Question 1035331
{{{f (x) = xe^(-0.2x)}}}  ==> {{{df(x)/dx = (1-0.2x)e^(-0.2x)}}}

Let {{{df(x)/dx = (1-0.2x)e^(-0.2x) = 0}}}

==> {{{1-0.2x = 0}}}.  Note that {{{e^(-0.2x)}}} is always greater than 0.

==> x = 5.  So there is a critical point at this x-value.

Now {{{d^2f(x)/dx^2 = (-0.4 + 0.04x)e^(-0.2x)}}}, and {{{d^2f(5)/dx^2 = (-0.4 + 0.04*5)e^-1 < 0}}},

hence there is a local max, which is also an absolute maximum at x = 5.

The maximum value is {{{f (5) = 5/e}}}.