Question

Find the shortest curve surface of a cylinder that can be inscribed inside a sphere of radius $R$ ?

Answer 1

Given the height $h$,we can find the radius of the cylinder in terms of $r$ using the Pythagorean Theorem.

Note that $h$ refers to half of the total height of the cylinder.I chose to use $h$ instead of $\frac{h}{2}$ to simplify things later on.

To find the volume of our cylinder,we need to multiply the area of the top by the total height of the cylinder.In other words;

$V=\pi {\left(radiusofcylinder\right)}^2\ast \left(heightofcylinder\right)$

$V=\pi {\left(\sqrt{r^2-h^2}\right)}^2\left(2h\right)$

$V=2\pi h(r^2-h^2)$

This is our volume function.Next we take the derivative of the volume function and set it equal to zero.If we move the h inside the parenthesis,we only need to use the power rule to the derivative.

$V=2\pi (r^{2}h-h^3)$

$\frac{d}{dx}V\left(h\right)=2\pi (r^2-3h^2)=0$

The $2\pi$ divides out and we are left with;

$(r^2-3h^2)=0$

After some rearranging;

$h^2=\frac{r^2}{3}$

Take the square root of both sides.

$h=\frac{r}{\sqrt{3}}$

This is our optimized height.To find the optimized volume,we need to plug this into the volume function.

$V=2\pi h(r^2-h^2)=2\pi \left(\frac{r}{\sqrt{3}}\right)(r^2-{\left(\frac{r}{\sqrt{3}}\right)}^2)$

$V=\frac{2\pi r}{\sqrt{3}}(r^2-\frac{r^2}{3})$

$V=\frac{2\pi r}{\sqrt{3}}\left(\frac{{3r}^2-r^2}{3}\right)$

$V=\frac{2\pi r}{\sqrt{3}}\left(\frac{{2r}^2}{3}\right)$

$V=\frac{4\pi r^3}{3\sqrt{3}}$

$V=\frac{4\sqrt{3}\pi r^3}{9}$

This is the optimized volume for the cylinder.Its a good check to notice that $V$ is in terms of $r^3$ since volume should have cubic units.In other words,if our radius was given in term of meters,our volume units would be $m^3$.