A right triangle is drawn in a semicircle of radius  $\frac{1}{2}$  with one of its legs along the diameter. If the maximum area of the triangle is M, then the value of   $32\sqrt{3}Mc$   is ______

$BC\times CE=AC\times CD\therefore \left(BC\right)\left(CE\right)=x(1-x)$

$\Rightarrow \text{But }BC=CE\therefore BC=\sqrt{x(1-x)}$

$\therefore \text{Area }\mathrm{\Delta }=\frac{x\sqrt{x-x^2}}{2}$   or 

${\mathrm{\Delta }}^2=\frac{x^3-x^4}{2}$ or

$\frac{d{\mathrm{\Delta }}^2}{dx}=\frac{3x^2-4x^3}{2}$

If  $\frac{d{\mathrm{\Delta }}^2}{dx}=0$,  then  $x=\frac{3}{4},$  which is the point of maxima. Hence maximum area is  $\frac{3\sqrt{3}}{32}$