11. Find a point inside a triangle such that the product of its distances to the sides of the triangle is maximized.
11. Let $x, y, z$ be the distances from a point to the sides of a triangle, then $a x + b y + c z = 2 S$, where $S$ is the area of the triangle. $x y z$ reaches its maximum value when $a x b y c z$ does, and therefore $a x = b y = c z = \frac{2}{3} S$.

$x = \frac{\boldsymbol{h}_{a}}{3} ; y = \frac{\boldsymbol{h}_{b}}{3} ; z = \frac{\boldsymbol{h}_{c}}{3}$. The desired point is the centroid of the triangle.