Answer
The maximum velocity occurs at ${\bf{F}} = \left( {0,20} \right)$ with maximum speed $20$ m/s.
Work Step by Step
We have the velocity vector field: ${\bf{F}} = \left( {\frac{{ - x}}{{20}},20 - \frac{{{x^2}}}{{1000}}} \right)$. So, the speed is
$v = ||{\bf{F}}|| = \sqrt {\left( {\frac{{ - x}}{{20}},20 - \frac{{{x^2}}}{{1000}}} \right)\cdot\left( {\frac{{ - x}}{{20}},20 - \frac{{{x^2}}}{{1000}}} \right)} $
$ = \sqrt {\frac{{{x^2}}}{{400}} + {{\left( {20 - \frac{{{x^2}}}{{1000}}} \right)}^2}} = \sqrt {\frac{{{x^2}}}{{400}} + 400 - \frac{{{x^2}}}{{25}} + \frac{{{x^4}}}{{{{10}^6}}}} $
$v = \sqrt {\frac{{{x^4}}}{{{{10}^6}}} - \frac{{3{x^2}}}{{80}} + 400} $
To find the maximum speed, we solve for the critical points of $v$:
$\frac{{dv}}{{dx}} = \frac{1}{2}{\left( {\frac{{{x^4}}}{{{{10}^6}}} - \frac{{3{x^2}}}{{80}} + 400} \right)^{ - 1/2}}\left( {\frac{{4{x^3}}}{{{{10}^6}}} - \frac{{3x}}{{40}}} \right) = 0$
So,
$\frac{{4{x^3}}}{{{{10}^6}}} - \frac{{3x}}{{40}} = 0$
$x\left( {\frac{{4{x^2}}}{{{{10}^6}}} - \frac{3}{{40}}} \right) = 0$
$x=0$, ${\ \ \ \ \ }$ $\frac{{4{x^2}}}{{{{10}^6}}} - \frac{3}{{40}} = 0$
The last equation gives $x = \pm 25\sqrt {30} $.
So, the critical points occur at $x=0$, $x = \pm 25\sqrt {30} $.
We plot the speed $v = \sqrt {\frac{{{x^4}}}{{{{10}^6}}} - \frac{{3{x^2}}}{{80}} + 400} $ versus $x$-coordinate and see the maximum occurs at $x=0$ (please see the figure attached). Notice that the $x$-values: $x = \pm 25\sqrt {30} $ is not part of the region, which is $ - 100 \le x \le 100$.
Substituting $x=0$ in $v$ and ${\bf{F}}$ give ${v_{\max }} = 20$ and ${\bf{F}} = \left( {0,20} \right)$, respectively.
So, the maximum velocity occurs at ${\bf{F}} = \left( {0,20} \right)$ with maximum speed $20$ m/s.
