Question

The opening of the tunnel can be modeled by the graph of the equation y=-0.18x^2+4.4x-12 where x and y are measured in feet. Find the maximum height of the tunnel.

Answer 1

$\textit{Derivative}$
$y= -0,18x^2+4,4x-12$
$\textit{Find Relative Extrema.}$

$\textit{Relative Extrema}$
$y'= -0,36x+4,4$
$y' = 0$
$0 = -0,36x+4,4$
$x = \frac{440}{36}$
$x = \frac{110}{9}$

$y(\frac{110}{9})=-\frac{18}{100}\cdot\frac{12100}{81}+\frac{44}{10}\cdot\frac{110}{9}-12$
$y(\frac{110}{9})=-\frac{242}{9}+\frac{484}{9}-12$
$y(\frac{110}{9})=\frac{242}{9}-\frac{108}{9}$
$y(\frac{110}{9})= \frac{134}{9}$