Single Variable Calculus, Chapter 7, Review Exercises, Section Review Exercises, Problem 116

What is the area of the largest rectangle in the first quadrant with two sides on the axes and one vertex on the curve $y = e^{-x}$.







$\displaystyle A = xy = xe^{-x} = \frac{x}{e^x}$

If we want to maximize the area we set $A'(x) = 0$ and solve for the dimensions of the box. So,


$
\begin{equation}
\begin{aligned}

\text{If } A =& \frac{x}{e^x}, \text{ then by Quotient Rule}
\\
\\
A'(x) =& \frac{e^x (1) - (x) e^x}{(e^x)^2}
\\
\\
A'(x) =& \frac{e^x - xe^x}{(e^x)^2}
\\
\\
\text{If } A'(x) =& 0
\\
\\
0 =& e^x (1 - x)

\end{aligned}
\end{equation}
$


We have,

$1 - x = 0 $ and $ e^x = 0 $ (we don't have solution for this)

So,

$x = 1$

Therefore,

the largest area is $\displaystyle A = \frac{x}{e^x} = \frac{1}{e^1} = \frac{1}{e} $ Square units