Question: A cylinder is obtained by revolving a rectangle about the $x$-axis, the base of the rectangle lying on the $x$- axis and the entire rectangle lying in the region between the curve: $y=\dfrac{x}{x^2+1}$ and the $x$-axis. Find the maximum possible value of the cylinder so formed.
So I proceeded by first graphing out the function to get a visual:
Now, we call the length of the side of the rectangle parallel to the $y$ axis as $y$, and the length parallel to the $x$ axis as $x$. The volume of the cylinder formed with these parameters would be
$$V= πy^2x$$
If I find a relation between $x$ and $y$, I can differentiate the expression for volume, and arrive at the maximum value. This is where I'm exactly stuck. I can't find such a relation. Any help would be appreciated.
We consider the line $y=c$ and we impose the condition
$c=\frac{x}{x^2+1}$
$cx^2-x+c=0$
that permit us to find two solutions $x_1,x_2$ in the definition domain of the intersection between the line and the curve, that corresponds in this case to $1-4c^2>0$
Moreover we obtain
$x_2-x_1=\frac{\sqrt{1-4c^2}}{c}$
But now
$V(c)=\pi c^2(x_2-x_1)=\pi c \sqrt{1-4c^2}$
What is the maximum value of $V(c)$? The maximum point $c$ for $V(c)$ is exactly also the maximum point for $T=V^2(c)$, that is simple to study:
$T(c)=V^2(c)=\pi^2 c^2(1-4c^2)$
and so
$T’(c)=2\pi^2 c(1-4c^2)-8\pi^2c^3\geq 0$
if and only if
$8c^3\leq c$ so $c^2\leq \frac{1}{8}$
Thus the maximum point for $V$ will be
$c=\frac{\sqrt{2}}{4}$ and $V(\frac{\sqrt{2}}{4})=\frac{\pi}{4}$