The largest rectangle in a semicircle

Area. The rectangle has width $2x$ and height $\sqrt{R^2-x^2}$, so
$$A(x)=2x\sqrt{R^2-x^2},\qquad A(x)^2=4x^2(R^2-x^2)=4R^2x^2-4x^4.$$

Maximising. As $A>0$, $A$ and $A^2$ have the same maximiser. The derivative of $A^2$ is $8R^2x-16x^3=8x(R^2-2x^2)$, positive for $x<\frac R{\sqrt2}$ and negative after. So the maximum is at $x=\dfrac R{\sqrt2}$, where the height equals $\dfrac R{\sqrt2}$ too.

Answer.
$$A_{\max}=2\cdot\frac R{\sqrt2}\cdot\frac R{\sqrt2}=R^2.$$

Remark. The optimal rectangle is twice as wide as it is tall. Its area is $\frac{R^2}{\pi R^2/2}=\frac2\pi\approx64\%$ of the half-disc.