How could we find the largest number in the sequence $ \sqrt{50},2\sqrt{49},3\sqrt{48},\cdots 49\sqrt{2},50$?

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 $n=34$.

I like the AM-GM argument a lot, but here's another more down-to-earth solution.

Write $a_n=n\sqrt{51-n}$, $1\le n\le 50$. Everything in sight is positive, so $a_n\le a_{n+1}$ if and only if $$ a_n^2\le a_{n+1}^2\Leftrightarrow n^2(51-n)\le (n+1)^2(50-n). $$ This latter inequality simplifies to the quadratic inequality $-3n^2+99n+50\ge0$. The plot of this function is a parabola opening downwards. Therefore the inequality holds between the zeros $n_1\approx-0.5$ and $n_2\approx 33.5$.

We have shown that $a_{n+1}$ is larger than $a_n$, when $1\le n\le 33$, and that $a_{n+1}$ is smaller than $a_n$, when $n\ge 34$. Therefore we can conclude that $a_{34}$ is the largest of this lot.

I would like to apply calculus whenever it is possible. So, here is my trying (IMO this is not a better solution though):

Consider the function $f(x)=x^2(51-x)$ over $[1,50]$. Then as usual, $f'(x)=0\Rightarrow x=0,34$ and $f''(0)>0,f''(34)<0$ implies $f$ has a unique global maximum at $x=34$ and global minimum at $x=0$. So,...