6\\ \Rightarrow 0-0.03{\left(x-14\right)}^2+6\\ \Rightarrow {\left(x-14\right)}^2=\frac{6}{0.03}\\ \Rightarrow {\left(x-14\right)}^2=200\end{array}$

Taking the square root on both sides:

  $\begin{array}{l}x-14=\pm \sqrt{200}\\ \Rightarrow x=14\pm 10\sqrt{2}\end{array}$

Hence, the root values are:

  $R_1=14-10\sqrt{2}{\text{, R}}_2=14+10\sqrt{2}$

Now, the Kangaroo’s jump length is simply the difference of these root values:

  $\begin{array}{l}{\text{L= R}}_2-R_1\\ \Rightarrow \text{L= }14+10\sqrt{2}-14+10\sqrt{2}\\ \Rightarrow \text{L= }20\sqrt{2}\end{array}$

So, the kangaroo’s maximum height is $20\sqrt{2}\text{ feet}\text{.}$

Now, for the function $y=-0.03{\left(x-14\right)}^2+6,$ the first term $-0.03{\left(x-14\right)}^2$ would be non-positive. Hence, the function attains maximum value when the first term $-0.03{\left(x-14\right)}^2$ is zero. Therefore, the function will attain the maximum value at $x=14.$

Now, at $x=14$ :

  $\begin{array}{l}y=-0.03{\left(x-14\right)}^2+6\\ y=0+6\\ y=6\end{array}$

Hence, maximum value of the function is $6.$ Therefore, the kangaroo’s maximum height is $6\text{ feet}\text{.}$