Find the maximum area of triangle DEM We have triangle ABC. On line AB is point M. The line which is paralel with AB cut AC in point D, and CB in E. In which distance "x" we should put line DE to get a maximum area of triangle DEM.
 A: Referring to the diagram below:-

In $\triangle CAB$, let the length of the altitude through $C$ be $1$ unit.
In $\triangle CDE$, let the length of the altitude through $C$ be $x$ unit. 
$\triangle CAB \sim \triangle CDE \implies AB : DE = 1 : x$
Area of $\triangle DEM$
$=\frac{1}{2}.DE. (1 – x)$
$=\frac{1}{2}.x. AB. (1 – x)$
Max(Area of $\triangle DEM$)
$= Max \frac{1}{2}.x. AB. (1 – x)$
$= \frac{AB}{2} Max(x – x^2)$      [$AB$ is given as constant]
For the quadratic function $(x – x^2)$, maximum occurs when $x = \frac{-(1)}{2(-1)} = \frac{1}{2}$
That is, $DE$ should be $0.5$ units from $C$ (on the altitude of $\triangle CAB$ through $C$ and its length is $1$ unit).
A: In mick's example, if the height were $h$
$$[DEF] = \frac12\cdot \frac{AB}{h}\cdot{hx - h^2}$$
The $\frac12\cdot \frac{AB}{h}$ is constant and we want to maximise $hx - x^2$. The maximum value of quadratic equation $ax^2 + bx + c$ is at $x = \frac{-b}{2a}$.
So the maximum of $hx - x^2$ will be when $x = \frac{-h}{2\cdot -1} = \boxed{\frac{h}{2}}$.
So the line $DE || AB$ must pass through half the height from A.
