Sorry, one more extrema

[sigma]

Find the area of the largest rectangle that can be drawn with one side along the x-axis and two vertices on the curve with equation \(\displaystyle \
\L\
y = \frac{1}{{x^2 + 1}}
\\)

Just need somebody to check over my work once more. Just want to make sure I'm doing it right.



\(\displaystyle \
\L\
\begin{array}{l}
A = 2xy{\rm but }y = \frac{1}{{x^2 + 1}}{\rm note: Domain:}\left( {0,\infty } \right) \\
A = 2x\left( {\frac{1}{{x^2 + 1}}} \right) \\
A = \frac{{2x}}{{x^2 + 1}} \\
\frac{{dA}}{{dx}} = \frac{{2(x^{\rm 2} + 1) - 2x(2x)}}{{(x^{\rm 2} + 1)^2 }} \\
0 = \frac{{2 - 2x^2 }}{{(x^{\rm 2} + 1)^2 }} \\
0 = - 2x^2 + 2 \\
0 = - 2(x^2 - 1) \\
x = \pm 1{\rm but not - 1} \\
y = \frac{1}{{(1)^2 + 1}} = \frac{1}{2} \\
A = 2(1)\left( {\frac{1}{2}} \right) = \frac{2}{2} = 1 \\
{\rm Therefore the largest area is 1} \\
\end{array}
\\)

 

[Daniel_Feldman]

It's correct, but you need to verify that maximum area occurs at x=1.

You can do that one of two ways:

1. Preform a factor line analysis test on the first derivative and make sure that it changes from positive to negative at x=1.

Or

2. Find the second derivative at x=1, and if it's negative, there's a relative maximum at x=1.