Answer
a. $\frac{20}{3}, \frac{40}{3}$
b. $\frac{1}{4}, \frac{79}{4}$
Work Step by Step
a. Assuming the two nonnegative numbers are $x,y$, we have $x+y=20$ and the product of one number and the square root of the other is given by $z=x\sqrt y=x\sqrt {20-x}$. Take the derivative to get $z'=\sqrt {20-x}-\frac{x}{2\sqrt {20-x}}$; letting $z'=0$, we get $x=\frac{40}{3}$. Checking the signs of $z'$, we have $(0)..(+)..(\frac{40}{3})..(-)..(20)$. Thus the maximum of $z$ happens when the two numbers are $\frac{20}{3}, \frac{40}{3}$
b. In this case, $z=\sqrt x+y=\sqrt x+20-x$. Take derivative to get $z'=\frac{1}{2\sqrt x}-1$. Letting $z'=0$, we get $x=\frac{1}{4}$. Checking the signs of $z'$, we have $(0)..(+)..(\frac{1}{4})..(-)..(20)$. Thus the maximum of $z$ happens when the two numbers are $\frac{1}{4}, \frac{79}{4}$
