Answer
a. $0,36$
b. $18,18$
Work Step by Step
a. Assuming the two nonnegative numbers are $x,y$, we have $x+y=36$ and the difference of their square roots is given by $z=\sqrt x-\sqrt y=\sqrt x-\sqrt {36-x}$. Take derivative to get $z'=\frac{1}{2\sqrt x}+\frac{1}{2\sqrt {36-x}}\gt0$, and the only critical points are $x=0,36$ which gives $z=-6,6$. Thus the largest difference of square roots is $6$, with the two numbers as $0,36$
b. In this case, $z=\sqrt x+\sqrt y=\sqrt x+\sqrt {36-x}$. Take derivative to get $z'=\frac{1}{2\sqrt x}-\frac{1}{2\sqrt {36-x}}$; letting $z'=0$, we get $x=y=18$, which gives $z=6\sqrt 2\gt6$ (endpoints). Thus the largest sum of square roots is $6\sqrt 2$, with the two numbers as $18,18$
