Find the maximum area of a rectangle that can be drawn inside a circle's quarter How can I find the maximum area of the rectangle here? Given that the circle's radius is 6.

Any hint please?
 A: Choosing one vertex to be $(r \cos\theta, r\sin\theta)$ then another is $(r (\cos\theta- \sin\theta),0)$, a third is $(0,r (\cos\theta- \sin\theta))$ and the fourth is $(r \sin\theta, r\cos\theta)$ 
The area is then $2r^2 \sin\theta(\cos\theta- \sin\theta) = \sqrt{8}r^2 \sin(\theta) \sin\left(\frac\pi4-\theta\right) $ which is maximised when $\theta= \frac\pi8$ with the maximum area being $r^2(\sqrt{2}-1)$ 
Added to illustrate comment:  If the arrangement is not symmetric, you will not get two vertices on the arc and two on the axes.  Instead you will get coomething like this 

A: I would compute the coordinates of the given points:
$$A(x_A,0),B(x_B,\sqrt{6^2-x_B^2}),C(x_C,\sqrt{6^2-x_C^2}),D(0,y_B)$$ Now it must be
$$\vec{AB}=[x_B-x_A,\sqrt{6^2-x_B^2}]$$ and $$\vec{AD}=[-x_A,y_D]$$ and it must be
$$\vec{AB}\cdot \vec{AD}=0$$
Can you proceed?
A: Let $ABCD$ be our rectangle, where $A$ be placed on the perpendicular radius and $D$ be placed on the horizontal radius.
Also, let $OE$ be a radius of the circle such that $OE\perp AD$, $OE\cap BC=\{F\}$, $OE\cap AD=\{G\}$ and let $HE$ be a diameter of the circle. 
Now, since $G$ is a midpoint of $BC$, we obtain that $G$ is a midpoint of $AD$.
Thus, $OG=AG=DG$ and let $OG=x$, $DC=y$.
Now, since $\measuredangle ECH=90^{\circ},$ we obtain:
$$CE^2=EG\cdot GH$$ or
$$x^2=(6-x-y)(6+x+y)$$ or
$$(x+y)^2=36-x^2$$ or
$$y=\sqrt{36-x^2}-x.$$
Id est, $$S_{ABCD}=2x\left(\sqrt{36-x^2}-x\right).$$
Can you end it now?
