find the dimensions of the right circular cylinder of maximum volume that can be placed inside of a sphere…

Answer:

$r = \frac{\sqrt{2}}{3}R$, $h=\frac{2\sqrt{3}}{3}R$

Explanation:

Step1: Establish relationship

Let the radius of the cylinder be $r$ and height be $h$. Using Pythagorean - theorem in the cross - section, we have $r^{2}+\left(\frac{h}{2}\right)^{2}=R^{2}$, so $r^{2}=R^{2}-\frac{h^{2}}{4}$.

Step2: Write volume formula

The volume of a right - circular cylinder is $V=\pi r^{2}h$. Substitute $r^{2}=R^{2}-\frac{h^{2}}{4}$ into the volume formula, we get $V = \pi\left(R^{2}-\frac{h^{2}}{4}\right)h=\pi R^{2}h-\frac{\pi}{4}h^{3}$.

Step3: Find derivative

Differentiate $V$ with respect to $h$. $V^\prime=\pi R^{2}-\frac{3\pi}{4}h^{2}$.

Step4: Set derivative to zero

Set $V^\prime = 0$ to find critical points. $\pi R^{2}-\frac{3\pi}{4}h^{2}=0$. Then $R^{2}-\frac{3}{4}h^{2}=0$, so $h^{2}=\frac{4}{3}R^{2}$, and $h=\frac{2\sqrt{3}}{3}R$ (we take the positive value since $h$ represents a length).

Step5: Find $r$

Substitute $h = \frac{2\sqrt{3}}{3}R$ into $r^{2}=R^{2}-\frac{h^{2}}{4}$. $r^{2}=R^{2}-\frac{1}{3}R^{2}=\frac{2}{3}R^{2}$, so $r=\frac{\sqrt{2}}{\sqrt{3}}R=\frac{\sqrt{6}}{3}R$.