Answer
The maximum height of ${\bf{r}}\left( t \right)$ above the $xy$-plane is $4$.
The point where this occurs is ${\bf{r}}\left( 2 \right) = \left( {{{\rm{e}}^2},\sin 2,4} \right)$.
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},\sin t,t\left( {4 - t} \right)} \right)$. So, the coordinate functions are
$x\left( t \right) = {{\rm{e}}^t}$, ${\ \ }$ $y\left( t \right) = \sin t$, ${\ \ }$ $z\left( t \right) = t\left( {4 - t} \right)$
Now, the $z$-coordinate represents the height of ${\bf{r}}\left( t \right)$ above the $xy$-plane. Thus, we find the maximum of the function $z\left( t \right) = 4t - {t^2}$.
Taking the derivative of $z\left( t \right)$ with respect to $t$ we get $z'\left( t \right) = 4 - 2t$. The critical point can be found by solving the equation $4 - 2t = 0$. So, we obtain $t=2$. Since the second derivative: $z{\rm{''}}\left( t \right) = - 2 < 0$, the function has a maximum. Substituting $t=2$ in $z\left( t \right)$ and ${\bf{r}}\left( t \right)$ give $z\left( 2 \right) = 4$ and ${\bf{r}}\left( 2 \right) = \left( {{{\rm{e}}^2},\sin 2,4} \right)$, respectively. Thus, the maximum height of ${\bf{r}}\left( t \right)$ above the $xy$-plane is $4$.
