Question

A right cylinder is inscribed in a sphere of radius r. Find the largest possible surface area.

Answer 1

then are = \[\pi(r ^{2}h)\]

Answer 2

\[\pi(r ^{2}h)=\pi(r ^{2}2y)\]
since \[x ^{2}+y ^{2}=r ^{2}\]
\[y=\sqrt{r ^{2}-x ^{2}}\]
area
\[\implies 2\pi(r \sqrt{r ^{2}-x ^{2}})\]

Answer 3

That gives you the area of the inscribed cylinder, but the problem wants you to find the largest possible value. How are you going to do that?

Answer 4

Here r =x
then \[area=2\pi(r ^{2}\sqrt{r ^{2}-x ^{2}})\]
becomes \[Area=2\pi(x ^{2}\sqrt{r ^{2}-x ^{2}})\]
taking first derivative
\[A'=2\pi(\frac{ x ^{2} }{\sqrt{r ^{2}-x ^{2}} }(-2x)+(2x)\sqrt{r ^{2}-x ^{2}})\]
Now A'=0
\[\frac{2 x ^{3} }{\sqrt{r ^{2}-x ^{2}} }=2x \sqrt{r ^{2}-x ^{2}}\]

Answer 5

\[\implies x ^{2}=r ^{2}-x ^{2}\]
\[2x ^{2}=r ^{2}\]

Answer 6

Set R as the radius of the cylinder, h as the height, and r as the radius of the sphere. and w/ Pythagorean Theorem, (h/2)^2 + R^2 = r^2. I then chose to solve for R and R^2. Plug that into the SA formula of the cylinder and take the derivative

Answer 7

Solve for another variable

Answer 8

and plug it in again. :D

Answer 9

and
\[x= \frac{ r }{\sqrt{2} }\]
since \[x ^{2}+y ^{2}=r ^{2}\]
\[y= \frac{r }{ \sqrt{2} }\]
so (x,y)=\[(\frac{ r }{ \sqrt{2} },\frac{ r }{ \sqrt{2} })\]
will yield maximum surface area

Answer 10

Thanks! ^_^

Answer 11

Welcome