Rectangle of maximum area with a side parallel to a leg

Text

A right triangle has leg $AB$ of length $3$ and leg $AC$ of length $4$ (right angle at $A$). Draw the parallel to $AC$, which meets the other leg at $S$ and the hypotenuse at $R$. From $R$ drop the perpendicular to $AC$, meeting it at $T$. Determine the sides of the rectangle $ASRT$ so constructed so that its area is maximum.

Solution

Place $A(0;0)$, $B(3;0)$ (leg $AB$ on the $x$-axis) and $C(0;4)$ (leg $AC$ on the $y$-axis). The parallel to $AC$ is a vertical line $x=s$: it meets $AB$ at $S(s;0)$ and the hypotenuse $BC$ at $R$. The perpendicular to $AC$ from $R$ is horizontal and meets $AC$ at $T(0;y_R)$; rectangle $ASRT$ has area $\mathcal A=s\cdot y_R$. Hypotenuse $BC$: $\dfrac{x}{3}+\dfrac{y}{4}=1\Rightarrow y=4\!\left(1-\dfrac{s}{3}\right)$. Hence $\mathcal A(s)=s\cdot4\!\left(1-\frac{s}{3}\right)=4s-\frac43 s^2,$ a downward parabola, maximum at the vertex: $s=\frac32,\qquad y_R=2.$ Thus $AS=\dfrac32$, $AT=2$ and $\boxed{\mathcal A_{\max}=\frac32\cdot2=3.}$