Gemetry Revisited 3

Let xy be the coordinates of the top right vertex of the rectangle, and origin as the center of the semicircle.
$Area~f(A)=2*x*y, ~~\therefore ~\color{#3D99F6}{f(A_{max})=2y+2x\dfrac{dy}{dx}=0.}\\ But ~x^2+y^2=5, ~ \implies~2x+2y\dfrac {dy}{dx}=0, ~\implies ~\dfrac {dy}{dx}= - \dfrac x y.\\ \therefore ~2y+2x*(- \dfrac x y)=0, \implies~x^2=y^2. \\ \text{Substituting in the equation of the semicircle, since y>0, we get } y=\sqrt{\dfrac 5 2}, x=\pm \sqrt{\dfrac 5 2}.\\ f(A_{max})=2*x*y=\Large ~~~\color{#D61F06}{ 5 }$

For simplicity, let's do most of the calculation with radius equal $1$ , then scale up. The height of the rectangle is $sin(\theta)$ , half the width is $cos(\theta)$ , so the area of the rectangle is $A=2 sin(\theta)cos(\theta)=sin(2\theta)$ .

The function $sin(2\theta)$ has a maximum at $\theta=45^\circ$ so the height of the rectangle will be $\frac{1}{\sqrt{2}}$ and the width twice that or $\sqrt{2}$ giving us area of $\frac{1}{\sqrt{2}}\times\sqrt{2}=1$ .

Scaling will multiply all distances by $\sqrt{5}$ , so the area will be $5$ times more, or 5.