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...
 A: Another  way: WLOG $x=\sin t,y=\cos t$ where $0\le t\le\pi$
$$xy+y^2=\dfrac{\sin2t+\cos2t+1}2=\dfrac{\sqrt2\sin(2t+\pi/4)+1}2$$
Clearly, the maximum value occurs if $2t+\dfrac\pi4=2m\pi+\dfrac\pi2\iff t=m\pi+\dfrac\pi8$
A: 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 $(\sqrt{2}-1)^2=3-2\sqrt{2}$.
A: Here is another approach:
Let $A=\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$ and  write the problem as 
$\max \{ {1 \over 2} x^T Ax | \|x\| = 1, x_1 \ge 0 \}$ (where $x=(x_1,x_2)^T$ for notational convenience). Note that $A$ is real and symmetric.
Note that the constraint $x_1 \ge 0$ is unnecessary since if $x$ solves
$\max \{ {1 \over 2} x^T Ax | \|x\| \le 1 \}$, then $(\operatorname{sgn} x_1)x$
is also a solution and solves the original problem.
Hence we can look for solutions of $\max \{ {1 \over 2} x^T Ax | \|x\| \le 1 \}$.
Since $A$ is symmetric, it has an orthonormal basis of eigenvectors $v_1,v_2$ and
with $V=[v_1 \ v_2]$, and $y=V^{-1} x$ we have
$x^T Ax = \lambda_1 y_1^2+\lambda_2 y_2^2$ and $\|x\| = \|y\|$.
Without loss of generality, take $\lambda_1 \ge \lambda_2$.
It is straightforward to check that $\lambda_1= 1+ \sqrt{2}>0$,
and that
$\lambda_1 \|y\|^2 =\lambda_1 \ge \lambda_1 y_1^2+\lambda_2 y_2^2$,
and so
$\max \{ {1 \over 2} (\lambda_1 y_1^2+\lambda_2 y_2^2) | \|y\| \le 1 \} = {1 \over 2} \lambda_1$ and a (in fact, the) maximising $y$ is $(1,0)^T$.
Hence the problem reduces to finding a unit eigenvector corresponding
to $\lambda_1$, and it is easy to check that
$x= ({ 1\over \sqrt{2} \sqrt{2+ \sqrt{2}}}, { 1+ \sqrt{2}\over \sqrt{2} \sqrt{2+ \sqrt{2}}} )^T$ is a solution.
