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).