Maximum of $xy+y^2$ subject to right-semicircle $x\ge 0,x^2+y^2\le 1$ Maximum of:
$$
xy+y^2
$$
Domain:
$$
x \ge 0, x^2+y^2 \le1
$$
I know that the result is:
$$
\frac{1}{2}+\frac{1}{\sqrt{2}}
$$
for
$$
(x,y)=\left(\frac{1}{\sqrt{2(2+\sqrt{2})}},\frac{\sqrt{2+\sqrt{2}}}{2}\right)
$$
But I don't know how to get this result.
I know that:
$$
xy+y^2 \le \frac{1}{2}+y^2
$$
so:
$$
xy \le \frac{1}{2}
$$
And also:
$$
xy+y^2 \le xy+1-x^2 \equiv 1+x(y-x)
$$
But I don't know what to do next...

Hint$$x^2+y^2=x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 $$
Now notice $$x^2+(3-2\sqrt{2})y^2+(2\sqrt{2}-2)y^2 \ge 2(3-2\sqrt{2})^{\frac{1}{2}}xy+(2\sqrt{2}-2)y^2 (\because \text{AM-GM})$$Now note $(\sqrt{2}-1)^2=3-2\sqrt{2}$.
