Rectangle of maximum area inside a right triangle

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Triangle $ABC$ is right-angled at $B$, with hypotenuse $\overline{AC}=5$ and leg $\overline{AB}=4$. A rectangle $MNPQ$ is inscribed with $M,N$ on the hypotenuse $AC$, $P$ on $BC$ and $Q$ on $AB$. (a) Find the missing leg, the angles of the triangle and the altitude on the hypotenuse. (b) Derive the relation between the base and the height of the rectangle. (c) Find the dimensions of the rectangle of maximum area and show that this area equals $3$.

Solution

(a) By the Pythagorean theorem $\overline{BC}=\sqrt{5^2-4^2}=\sqrt{9}=3$ (the $3$-$4$-$5$ triple). Angles: $\cos A=\dfrac{\overline{AB}}{\overline{AC}}=\dfrac45\Rightarrow A\approx 36.87^\circ$, hence $C\approx 53.13^\circ$ and $B=90^\circ$. Altitude on the hypotenuse: $h=\dfrac{\overline{AB}\cdot\overline{BC}}{\overline{AC}}=\dfrac{4\cdot3}{5}=\dfrac{12}{5}=2.4$.

(b) Let $y$ be the height of the rectangle, measured from the hypotenuse toward vertex $B$, and $b$ its base on $AC$. The small triangle cut off above the rectangle is similar to $ABC$: as $y$ runs from $0$ to $h$ the base goes from $\overline{AC}$ to $0$, so $b=\overline{AC}\left(1-\frac{y}{h}\right)=5\left(1-\frac{y}{2.4}\right).$

(c) Area as a function of $y$: $A(y)=b\,y=5y\left(1-\frac{y}{2.4}\right)=5y-\frac{5y^2}{2.4}.$ Differentiating and setting to zero: $A'(y)=5-\frac{10y}{2.4}=0\ \Rightarrow\ y=1.2=\frac h2.$ Since $A''(y)=-\dfrac{10}{2.4}<0$, this is a maximum. Then $b=5\left(1-\frac{1.2}{2.4}\right)=2.5=\frac{\overline{AC}}{2},\qquad A_{max}=2.5\cdot1.2=3.$ $\boxed{\;b=2.5,\quad y=1.2,\quad A_{max}=3\;}$