Rectangle of maximum area in an isosceles triangle (base 5, height 8)

Text

You cut an isosceles triangle $A\hat{B}C$ with base $\overline{AB}=5$ and height $\overline{CH}=8$ with a line parallel to the base, meeting $AC$ at $D$ and $BC$ at $E$. From $D$ and $E$ you drop perpendiculars to the base, meeting it at $F$ and $G$, forming the rectangle $DEGF$. Letting $x$ be the height $EG$ of the rectangle, determine the value of $x$ that maximizes the area.

Solution

At height $x$ above the base, by similarity the width $DE$ is $\frac{DE}{AB}=\frac{8-x}{8}\ \Longrightarrow\ DE=5\!\left(1-\frac{x}{8}\right).$ The rectangle’s area is $\mathcal A(x)=DE\cdot x=5x-\frac{5}{8}x^2,$ a downward parabola, maximum at the vertex: $x=4,\qquad DE=\frac52.$ $\boxed{x=4,\qquad \mathcal A_{\max}=10.}$