If a right triangle is drawn in a semicircle of radius $1 / 2$ with one leg (not the hypotenuse) along the diameter, what is the triangle's maximum possible area?
Solution: It is easy to see that we will want one vertex of the triangle to be where the diameter meets the semicircle, so the diameter is divided into segments of length $x$ and $1-x$, where $x$ is the length of the leg on the diameter. The other leg of the triangle will be the geometric mean of these two numbers, $\sqrt{x(1-x)}$. Therefore the area of the triangle is $\frac{x \sqrt{x(1-x)}}{2}$, so it will be maximized when $\frac{d}{d x}\left(x^{3}-x^{4}\right)=3 x^{2}-4 x^{3}=0$, or $x=3 / 4$. Therefore the maximum area is $\frac{3 \sqrt{3}}{32}$.