How could we find the largest number in the sequence $ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$? How to find the largest number in the sequence$$ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$$
I am interested in a "calculus-free" approach.
Thanks,

The $n$-th term in the sequence is $n\sqrt{51-n}=\sqrt{n^2(51-n)}$. So the question is: for which $n$ ($1\le n\le 50$), does $n^2(51-n)$ become the largest?
If you want to avoid calculus, you could use the AM-GM inequality:
if $x,\,y,\,z\ge 0$, then $$\frac{x+y+z}{3}\ge\sqrt[3]{xyz},$$
with equality if and only if $x=y=z$.
If we set $x=y=n/2$ and $z=51-n$, we obtain: $$\frac{51}{3}\ge \sqrt[3]{\frac{n}{2}\cdot\frac{n}{2}\cdot (51-n)},$$
with equality if and only if $n/2=51-n$ or $n=34$.
It follows that $n^2(51-n)\le 4\cdot 17^3$, or $\sqrt{n^2(51-n)}\le 2\cdot 17^{3/2}$, where equality holds for $n=34$.
