$ 5r + 4s + 3t + 6u = 100, \:\: r \ge s \ge t \ge u \ge 0 $ maximum and minimum possible of $r + s + t + u$? We have
$$ 5r + 4s + 3t + 6u = 100, \:\: r \ge s \ge t \ge u \ge 0 $$
What is the sum of the maximum and minimum possible  of $r + s + t + u$?

Attempt:
Assume that $r'+s'+t'+u'$ is the maximum. Now if $u > 0$, then we can have
$$ (r' + K) + s' + t' + (u' - \frac{5}{6} K),  \:\: K > 0 $$ 
is bigger than the claimed maximum, with ($r=r'+K, s=s', t=t', u = u' - \frac{5}{6}K$). We have a contradiction. So $u'=0$.
Now see
$$ 5r + 4s + 3t = 100 $$
Similarly, let $r^{*} + s^{*} + u^{*}$ maximum. Then
$$ r^{*} + (s^{*} - \frac{3}{4}C) + (t^{*} + C),  \:\: C > 0 $$
is bigger then the claimed maximum, provided that $s = s^{*} -(3/4)C \ge t = t^{*} + C$
with 
$$ C \le \frac{4}{7}(s^{*}-t^{*})$$
After this I have no idea.


Just to elaborate on one of the bounds in @Dr. Sonnhard Graubner's answer. Using Chebyshev's inequality, because $5>4>3$ and $r\geq s \geq t$
$$100=5r + 4s + 3t + 6u\geq \frac{5+4+3}{3}\cdot(r+s+t) + 6u=\\
4\cdot(r+s+t) + 6u\geq 
4\cdot(r+s+t) + 4u$$
and $$r+s+t+u\leq 25$$
