If $p$ and $q$ are positive real number such that $p^2+q^2=1$,

then the maximum value of $p+q$, is

Let $x=p+q.$ Then,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}x=p+\sqrt{1-p^2}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\left[\because p^2+q^2=1\right]\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dx}{dp}=1-p{\left(1-p^2\right)}^{12}\text{ and }\frac{d^{2}x}{d{p}^2}=-\frac{1}{{\left(1-p^2\right)}^{3/2}}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{dx}{dp}=0\Rightarrow p=\frac{1}{\sqrt{2}}.\end{array}$

Clearly, $\frac{d^{2}x}{d{p}^2}<0\text{ for }p=\frac{1}{\sqrt{2}}$

So, $x=p+q$ is maximum when $p=\frac{1}{\sqrt{2}}$.

For  $p=\frac{1}{\sqrt{2}}$, we get $q=\frac{1}{\sqrt{2}}$                                                     $\left[\because p^2+q^2=1\right]$

$\therefore p+q=\sqrt{2}$