2022/073) Maximize $x \sqrt{1-y^2} + \sqrt{1-x^2}$

 In the above expression x  not dependent on y term so let us maximize $\sqrt{1-y^2}$ and this is {maxium at y= 0 and the value is 1.

Putting the same we get $x  + \sqrt{1-x^2}$

As we have $\sqrt{1-x^2}$ we can put $x=\sin\,t$ ( $0 \le t \le \frac{\pi}{4}$ to get

$x  + \sqrt{1-x^2}= \sin\, t + \cos \, t$

We need to maximize $\sin\, t + \cos \, t$ so let us get in form $r \cos\, w \sin \, t + r \sin\, w \cos  \, t$

which is $r \sin ( t+w)$ and maximum value is |sqrt{2}$

So let $r\cos\, t = 1$ and $r \sin\, t = 1$

By squaring both and adding 

$r^2 = 2$ or $r=\sqrt{2}$

Hence largest valus is $\sqrt{2}$