Rectangle of maximum area inscribed in an isosceles triangle

Text

You cut an isosceles triangle $A\hat{B}C$ with base $\overline{AB}=4$ and height $\overline{CH}=6$ 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 a height $x$ above the base, the width $DE$ of the triangle follows by similarity: the upper triangle, similar to $ABC$, has height $6-x$, so $\frac{DE}{AB}=\frac{6-x}{6}\ \Longrightarrow\ DE=4\!\left(1-\frac{x}{6}\right).$ The rectangle’s area is $\mathcal A(x)=DE\cdot x=4x-\frac{2}{3}x^2,$ a downward parabola, maximum at the vertex: $x=3,\qquad DE=2.$ $\boxed{x=3,\qquad \mathcal A_{\max}=6.}$