Optimization, cylinder in sphere

[Tclack]

Homework Statement

Find the dimensions(r and h) of the right circular cylinder of greatest Surface Area that can be inscribed in a sphere of radius R.

Homework Equations

SA=2\pi r^2+2\pi rh
r^2 + (\frac{h}{2})^2 = R^2 (from imagining it, I could also relate radius and height with r^2 = h^2 +2R^2)

The Attempt at a Solution

SA=2\pi r^2+2\pi rh

r^2 + (\frac{h}{2})^2 = R^2

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

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

\frac{dSA}{dr}=4\pi r+4\pi (\sqrt{R^2-r^2}+\frac{-2r^2}{2\sqrt{R^2-r^2}})

I tried setting that equal to zero, but I wasn't coming up with the right answer

The answer in the book(not mine): r=\sqrt{\frac{5+\sqrt{5}}{10}}R
h=2\sqrt{\frac{5-\sqrt{5}}{10}}R

Can anyone see my error, or did I make one?

 

[Dick]

You haven't made any errors yet. I guess the error is in the part you didn't show us. How did you get a different answer from the book?

 

[Tclack]

Well, from
\frac{dSA}{dr}=4\pi r+4\pi (\sqrt{R^2-r^2}+\frac{-2r^2}{2\sqrt{R^2-r^2}})=0
4\pi r+4\pi (\sqrt{R^2-r^2}-\frac{r^2}{\sqrt{R^2-r^2}})=0
4\pi r+4\pi (\frac{R^2-r^2-r^2}{\sqrt{R^2-r^2}})=0
-4\pir=4\pi (\frac{R^2-2r^2}{\sqrt{R^2-r^2}})
-r{\sqrt{R^2-r^2}={R^2-2r^2}

I seem to be going nowhere, I could square both sides

r^2(R^2-r^2)=R^4-4R^2r^2+4r^4
5r^2R^2-3r^4=R^4
But it doesn't clarify anything. Plus even, If I plug in \sqrt{\frac{5+\sqrt{5}}{10}}R for r, I'm getting nothing.

 

[ideasrule]

r^2(R^2-r^2)=R^4-4R^2r^2+4r^4
5r^2R^2-3r^4=R^4

You didn't simplify correctly. Aside from that, all you need to do now is use the quadratic equation to calculate r^2 (since r^2 = r^2 squared).