Inscribed rectangle of maximum area in a right triangle

Text

In triangle $A\hat{C}B$, right-angled at $\hat{C}$, we have $\overline{AC}=3$ and $\overline{BC}=4$. Take a point $P$ on the hypotenuse $AB$ and drop the perpendiculars to the legs, meeting $AC$ at $R$ and $BC$ at $S$. Determine where to place $P$ so that rectangle $PRCS$ has maximum area.

Solution

Place the right-angle vertex at the origin: $C(0;0)$, $A(3;0)$, $B(0;4)$. Let $P(x;y)$ be on the hypotenuse, with $R(x;0)$ on $AC$ and $S(0;y)$ on $BC$; rectangle $PRCS$ has area $\;\mathcal A=x\,y$. The hypotenuse $AB$ has equation $\dfrac{x}{3}+\dfrac{y}{4}=1$, so $y=4\!\left(1-\dfrac{x}{3}\right)$. Then $\mathcal A(x)=x\cdot4\!\left(1-\frac{x}{3}\right)=4x-\frac{4}{3}x^2,$ a downward parabola; the maximum is at the vertex: $x=-\frac{4}{2\cdot(-\tfrac43)}=\frac{3}{2},\qquad y=2.$ Thus $P$ is the midpoint of the hypotenuse, $R$ the midpoint of $AC$, and $\boxed{\mathcal A_{\max}=\frac32\cdot2=3.}$