Question

Set up an equation of a tangent to the graph of the following function.
Given a sphere of radius r. Inscribe in it a cone which has the greatest lateral area.Find that area.

Answer 1

First we want to write $h$ as a function of $r$. From the diagram be see that we have a right triangle with hypotenuse of length $R$ and length of length $r$ and $\frac{h}{2}$ (since the second leg only goes to the center of sphere, not the full length of cylinder). Therefore

$R^2=r^2+\frac{h^2}{4}$

$\Rightarrow h=2\sqrt{R^2-r^2}$

Then, we are given the lateral surface area $A$ is given by formula

$A=2\pi rh$

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

$=4\pi r\sqrt{R^2-r^2}$

Taking the function $f\left(r\right)$ and differentiating we find

$f^1\left(r\right)=4\pi \sqrt{R^2-r^2}-\frac{4\pi r^2}{\sqrt{R^2-r^2}}$

Setting this equal to $0$

$f^1\left(r\right)=0$

$\Rightarrow \frac{4\pi r^2}{\sqrt{R^2-r^2}}=4\pi \sqrt{R^2-r^2}$

$\Rightarrow r^2={\left(\sqrt{R^2-r^2}\right)}^2$

$\Rightarrow {2r}^2=R^2\Rightarrow R=\frac{R}{\sqrt{2}}\Rightarrow \frac{h}{2}=\sqrt{R^2-\frac{R^2}{2}}=\frac{R}{\sqrt{2}}$

Since $f^1\left(r\right)<0$ when $r<\frac{R}{\sqrt{2}}$

$f^1\left(r\right)>0$ when $r>\frac{R}{\sqrt{2}}$

this point is a minimum. Here the lateral surface area is minimam

When $r=\frac{h}{2}=\frac{R}{\sqrt{2}}$