Single Variable Calculus, Chapter 4, 4.7, Section 4.7, Problem 26

Determine the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 3cm and 4cm if two sides of the rectangle lie along the legs.




By using Two Point Form, we can determine the equation of the line that intersect the vertex of the rectangle.

$
\begin{equation}
\begin{aligned}
y - y_1 &= \frac{y_2 - y_1}{x_2 - x_1} (x - x_1)\\
\\
y -0 &= \frac{4-0}{0-(-3)} (x -(-3))\\
\\
y &= \frac{4}{3} ( x+3)\\
\\
y &= \frac{4}{3} x + 4
\end{aligned}
\end{equation}
$

The area of the rectangle is $A = -xy$
when $\displaystyle y = \frac{4}{3} + 4$,
$\displaystyle A = -x \left( \frac{4}{3}x + 4 \right) = -\frac{4}{3} x^2 - 4x$
By taking the derivative of the area,
$\displaystyle A' = \frac{-8}{3}x - 4$

$
\begin{equation}
\begin{aligned}
\text{when } A' &= 0 \\
\\
0 &= \frac{-8}{3}x - 4\\
\\
4 &= \frac{-8}{3}x\\
\\
x &= \frac{-12}{8}\text{cm}
\end{aligned}
\end{equation}
$


Then, the critical number is $\displaystyle x = \frac{-3}{2}$cm
when $\displaystyle x = \frac{-2}{2}$, then
$\displaystyle y = \frac{4}{3} (x) + 4 = \frac{4}{3} \left( \frac{-3}{2} \right) +4 = -2 + 4 = 2$cm

Therefore, the largest rectangle that can be inscribed in the right triangle is $A = -xy = - \left( \frac{-3}{2} \right)(2) = 3\text{cm}^2$